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QUESTION BANKPHYSICS
Gujarat Secondary and HigherSecondary Education Board,
Gandhinagar
Price : 85.00Published by :
SecretaryGujarat Secondary and Higher Secondary Education Board,
Gandhinagar
I
Copyright of this book is reserved by Gujarat Secondary and Higher Secondary EducationBoard, Gandhinagar. No reproduction of this book in whole or in part, or in any form ispermitted without written permission of the Secretary, Gujarat Secondary and Higher
Secondary Education Board, Gandhinagar.
Contribution
1 Dr. Hasmukh Adhiya (IAS) Principal Secretary , Education Department Gandhinagar
2 Shri R. R. Varsani (IAS) Chairman , G.S&H.S.E. Bord, Gandhinagar
3 Shri H. K. Patel (G.A.S) Dy. Chairman, G.S&H.S.E. Bord, Gandhinagar
4 Shri M. I. Joshi (G.E.S) Secretary , G.S&H.S.E. Bord, Gandhinagar
Coordination
1 Shri B. K. Patel O.S.D., G.S&H.S.E. Bord, Gandhinagar
2 Shri D. A.Vankar Assistant Secretary (Retd.), G.S&H.S.E. Bord, Gandhinagar
5 Shri G. M. Rupareliya Assistant Secretary, G.S&H.S.E. Bord, Gandhinagar
Expert Teachers
1. Shri J. M. Patel Shree J. M. Chaudhary Sarvajanik Vidhyalaya, Mehsana
2. Shri K. D. Patel J. N. Balika Vidhyalaya, Saraspur
3. Shri Mayur M. Raval P. J. Vakharia High School, Kalol
4. Shri S. G. Patel Sarkari Schook, Sector-12, Gandhinagar
5. Shri J. P. Joshi Diwan Ballubhai High School, Ahmedabad
6. Shri Vasudev B. Raval Vidhya Mandir High School, Palanpur
7. Shri Surendrabhai M. Rajkutir Convent of Jesus And Merry
8. Shri Sureshchandra H. Patel Alambic Vidhyalaya, Vadodara
9. Shri C. D. Patel Lalbahadur Shastri Vidhyalaya, Vadodara
10. Shri Mukesh N. Gandhi New English School, Nadiad
11. Shri Dineshbhai V. Suthar Retired Teacher
12. Shri S. S. Patel J. M. Chaudhary Sarvajanik Vidhyalaya, Mehsana
13. Shri Jayesh M. Purohit Ankur Vidhyalaya, Ahmedabad
14. Smt. Asha M. Patel Shree M.B. Vamdot Sarvajanik High School, Bardoli
15. Shri Maheshbhai Dhandhla Bhavnagar
16. Shri Mukesh M. Bhatt Bhavnagar
17. Shri Anilkumar Trivedi Anand
18. Shri Anand Thakkar Navchetan High School, Ahmedbad
19. Shri Sudhirkumar G. Patel Nutan High School, Visnagar
20. Smt. Anita Pillai Surat
II
P R E FA C E
Uptil now , the Students had to appear in various entrance examinations forengineering and medical courses after std-12. The burden of examinations on the side of thestudents was increasing day-by-day. For alleviating this difficulty faced by the students,from the current year, the Ministry of Human Resource Development , Government of India,has Introduced a system of examination covering whole country. For entrance to engineeringcolleges, JEE(Main) and JEE(Advanced) examinations will be held by the CBSE. TheGovernment of Gujarat has except the new system and has decided to follow the examinationsto be held by the CBSE.
Necessary information pertaining to the proposed JEE (Main) andJEE(Advanced) examination is available on CBSE website www.cbse.nic.in and it is requestedthat the parents and students may visit this website and obtain latest information – guidanceand prepare for the proposed examination accordingly. The detailed information about thesyllabus of the proposed examination, method of entrances in the examination /centers/places/cities of the examinations etc. is available on the said website. You are requested togo through the same carefully. The information booklet in Gujarati for JEE( Main) examinationbooklet has been brought out by the Board for Students and the beneficieries and a copy ofthis has been already sent to all the schools of the state. You are requested to take fulladvantage of the same also However, it is very essential to visit the above CBSE websitefrom time to time for the latest information – guidance . An humble effort has been made bythe Gujarat secondary and Higher Secondary Education Boards, Gandhinagar for JEE andNEET examinations considering the demands of the students and parents , a question bankhas been prepared by the expert teachers of the science stream in the state. The MCQ typeObjective questions in this Question Bank will provide best guidance to the students and wehope that it will be helpful for the JEE and NEET examinations.
It may please be noted that this “Question Bank” is only for the guidance of theStudents and it is not a necessary to believe that questions given in it will be asked in theexaminations. This Question Bank is only for the guidance and practice of the Students. Wehope that this Question Bank will be useful and guiding for the Students appearing in JEE andNEET entrance examinations. We have taken all the care to make this Question Bank errorfree, however, if any error or omission is found, you are requested to refer to the text –books.
M.I. Joshi R.R. Varsani (IAS)Date: 02/ 01/ 2013 Secretary Chairman
III
INDEX
IV
Unit Unit Name GSEB NEET JEE PageNo. No.
PART - I
1 Phisycs & Measurement Y Y Y 1
2 Kinematics Y Y Y 21
3 Laws of Motion Y Y Y 51
4 Work, Energy and Power Y Y Y 78
5 Rotationl Motion Y Y Y 107
6 Gravitation Y Y Y 147
7 Properties of Solids and Liquids Y Y Y 195
8 Thermodynamics Y Y Y 258
9 Kinetic Theory of Gases Y Y Y 288
10 Oscilations and Waves Y Y Y 321
1
Unit-1Important Formula
2
SUMMARY• Measurement of large distance (Parallax Method)
equation D = b
where D = distance of the planet from the earth.
where = parallax angle.b = distance between two place of observation.
• Measurement of the size of a planet or a star.
equation d D
where D = distance of planet from the earth,
d = diameter of planet. angular diameter of planet.
• Measurement of massThe gravitational force on an object, of mass m, is called the weight of the object.1 amu = 1.66 1027 kg = 1u
• Estimation of ErrorAbsolute Error - Suppose the values obtained in several measurement of physical quantitya are a1, a2, ............... an If their arithmetic mean is a
then 1 2
1
.... 1 n
ni
i
a a + aa an n
a1 = a - a1, a2 = a - a2 ,---------- an = a - an
a2, a2 ------- an are called absolute error
• Average absolute error
1 2
1
... 1
nn
ii
a a aa a
n n
• Fractional Error = aa
a
• Percentage Error
Percentage error = a 100 % = a 100 %
a
• Combination of errorsAddition Z = A + B Z A B
Substraction Z = A – B Z A B
Division Z = AB
Z A B
Z A B
Multiplication Z = A • B Z A B
Z A B
Power Z = nA Z An
Z A
3
• Rule for determining number of significant figures
• All the non - zero digits are significant
• All the zeros between two non zero digits are significant no matter where the decimalpoint is it at all.
• If the number is less then 1 then zeros on the right of decimal point but to the left ofthe first non - zero digit are not significant.
• In a number without decimal point the zeros on the right side of the last non zero digitare not significant.
• Dimensions and Dimensional formulas.
• The expression of a physical quantity with appropriate powers of M, L, T, K, A etcis called the dimensional formula of that physical quantity.
• The power of exponents of M, L, T, K, A are called dimensions of that quantity.
• • Some important units of distance–151fermi (fm) 10 m
o–101 A 10 m
111AU 1.496 10 m 151light year 9.46 10 m
161parsec 3.08 10 m
4
MCQ QuestionsFor the answer of the following questions choose the correct alternative from among thegiven ones.Physics - scope and Excitement
- Physics, Technology and society.- Fundamental sources of nature.- Nature of Physical laws
1. Physics is one of the basic disciplines in the category of ............... sciences.(A) Astro (B) Natural (C) Space (D) Genetic
2. ‘Physics’ comes from a ............... word meaning nature(A) Hindi (B) German (C) Greek (D) Sanskrit
3. Mechanics and newton’s motion laws as ............... laws dependad.(a) liner momentum (b) Energy conservation(c) Gravitational (d) Charge conservation
4. What is the approximate value of the Radious of a nucleus ?(a) –1410 m (b) –3110 m (c) –1910 m (d) –1510 m
5. The scope for ratio of length is in order to ...............(a) –4010 (b) 4010 (c) 2010 (d) 3010
6. The range of time scale is about ...............
(a) –10 2610 sec to10 sec (c) –15 1510 sec to10 sec
(b) –22 1810 sec to10 sec (d) 20 2510 sec to10 sec
7. Birth, evolution and death of stars etc. are studid in branch of physics known as ...............(a) Thermodynamics (c) Astro physics(b) Quantam physics (d) Electronics
8. ............... is a branch of physics in wich heat engine and refrigeratior efficiency is studied.(a) optics (b) Thermodynamics (c) Mechanics (d) Quantom physics
9. What is full name of LHC(a) Large hadron collider (c) Large heavy cullent(b) Large hadron cullent (d) Light heavy cullent
10. The range of mass varies from ...............
(a) –15 2610 kg to10 kg (b) –20 2810 kg to10 kg (c) –30 5510 kg to10 kg (d) –20 2010 kg to10 kg
11. Length of Galaxies is in order of ...............
(a) 2610 m (b) 3610 m (c) 2810 m (d) –1410 m
12. The approximate value of charge of an electron is ...............
(a) –1810 c (b) 1510 c (c) –3810 c (d) –1910 c
13. The universe is made up of ...............(a) matter only (b) radiation only (c) vaccum (d) matter and radiation
5
14. Nucleus of molecule is made up of wich fundamental constituents ?(a) only Electron (c) Electron and Proton(b) Proton and neutron (d) Electron and neutron
15. In the development of nenotechnology and biotechnology ............... have played a vital role.(a) ECG (b) ESR (c) NMR (d) AFM
16. What is full form of AFM ?(a) Atomatic force mioroscope (c) Atomatic fire microscope(b) Atomic force mirror (d) Atomic force microscope
17. What is full name of ECG ?(a) Electron cardiograph (c) Electron colour gram(b) Electro cardiograph (d) Electric colour graph
18. What is full name of ESR ?(a) Electric space Radar (c) Electron spin Resonance(b) Electron space Range (d) Electric spin Resonance
19. What is full name of NMR ?(a) Nuclear magnetic Resonance (c) Nuclear mega Radar(b) Neutron mega Resonance (d) Nuclear micro Radar
20. ............... deals with electric charge and magnatic phenomenna(a) Dynamics (b) Electro dynamic (c) Themodynamic (d) Mechanis
21. At present state, there are .............. fundamental forces in nature.(a) six (b) four (c) two (d) five
22. When charges are at rest the force is given by ............... law.(a) coulomb’s (b) Newton’s (c) Ampere’s (d) Faraday’s
23. The ................ force is the force of mutual attraction between any two objects by virtue of theirmasses.(a) Weak (b) Electromagnetic (c) Nuclear (d) Gravitational
24. The ............... force is the strongest of all fundamental forces.(a) nuclear (b) Electromagnetic (c) Gravitational (d) Weak nuclear
25. Electromagnetic force is ...............(a) attractive force only (c) repulsive force only(b) attractive and repulsive force (d) a short range force
26. Which of the following force binds The particle in the nucleons ?(a) Electromagnetic force (b) Strong force (c) Gravitational force (d) Weak force
27. Electromagnetic force is ............... range force(a) Short (b) long (c) medium (d) very short
28. Quarks - Quarks force is produced between -(a) Proton - neutron (b) proton - proton (c) neutron - neutron (d) (a),(b), (c) are true
6
29. Which partical are emitted during the decay from the nucleus ?
(a) neutron and proton (c) electron and neutrino(b) electron and neutron (d) electron and proton
30. ............... and ............... law’s are called inverse square law(a) Gravitation and weak (c) Coulomb’s and strong(b) Gravitation and coulomb’s (d) Electromagnetic and coulomb’s
31. Which property of object is responsible for the electric force ?(a) electric charge (b) pressure (c) volume (d) mass
32. Which property of object is responsible for the Gravitational force.(a) electric charge (b) mass (c) pressure (d) volume
33. How much times is the strong nuclear force stronger then weak nuclear force ?
(a) 1310 (b) 210 (c) –1310 (d) –210
34. How much times is the strong nuclear force stronger then electro magnatic force ?
(a) 1310 (b) 210 (c) –1310 (d) –210
35. How much times is the electromagnatic force stronger then Gravitational force
(a) 1310 (b) –1310 (c) 3610 (d) –3610
36. Who has unified electromagnetism and optics ?(a) Newton (b) Maxwell (c) Coulomb (d) Faraday
37. Who has unified terrestrial and celestial domains under a common law of Gravitational(a) Newton (b) Maxwell (c) Coulomb (d) Farady
38. The weak nuclear force, Gravitational force and electromagnatic force are A, B and C Respectivelythen ...............(a) C > A > B (b) C > A < B (c) B > A > C (d) C < A < B
39. Range of weak nuclear force is ...............
(a) –1510 km (b) –1410 km (c) –1810 km (d) –2010 km
40. Strong nuclear force close not exist on ...............(a) Proton (b) nuclear (c) neutron (d) electron
41. The force acting between two point charges kept at a certain distance is F1 Now magnitudeof charge are double and distance between them is double. The force acting between them isF2 find out the ratio of F2/F1 = ...............(a) 16 : 1 (b) 1: 16 (c) 1: 1 (d) 1: 8
42. If the resulting external force acting on system is zero then ............... of the system is constantand if the resultant external torque acting on a system is zero then ............... of the system isconstart.(a) total energy, angularmomentum (c) linermomentam, energy(b) liner momentam, angularmomentum (d) angular and linear momentam
7
43. Space is homogeneous and isotropic so ............... law of servation is the result of this(a) linear and angular momentum (c) energy and charge(b) angular and linear momentum (d) charge and energy
44. Time is homogeneous so ............... law of conserbation is the result of this(a) angular momentum (b) linear momentum (c) energy (d) charge
45. The basic reason behind existance of which conseration of law is still not known ?(a) angular momentum (c) energy(b) linear momentum (d) charge
46. The Gravitational force between any two body charges with distance as nF r where n = ..........
(a) –1 (b) 2 (c) –3 (d) –247. Match the column
Column - I Column - II(1) space is isotropic (P) conservation of linear momentum(2) space is homogeneous (Q) conservation of energy(3) Time is homogeneous (R) conservation of charge still not known(4) Time is isotropic (S) conservation of angular momentam(a) 1- (S), 2-(P), 3-(R), 4-(Q) (c) 1-(P), 2-(S), 3-(R), 4-(Q)(b) 1-(S), 2-(P), 3-(Q), 4-(R) (d) 1-(R), 2-(Q), 3-(P), 4-(S)
Measurement and system of units• Units of physical quantities, system of units, SI system of units, fundamental or Base units. precision
in measurement. Error in measurement and significant figures.• Dimensions and Dimensional formula, Dimensional analysis and its uses.48. Which of the following unit is not of length ?
(a) light year (b) fermi (c) oA (d) becquerel
49. becquerel is a ............... unit and its symbol is ...............(a) supplementary, Bq (b) fundamental, Bq (c) derived, Bq (d) derived, Bv
50. How many fundamental units are there in SI system ?(a) 5 (b) 7 (c) 6 (d) 4
51. Which of the following physical quantity is fundamental ?(a) viscosity (b) velocity (c) force (d) time
52. Poise is the unit of(a) viscosity (b) velocity (c) force (d) time
53. Which unit of physical quantity remains same for all unit system ?(a) meter (b) second (c) ampere (d) kilogram
8
54. Which of the following system of unit is not based on only units of mass length and time.(a) SI (b) MKS (c) CGS (d) FPS
55. Which of the following symbol of unit does not follow practical norms for the use of SI system ?(a) Kg (b) kg. (c) k (d) A
56. Why derive luminous intensity simbol form of SI system ?(a) cd (b) Cd (c) cd. (d) CD
57. What is the ratio of 10 micron to 1 nenometer ?
(a) 410 (b) 310 (c) 1610 (d) 1510
58.1 fem tom eter
100 nenom eter = ...............
(a) –610 (b) –810 (c) 2410 (d) –2410
59. If value of gravitational constant in MKS is 2
–112
Nm6.67 10kg
then value of G in
CGS = ............... 2
2
dyn cmgm
(a) –96.67 10 (b) –76.67 10 (c) –86.67 10 (d) –56.67 10
60. A partical has an acceleration of 72 km / min2 find acceleration in SI system.
(a) 20.5 m/ s (b) 230 m/s (c) 218 m/ s (d) 220 m/s
61. 950 dyne = ............... newton
(a) –39.5 10 (b) –595 10 (c) –7950 10 (d) –49.5 10
62. 100 picometer = ...............
(a) –810 cm (b) –710 m (c) –610 10 m (d) –810 10 m
63. 100 walt hour = ............... joule.
(a) 53.6 10 J (b) 63.6 10 J (c) 536 10 J (d) 636 10 J
64. If x meter is a unit of length then area of 1m2 = ...............(a) x (b) x2 (c) x–2 (d) x–1
65. 1 Mev = ............... ev
(a) 710 (b) 410 (c) 510 (d) 610
66. Wave length of light radiation 0.000015 m = ...............(a) 15 micron (b) 1.5 micron (c) 150 micron (d) 0.15 micron
67. 01 = ...............
(a) 600'' (b) 3600'' (c) 180'' (d) 3600'
68. 1 rad = ...............
(a) 0180 (b) 03.14 (c) 0180
(d)
0
180
9
69. 1 g = ............... amu
(a) 236.02 10 (b) –236.02 10 (c) –271.66 10 (d) 271.66 10
70. 1 parsec = ...............
(a) –1510 m (b) 111.496 10 m (c) 151.496 10 m (d) 163.08 10 m
71. Which of the following unit does not represent the unit of power ?(a) ampere/volt (c) (ampere)2 ohm(c) joule/second (d) ampere volt
72. Write the unit of angular acceleration in the SI system.
(a) N.Kg (b) 2rad / (sec) (c) m/sec (d) N/kg
73. unit of universal gravitational constant is ...............
(a) –1kg m sec (b) –1N m sec (c) 2 –2N m kg (d) –1N m kg
74. The unit of stefen Boltzman constant ( ) is ...............
(a) 2 –2 –1w m k (b) 2 –3w m k (c) –2 4w m k (d) –2 –4w m k
75. Unit of momentum physical quantity ?(a) newton - second (b) newton/second (c) Jule (d) Jule/second
76. Light year is a unit of ...............(a) Mass (b) volume (c) density (d) Distance
77. Joule/seed is the unit of ...............(a) Work (b) angular momentum (c) Pressure (d) Energy
78. The SI unit of momentum is ...............
(a) kg newton (b) –2 2kg m s (c) –1kg m (d) –1kg ms
79. Volt/meter is the unit of ...............(a) Work (b) viscosity (c) Electric fild intensity (d) velosity
80. The force F is represented by equation F = 1P + Q , where is the length. The unit ofP is same as that of ...............(a) Surface tension (b) velocity (c) force (d) momentum
81. Write the unit of surface tension in SI system.
(a) 2Nm
(b) Nm
(c) 2dynecm
(d) dynecm
82. Which physical quantity has unit of pascal - secod ?(a) Velocity (b) viscocity (c) energy (d) coefficient of viscocity
83. Which physical quantity has unit of joule - second ?(a) velocity (b) plank’s constant (c) energy (d) vescocity
10
84. What is the least count of vernier callipers ?(a) –410 m (b) –510 m (c) –210 m (d) –310 m
85. What is the least count of screw gauge ?(a) –410 m (b) –510 m (c) –210 m (d) –610 m
86. For measurement of astronomical distance ............... is used.(a) vernier callipers (b) spherometer (c) screwgauge (d) indirect method
87. Which mictoscope is used to measure the dimension of particle having dimension less than04000 A ?
(a) electron microscope (b) simple microscope (c) optical microscope (d) none of above88. In electron microscope electron behave like ...............
(a) charge (b) mass (c) particles (d) wave89. Which wave length of light is used in an optical microscope ?
(a) radiowave (b) X - ray (c) infrared (d) visible90. The intercepted area of the spherical surface about the center is 0.25m2 having diameter 50
cm what will be solid angle ?
(a) –14 10 sr (b) 31 10 sr (c) –110 sr (d) –15 10 sr
91. One planet is observed from two diametrically opposite point A and B on the earth the anglesubtended at the planet by the two directions of observations is 1.8o. Given the diameter ofthe earth to be about 71.276 10 m . What will be distance of the planet from the earth ?
(a) 840.06 10 m (b) 84.06 10 m (c) 13400.6 10 m (d) 811 10 m
92. Find the distance at which 4 AU would subtend an angle of exactly 1" of arc.11 16[1AU 1.496 10 m,1" 4.85 10 rad]
(a) 51.123 10 m (b) 511.23 10 m (c) 171.123 10 m (d) 1711.23 10 m
93. The percentage error in the distance 100 5 cm is ...(a) 5 % (b) 6% (c) 8 % (d) 20 %
94. In an experiment to determine the density of a cube the percentage error in the measurementof mass is 0.25 % and the percentage error in the measurement of length is 0.50 % what willbe the percentage error in the determination of its density ?(a) 2.75 % (b) 1.75 % (c) 0.75 % (d) 1.25 %
95. If 4A b the fractional error in A is ...............
(a) 4bb (b)
bb
(c) b4
b
(d) 4b
96. If 2
3A BPC
where percentage error in A , B and C are respectively 2% 3% and 5% then
total percentage error in measurement of p(a)18 % (b) 14 % (c) 21 % (d) 12 %
11
97. In the experiment of simple pendulum error in length of pendulum ( ) is 5 % and that of gis 3 % then find percentage error in measurement of periodic time for pendulum
(a) 4.2 % (b) 1.2 % (c) 2 % (d) 4 %
98. Acceleration due to gravity is given by 2GMgR
what is the equation of the fractional error g / g
in measurement of gravity g ? [G & M constant]
(a) R–
R
(b) R2
R
(c) R–2
R
(d) 1 R2 R
99. The period of oscillation of a simple pendulum is given by T 2g
what is the equation of
the relative error T
T
in measurement of period T ?
(a) 12
(b) 2
(c) 14
(d) 4
100. The length of a rod is (10.15 0.06) cm what is the length of two such rods ?(a) (20.30 0.06) cm (b) (20.30 1.6) cm (c) (10.30 0.12) cm (d) (20.30 0.12) cm
101. For a sphere having volume is given by 34V r3
What is the equation of the relative error
VV
in measurement of the volume V ?
(a) r3
r
(b) r4
r
(c) 4 r3 r
(d) 1 r3 r
102. Kinetic energy K and linear momentum P are related as 2pK
2m . What is the equation of the
relative error k
k
in measurement of the K ? (mass in constant)
(a) pp (b)
p2p
(c) p
2 p (d) p4
p
103. Heat produced in a current carrying conducting wire is H = I2Rt it percentage error in I, Rand t is 2 % , 4 % and 2 % respectively then total percentage error in measurement of heatenergy ...............(a) 8 % (b) 15 % (c) 5 % (d) 10 %
104. The resistance of two resistance wires are 1R (100 5) and 2R (200 7) are connectedin series. find the maximum absolute error in the equivalent resistance of the combination.(a) 35 (b) 12 (c) 4 (d) 9
105. The periodic time of simple pendulum is T 2g
relative error in the measurement of T and
are a and b respectively find relative error in the measurement of g(a) a + b (b) 2b + a (c) 2a + b (d) a - b
12
106. A physical quantity x is given by x = 4
3
14 4
3
A BC D
due to which physical quantgity produced the
maximum percentage error in x(a) B (b) C (c) A (d) D
107. The resistance VRI
where V 100 5 volts and I 10 0.3 anperes calculate the percentageerror in R.(a) 8 % (b) 10 % (c) 12 % (d) 14 %
108. The number of significant figures in 0.000150 is ...............(a) 3 (b) 5 (c) 2 (d) 4
109. Which of the following numerical value have significant figure 4 ?(a) 1.011 (b) 0.010 (c) 0.001 (d) 0.100
110. What is the number of significant figures in 35.50 10 ?
(a) 2 (b) 7 (c) 3 (d) 4111. The mass of substance is 75.5 gm and its volume is 25 cm2. It’s density up to the correct
significant figure is ...............
(a) 33.02 gm/ cm (b) 33.200 gm/ cm (c) 33.02 gm/ cm (d) 33.1gm/ cm
112. The area of a rectangle of size 1.25 2.245 cm in significant figure is ...............
(a) 22.80625 cm (b) 22.81cm (c) 22.806 cm (d) 22.8062 cm
113. The significant figures in 500.5000 are ...............(a) 5 (b) 3 (c) 7 (d) 6
114. Addition of measurement 15.225 cm, 7.21 cm and 3.0 cm in significant figure is ...............(a) 25.43 cm (b) 25.4 cm (c) 25.435 cm (d) 25.4350 cm
115. Substract 0.2 J from 7.36 J and express the result with correct number of significant figures.(a) 7.160 J (b) 7.016 J (c) 7.16 J (d) 7.2 J
116. After rounding of the number 9595 to 3 significant digits the value becomes ...............(a) 9600 (b) 9000 (c) 9590 (d) 9500
117. How many significant numbers are there in 5(2.30 4.70) 10 ?
(a) 3 (b) 4 (c) 2 (d) 5118. The radius of circle is 1.26 cm. According to the concept of significant figures area of it can
be represented as -
(a) 24.9850 cm (b) 24.985 cm (c) 24.98 cm (d) 29.98 cm
119. If A = 3.331 cm B = 3.3 cm then with regard to significant figure A + B = ...............(a) 6.6 cm (b) 6.31 cm (c) 6.631 cm (d) 6 cm
13
120. If the length of rod A is (2.35 0.01) cm and that of B is (5.68 0.01) cm then the rod B islonger than rod A by ...............
(a) (2.43 0.00) cm (b) (3.33 0.02) cm (c) (2.43 0.01) cm (d) (2.43 0.001) cm
121. In acceleration, The dimensions for mass ............... for length .. and for time(a) 0,1,–2 (b) 1,0,–2 (c) –2,0,1 (d) –2,1,0
122. Dimensional formula for power is ...............
(a) 2 –2 –3M L T (b) 1 2 –2M L T (c) 1 3 –1M L T (d) 0 2 –2M L T123. Dimensional formula for calories is ...............
(a) 1 1 –2M LT (b) 2 1 –2M LT (c) 1 2 –2M L T (d) 2 2 –2M L T124. Dimensional formula for thermal conductivity (k) is ..
(a) 2 1 –2 –1M LT K (b) 1 1 –2 1M LT K (c) 1 0 –3 –1M L T K (d) 1 1 –3 –1M LT K125. Dimensional formula for Resistance (R) is ...............
(a) 1 1 –3 –1M LT A (b) 1 1 0 –1M LT A (c) 1 2 –3 –2M L T A (d) 1 0 –3 –1M L T A126. Dimensional formula for conductance is ...............
(a) –1 2 –3 2M L T A (b) 1 2 –2 1M L T A (c) 1 –2 3 2M L T A (d) –1 –2 3 2M L T A
127. Which physical quantity is represented by 1 3 –3 2M L T A ?(a) Resistivily (b) Resistance (c) conductance (d) conductivity
128. Which physical quantity is represented by –1 –3 3 2M L T A ?(a) Resistivity (b) Resistance (c) conductance (d) conductivity
129. Which physical quantity is represented by 1 1 –3 –1M LT A ?(a) Stress (b) Resistance (c) Electricfield (d) potential Difference
130. The dimensional formula of plank’s constant is ...............
(a) 3 2 –1M L T (b) 1 2 –1M L T (c) 2 1 –1M LT (d) 1 2 –3M L T
131. Dimensional formula of latent heat is ...............
(a) 0 2 –2M L T (b) 2 0 –2M L T (c) 1 2 –1M L T (d) 2 2 –1M L T132. Dimensions of impulse are.
(a) –1 –1 1M L T (b) 1 1 –1M LT (c) 1 1 1M LT (d) 1 2 –2M L T133. Write dimensional formula of coefficient of viscosity
(a) 1 2 –1M L T (b) –1 1 1M LT (c) 1 –1 –1M L T (d) 1 1 –1M LT134. Dimensional formula for torque is
(a) 2 2 –3M L T (b) 2 1 –2M LT (c) 1 1 –2M LT (d) 1 2 –2M L T135. Dimensional formula for capisitance (C)
(a) –1 –2 4 2M L T A (b) 1 –2 4 2M L T A (c) –1 –2 3 1M L T A (d) 3 1 –1 –2M LT A
14
136. Dimensional formula for Boltzmann’s constant is ...............
(a) 1 1 –2 –1M LT K (b) 2 1 –2 –1M LT K (c) 1 2 –2 –1M L T K (d) 2 2 1 –2M L T K137. Dimensional formula for electromotive force (emf)
(a) 2 1 –1 –3M LT K (b) 1 2 –3 –1M L T K (c) 1 1 –3 –1M LT K (d) 1 2 3 –1M L T K138. Which physical quantity has dimensional formula as CR where C - capisitance and R - Resistance ?
(a) Frequency (b) current (c) Time period (d) acceleration139. Write the dimensional formula of the ratio of linear momentum to angular momentum.
(a) 0 –1 0M L T (b) 1 1 0M LT (c) 0 1 0M LT (d) 0 1 1M LT140. If L and R are respesented as the inductance and resistance respectively then the dimensional
formula of RL
will be ...............
(a) –2 1 –2 1M LT A (b) 0 0 –1 0M L T A (c) 1 –1 0 1M L T A (d) 1 3 1 0M L T A141. Write the dimensional formula of r.m.s (root mean square) speed.
(a) 1 2 –2M L T (b) 0 2 –2M L T (c) 0 1 –1M LT (d) 1 0 –1M L T
142. One physical quantity represented by an equation as (p – q)c2
where p, q and c are length
then quantity is ..(a) length (b) velocity (c) Area (d) volume
143. The dimensional formula of magnetic flux is ...............(a) 1 2 –2 –1M L T A (b) 1 2 1 2M L T A (c) 1 2 –2 2M L T A (d) –1 –2 1 2M L T A
144. Which physical quantity has unit of pascal - second ?(a) Force (b) Energy (c) Coefficient of viscocity (d) velocity
145. Dimensional formula of CV ? where C - capacitance and V - potential different
(a) 1 –2 4 2M L T A (b) 1 2 –3 1M L T A (c) 0 0 1 –1M L T A (d) 0 0 1 1M L T A
146. The equation of a wave is given by Y A sin –x kv
where is the angular velocity and
v is the linear velocity. Write the dimensional formula of K
(a) 0 0 1M L T (b) 1 0 –1M L T (c) 0 1 1M LT (d) 1 –1 1M L T147. If P and q are diffrent physical quantities then which one of following is only possible dimensionally ?
(a) p + q (b) pq (c) p – q (d) p = q
148. From 2
ap v – bv
constant equation is dimensionally correct find the dimensional formula
for b ? where P = preasure V = volume
(a) 0 3 0M L T (b) 1 3 0M L T (c) 0 1 3M LT (d) 1 1 –1M LT
15
149. Pressure P = A cosBx + c sinDt where xin meter and t in time then find dimensional formula
of DB
(a) 1 1 –1M LT (b) 0 1 –1M LT (c) 1 1 0M LT (d) –1 0 1M L T150. Find the dimensional formula for energy per unit surface area per unit time
(a) 1 0 –2M L T (b) 0 1 –1M LT (c) 1 0 –3M L T (d) 1 –1 1M L T
151. Equation of force 2F at bt where F is force in Newton t is time in second, then write unitof b.
(a) –1Nm (b) 2Nm (c) Nm (d) –2Nm
152. Pressure 2atP
bx where x = distance, t= time find the dimensional formula for
ab
(a) 1 0 –4M L T (b) 1 1 –1M LT (c) 1 0 –2M L T (d) 1 0 –2M L T
153. 2–Bxt0F A (1 – e ) where F is force and x is desplacement. write the dimension formula of B
(a) 2 1 –1M LT (b) 0 –1 –2M L T (c) 1 0 –2M L T (d) 1 2 –1M L T
154. Equation of physical quantity 2at bt v where v = velocity t = time so write the dimensionalformula of a in this equation
(a) 0 1 –1M LT (b) 1 1 –1M LT (c) 0 1 –2M LT (d) 1 2 0M L T
155. Density of substance in CGS system is 3.125 3gm / cm what is its magnitude is SI system ?
(a) 0.3125 (b) 3.125 (c) 31.25 (d) 3125
156. The resistivity of resistive wire is ARL
where L = length of wire A = Area of wire and
R is resistance of wire find dimension formula of
(a) 1 3 –3 –2M L T A (b) 1 2 –3 –2M L T A (c) 2 3 1 2M L T A (d) 2 3 –3 –2M L T A157. A cube has numerically equal volume and surface area calculate the volume of such a cube.
(a) 2000 Unit (b) 216 Unit (c) 2160 Unit (d) 1000 Unit158. Which out of the following is dimensionally correct.
(a) p2 = hg (b) p = h2g (c) p = hg (d) p = h2g
159. If energy p q rE G h c where G is the universal gravitational constant. h is the plank’s constantand c is the velocity of light, then the values of p, q and r are respectively
(a) 1 1 5– , ,2 2 2
(b) 1 1 5, ,2 2 2
(c) 5 1 1, , –2 2 2
(d) 1 1 5, – ,2 2 2
160. If the centripetel force is of the form mavbrc find the values of a, b and c(a) 1,2,1 (b) 1,2,–1 (c) 1,3,–2 (d) –1,3,–1
16
161. equation of 0 2 1[1 ( – )] t T T find out the dimensions of the coefficient of linear expansion suffix.
(a) 0 0 1 1M L T K (b) 0 1 1 1M LT K (c) 1 1 0 1M LT K (d) 0 0 0 –1M L T K162. Test if the following equation are dimensionally correct (S = surface tension = density
P = pressure v = volume n = coefficient of viscocity r = redious)
(a) 2Scosh
rg
(b)
pv (c)
4pr tv8n
(d) all correct
163. Match list - I with list - IIList - I List - II(1) Joule (a) henry ampere/sec(2) Walt (b) coulomb volt(3) volt (c) metre ohm(4) Resistivity (d) (ampere)2 ohm
(a) b,d,c,a (b) c,a,b,d (c) b,d,a,c (d) b,c,a,d
164. Match column - I with column - IIColumn -I Column - II
(1) capacitance (a) 1 1 –3 –1M LT A
(2) Electricfield (b) 1 2 –1M L T
(3) planck’s constant (c) –1 –2 4 2M L T A
(4) Angular momentum (d) 1 2 –1M L T
(a) a,c,b,d (b) c,a,d,b (c) c,a,b,d (d) a,b,d,c
165. In the relation –
P e ,
B
zk P is pressure, z is distance, k is boltz mann constant and is
the temperature. The dimensional formula of B will be
(a) 0 2 0M L T (b) 1 0 1M L T (c) 1 1 –1M LT (d) 1 1 0M LT
17
1(B) 26(B) 51(D) 76(D) 101(A) 126(D) 151(D)
2(C) 27(B) 52(A) 77(B) 102(B) 127(A) 152(A)
3(C) 28(A) 53(B) 78(D) 103(D) 128(D) 153(B)
4(A) 29(C) 54(A) 79(C) 104(B) 129(C) 154(C)
5(B) 30(B) 55(B) 80(A) 105(C) 130(B) 155(D)
6(B) 31(A) 56(A) 81(B) 106(C) 131(A) 156(A)
7(C) 32(B) 57(A) 82(D) 107(A) 132(B) 157(B)
8(B) 33(A) 58(B) 83(B) 108(A) 133(C) 158(C)
9(A) 34(B) 59(C) 84(A) 109(A) 134(C) 159(A)
10(C) 35(C) 60(D) 85(B) 110(C) 135(D) 160(B)
11(A) 36(B) 61(A) 86(D) 111(D) 136(A) 161(D)
12(D) 37(A) 62(C) 87(A) 112(B) 137(B) 162(D)
13(D) 38(A) 63(A) 88(D) 113(C) 138(C) 163(C)
14(B) 39(C) 64(C) 89(D) 114(B) 139(A) 164(B)
15(D) 40(D) 65(D) 90(A) 115(D) 140(B) 165(B)
16(D) 41(C) 66(A) 91(B) 116(C) 141(C)
17(B) 42(B) 67(B) 92(C) 117(A) 142(C)
18(C) 43(A) 68(C) 93(A) 118(C) 143(A)
19(A) 44(C) 69(A) 94(B) 119(A) 144(C)
20(B) 45(D) 70(D) 95(C) 120(B) 145(D)
21(B) 46(D) 71(A) 96(C) 121(A) 146(A)
22(C) 47(B) 72(B) 97(D) 122(B) 147(B)
23(D) 48(D) 73(C) 98(B) 123(C) 148(A)
24(A) 49(A) 74(D) 99(A) 124(D) 149(B)
25(B) 50(B) 75(A) 100(D) 125(C) 150(C)
KEY NOTE
18
HINT91 01.8 0.01 rad
7b 1.27 10 m
8bD 4.06 10 m
94 3
mass mdensityvolume l
( )( ) =
( )
percentage error in density
= M l[ + 3 ] 100
M l
= 1.75 %
962
3
A BPC
P A B C% [2 3 ]P A B C
= 21 %
97lT 2g
T 1 l 1 g100 [ ] 100T 2 l 2 g
= 4 %100 length of two rods 2l
= 2(10.15 0.06)
= (20.30 0.12) cm
103 2heat energy H = I RT
H I R T100 [2 ] 100H I R T
= 10 %
105 22g 4
T
l
g l T2g l T
= b + 2q
33Strong nuclear force
Electronmagnaticforce = 2
110 = 102
34Strong nuclear forceWeak nuclear force = -13
110 = 1013
35Electronmagnetic force
Gravational force = -2
-38
1010
= 1036
41 1 21 2
1
kq qFr
' '1 2
2 22
kq qFr
49–6
4–9
10 10 1010
58–15
–8–9
10 10100 10
60 2
72 100072 20min 3600 (sec)km m
2( )
64 Area = 2
A = 2 2x m
2 –22 2
A 11 m xx x
69 1 amu = –271.66 10 kg
= –241.66 10 gm231gm 6.023 10 amu
80 1 –1F p q |1p F
F N(Neuton)P surface tension(meter)
90 2A 0.025 m
2r 0.5m
2ASolid angle 0.4 Sr
r
=
= –14 10 Sr
19
107VResis tan ce RI
R V IR V I
R 5 0.3R 100 10
R 8R 100
R % 8%R
111mass 75.5density
volume 25 =
= 3.02 = 33.1g / cm
128 Resis tan ce AreaResistivitylength
129 forceElectricfieldelectric ch arg e
1302mass (dis tan ce)plank 's cons tan t
time
131heat energylatent heat Q
mass( ) =
133 2Force timecoefficientof viscos ity
(length)
141 2Urms u root mean square speed
142 If p = q = c = Lthen (p - q)c = L2 = Area
144dIf F = nAdx
Fn = pascal secondddx
v
vA =
146xy A sin ( – k) v
x k v
0 0 1xk M L T v
148 2
aP (v – b) cons tan tv
2PV – Pb – constanta abv v
PV – Pb
0 3 0V b M L T
149 cosBx dimensionl less
0 0 0Bx M L T
0 0 00 –1 0M L TB M L T
X
Same as0 0 –1D M L T
0 1 –1D M LTB
151 2F at bt
2F bt at
2 2F Nbt m
153 2– BxtF A (1 – e )
2Bxt dimensional less
0 0 00 –1 –2
2M L TB M L T
xt
155 3gmDensity 3.125cm
=
= –3
–6 33.125 10 kg
10 m
= 3125 3kg / m
157 3volume of cube V a2total surface area of cube A = 6a
V A 3 2a 6a
a 63V (6) 216 unit
20
159 P q rE G h c1 2 –2E M L T–1 3 –2G M L T
1 2 –1h M L T0 1 –1c M LT take it
1 2 –2 –1 3 –2 p 1 2 –1 q(M L T ) (M L T ) (M L T )0 1 –1 c(M LT )
= –p q 3p 2q r –2p–q–rM L T
1 1 5P , q , r2 2 2
160 a b cF m r v1 1 –2F M LT
0 1 –1M LTv0 1 0r M LT
1 0 0m M L T take it1 1 –2 1 a 1 –1 b 1 c(M LT ) (M ) (LT ) (L )
= a b c –bM L T
a 1, b 2, c –1
165 0 0 0z M L TkB
kBz
and P =
u
kBp pz
0 2 0M L T
21
Unit - 2Kinematics
22
SUMMARY• speed =
distance xtime t
Total distanceAverage speed =Total time
•t 0
xInstantaneous speed = limt
•displacement rVelocity =
time tv
t 0
r rIns tan eous velocity υ limt dt
• Average acceleration Gavet
•t
υ dυIns tan taneous acceleration a limt dt
• Equation for Uniformally accelerated motion
(1) υ = υ 0 + at (3) 2o
1d = t + at2
(2) Vo Vs t
2
(4) V2 = Vo2 + 2ad
• thn o
aDistance covered in n SecondS V (2n –1)2
• About Vectors
A . B
= AB cosq A ´ B
= AB sinq n
A . A
= 2| |A
A ´ A
= 0
i . i = j . j = k . i = 1 i ´ i = j ´ j = k ´ k = 0
i . j = j . k = k . k = 0 i ´ j = K j ´ k = i k ´ i = j
cosq =. BA
AB
A ´ B
=
ˆˆ ˆ
A A AB B B
i j kx y zx y z
23
A^ B
then A . B
= 0 A^ B
then | A ´ B
| = AB
A || B
then A . B
= AB A || B
then A ´ B
= 0
| A | = | B
| and A and B
is Q the angle between
(1) θ =0 then | A + B
| = 2A
(2) θ =180 then | A + B
| = 0
(3) θ =90 then | A + B
| = 2 A
(4) θ =60 then | A + B
| = 3 A
(5) θ =120 then | A + B
| = A
For projectile
- Time to reach the highest point oυ sintmg
- Maximum height H = 20
2υ sin2
θg
- Range R = 20υ sin2θ
g
- Maximum Range 20υR
g
- o2υ sinFlight time Tg
- Equation of trajectory 2
2 2o
gxy x tan –2 cos
- R 4H cot
24
MCQFor the answer of the following questions choose the correct alternative from among the given ones.(1) A branch of physics dealing with motion without considering its causes is known as ....
(A) Kinematicas (B) dynamics(C) Hydrodynemics (D) mechanics
(2) Mechanics is a branch of physics. This branch is ...(A) Kinematics without dynamics (B) dynamics without Kinematics(C) Kinematics and dynamics (D) Kinematics or dynamics
(3) To locate the position of the particle we need ...(A) a frame of referance (B) direction of the particle(C) size of the particle (D) mass of the particle
(4) Frame of reference is a ... and a ... from where an obeserver takes his observation,(A) place, size (B) size, situation(C) situation, size (D) place, situation
(5)
-2
B
-1
0
1 2 3
A
(m)
As shown in the figure a particle moves from 0 to A, and then A to B. Find pathlength anddisplacement.
(A) 2m, –2m (B) 8m, –2m (C) 2m, 2m (D) 8m, –8m
(6) A particle moves from A to B and then it moves from B to C as shown in figure. Calculatethe ratio between path lenghth and displacement.
(A) 2 (B) 1 (C) 12
(D)
(7) A particle moves from A to P and then it moves from P to B as shown in the figure. Findpath length and dispalcement.
600P
(A) 2
3l
, l (B) 3l
, l (C) 2l, l (D) l, 3
2l
25
(8) A car goes from one end to the other end of a semicircular path of diameter ‘d’. Find theratio between path legth and displacement.
(A) 32
(B) (C) 2 (D) π2
(9) A particle goes from point A to B. Its displacement is X and pathlength is y. So xy
.....
(A) > 1 (B) < 1 (C) 1 (D) 1(10) As shown in the figure a partricle statrs its motion from 0 to A. And then it moves from
A to B. AB is an arc find the Path length
Or
3
(A) 2r (B) r3
(C)
πr 1 +3
(D) r 13
(11) Here is a cube made from twelve wire each of length l. An ant goes from A to G throughpath A-B-C-G. Calculate the displacement.
G
E
F
H
(A) 3l (B) 2l (C) 3l (D) l3
(12) As shown in the figure particle P moves from A to B and particle Q moves from C to D.Desplacements for P and Q are x and y respectivey then
5
4
3
2
1
1 2 3 4 5O
(A) x > y (B) x < y (C) x = y (D) x y
26
(13) Shape of the graph of position time given in the figure for a body shows that
t
x
o
(A) The body moves with constant acceleration(B) The body moves with zero velocity(C) The body returns back towards the origin(D) nothing can be said
(14) The graph of position time shown in the figure for a particle is not possible because ...
t
x
o
(A) velocity can not have two values on one time(B) Displacement can not have two values at one time(C) Acceleration can not have two values at one time(D) A, B and c are true
(15) An ant goes from P to Q on a circular path in 20 second Raidus OP = 10m. What is theaverage speed and average velocity of it ?
1200
p
O
R
(A) –1 –1ms , 3 ms6
(B) 3 –1 –13ms , ms
2
(C) –1 –1ms , 3 ms3
(D) –1 –1ms , 6 ms
(16) A particle is thrown in upward direction with initial velocity of 60 m/s. Find average speed andaverage velocity after 10 seconds. [g = 10 ms–2]
(A) 26ms–1, 16ms–1 (B) 26ms–1, 10ms–1
(C) 20ms–1, 10ms–1 (D) 15ms–1, 25ms–1
27
(17) The ratio of pathlength and the resepective time interval is(A) Mean Velocity (B) M ean speed(C) intantaneous velocity (D) intantaneous speed
(18) A car moving over a straight path covers a distance x with constant speed 10 ms–1 and thenthe same distance with constant speed of V2. If average speed of the car is 16ms–1, then V2 = ....
(A) 30 ms–1 (B) 20 ms–1 (C) 40 ms–1 (D) 25 ms–1
(19) A bus travells between two points A ans B. V1 and V2 are it average speed and averagevelocity then(A) v1 > v2 (B) v1 < v2 (C) v1 = v2 (D) depends on situation
(20) A car covers one third part of its straight path with speed V1 and the rest with speed V2. Whatis its average speed ?
(A) 1 2
1 2
32
v vv v (B)
1 2
1 2
23
v vv v (C) 1 2
1 2
32
v vv v
(D) 1 2
1 2
32 2
v vv v
(21) Rohit completes a semicirular path of radius R in 10 seconds. Calculate average speed and averagevelocity in ms–1.
(A) 2 R 2R,10 10 (B) R R,
10 10 (C) πR 2R,
10 10(D) 2 R R,
10 10
(22) A particle moves 4m in the south direction. Then it moves 3m in the west direction. The timetaken by the particle is 2 second. What is the ratio between average speed and average velocity
?
(A) 57 (B)
75 (C)
145 (D)
514
(23) A particle is projected vertically upwards with velocity 30ms–1. Find the ratio of average speedand instantaneous velocity after 6s. [g = 10ms–1]
(A) 12 (B) 2 (C) 3 (D) 4
(24) The motion of a particle along a straight line is described by the function x = (3t – 2)2. Calculatethe acceleration after 10s.
(A) 9ms–2 (B) 18mls (C) 36ms– (D) 6ms–(25) Given figure shows a graph at acceleration time for a rectilinear motion. Find average
acceleration in first 10 seconds.
ot15105
102
maS
(A) 10ms–2 (B) 15ms–2 (C) 7.5ms–2 (D) 30ms–2
(26) A body starts its motion with zero velocity and its acceleration is 3m/s2. Find the distance travelledby it in fifth second.
(A) 15.5m (B) 17.5m (C) 13.5m (D) 14.5m
28
(27) A body is moving in x direction with constant acceleration . Find the difference of thedisplacement covered by it in nth second and (n–1)th second.
(A) α (B) 2
(C) 3 (D) 32
(28) What does the speedometer measure kept in motorbike ?(A) Average Velocity (B) Average speed(C) intantaneous speed (D) intantaneous Velocity
(29) The displacement of a particle in x direction is given by x = 9 – 5t + 4t2. Find the Velocityat timt t = 0
(A) –8 ms–1 (B) –5 ms–1 (C) 3 ms–1 (D) 10 ms–1
(30) A freely falling particle covers a building of 45m height in one second. Find the height of thepoint from where the particle was released. [g = 10ms–2](A) 120m (B) 125m (C) 25m (D) 80m
(31) The distance travelled by a particle is given by s = 3 + 2t + 5t2 The initial velocity of the particleis ...(A) 2 unit (B) 3 unit (C) 10 unit (D) 5 unit
(32) A particle is thrown in upward direction with Velocity V0. It passes through a point p of heighth at time t1 and t2 so t1 + t2 = ....
(A) 0
gv (B)
v02g
(C) 2hg
(D) h
2g(33) A particle is thrown in upward direction with initial velocity V0. It crosses point P at height h
at time t1 and t2 so t1t2 = _______
(A) 2hg (B)
20V
2g (C) 2
02Vg (D)
h2g
(34) Ball A is thrown in upward from the top of a tower of height h. At the same time ball B startsto fall from that point. When A comes to the top of the tower, B reaches the ground. Find thethe time to reach maximum height for A.
(A) hg
(B) 2hg
(C) h2g
(D) 4hg
(35) In the figure Velocity (V) position graph is given. Find the true equation.
xoX
Vo
V
(A) 00
0
x –xvv v (B) 0
00
– xxvv v (C) 0
00
– x –xvv v (D) vv v0
00
= x +x
29
(36) In the figure there is a graph of a x for a moving particle. Hence da =dt
.... V
xoX
a
ao
o
(A) 0
0
xa
(B) 0
0
–xa
(C) 0
0
–ax
(D) 0
0
ax
(37) A particle is moving in a straight line with intial velocity of 10 ms–1. A graph of acceleration time of theparticle is given in the figure. Find velocity at t = 10 s.
o 10
5a
(A) 25 ms–1 (B) 35 ms–1 (C) 45 ms–1 (D) 15 ms–1
(38) A graph of moving body with constant acceleration is given in the figure. What is the velocity after time t?
o t
A
V
D
B C
E
t1
(A) DC0A + 0EBC
(B) DC0A + DEBC
(C) BCAB + 0EDC
(D) DC0A + ADBC
(39)
t
V
o
The graph given in the figure shows that the body is moving with .....(A) increasing acceleration (B) decreasing acceleration(C) constant velocity (D) increasing velocity
30
(40) Slope of the velocity-time graph gives of a moving body..(A) displacement (B) acceleration (C) initial velocity (D) final velocity
(41) The intercept of the velocity-time graph on the velocity axis gives.(A) initial velocity (B) final velocity (C) average velocity(D) instanteneous velocity
(42) Here are the graphs of velocity time of two cars A and B, Find the ratio of the accelerationafter time t.
t
V
130
60
B
A
(A) 13
(B) 13
(C) 3 (D) 3
(43) Here is a velocity - time graph of a motorbike moving in one direction. Calculate the distancecovered by it in last two seconds.
o
V
t1 2 3 4 5
C10m
5
(A) 5 m (B) 20 m (C) 50 m (D) 25 m
(44) a
t
In the above figure acceleration (a) time (t) graph is given. Hence V .....(A) a (B) a (C) a2 (D) a3
(45)
t
X
o
A
B
The graph of displacent (x) time (t) for an object is given in the figure. In which part ofthe graph the acceleration of the particle is positive ?(A) OA (B) AB(C) 0 - A - B (D) acceleration is not positive at any part.
31
(46) In a uniformly accelerated motion the slope of velocity - time graph gives ....(A) The instantaneous velocity (B) The acceleration(C) The initial velocity (D) The final velocity
(47) The area covered by the curve of V – t graph and time axis is equal to magnitude of ....(A) change in velocity (B) change in acceleration(C) displacement (D) final velocity
(48) An object moves in a straight line. It starts from the rest and its acceleration is 2ms–2. After reaching a cer-tain point it comes back to the original point. In this movement its acceleration is -3ms-2. till it comes to rest.The total time taken for the movement is 5 second. Calculate the maximum velocity.(A) 6 ms–1 (B) 5 ms–1 (C) 10 ms–1 (D) 4 ms–1
(49) The relation between time and displacement of a moving particle is given by 2t 2 x where is aconstant. The shape of the graph x y is ...(A) parabola (B) hyperbola (C) ellips (D) circle
(50) Here are the graphs of x t of a moving body. Which of them is not suitable ?
(A)
t
x
o
(B)
t
x
o
(C)
t
x (D)
t
x
(51) Here are the graphs of v t of a moving body. Which of them is not suitable ?
(A)
t
V(B)
t
V
(C)
t
V (D)
t
V
Comprehension type questions
t
V
A
o D
C
B
In the figure there is a graph of velocity time for a particle.
32
(52) Which area shows the displacement covered by the particle after time t(A) closed fig AODCA (B) closed fig. ABCA(C) closed fig. AODCBA (D) none of above
(53) Which part shows initial velocity of the particle ?(A) OA (B) AB (C) AC (D) AOA
(54) How will you calculate the acceleration of the particle ?(A) taking length of AB (B) taking magnitude of BC(C) taking slope of AC (D) taking slope of AB
(55)
o
X
t2 84 6
Given graph shows relation between position and time. Find correct graph of acceleration time
(A) a
t2 84 6
(B)
o
a
t2 84 6
(C)
o
X
t2 84 6
(D)
o
X
t2 84
(56)
t
X
130
60
B
A
Here are displacement time graphs of particle A and B. If VVA and VB are velocities of the particles
respectively, then A
B
VV
= .....
(A) 13
(B) 3 (C) 13
(D) 3
33
(57) X
t2 84 6
Given graph shows relation between position (x) time (t) Find the correct graph of velocity time.
(A)
o
V
t2 84 6
(B) V
t2 84 6
(C) V
t2 84 6
(D) V
t2 84 6
(58) Particles A and B are released from the same height at an interval of 2 s. After some time tthe distance between A and B is 100m. Calculate time t.(A) 8 s (B) 6 s (C) 3 s (D) 12 s
(59) As shown in the figure a particle is released from P. It reachet at point Q at time t1
and reaches at point R at time t2 so 1
2
tt
= ....
(A) 13
(B) 12
(C) 21
(D) 41
(60) A particle moves in stright line. Its position is given by x = 2 + 5t – 3t2.Find the ratio of intial velocity and initial acceleration.
(A) 56
(B) 5–6
(C) 65
(D) 6–5
(61) A particle is moving in a circle of radius R with constant speed. It coveres an angle in sometime interval. Find displacement in this interval of time.
(A) 2R cos2 (B) θ2Rsin
2(C) 2Rcos (D) 2Rsin
(62) A particle is moving in a straight line with initial velocity of 200 ms–1 acceleration of the particleis given by a = 3t2 – 2t. Find velocity of the particle at 10 second.(A) 1100ms–1 (B) 300ms–1 (C) 900ms–1 (D) 100ms–1
34
(63) Angle of projection, maximum height and time to reach the maximum height of a particle are , H and tmrespectivley. Find the true relation.(A)
mHt2g
(B)
m2Ht =g
(C) m
4Ht =g
(D) m
Ht =4g
(64) Particle A is projected vertically upward from a top of a tower. At the same time particle B is droppedfrom the same point. The graph of distance (s) between the two particle varies with time is.
(A)
t
S (B)
t
S
(C)
t
S (D)
t
S
(65) A car is moving with speed 30m. Due to application of brakes it travells 30m before stopping.Find its acceleration.
(A) 2
m15s
(B) 2
m–15s
(C) 2
m30s
(D) 2
m10s
(66) A particle moves with a constant acceleration 2m/s2. Its intial velocity is 10m/s. Find velocity after t second.(A) (10 + t) ms–1 (B) 5(2 + t)ms–1 (C) 2 (5 + t)ms–1 (D) (10 + t2) ms–1
(67) A particle moves in a straight lime with constant acceleration. At t = 10s velocity and displacementof the particle are 16ms–1 and 39m respectively. What will be the velocity after 10 s ...(A) 22 ms–1 (B) 18 ms–1 (C) 20 ms–1 (D) 28 ms–1
(68) A particle moves with constant acceleration 2m/s2 in x direction. The distance travelled in fifthsecond is 19 m. Calculate the distance travelled after 5 second.(A) 50 m (B) 75 m (C) 80 m (D) 70 m
(69) Two bodies of masses m1 and m2 are dropped from heights H and 2H respectively. The ratio of time takenby the bodies to touch the ground is ...
(A) 12
(B) 2 (C) 12
(D) 21
(70) A freely falling stone crashes through a horizontal glass plate at time t and losses half of its velocity. Af-
ter time t2
it falls on the ground. The glass plate is 60 m high from the ground. Find the total distance travelled
by the stone. [g = 10ms–2](A) 120 m (B) 80 m (C) 100 m (D) 140 m
(71) A freely falling object travells distance H. Its velocity is V. Hence, in travelling further distance of 4H itsvelocity will become ....(A) 3V (B) 5V (C) 2V (D) 3V
h
3h
R
Q
P
35
(72) A ball is thrown vertically upward direction. Neglacting the air resistance velocity of the ball inair will(A) zero (B) decrease when it is going up(C) decrease when it is coming down (D) remain constant
(73) Two particles P and Q get 5 m closer each second while travelling in opposite direction. Theyget 1 m closer each second while travelling in same direction. The speeds of P and Q arerespectively ...(A) 5 ms–1, 1 ms–1 (B) 3 ms–1, 4 ms–1 (C) 3 ms–1, 2 ms–1 (D) 10 ms–1, 5 ms–1
(74) Motion of a porticle is described by an equation 12υ = A + y where v, y and A are velocity
distance and a constant respectively. Find the acceleratrion of the particle.
(A) 1 unit (B) 2 unit (C) 1 unit2
(D) 3 unit
(75) The minumum distance in which a car can be stopped is x. The velocity of the car is V. Ifthe velocity is 2V then find the stopping distance.
(A) 2x (B) 4x (C) 3x (D) 1 x2
(76) A particle moves in one direction with acceleration 2 ms–2 and initial velocity 3 ms–1.After whattime its displacement will be 10 m ?(A) 1 s (B) 2 s (C) 3 s (D) 4 s
(77) A goods train is moving with constant acceleration. when engine passes through a signal its speedis U. Midpoint of the train passes the signal with speed V. What will be the speed of the lastwagon ?
(A) 2 2V – U
2(B)
2 2V U2
(C) 2 22V U2
(D) 2 22V – U
(78) Displacement of a particle in y direction is given by y = t2 – 5t + 5 where t is in second.Calculate the time when its velocity is zero.(A) 5 s (B) 2.5 s (C) 10 s (D) 3 s
(79) The area under acceleration versus time graph for any time interval represents...(A) Intial velocity (B) final velocity(C) change invelocity in the time interval(D) Distance covered by the particle
(80) A ball is thrown vertically upward. What is the velocity and acceleration of the ball at the maximumheight ?(A) –gt ms–1, 0 (B) 0, –9 ms–2 (C) g ms–1, 0 (D) 0, –gt ms–2
(81) The relation between velocity and position of a particle is V = Ax + B where A and B are constants.Acceleration of the particle is 10 ms–2 when its velocity is V, How much is the acceleration whenits velocity is 2V.(A) 20 ms–2 (B) 10 ms–1 (C) 5 ms–2 (D) 0
36
(82) A particle moves on a plane along the path y = Ax3 + B in such a way that dx cdt
. c, A,
B are constans. Calculate the acceleration of the particle.
(A) ^
–23Axc jms (B) ^
2 –25Axc jms
(C) ^
2 –23Axc jms (D) ^ ^
2 –2c i 3Axc j ms
(83) The relation between velocity and position of a particle is given by V – x . Its initial velocity
is zero. Find its velocity at time 1tB
(A) e ms–1 (B) 0 ms–1 (C) –11 mse
(D) e2 ms–1
(84) An object moves in x - y plane. Equations for displacement in x and y direction are x = 3sin2tand y = 3cos2t Speed of the particle is(A) zero (B) constant and nonzero(C) increasing with time t (D) decreasing with time t
(85) Motion of a particle is decribed by x = (t – 2)2 Find its velocity when it passes through origin.(A) 0 (B) 2 ms–1 (C) 4 ms–1 (D) 8 ms–1
(86) To introduce a vector quantity ....(A) it needs magnitude not direction (B) it needs direction not magnitude(C) it need both magnitude and direction (D) nothing is needed
(87) Which pair of two vectors is antiparallel.
(A) AB
(B) AB
(C) A
B
(D) A
B
(88) In the above figure P and Q
are two vectors. What from followings is true
P
Q
(A) P
and Q
are equal (B) P
and Q
are perpendicular
(C) P
and Q
are antiparallel (D) P
and Q
are in same direction(89) Which from the following is a scalar ?
(A) Electric current (B) Velocity (C) acceleration (D) Electric field(90) P
and Q
are equal vectors what from the followings is true.
(A) P
and Q
are antiparallel (B) P
and Q
are parallel
(C) P
and Q
may be perpendicular (D) P
and Q
may be free vectors
37
(91) P
= Q
is true, if ...(A) their magnitudes are equal(B) they are in same direction(C) their magnitudes are equal and they are in same direction(D) their magnitudes are not equal and they are not in same direction
(92) A
and B
are in opposite direction so they are(A) parallel vectors (B) anti parallel vector(C) equal vector (D) perpendicular vector
(93)^ ^ ^
A 1 2 j 2k
. Calculate the angle between A
and Y axis.
(A) –1 5sin3
(B) –1 1sin
3 (C) –1 10sin3
(D) –1 5cos3
(94) A
and B
are nonzero vectors. Which from the followings is true ?
(A) 2 2 2 2A + B – A – B = 2 A + B
(B) 2 2 2 2A B – A – B 2 A – B
(C) 2 2 2 2A B – A – B A B
(D)
2 2 2 2A B – A – B A – B
(95) C A B
and A = B = C Find the angle between A
and B
(A) 3
(B) 6
(C) 23
(D) 0
(96) The resultant of two vectors is maximum when they.(A) are at right angles to each other (B) act in oppsite direction(C) act in same direction (D) are act 1200 to each other
(97) The resultant of two veetors A
and B
(A) can be smaller than A – B in magnitude(B) can be greater than A + B in magnitude(C) can’t be greater than A + B or smaller than A – B in magnitude(D) none of above is true
(98) The resultant of two forces of magnitude 2N and 3N can never be.
(A) 4N (B) 1N (C) 2.5N (D) 1 N2
(99) The sum of P
and Q
is at right agnles to their difference then(A) A = B (B) A = 2B (C) B = 2A (D) A = B = A – B
(100) Magnitudes of A, B and C
are 41, 40 and 9 respectively, A B C
Find the angle between A and B
(A) –1 9sin40
(B) –1 9sin41
(C) –1 9tan41
(D) –1 41tan40
(101) If A 3i 4 j 9k is multiplied by 3, then the component of the new vector along z direction is ...
(A) –3 (B) + 3 (C) –27 (D) + 27
38
(102) A B
is perpendicular to A
and B 2 A B
What is the angle between A and B
(A) 6
(B) 5π6 (C)
2π3 (D)
π3
(103) Out of the following pairs of forces, the resultant of which can not be 18N(A) 11 N, 7 N (B) 11 N, 8 N (C) 11 N, 29 N (D) 11 N, 5 N
(104) A 3i 2 j – 5k and B i j 2k
Find A 2B
(A) 40 (B) 42 (C) 39 (D) 2
(105) What is the angle between Q
and the resultant of P Q
and Q – P
(A) 900 (B) 600 (C) 0 (D) 450
(106) A 2i 2j – k and B 2i – j – 2k
Find 3A – 2B
(A) 2i 7 j k (B) 2i 8j – k
(C) 2i + 8j + k (D) i + 7 j + k
(107) Linear momentajm of a particle is 3i + 2j – k kgms–1. Find its magnitude.
(A) 14 (B) 12 (C) 15 (D) 11
(108) A × B = C, Then C
is perpendicular to
(A) A
only (B) B
only
(C) A
and B
both when the angle between them is ...(D) A
and B
both whatever to be the angle between them
(109) If A B 0
then
(A) A
must be zero (B) B
must be zero
(C) either A 0, B 0
or 0 (D) either A = 0, B = 0
or πθ =2
(110) y component of A B
is ....
A Axi + Ay j Azk
B Bxi + Byj Bzk
(A) AxBy – AyBx (B) AzBx – AxBz (C) AxBz – AzBx (D) AzBy – AyBz
(111) If A B 0
then A B
= ...
(A) AB (B) AB
(C) 1 AB2
(D) 0
(112) If A 4i + 3j – 2k and B 8i + 6j – 4k
the angle between A
and B
is(A) 450 (B) 0 (C) 60 (D) 90
39
(113) A 2i – 3j k and B 8i + 6j – 4k
then A B
= ....(A) 28 (B) 14 (C) 0 (D) 7
(114) If A 2i + 5j – k and B 3i – 2 j – 4k
the angle between A
and B
is ...
(A) 0 (B) π2
(C) π4
(D) π6
(115) Which statement is true ?(A) A B B A
(B) A × B = – B × A
(C) A B – B A
(D) A B AB
(116) Which from the following is true ?
(A) A B
cosAB
(B) A BsinAB
(C) A × B
tanθ =A - B
(D) ABcot
A B
(117) A B A B
....
(A) 0 (B) A2 + B2 (C) 2 2A B (D) A2B2
(118) The angle between i + j and z axis is ....(A) 0 (B) 45 (C) 90 (D) 180
(119) A 2i – j 2k and B – i – 2j 2k
what is the angle between A
and B
(A) cos–1 0.8888 (B) cos–1 0.4444 (C) sin–1 0.4444 (D) sin–1 0.8888
(120) A Pi – 2Pj – k and B – 3i + 2j 14k
are perependcular to each other. Then p = ...(A) 3 (B) 4 (C) 2 (D) 1
(121) In the above triangle AC = 5, BC = 8 and B = 6 Find the value of angle A.
8
5
6
(A) sin–1 0.6 (B) sin–1 0.8 (C) sin–1 0.12 (D) sin–1 0.4(122) A – l + j – 2 k and B 2i – j k
Find the unit vectior in direction of A B
(A) 1 –i – 5j – 2k23
(B) 1 –i + 5j – 3k35
(C) 1 –i – 5j – 3k29
(D) 1 –i – 5j – 3k35
40
(123) What is unit vector along i + j ?
(A) i + j2
(B) i + j
2
(C) i + j
3
(D) i – j
2
(124) Unit vector of A B
is k . Unit vector of A
is i Then what is the unit vector of B
(A) j (B) – j
(C) any unit vector in xy plane (D) any unit vector in xz plane
(125) Find a unit vector in direction of i + 2 j – 3k
(A) 1 i + 2j – 3k7 (B) 1– i + 2 j – 3k
2
(C) 1 i + 2j – 3k14
(D) 1 i + 2 j – 3k5
(126) Find a unit vector perpenduicular to both A
and B
(A) A BAB
(B) A × B
ABsinθ
(C) A × B
ABcosθ
(D) A B
ABsinθ
(127) If resultant of A 2i + j – k , B i – 2j + 3k
and C
is unit vector in y direction, then C
is
(A) – j (B) 3i – 2j + 2k (C) j (D) 2i + 3k
(128) A
and B
are two vectors A BU = U Now find the true option.
(A) A
and B
equal vectors (B) A and B
are in opposite direction
(C) A
and B
are in same direction (D) A B
(129) Unit vector in direction of A
is
(A) A
(B) AA
(C) A A
(D) A
A
(130) 2 2i + j + pk3 3 is a unit vector so p = .....
(A) 23
(B) 1–3
(C) 1 (D) 19
(131) Find a unit vector from the followings.
(A) i + j (B) i – j (C) 1 1i + j2 2 (D)
1 1i – j22
(132) Train A is 56 m long and train B 54 m long. They are travelling in opposite direction with velocity
m15s
and m5s
respectively. The time of crossing is.
(A) 12 s (B) 6 s (C) 3 s (D) 18 s
41
(133) Graphs of velocity time for two cars A and B moving in a straight line are given in the fig.The area covered by PQRS gives.
t
AV
PB
R
t2
S
Q
t1
(A) distance from A to B at time t2
(B) distance from A to B at time t1
(C) distance from A to B in time interval t2 – t1
(D) change in distance from A to B in time interval t2 – t1
(134)
o
AV B
t
Graphs velocity time is given for cars A and B moving in a straight line in same direction.At time t = 0 they are moving in the direction from A to B, then.(A) They will meet once (B) They will never meet(C) They will meet twice (D) none of above is true
(135) Velocity of particle A with respect to particle B is m4s
while they are moving in same direction.
And it is m10s
while they are in opposite direction. What are the velocities of the particles with
respect to the stationary frame of reference.(A) 7 ms–1, 3 ms–1 (B) 4 ms–1, 5 ms–1
(C) 7 ms–1, 4 ms–1 (D) 10 ms–1, 4 ms–1
(136) Stone A is thrown in horizontal direction with velocity of 10 ms–1 at the same time stone B freelyfalls vertically in downword direction. Calculate the velocity of B with respect to A after 10 second.
(A) 10 ms–1 (B) –1101 ms (C) –110 101ms (D) 0
(137) A car moves horizontally with a speed of 3 ms–1. A glass wind screen is kept on the front sideof the car. Rain drops strike the screen vertically. With the Velocity of 5 ms-1 Calculate the velocityof rain drops with respect to a ground.(A) 6 ms–1 (B) 4 ms–1 (C) 3 ms–1 (D) 1 ms–1
42
(138) A man crosses a river through shortest distance D as given in the figure. RV
is velocity of
water and mV
is velocity of man in still river water. If mRV
is relative velocity of man w.r.t.river, then find the angle made by swimming man with the shortest distance AB
A
B
D
(A) –1 R
m R
VtanV – V
(B) 2 2
–1 m
R m
VtanV – V
(C) 2 2
–1 R
m R
VtanV – V
(D) 2 2
–1 R
R m
VtanV – V
(139) A particle has initial velocity –12i + 3j ms and has acceleration –2i + j ms . Find the velocity
of the particle after 2 second.
(A) –13i + 5j ms (B) –14i + 5j ms
(C) –13i + 2 j ms (D) –15i + 4 j ms
(140) Motion of a body in x - y plane is described by 2r 2ti 5t 6t j m Then find velocity
of the body at t = 1 second.(A) –1144ms (B) –1148ms (C) –1150ms (D) –1260ms
(141) A particle moves in x - y plane. The position vector of the particle is given by 2r 3ti – 2t j m Find
the rate of change of at t = 1 second. Where is the anghe betheen direction of motion and x
(A) 1625
(B) 1225
(C) 12–25
(D) 169
(142) x and y co-ordinates of a particle moving in x-y plane at some instant are 2 23x 2t and y t2
Calculate y co-ordinate when its x coordinate is 8m.(A) 3 m (B) 6 m (C) 8 m (D) 9 m
(143) A particle in xy plane is governed by x = A cos t, y = A (1 – sin t). A and are constants.What is the speed of the particle.
(A) Aω t (B) Aω 2t (C) A cos t (D) 2 tA sin2
43
(144) A particle is moving in a xy plane with y = 2x and Vx = 2 – t. Find Vy at time t = 3 second.(A) 2 ms–1 (B) –3 ms–1 (C) +3 ms–1 (D) –2 ms–1
(145) t1 and t2 are two values of time of a projectile at the same height t1 + t2 = ...(A) Time to reach maximum height (B) fight time for the projectile
(C) 34
time of the fight time. (D) 32
time of the fight time.
(146) Eequetion of a projectile is given by y = Ax – Bx2. Find the range for the particle.
(A) AB (B)
A4B (C)
A2B (D)
2AB
(147) Angle of projection of a projectile with horizonal line is at time t = 0, After what time theangle will be again ?
(A) V cosg
(B) Vsing (C) 0V sin
2g
(D) 02V sinθg
(148) A particle is projected with initial speed of V0 and angle of . Find the horizontal displacementwhen its velocity is perpendicular to initial velocity.
(A) 2
0Vgtanθ (B)
20V
gsinθ(C) 0V sin
g
(D) 2
0Vtan
(149) Intial anlge of a projectile is and its initial velocity is V0. Find the angle of velocity with horizontalline at time t.
(A) –1
0
gsin 1 – tV cos
(B) –1
0
gtan 1 – tV cos
(C)
–1
0
gtan tanθ – tV cosθ
(D) –1
0
gsin tanθ – tV cosθ
(150) A stone is projected with an angle θ and velocity VV0 from point P. It strikes the ground at pointQ. If the both P and Q are on same horizontal line, then find average velocity.
(A) 0V cosθ (B) 0V sinθ (C)
0θV cos2
(D) 0
θV sin2
(151) An object is projected with initial velocity of 100 ms–1 and angle of 60. Find the verticle velocitywhen its horizontal displacement is 500 m. (g = 10 ms–1)(A) 93.35 ms–1 (B) –93.35 ms–1 (C) –8.65 ms–1 (D) 98 ms–1
(152) Angle of projection of a projectile is changed, keeping initial velocity constant. Find the rate ofchange of maximum height. Range of the projectile is R.
(A) R4 (B)
R3 (C)
R2 (D) R
(153) A cartasian equation of a projectile is given by y = 2x – 5x2. Calculate its initial velocity.(A) –110ms (B) –15ms (C) –12ms (D) 4 ms–1
44
(154) A body travelling in a circle at constant speed.(A) has a constant velocity (B) is not accelerated(C) has an outward radial acceleration (D) has an inward radial acceleration
(155) The speed of a particle moving in a circle of radius r = 4 meter is 10 ms–1. What is its radial acceleration ?(A) 25 ms–2 (B) 20 ms–2 (C) 10 ms–2 (D) 15 ms–2
(156) If f is the frequency of a body moving in a circular path with constant speed. a is its centrifugal accelera-tion, so.
(A) a f (B) 2a α f (C) 3a α f (D) 1a αf
• Following question is Assertion - Reason type question. choose(A) If both Assertion - Reason are true, reason is correct explanation of Asserton.(B) If both Assertion - Reason are true but reason is not correct explanation of Asserton. (C) IfAsserton is true but Reason is false.(D) If Reason is true but Asserton is false.
(157) Asserton :At the highest point of projectile motion the velocity is not zero.Reason : Only the verticle component of velocity is zero. Where as horizontal component still exists.(A) a (B) b (C) c (D) dMatch coloum type question.
(158) A balloon rise up with constant net acceleration of 10 ms–2. After 2 second a particle drops from theballoon. After further 2 s match the following (take g = 10 ms–2)Table - 1 Table 2(A) Hight of the particle from ground (P) zero(B) Speed of particle (Q) 10 SI unit(C) Displacement of particle (R) 40 SI unit(D) Accelereation of the particle (S) 20 SI unit(A) A - P, B - Q, C - R, D - S(B) A - Q, B - R, C - S, D - P(C) A - R, B - P, C - S, D - Q(D) A - R, B - S, C - P, D - QComprehensions type questions.A particle is moving in a circle of radius R with constant speed. The time period of the particle is T Now
after time t = T6
.
(159) Average speed of the particle is
(A) R6T (B) 2 R
3T (C) 2 R
T (D) R
T(160) Average velocity of the particle is
(A) 3RT (B)
6RT (C)
2RT (D)
4RT
(161) Range of a projectile is R and maximum height is H. Find the area covered by the path of the projectileand horizontal line.
(A) 2 RH3 (B)
5 RH3 (C)
3 RH5 (D)
6 RH5
45
KEY NOTE1 A 32 B 63 B 94 A 125 C2 C 33 A 64 A 95 C 126 B3 A 34 C 65 B 96 C 127 B4 D 35 D 66 B 97 C 128 C5 B 36 C 67 C 98 D 129 B6 A 37 C 68 B 99 A 130 B7 A 38 A 69 C 100 B 131 C8 D 39 B 70 D 101 C 132 B9 C 40 B 71 B 102 B 133 C
10 C 41 A 72 B 103 D 134 A11 C 42 B 73 C 104 B 135 A12 A 43 C 74 C 105 C 136 C13 C 44 C 75 B 106 C 137 B14 D 45 D 76 B 107 A 138 C15 B 46 B 77 D 108 D 139 B16 B 47 C 78 B 109 D 140 D17 B 48 A 79 C 110 B 141 C18 C 49 B 80 B 111 A 142 B19 D 50 C 81 A 112 B 143 C20 A 51 D 82 B 113 C 144 D21 C 52 C 83 C 114 B 145 B22 B 53 A 84 B 115 B 146 A23 A 54 D 85 A 116 C 147 D24 B 55 D 86 C 117 A 148 A25 C 56 A 87 D 118 C 149 C26 C 57 A 88 B 119 B 150 A27 A 58 B 89 A 120 C 151 B28 C 59 B 90 B 121 B 152 C29 B 60 B 91 C 122 D 153 B30 B 61 B 92 B 123 B 154 D31 A 62 A 93 A 124 C 155 A
156 B157 A158 C159 C160 B161 A
46
• • •
HINT(9) Pathlenth is always greater or equal to displacement
(18) 1 2
1 2
2V VAverage speed =V + V
(23)2 2
0 0V 2Vh = speed =
2g 2g × 6
(24)dx dvV = and a =dt dt
(25)change in velocityFind
timechange in velocity is the area covered by the graph.
(30) 2 22 1
1 1h – h = g (t +1) – gt = 452 2
(31)dsV = = 2 + 10tdt
(32,33) 20
1h = V t – gt2
20
1 gt – V t + h = 02
2 02V 2ht + + = 0g g
There are two real values of t.
(35) slope is 0
0
Vx
(37) Aera covered by a t graph gives change in velocity..(42) acceleration = slope = tanθ
(44)da = constantdt
da dv = kdv dt
da k=dv a
ada = kdv 2a = kv
2
47
(45)2
2
d xa =dt
(48) If maximum velocity is V.V = a1t1 and V = a2t2
1 21 2
v vT = t + t = +a a
1 2
1 2
a a TV =a + a
(59) 2 21 2
1 1h = gt and (h + 3h) = gt2 2
(61)
Q
R
R
S
S 2 2 2R R 2R cosθ
2 2 2= 2R -2R Cosθ 2R (1-Cosθ)θ = 2Rsin2
s
(65)2
0Vstopping distance ds =2g
(66) V = V0 + at
20
1d = V t + at2
(69) 2 21H = gt H α t2
(70) At time t velocity V = gt
(70) 20
1y = V t + gt2
2t 1 ty = gt × – g2 2 2
(73) V1 + V2 = 5V1 – V2 = 1 So 2V1 = 6
V1 = 3m/s
(74)12V = (A + y)
48
dv dv dya =dt dy dt
(75)2vs =
2a(81) V = Ax + B
a = Ava α V
(82)2
2
dx dx= c and = 0dt dt
y = Ax3 + B
2dy dx= 3Axdt dt
2
2
d y dx dxand = 6Ax ×dt dt dt
22
2
d y = 6Axcdt
2 2^ ^
2 2
d x d yNow a = i + jat dt
(95) C A B
2 2 2C = A + B + 2AB cosθ
(116) A – B = ABcosθ
A B = ABsinθ
(127)^
A + B + C = j
(130) 24 4+ + p = 19 9
(131) Velocity of A wrt ground^
AGV = 10 l
Velocity of B w.r.t. ground^
BGV = –gt j
BA BG GAV = V V
49
(138) BωV
MωVMBV
MW WB MBV + V = V
MWV = Velocity of man w.r.t. water
WBV = Velocity of water w.r.t. Bank
MBV = Velocity of man w.r.t. Bank
WB
MB
Vtanθ =V
(141) 2r = 3ti – 2t j
drV = = 3i – 4t jdt
–4tanθ = t3
(142) x = 2t2 = 8t = 2s
2 23 3y = t = (2) = 6m2 2
(143) x = Acoswt y = A (i – sinwt)
dxVα = = –Aωsinwtdv Vy = –Aω coswt
2 2V = Vx + Vy
(145)
o
A B
C
t1 = t (0A) = t (BC)t2 = t (0B)t1 + t2 = t (0A) + t (0B)t1 + t2 = t (BC) + t (0B)
50
(146) y = Ax – Bx2
dy = A – 2Bxdx
At maximum height A – 2Bx = 0
Ax =2B
(147)
Q
(148) 0 0 0V = V cosθi + V sinθ j
0 0V = V cosθi + V smθ – gt j
0V V = 0
0Vt = Now find xgsinθ
(149) At time t V x = V0 = 0V cosθ
yV = Visinθ – gt
0
0
V sinθ – gtVytanα = =Vx V cosθ
(150) Range 2
0V sin2θR =g
0f
2V sinθFlight time t =g
f
RAverage velocity =t
(156) Cenrifugal acceleration22
2 2r
v 2πr 1a = = = 4π rfr T r
(161)R
0
A = ydx2
0
gxtake y = xtanθ –2V xcosθ
51
Unit - 3Laws of Motion
52
SUMMARY
Important Points The law of inertia given by Galileo was represented by Newton as the first law of motion :" If no external
force acts on a body, the body at rest remains at rest and a body in motion continues to move with thesame velocity."
This law gives us definition of force.
The momentum of a body
vmp is a vector quantity.It gives more information than the velocity.It's
unit is 1kgms or Ns and dimensional formula M1L1T-1.
Newton's second law of motion : The time-rate of change in momentum of a body is equal to theresultant external force applied on the body and is in the direction of the exterrnal force.
F =dPdt
ma
is the vector relationship
The SI unit of force is newton (=N) . 1 N = 1 kg ms–2 . This law gives the value of force.It is consistent
with the first law.(
F =0 indicates that a
=0). In this equation the acceleration of the body a
is that
which it has when the force is acting on it.(Not of the Past!) .
F is only the resultant external force.
The impulse of force is the product of force and the time for which it acts. when a large force acts for
a very small time, it is difficult to measure
F and t but the change in momentum can be measured,
which is equal to the impulse of force (
F t )
Newton's third law of motion: " To every action there is always an equal and opposite reaction."
Forces always act in pairs, and,
FAB= –
FBA. The action and the reaction act simultaneously.They act on
different bodies,hence they cannot be cancelled by adding. But the resultant of the forces betweendifferent parts of the same body becomes zero.
The law of conservation of momentum is obtained from Newton's second law and the third law.It iswritten as-"The total momentum of an isolated system remains constant."
The concurrent forces are those forces of which the lines of action pass through the same point. For
equilibrium of the body, under the effect of such forces,
F must be = 0 Moreover, the sum of thecorresponding components also should be zero. ( F
x = 0, F
y=o, F
z = 0)
- Friction is produced due to the contact force between the surfaces in contact.It opposes the impending orthe real relative motion.
Static frictional force fs f
s (max) =
s N and the kinetic friction is f
k =
k N
53
Sµ coefficient of static friction
kµ coefficient of kinetic friction and k s
* The reference frame , in which Newton's first law of motion is obeyed is called the inertial frame ofreference and the one in which it is not obeyed is called non-inertial frame of reference. The frame ofreference with constant velocity is an inertial frame of reference and one which has acceleration is non-inertial frame of reference.
* On a body performing uniform circular motion a force equal to mv2 / r acts towards the centre of thecircular path. This is called the centripetal force.
The maximum safe speed on level curved road is vmax
= rgµs
The maximum safe speed on a banked curved road is vmax
=
tanθµ1tanθµrg
s
s
* Motion of a body on a friction less inclined plane:
As shown in figure body of mass m is placed on a smooth inclined plane making an angle θ with thehorizontal.
Here, Ncosmg1 †ÃÜå masinmg2
acceleration of body θ = gsinθ .
* Motion of a body on a Rough inclined plane:(1) Body moving down : If body moving down
with acceleration a then,
mg sinθ – maNµs and N = mg cosθ
macosθmgµsinθmg s
a cosθµsinθg s
(2) Body moves up : If body is moving upwards with acceleration a then sg sin θ μ cos θ .a
Pseudo Force : In non-inertial frame of reference due to acceleration one more additional force acting ona body in the opposite direction of acceleration of frame of reference is called pseudo force (F
P) .
when a man of weight m climbs on the rope with acceleration a then tension in the rope is T = m(g + a). When man sliding down with acceleration a then tension in the rope is T = m(g – a).
54
* When masses in contact:In figure masses m1 ,m2 and m3
are placed in contact on a surface.
acceleration1 2 3
Fam m m
Force on m1 : F1 = F
Force on m2 : F2 = (m2 + m3) 2 3
1 2 3
m m Fa
m m m
Force on m3 :
3
3 31 2 3
m FF m am m m
* Masses connected by strings:
Consider two masses 1m and 2m placed on a frictionless horizontal surface connected by a light
inextensibule string.
If 1m is pulled by a force F then a tension T is developed,
in the string.
For 1m : For1 2 2F – T = m a , m : T = m a
Tension in string 2
1 2
,m FTm m
acceleration
1 2
.Fam m
a
2m 1mT T F
F a1m
2m 3m
55
MCQFor the answer of the following questions choose the correct alternative from among the given ones.
1. The velocity of a body of mass 20 kg decreases from 20 ms–1 to 5 ms–1 in a distance of 100 m. Force
on the body is
(A) -27.5 N (B) -47.5 N (C) -37.5 N (D) -67.5 N
2. A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves
0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the
force. (Consider g = 10 ms–2)
(A) 16 N (B) 20 N (C) 22 N (D) 4 N
3. Formula for true force is
(A) F ma (C) d mv
Fdt
(B) dvF mdt
(D) 2
2
d xF = mdt
4. A particle moves in the X–Y plane under the influence of a force such that its linear momentum is^ ^
P (t) A[ i cos (kt) j sin(kt)]
where A and k are constants. The angle between the force and
momentum is
(A) 00 ` (B) 300 (C) 450 (D) 900
5. Force of 5 N acts on a body of weight 9.8 N. what is the acceleration produced in ms-2
(A)49.00 (B) 5.00 (C) 1.46 (D) 0.51
6. A lift is going up. The total mass of the lift and the passenger is 1000 kg The variation in the speed of
the lift is as given in the graph. The tension in the rope pulling the lift at t= 10.5 sec will be
(A) 8000 N
(B) Zero
(C) 12000 N
(D) 17400 N
7. Same force acts on two bodies of different masses 2 kg and 4 kg initially at rest. The ratio of timesrequired to acquire same final velocity is
(A) 2:1 (B) 1:2 (C) 1:1 (D) 4:16
8. Which of the following quantities measured from different inertial reference frames are same
(A) Force (B) Velocity (C) Displacement (D) Kinetic Energy
2 10 12
3.6
1
t (sec)
56
9. 10,000 small balls, each weighing 1 g strike one square cm of area per second with a velocity100 ms–1 in a normal direction and rebound with the same velocity. The value of pressure on the surfacewill be
(A) 23 Nm102 (B) 25 Nm102
(C) 27 Nm10 (D) 27 Nm102
10. When the speed of a moving body is doubled
(A) Its acceleration is doubled
(B) Its momentum is doubled
(C) Its kinetic energy is doubled
(D) Its potential energy is doubled
11. A particle moves in the XY Plane under the action of a force F such that the components of its linearmomentum P at any time t are Px = 2 cost, Py = 2 sint. The angle between F and P at time t is
(A) 900 (B) 00 (C) 1800 (D) 300
12. A player caught a cricket ball of mass 150 g moving at the rate of 20 ms-1 . If the catching process becompleted in 0.1 s the force of the blow exerted by the ball on the hands of player is
(A) 0.3 N (B) 30 N C) 300 N (D) 3000 N
13. A body of mass 5 kg starts from the origin with an initial velocity ^ ^
-1u=30 i +40 j ms
. If a constant
Force –^ ^
F= i +5 j N
acts on the body, the time in which the y-component of the velocity becomes
zero is
(A) 5 s (B) 20 s (C) 40 s (D) 80 s
Third law of motion
14. Swimming is possible on account of
(A) First law of motion (B) second law of motion
(C) Third law of motion (D) Newton's law of gravitation
15. A cold soft drink is kept on the balance.When the cap is open, then the weight
(A) Increases (B) Decreases (C) First increase then decreases (D) Remains same
57
Conservation of linear momentum and impulse
16. A wagon weighing 1000 kg is moving with a velocity 50 km h-1 on smooth horizontal rails. A massof 250 kg is dropped into it. The velocity with which it moves now is
(A) 1hkm5.2 (B) 1hkm20
(C) 1hkm40 (D) 1hkm50
17. The Figure shows the Position-time (x-t) graph of one dimensional motion of a body of mass 0.4 kg .The magnitude of each impluse is
(A) 0.2 Ns (B) 0.4 Ns
(C) 0.8 Ns (D) 1.6 Ns
Equllibrium of Forces
18. Three Forces F1, F2 and F3 together keep a body in equilibrium. If F1
= 3 N along the positive X- axis,F2
= 4N along the positive Y-axis ,then the third force F3 is
(A) 5 N -making an angle θ = tan–1 43 with negative y-axis
(B) 5 N - making an angle θ = tan–1 34 with negative y-axis
(C) 7 N - making an angle θ = tan–1 43 with negative y-axis
(D) 7 N - making an angle θ = tan–1 34 with negative y-axis
19. A solid sphere of mass 2 kg is resting inside a cube as shown in the figure. The cube is moving with
a velocity –
^ ^1v = 5t i+ 2t j ms
Here t is the time in second. All surfaces are smooth. The sphere
is at rest with respect to the cube. what is the total force exerted by the sphere on the cube
(g = 10 ms–2)
(A) N29 (B) 29 N
(C) 26 N (D) N89
y
0 x
A B
D C
x (m)
0 2 4 6 8 10 12 14 16t (s)
58
20. A particle of mass 2 kg is initially at rest. A force acts on it whose magnitude changes with time. Theforce time graph is shown below.
The velocity of the particle after 10s is
(A) 10 ms-1 (B) 20 ms-1
(C) 75 ms-1 (D) 50 ms-1
21. A block of mass 4 kg is placed on a rough horizontal plane. A time dependent force F = Kt2 acts on
a block , where k = 2 2sN , co-efficient of friction 0.8µ .Force of friction between the block and the
plane at t = 2 S is.....
(A) 32 N (B) 4 N (C) 2 N (D) 8 N
22. A 7 kg object is subjected to two forces (in newton)^ ^
1F =20i+30 j
and ^ ^
2F = 8i-5 j
The magnitude of
resulting acceleration in ms–2 will be
(A) 5 (B) 4 (C) 3 (D) 223. A car travelling at a speed of 30 km/h is brought to a halt in 8 metres by applying brakes. If the same
car is travelling at 60 km/h it can be brought to a halt with the same breaking power in
(A) 8 m (B) 16 m (C) 24 m (D) 32 m
24. A given object takes n times more time to slide down 450 rough inclined plane as it takes to slide downa perfactly smooth 450 incline. The coefficient of kinetic friction between the object and the incline is
(A) 2
12 n (B) 2
11 n (C) 211 n (D) 2
11 - n
25. Two bodies of equal masses revolve in circular orbits of radii R1 and R2 with the same period Theircentripetal forces are in the ratio.
(A) 2
1
2
RR
(B)
2
1R
R(C)
2
2
1R
R
(D) 1 2R R
26. Two masses M and 2M are joined together by means of light inextensible string passed over a
frictionless pulley as shown in fig. When the bigger mass is released, the small one will ascend with anacceleration
(A) 3g (B)
32g
(C) g (D) 2g
0
10
20
F (N)
t (s)
2M
M
59
27. A 0.5 kg ball moving with a speed of 12 ms-1 strikes a hard wall at an angle of 300 with the wall.It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for0.025 S the average force acting on the wall is
(A) 96 N (B) 48 N (C) 24 N (D) 12 N
28 A shell of mass 200g is ejected from a gun of mass 4 kg by an explosion that generates 1.05 KJ ofenergy. The initial velocity of the shell is
(A) 100 ms-1 (B) 80 ms-1 (C) 40 ms-1 (D) 120 ms-1
29. A gramophone record is revolving with an anguler velocity A coin is placed at a distance r from thecentre of the record. The coefficient of static friction is µ . The coin will revolve with the record if
(A) r = µ g2 (B) 2
r µg (C) 2
µgr
(D) 2
µgr
30. A stone of mass 2 k g is tied to a string of length 0.5 m It the breaking tension of the string is 900N,then the maximum angular velocity the stone can have in uniform circular motion is
(A) 30 rad/s (B) 20 rad/s (C) 10 rad/s (D) 25 rad/s
31. A body of mass 6 kg is hanging from another body of mass 10 kg as shown in fig. This conbination isbeing pulled up by a string with an acceleration of 2 ms–2. the tension T1
is (g = 10 ms–2 )
(A) 240 N
(B) 150 N
(C) 220 N
(D) 192 N
32. A sparrow flying in air sits on a stretched telegraph wire. If the weight of the sparrow is W ,which of thefollowing is true about the tension T produced in the wire?
(A) T = W (B) T < W (C) T = 0 (D) T >W
33. Fig. shows the displacement of a particle going along X-axis as a function of time. The force acting onthe particle is zero in the region
(A) AB
(B) BC
(C) CD
(D) None of these
T1
10 kg
6 kg
aT2
XAD
ispla
cem
ent B C
D
Time
Y
60
34. A Force (F) varies with time (t) as shown in fig. Average force over a complete cycle is-
(A) Zero (B) 2F0
(C) 0F (D) 0F2
35. A body of mass 0.05 kg is falling with acceleration 9.4 ms–2 . The force exerted by air opposite tomotion is N (g=9.8 ms–2)
(A) 0.02 (B) 0.20 (C) 0.030 (D) Zero
36. The average force necessary to stop a hammer with 25 NS momentun in 0.04 sec is ___________N
(A) 625 (B) 125 (C) 50 (D) 2537. Newton's third law of motion leads to the law of consrevation of
(A) Angular momentum (B) Energy (C) mass (D) momentum
38. A ball falls on surface from 10 m height and rebounds to 2.5 m. If duration of contact with floor is0.01 sec. then average aceleration during contact is _______________ms-2
(A) 2100 (B) 1400 (C) 700 (D) 400
39. A vehicle of 100 kg is moving with a velocity of 5 sm . To stop it in 10
1 sec, the required force
in opposite direction is _______________N
(A) 50 (B) 500 (C) 5000 (D) 1000
40. The linear momentum P of a particle varies with the time as follows. 2btaP Where a and b a reconstants. The net force acting on the particle is _____________
(A) Proportional to t (B) Proportional to t2 (C) Zero (D) constant
41. A vessel containing water is given a constant acceleration a towards the right, along a straight horizontalpath. which of the following diagram represents the surface of the liquid ?
(A) (B) (C) (D)
42. A body of 2 kg has an initial speed 5 m/s. A force act on it for some time in the directine of motion.
The force ( )F
---------time (t) graph is shown in figure. The final speed of the body is _________
(A) 9.25 ms-1 (B) 5 ms-1
(C) 14.25 ms-1 (D) 4.25 ms-1
2 4 4.5 6.5
F (N)
4
2.5
t (s)
F
t0
F0
a a a a
61
43. which of the following statement is correct?
(A) A body has a constant velocily but a varying speed.
(B) A body has a constant speed but a varying value of acceleration.
(C) A body has a constant speed and zero accelaration.
(D) A body has a constant speed but velocity is zero.
44. A force of 8 N acts on an object of mass 5kg in X- direction and another force of 6 N acts on it in Y -direction. Hence, the magnitude of acceleration of object will be
(A) 1.5 ms–2 (B) 2.0 ms–2 (C) 2.5 ms–2 (D) 3.5 ms–2
45. Three forces are acting simultaneously on a particle
moving with velocity
V .These forces are represented
in magnitude and direction by the three sides of atriangle ABC. The particle will now move with velocily__________
(A) Less than v
(B) greater than v
(C) v
in the direction of the largest force BC (D) v
remaining unchanged.
46. A plate of mass M is placed on a horizontal frictionless surface and a body of mass m is placed on thisplate, The coefficient of dynamic friction between this body and the plate is . If a force 2mg. isapplied to the body of mass m along the horizonal direction the acceleration of the plate will be________________
(A) gMµm
(B) gmMµm
(C) gM
m2(D) gmM
m2
47. On the horizontal surface of a truck 6.0µ , a block of mass 1 kg is placed. If the truck is
accelerating at the rale of 2s/m5 then frictional force on the block will be_____________N
(A) 5 (B) 6 (C) 5.88 (D) 848. Two blocks of nass 8 kg and 4 kg are connected a heavy string Placed on rough horizontal Plane, The
4 kg block is Pulled with a constant force F.The co-efficient of friction between the blocks and theground is 0.5, what is the value of F, So that the tension in the spring is constant throughout duringthe motion of the blocks ? (g=10 ms–2)
(A) 40 N (B) 60 N
(C) 50 N (D) 30 N
C
A B
M 2mg
m
8 kg 4 kg F
62
49. Seven blocks, each of mass 1 kg are arranged one above the other as
shown in figure. what are the values of the contact forces exerted onthe third block by the forth and the second block respectively ?
( g = 10 ms-2 )
(A) 40 N, 50 N (B) 50 N, 40 N
(C) 40 N, 20 N (D) 50 N, 30 N
50. A man is standing on a spring balance. Reading of spring balance is 60 kgf. If man jumps outside
balance, then reading of spring balance____________
(A) First increase than decreases to zero (B) Decreses
(C) Increases (D) Remains same
51. A car turns a corner on a slippery road at a constant speed of 10 m/s. If the coefficient of friction is
0.5, the minimum radius of the arc at which the car turns is_____________meter.
(A) 20 (B) 10 (C) 5 (D) 4
52. A person standing on the floor of a lift drops a coin. The coin reaches the floor of the lift in time to if
the lift is stationary and the time t2 if it is accelerated in upward direction. Than
(A) 21 tt (C) 21 tt
(B) t1 > t2 (D) Cannot say anything
53. A lift of mass 1000 kg is moving with an acceleration of 1 ms–2 in upward direction Tension developedin the rope of lift is _________________N (g = 9.8 ms-2 )
(A) 9800 (B) 10,000 (C) 10,800 (D) 11,000
54. Three blocks of masses m1,m2 and m3
are connected by massless strings as shown in figure, on a
frictionless table. They are Pulled with a force N40T3 . If 1 10m kg , kg6m2 and
kg4m3 the tension T2 will be =________________N
(A) 20 (B) 40
(C) 10 (D) 32
55. If the surfaccs shown in figure are frictionless, the ratio of 1T and 2T is __________
(A) 2:3 (B) 3:1
(C) 5:1 (D) 1:5
76
4
2
m1 m2 m3
T1 T2 T3= 40 N
3kg 15kgT2 T1
12kg 300
F
63
56. Three masses 1 kg, 6 kg and 3 kg are connected to eachother with threads and are placed on a table as shown infigure. The alcceleration with which the system is movingis ______ ms-2 (g=10ms-2)
(A) zero (B) 1(C) 2 (D) 3
57. A rope which can withstand a maximum tension of 400 N hangs from a tree. If a monkey of mass 30 kgclimbs on the rope in which of the following cases-will the rope break?
(take g =10 ms-2 and neglect the mass of rope)
(A) When the monkey climbs with constant speed of 5 ms–1
(B) When the monkey climbs with constant acceleration of 2 ms–2
(C) When the monkey climbs with constant acceleration of 5 ms–2
(D) When the monkey climbs with the constant speed of 12 ms–1
58. An object of mass 3 kg is moving with a velocity of 5 m/s along a straight path. If a force of 12 N isapplied for 3 sec on the object in a perpendicuiar to its direction of motion.The magnitude of velocityof the particle at the end of 3 sec is____________m/s.
(A) 5 (B) 12 (C) 13 (D) 459. Same forces act on two bodies of different mass 2 kg and 5 kg initialy at rest. The ratio of times
required to acquire same final velocity is ____________
(A) 5:3 (B) 25:4 (C) 4:25 (D) 2:5
60. A body of mass 5 kg starts motion form the origine with an initial veiocily ^ ^
0 =30 i+40 j m/s
If a
constant force ^ ^
F = – ( i +5 j)N
acts on the body, than the time in which the Y-component of the
velocity becomes zero is __________________
(A) 5 s (B) 20 s (C) 40 s (D) 80 s
61. A Block of mass 300 kg is set into motion on a frictionlesshorizontal surface with the help of frictionless pulley anda rope system as shown in figure. What horizontal forceF should be applied to produce in the block an aecelerationof 1 ms-2 ?(A) 150 N (B) 100 N
(C) 300 N (D) 50 N
6 kg T2T1
T1 T2
1 kg 3 kg
F a=1 ms-2
64
62. A body of mass m rests on horizontal surface. The coefficient of friction between the body and thesurface is µ . If the body is Pulled by a force P as shown in figure, the limiting friction between bodyand surface will be__________
A) µmg (B)
2Pmgµ
(C)
2Pmgµ (D)
2P3mg
63. Three blocks A , B, and C of equal mass m are placed one overthe other, one on a smooth horizontal ground as shown infigure. Coefficient of friction between any two blocks of A, Band C is 0.5. What would be the maximum value of mass ofblock D so that the blocks A, B and C move without slippingover each other ?
(A) 3 m (B) 5 m
(C) 6 m (D) 4 m
64. A train is moving along a horizontal track. A pendulum suspended from the roof makes an angle of 40
with the vertical, The acceleration of the train is ___________ms-2 (g = 10 ms-2 )
(A) 0.6 (B) 0.7 (C) 0.5 (D) 0.2
65. A bag of sand of mass m is suspended by rope. a bullet of mass 30m
is fired at it with a velocity υ
and gets emmbedded into it. The velocity of the bag finally is _________
(A) 3031υ
(B) 3130υ
(C) 31υ
(D) 30υ
66. Three blocks having equal mass of 2 kg are hanging on a string passing over apulley as shown in figure. what will be the tension produced in a string connectingthe blocks B and C
(A) zero (B) 13.1 N
(C) 3.3 N (D) 19.6 N
67. A partly hanging uniform chain of length L is resting on a rough horizontal table. is the maximumpossible length that can hang in equilibrium The coefficient of friction between the chain and table is____________
(A) L
L+l
l (B) Ll (C) L
l(D) L
ll
T2
T1
C
BA
ABC
D
65
68. As shown in figure,the block of 2 kg at one end and the other of 3 kg at the other end of a light stringare connected. It the system remains stationary find the magnitude and direction of the frictional force(g = 10 ms-2 )
(A) 20 N, downward on slope (B) 20 N, upward on slope
(C) 10 N Downward on slope (D) 10 N upward on slope
69. A particle is resting over a smooth horizontal floor, At t = 0, a horizontal force starts acting on it.Magnitude of the force increses with time according to law F = t , where = is constant Match thecolumn after seeing the figure.
Column-1 Column-2(a) curve (i) shows. (p) velocity against time
(b) curve (ii) shows (q) velocity against acceleration
(r) acceleration against time
(A) (i)-p (ii)-q (B) (i)-q, (ii)-r (C) (i)-r, (ii)-p (D) (i)-q, (ii)-p
70. A car of mass 1000 kg travelling at 32 m/s clashes into a rear of a truck of mass 8000 kg moving in thesame direction with a velocity of 4 m/s. After the collision the car bounces with a velocity of8 ms–1. The velocity of truck after the impact is ____________m/s
(A) 8 (B) 4 (C) 6 (D) 971. A Block of mass m = 2 kg is resting on a rough inclined plane of inclination 300 as shown in fignre.
The coefficient of friction between the block and the plane is µ =0.5 . What minimum force F shuldbe applied perpendicular to the plane of block so that block does not slip on the plane ? (g=10ms–2)
(A) zero (B) 6.24N(C) 2.68 N (D) 4.3 N
72. The upper half of an inclined plane of inclination θ is perfectly smooth while the lower half is rough Abody starting from the rest at top come back to rest at the bottom, then the coefficient of friction for thelower half is given by____________
(A) sμ sinθ (B) sμ cot θ (C) sμ 2cosθ (D) sμ 2 tan θ
73. A Block of mass m = 4 kg is placed over a rough inclined plane as shown in figure, The coefficient of
friction between the block and plane is sµ = 0.6. A force F = 10 N is applied on the block of an
angle at 300. The contact force between the block and the plane is ___________
(A) 27.15 N (B) 16.32 N
(C) 10.65 N (D) 32.16 N
2 kg
3 kg
300
y
x
(ii)
(I)
450
300F
300
N = F + mg cos 300
66
74. The motion of a particle of a mass m is describe by 2gt21uty . Find the force acting on the
particle.
(A) F = ma (B) F = mg (C) F = 0 (D) None of these75. A balloon has a mass of 10 g in air, The air escapes from the balloon at a uniform rate with a velocity
of 5 scm and the balloon shrinks completely in 2.5 sec. calculate the average force acting on the
balloon.
(A) 20 dyne (B) 5 dyne (C) 0 dyne (D) 10 dyne
76. Two bodies A and B each of mass m are fixed together by a massless spring A force F acts on the massB as shown in figure. At the instant shown, a body A has an acceleration a. what is the accelaration ofB?
(A)
a
mF
(B) F–T
(C)
mFa (D) a
77. With what acceleration (a) should a box descend so that a block of mass M placed in it exerts a force
4Mg
on the floor of the box?
(A) 3g4
(B) 4g3
(C) 4g (D) 3g
78. A mass of 6 kg is suspended by a rope of length 2 m form the ceiling.A force of 50 N in the horizontal dircction is applied at the mid PointP of the rope- as shown in figure. what is the angle the rope makeswith the vertical in equilibrium ? (g = 10 ms-2) Neglect mass of therope.
(A) 400 (B) 300
(C) 350 (D) 450
79. The minimum force required to start pushing a body up a rough (cofficient of µ ) inclined plane is F1.While the minimum force needed to prevent it from sliding down is F2. If the inclined plane makes an
angle from the horizontal. such that tan 2 than the ratio 1
2
FF is
(A) 4 (B) 1 (C) 2 (D) 3
m m F
A B
T1
P
1m
w60 N
T2
50 N
67
80. When forces F1, F2, F3 are acting on a particle of mass m such that F2 and F3 are mutually perpendicular,then the particle remains stationary. If the force F1 is now removed than the acceleration of the
particle is
(A) 0Fbm (B) 1 2FF
m (C) mF-F 32 (D) 2F
m
Asseration and reason type question
Asseration and reason are given in following question. Each question have four options. One of them iscorrect select it.
(a) Asseration is true. Reason is true and reason is correct explanatin for Assertion.
(b) Asseration is true. Reason is ture but reason is not the correct explanatin of assertion.
(c) Asseration is ture. Reason is false.
(d) Asseration is false. Reason is true.
81. Asseration : Frictional forces are conservative forces.
Reason : Potential energy can be associated with frictional forces.
(A) a (B) b (C) c (D) d
82. Asseration : A body of mass 1 kg is moving with an accelaration of 1ms-1. The rate of change of its
momentum is 1 N.
Reason : The rate of change of momentum of body = force applied on the body.
(A) a (B) b (C) c (D) d
83. Asseration : A body falling freely under gravity becomes weightless.
Reason : R = m(g – a) = m(g – g) = 0(A) a (B) b (C) c (D) d
84. Asseration : It is difficult to move bike with its breaks on.
Reason : Rolling friction is converted into sliding friction, which is comparatively larger.
(A) a (B) b (C) c (D) d
Comprehension :-
According to newton's second low of mation, F= ma, where F is the force required to produce anaccelaration a in a body of mass m. If a = 0 than F = 0. If a force acts on a body for t seconds, the effect ofthe force is given by impulse = F t = change in linear momentum of the body.
68
with the help of the passage given above, choose the most appropriate alternative for each of thefollowing questions.
85 A cricket ball of mass 150 g. is moving with a velocity of 12 m/s and is hit by a bat so that the ballis turned back with a velocity of 20 m/s. If duration of contact between the ball and the bat is0.01 sec. The impulse of the force is
(A) 7.4 NS (B) 4.8 NS (C) 1.2 NS (D) 4.7 NS
86. Average force exerted by the bat is
(A) 480 N (B) 120 N (C) 1200 N (D) 840 N
87. The force acting on a body whose linear momentum changes by 20 kgms-1 in 10 sec is
(A) 2 N (B) 20 N (C) 200 N (D) 0.2 N
88. An impulsive force of 100 N acts on a body for 1 sec What is the change in its linear momentum ?
(A) 10 N-S (B) 100 N-S (C) 1000 N-S (D) 1 N-S
89. Match the column
Column - I Column - II
(a) Body lying on a horizontal surface (p) is a self adjusting force
(b) Static friction (q) is a maximum value of static friction
(c) Limiting friction (r) is less than limiting friction
(d) Dynamic friction (s) force of friction = 0(A) a-s, b-p, c-q, d-r (B) a-p, b-q, c-r, d-s
(C) a-s, b-r, c-q, d-p (D) a-r, b-q, c-p, d-s
90. A block B, placed on a horizontal surface is pulled with initial velocity V. If the coefficient of kineticfriction between surface and block is , than after how much time, block will come to rest ?
(A) gµv
(B) gµv (C)
gv (D)
vg
91. As shown in figure a block of mass m is attached with a cart. If the coefficient of static friction betweenthe surface of cart and block is than what would be acceleration of cart to prevent the falling ofblock ?
(A) mgαµ
(B) µmgα
(C) µgα (D)
gαµ
69
92. A particle of mass m is at rest at t = 0. The force exerting on it in x-direction is F(t) = Foe-bt . Which one
of the following graph is of speed V t t
(A) (B)
(C) (D)
KEY NOTE1 C 31 D 61 A 91 C2 C 32 D 62 C 92 B3 C 33 A 63 A4 D 34 A 64 B5 B 35 A 65 C6 A 36 A 66 B7 B 37 D 67 D8 A 38 A 68 A9 D 39 C 69 C10 B 40 A 70 D11 A 41 C 71 C12 B 42 C 72 D13 C 43 C 73 A14 C 44 B 74 B15 C 45 D 75 A16 C 46 A 76 A17 C 47 A 77 B18 C 48 B 78 A19 C 49 B 79 D20 D 50 A 80 A21 D 51 A 81 D22 A 52 C 82 A23 D 53 C 83 A24 B 54 D 84 A25 B 55 D 85 B26 A 56 C 86 A27 C 57 C 87 A28 A 58 C 88 B29 C 59 D 89 A30 A 60 C 90 A
0F bm
vt
t
0F bm
vt
t
0F bm
vt
t
0F bm
vt
t
70
HINT
1. F = ma = m 2 2
2Sv u
2. g2uH
2
max maxHg2u
This velocity is supplied to the ball by hand and initially the hand was at rest. It acquires this velocityin distance of 0.2 meter.
2
2uaS
So upword force F = m (g+a)
3. According to newton's second law force = rate of change of linear momentum.
4. ^ ^
P t A i cos kt j sin kt
)t(PdtdF
= Ak ^ ^i sin kt j cos kt
2F.P A k cos kt.sin kt sin kt.cos kt
0P.F
5. As weight W = mg = 9.8 N
m = 1 Kg
Fa m
6. At 10.5 sec the lift is moving upword with acceleration
2ms8.12
6.30a
Tension in rope T = m (g-a)
= 1000 (9.8 - 1.8)
= 8000 N
71
7. a1t
aVt (V is constant)
21
42
mm
aa
tt
2
1
1
2
2
1 [ m1a as F is constant]
9. n mv - -mvFP = =A A
2nmv=A
11.^ ^
P Pxi Py j
, dpFdt
Now o. 0 θ 90F P
12. Force exerted by ball, dvF mdt
13. vy = 40 ms–1 Fy = –5 N m = 5 kg
So ay = 11yF
msm
as y yv u at
0 = 40 –1t t = 40s14. Swimming is a result of pushing water in the opposite direction of the motion
15. Gas will cane out with sufficient speed in forward direction, So recetion of this forward force wil changethe reading of the spring balance.
16. According to the principle of conservation of linear momentum
km1000×50 = 1250×v \ v = 40 h
17. impulse = m f iP v v
19. As ^ ^
υ 5t i 2t j
^ ^a 5i 2 j
^ ^
ma x yF i m g a i
221 1 yF m ax g a
= 26 N
ay
ax
max
m(g+ )ay
72
20. Velocity after 10 sec.
21υ (Area enclosed between F - t graph) = 50 ms-1
21. Fmax = µ mg = 0.84 10 = 32 N
applied force at t = 2S is
F = kt2 = 2(2)2 = 8 N
The body fails to move. Hence the force of friction at t = 2s = applied force = 8 N
22. 21 FFF
Fa m
23. Here distance 2das υ is doubled, d becomes 4 times.
24. 21S ut at2
as u = 0 2Sta
for smooth plane a = gsinθFor rough plane 1a g (sinθ-μ cosθ)
1 2
sin θ-μ cos θSt
g
1 2sinθSt nt n
g
2n g sin θ-μ cos θ g sin θ
When θ = 45o sinθ = cosθ = 21 2n
1-1µ
25. 2 22 2
22 4m Rmv mRF mR
R R T T
As m and T are same for two bodies
F R
26. 1 2
1 2
m m ga
m m
MM 2 gMM 2
3g
73
27. As impulse = change in linear momentun
Ft = 02 sin30mv 2m sin30 Ft
28. 1 1 2 2m v +m v = 0 1 12 1
2
m vv = v = -m 20
Now 2 2
1 1 2 21 1E = m v + m v2 2
By solving weight 1v = 100 ms-1
29. The coin will revolve with the record if centrifugal force mrw2 < f
2mr <µR<µmg
2
µgr <
30. 2 30FF mr mr rads
31. Total mass pulled up m = 10 + 6 = 6 kg
agmT1
33. d-t graph AB is linier
Velocity is uniforce
a = 0 F = 0
35. agmFair
36. N62504.0
25tPF
38. 0 12v gh and 12v gh
2 1madt m v v
2 1v vadt
40. tFbt20dtdpF
42. 2 1.F t m v v and
F. t Area of lower part of graph F t
74
44. 22
21 FFF and M
Fa
46. Resultant force F = 2 µ mg-µ mg = µ mg
accleration in plate mFa
48. 0a gµ8T gµ4TF
51.2
µ µgvv rg r
52. 21
1d gt2
and 22
1d g a t2
By comparing both eqn.
2 21 2
1 1gt g a t2 2
21 tt
53. Tensile force T = m(g+a)
54. Tensile force
321
2132 mmm
mmTT
55. acceleration321 mmm
Fa
amT 12
ammT 211
56. acceleration 10631
13a
57. Calculate F = m (g+a)
58. P F t
m (vy–v0y ) = F t 12yv j
and resultant velocity of t = 3 Sec 2 2x yv v v
59. 1 1 0 1 1- υFt m v v m
2 2 0 2-Ft m v v m v
2
1
2
1
mm
tt
75
60. Fy = m
t
VV 0y
61. Tension T = 2 F ma = 2 F F = 2ma
62. N + P sinθ = mg N = mg - P sinθ = mg - 2P
limiting friction Fs = Nµs
63. gMµ(max)F Ass max
maxA
F ga 2m
and D
max DD
m ga M 3m3m m
OR WTTT 321
g.Mµmg3gmµ2µmg D m3m216 µm6MD
64. ma = mg sinθ a = 10 sin 4o a = 0.7 ms-2
65. According to the low of conservation of momentum
mm30 30m v v
31vv
66. for Block A T2 -2g = 2a ....(1)
for block B T1 +2g-T2= 2a ....(2)
for block C 2g - T1= 2a ....(3)
Adding (1) & (2) T1 = 4a ....(4)
from (3) & (4) 2g-4a = 2amm
30 30m v v
3ga
N1.133g4a4T1
67. LMµ
Weight of chain of lenght = gµ
Weight of a chain of lenght ( L ) = g-Lµ
s sµ R µ µ(L- ) gSf and µ gSf
sµ g µ .µ L- g
L
µ s
76
68. for 3 kg body m1a = m1g-T 1 1T m g m a......(1)
for 2 kg body T - m2 gsinθ = m2a
22
m gT m a ......(2)2
from (1) and (2) a = g = 10 ms-2
S 2= m a=20 Nf
69.dpF t t P tdtdt
2t
2mv
2α t v Which shows graph of 2tv is parabola so (ii) - (P)
now 1 (2 )2
dv m tdt
a α t Which shows graph of a - t is linier so (i)....R
70. according to eqn 1 1
1 1 2 2 1 1 2 2m v m v m v m v
12 9 mv s
71. N F mg cos30 and Nµ30sinmg s
72. For upper half 2 20v - v = 2 ad
22 µθsing20 2 sinv gl
For lower half 2 21 0v v 2a 'd
s0 glsin 2g sin cas .2
θtan2µ2
73. 45cosmg30sinFN
Nµmaxf ss †ÃÜå 22c sf N +f max
75. .dp d dmF mv vdt dt dt
dm 10 4gram
secdt 2.5 and cm5 sv
76. TF'ma maF'ma amF'a
77. As the box is descending
agmR
mg 3gmg ma a4 4
77
78. N 50θsinT1 and N 60θcosT1
5 θ 4tan 040θ
79. θcosµθsinmgF1
θcosµθsinmgF2
3µ-2µµµ2
µ-θtanµθtan
FF
2
1
80. Here 321 FFF aFa 1
90. 0 00v v at as v and v v
tVa maf maµmg
V Vµg tt µg
91. pseudo force exerted on block F = m α
force of friction f = μmαµN
when tallnotwillbolck,Wf
mgμmα mgα
92. F(t) = F0e–bt
ma = F0e–bt a = bt0 e
mF
dvdt bto e
mF dt btoFdv e dt
m
Integrating
0
tbtFov e dt
m
0
tbtoF ev
m b
1 btoFv emb
which is exponential function So graph (b) is true.
78
Unit - 4Work Energy
And Power
79
SUMMARY
The product of the magnitude of the displacement during the action of a force andthe magnitude of the component of the force in the direction of displacement isknown as work. Its unit in joule and its dimensional formula is M1L2T-2.
If angle between force and displacement is
(i) for = 0 W = Fd (ii) for = w = 0 (iii) for w = -Fd
If is an acute, work is positive and work is done by the force on the body. If is an obtuse angle, work is negative and work is done by the body aganist the force.
Work on a body lying on rough horizontalsurface acted by force. As there is nodisplacement along Y-axisN + Fsin = Mg N = Mg - F sinResponsible resultant force for displacementalong X-direction= F cos - N = F cos - (Mg - Fsin)Work W = [F cos - (Mg - Fsin]d = [F(cos - sin ) - Mg] d
The work done by a variable force is given by W = F . df
i
l
If a variable force and displacement due to it are in the same direction the areabelow the F - x graph gives value of work.
The ability of a body to do work by virtue of its motion is known as its Kineticenergy. If the velocity of a body of mass `m' is `v' its kinetic energy is
K = m2/pmv21 22 . It is scalar quantity..
Work Energy Theorem : Work done by a resultant force on a body is equal to
the change in its kinetic energy. W 220 0
1mv mv K2
When a body has the ability to do work due to its position in a force field or itsconfiguration. It is known as potential energy. It is scalar quantity.
If the gravitational potential energy, due to the gravitational field of Earth, israndomly taken to be zero on its surface the potential energy of a body of massm, at height h is mgh, where g is the gravitational acceleration. The value of `h' isnegligible compared to the radius of the earth.
The sum of the potential energy (U) and the kinetic energy (K) of a substance iscalled the mechanical energy. E = K + U.
80
Considering potential energy of a spring as zero in its normal state, if its langth is changed
by x, the potential energy of the spring is U = 2kx21
. Here k is the spring constant. Unit
of k is N/m and dimensional formula is M1L0T-2
The forces for which work done is independent of the path of motion of the body butdepends only on initial and final positions, are called conservative force. The force ofgravitation or the restoring force developed in a spring due to its compression or extensionare conservative forces.
The relation between the conservative force and the potential energy is F = –dUdx .
The time-rate of doing work is called power. Its unit is watt (J/s). Its dimensional formulais M1L2T-3.
The power P = W / t or P = F . v
1 horse power 746 wattUnit of electric energy for domestic use is 1 unit = 1 kWh = 3.6 106 J
If during collision of two bodies the kinetic energy is conserved the collision said to beelestic.
A body of mass m1 moving with velocity v1 undergoes elastic collision with a body of mass m2
moving with velocity v2 in the direction of v1 If after collision their velocities are '1v and '
2v .
'1v = 2
21
21
21
21 vmm
m2vmmmm
and '2v = 1 2 1
1 21 2 1 2
2m m mv vm m m m
In case of complete inelastic collision bodies colliding move together after collision with a
common velocity v. In this case
21
2211
mmvmvmv
A body of mass m1 moving with velocity v1 collides with a stationary body of mass m2
elastically. They move with velocities '1v and '
2v making angles 1 and 2 with direction of v1
then 22211111 cos'vmcos'vmvm
212111 sin'vmsin'vm0 and 222
211
211 'vm'vmvm
As Newton's low of impact, the co-efficient of restitution is,
e = 2 1
1 2
' ' , ' 'v v ev v
depends on the types of materials of bodies colliding.
For perfectly elastic collision e = 1 and for perfectly inelastic collision e = 0.Equations of velocities after collision of two bodies can be written as :
221
21
21
211 v
mmm)e1(v
mm)emm('v
and 1 1 22 1 2
1 2 1 2
(1 e)m (m e m )v' v vm m m + m
81
MCQ
For the answer of the following questions choose the correct alternative from among the given ones.1. How much is the work done in pulling up a block of wood weighing 2KN for a
length of 10m on a smooth plane inclined at an angle of 30o with the horizontal ?(A) 1.732 KJ (B) 17.32 KJ (C) 10 KJ (D) 100 KJ
2. A force of 7N, making an angle with the horizantal, acting on an object displacesit by 0.5m along the horizontal direction. If the object gains K.E. of 2J, what is thehorizontal component of the force ?(A) 2N (B) 4 N (C) 1 N (D) 14 N
3. A 60 kg JATAN with 10 kg load on his head climbs 25 steps of 0.20m height each.what is the work done in climbing ? (g = 10 m/s2)(A) 5 J (B) 350 J (C) 100J (D) 3500J
4. A ball of mass 5 kg is stiding on a plane with intial velocity of 10 m/s. If co-efficient of friction between surface and ball is 2
1 , then before stopping it willdescribe...... (g = 10 m/s2)(A) 12.5 m (B) 5 m (C) 7.5 m (D) 10 m
5. The relationship between force andposition is shown in the figure given (inone dimensional case) calculate thework done by the force in displacing abody from x = 0 cm to x = 5 cm
(A) 30 ergs (B) 70 ergs
(C) 20 ergs (D) 60 ergs
6. The force constant of a wire is K and that of the another wire is 3k when both thewires are stretched through same distance, if work done are W1 and W2, then...
(A) 22 1w 3w (B) W2 = 0.33W1 (C) W2 = W1 (D) W2 = 3W1
7. A ball is released from the top of a tower. what is the ratio of work done by forceof gravity in first, second and third second of the motion of the ball ?[hn (2n – 1)]
(A) 1 : 2 : 3 (B) 1 : 4 : 9 (C) 1 : 3 : 5 (D) 1: 5 : 3
•1
•2
•3
• 4
•5
•6
x (cm)
20
10
0
10
20
Forc
e (D
yne)
82
8. A spring of spring constatnt 103 N/m is stretched initially 4cm from theunstretched position. How much the work required to stretched it further by another5 cm ?
(A) 6.5 NM (B) 2.5 NM (C) 3.25 NM (D) 6.75 NM
9. The mass of a car is 1000 kg. How much work is required to be done on it to makeit move with a speed of 36 km/h ?
(A) 2.5 104 J (B) 5 103 J (C) 500 J (D) 5104 J
10. A body of mass 6 kg is under a force, which causes a displacement in it given by
S = 3t2 3 (in m). Find the work done by the force in first one seconds.
(A) 2 J (B) 3.8 J (C) 5.2 J (D) 24 J
11. A 8 kg mass moves along x - axis. Itsaccelerations as a function of itsposition is shown in the figure. What isthe total work done on the mass by theforce as the mass moves from x = 0 tox = 6cm ?
(A) 4810-3 J (B) 9810-3 J(C) 4.810-3 J (D) 9.810-3 J
12. The work done by a force actingon a body is as shown in the graph.what is the total work done incovering an intial distance of15m ?
(A) 50 J (B) 75 J(C) 100 J (D) 25 J
13. A spring gun of spring constant 290 10 N / M is compressed 4cm by a ball of mass16g. If the trigger is pulled, calculate the velocity of the ball.
(A) 60 m/s (B) 3 m/s (C) 90 m/s (D) 30 m/s
14. A uniform chain of length 2m is kept on a table such that a length of 50cm hangsfreely from the edge of the table. The total mass of the chain is 5kg. What is thework done in pulling the entire chain on the table. (g = 10 m\s2)
(A) 7.2 J (B) 3 J (C) 4.6 J (D) 120 J
Forc
e (N
)
20 •15 •10 •
5 •0 •
00 •
2 4 6• •a
(cm
/sec
)2
x (m)
83
15. A uniform chain of length L ans mass M is lying on a smooth table and one thirdof its is hanging vertically down over the edge of the table. If g is acceleration dueto gravity, the work required to pull the hanging part on to the table is(A) MgL (B) MgL / 3 (C) MgL/9 (D) MgL / 18
16. A cord is used to lower vertically a block of mass M by a distance d with constant
down-word acceleration 92 .work done by the cord on the block is
(A) - Mgd/2 (B) Mgd/4 (C) -3Mgd/4 (D) Mgd17. A block of mass 5 kg is resting on a smooth surface. At what angle a force of 20N
be acted on the body so that it will acquired a kinetic energy of 40J after moving4m(A) 300 (B) 450 (C) 600 (D) 1200
18. Natural length of a spring is 60 cm, and its spring constant is 2000N/m. A massof 20kg is hung from it. The extension produced in the spring is..... (g = 9.8 m/s2)(A) 4.9 cm (B) 0.49 cm (C) 9.8 cm (D) 0.98 cm
19. The potential energy of a body is given by U = A - Bx2 (where x is displacement).The magnitude of force acting on the particle is(A) constant (B) proportional to x(C) proportional to x2 (D) Inversely proportional to x
20. A uniform chain of length L and mass M is lying on a smooth table and ( 41 )th of
its length is hanging vertically down over the edge of the table. If g is accelerationdue to gravity, the work required to pull the hanging part on to the table is(A) MgL (B) MgL/9 (C) MgL/18 (D) MgL/ 32
21. If Wa, Wb, and Wc represent the work done inmoving a particle from X to Y along three differentpath a, b, and c respectively (as shown) in thegravitational field of a point mass m, find thecorrect relation between Wa, Wb and Wc(A) Wb > Wa > Wc (B) Wa< Wb < Wc(C) Wa > Wb > Wc (D) Wa = Wb = Wc
22. An open knife edge of mass m is dropped from a height h on a wooden floor. Ifthe blade penetrates upto the depth d into the wood, the average resistance offeredby the wood to the knife edge is,
(A) mg (B) mg (1+ dh ) (C) 2hmg (1 )d (D) mg (1 - d
h )
84
23. A toy car of mass 4 kg moves up a rampunder the influence of force F plottedagainst displacement x. The maximumheight attained is given by(Take g = 10 m/s2)(A) ymax = 5m(B) ymax = 10m(C) ymax = 15m(D) ymax = 20m
24. A particle of mass 0.5kg travels in a straight line with velocity v = 23ax ,
Where a = 5 1sm 21 . The work done by the net force during its displacement from
x = 0 to x = 2m is(A) 50 J (B) 45 J (C) 25 J (D) None of these
25. Velocity time graph of a particle of mass 2kgmoving in a straight line is as shown in figure.Wrok done by all forces on the particle is(A) 400 J(B) -400 J(C) -200 J(D) 200 J
26. A mass of M kg is suspended by a weight-less string, the horizontal force that isrequired to displace it until the string makes an angle of 60o with the intial varticaledirection is
(A) Mg / 3 (B) Mg . 2 (C) Mg / 2 (D) Mg . 327. Given below is a graph between a
variable force (F) (along y-axis)and the displaement (X) (along x-axis) of a particle in onedimension. The work done by theforce in the displacement intervalbetween 0m and 30m is(A) 275 J (B) 325 J(C) 400 J (D) 300 J
85
28. Force F on a particle moving in astraight line varies with distance das shown in the figure. The workdone on the particle during itsdisplacement of 12m.(A) 27 J (B) 24 J(C) 36 J (D)26 J
29. A force F = Ay2 + By + C acts on a body in the y-direction. The work done by thisforce during a displacement from y = -a to y = a is
(A) 3
Aa2 3
(B) ca23
Aa2 3
(C) ca2
Ba3
Aa2 23
(D) None of these.
(ENERGY AND CONSERVATVE, NONCONSERVATIVE FORCE)
30. A spring with spring constant K when streched through 2cm the potential energyis U. If it is streched by 6cm. The potential energy will be......
(A) 6U (B) 3U (C) 9U (D) 18U
31. If linear momentum of body is increased by 1.5%, its kinetic energy increasesby.......%
(A) 0% (B) 10% (C) 2.25% (D) 3%
32. With what velocity should a student of mass 40 kg run so that his kinetic energybecomes 160 J ?
(A) 4 m/s (B) 8 m/s (C) 16 m/s (D) 8 m/s
33. A body of mass 1 kg is thrown upwords with a velocity 20 m/s. It momentarilycomes to rest after a height 18m. How much energy is lost due to air friction.(g = 10 m/s2)
(A) 20 J (B) 30 J (C) 40 J (D) 10 J
34. Two bodies of masses m1 and m2 have equal kinetic energies. If P1 and P2 aretheir respective momentum, what is ratio of P2 : P1 ?
(A) m1: m2 (B) 12 m:m (C) 21 m:m (D) m12 : m2
2
86
35. A body having a mass of 0.5 kg slips along thewall of a semispherical smooth surface of radius20 cm shown in figure.What is the velocity ofbody at the bottom of the surface ? (g = 10 m/s2)
(A) s/m2 (B) 2 m/s
(C) 2 s/m2 (D) 4 m/s
36. Two bodies of masses m and 3m have same momentum. their respective kineticenergies E1 and E2 are in the ratio.....
(A) 1 : 3 (B) 3 : 1 (C) 1 : 3 (D) 1 : 6
37. What is the velocity of the bob ofa simple pendulum at its meanposition, if it is able to rise tovertical height of 18 cm(Take g = 10 m/s2)
(A) 0.4 m/s (B) 4 m/s
(C) 6 m/s (D) 0.6 m/s
38. A particle is placed at the origin and a force F = kx is acting on it (Where K ispositive constant) If Ucos = 0, which one of the following graph of U(x) versusx. (where U is the potential energy function)
(A) (B)
(C) (D)
R = 20cm
0.5 Kg
18 cm
U( )x
Y
X
U( )xY
X
Y
X
U( )x
Y
X
U( )x
87
39. A force time graph for a linearmotion is shown in figure where thesegments are circular. what is linearmomentum gained between zero and8 second ?
(A) -2 N S (B) 0 NS(C) 4 NS (D) -6 NS
40. A particle is dropped from a height-h. A constant horizontal velocity is given to theparticle. Taking g to be constant everywhere kinetic energy E of the particle withrespect to time t is correctly shown in....
(A) (B)
(C) (D)
41. Which of the following graph is correct between kinetic energy (E), potentialenergy (U) and height (h) from the ground of the particle.
(A) (B)
(C) (D)
• • • • • • • •2 4 6 8 t(s)
Forc
e (N
)
+2
0
–2
E
t
E
t
E
t
E
Ene
rgy
U
Height
E
Ene
rgy
U
Height
E
Ene
rgy
Height
U
Ene
rgy
U
Height
E
E
t
88
42. An engine pump is used to pump a liquid of density continuosly through apipe of cross-sectional area A. If the speed of flow of the liquied in the pipeis v, then the rate at which kinetic energy is being imparted to the liquid is
(A) 21
AV3 (B) 21
AV2 (C) 21
AV (D) AV
43. The velocity of a body of mass 400 gm is ˆ ˆ3i 4 j m/s. So its kinetic energy
is ......
(A) 5 J (B) 10 J (C) 8 J (D) 16 J
44. A particle is moving under the influence of a force given by F = kx, where kis a constant and x is the distance moved. What energy (in joule) gained by theparticle in moving from x = 1m to x = 3m ?
(A) 2 k (B) 3 k (C) 4 k (D) 9 k
45. A spring is compressed by 1 cm by a force of 4 N. Find the potential energyof the spring when it is compressed by 10 cm.
(A) 2 J (B) 0.2J (C) 20 J (D) 200 J
46. when 2kg mass hangs to a spring of length 50 cm, the spring stretches by 2 cm.The mass is pulled down until the length of the spring becomes 60 cm. What is theamount of elastic energy stored in the spring in this condition, if g = 10 m/s2
(A) 10 J (B) 2 J (C) 2.5 J (D) 5 J
47. The potential energy of a projectile at its highest point is th12 the value of its
initial kinetic energy. Therefore its angle of projection is ......
(A) 300 (B) 450 (C) 600 (D) 750
48. Two bodies P and Q have masses 5 kg and 20 kg respectively. Each one is actedupon by a force of 4 N. If they acquire the same kinetic energy in times tP and tQ
then the ratio tqtp = ..........
(A) 21 (B) 2 (C) 5
2 (D) 65
49. A particle of mass 0.1kg is subjectedto a force which varies with distanceas shown in figure. If it starts itsjounery from rest at x = 0. What isthe particle's velocity squre at x = 6cm ?
(A) 0 (m/s)2 (B) 240 2 (m/s)2
(C) 240 3 (m/s)2 (D) 480 (m/s)2
F(N)
x(m)
89
50. The potential energy of 2kg particle, free to move along x axis is given by
J2x
4x)X(U
24
. If its mechanical energy is 2 J, its maximum speed is.... m/s
(A) 23 (B) 2 (C) 2
1(D) 2
51. If the K.E. of a body is increased by 44%, its momentum will increase by.......
(A) 20 % (B) 22 % (C) 2 % (D) 120 %
52. A bullet of mass 0.10 kg moving with a speed of 100 m/s enters a wooden blockand is stopped after a distance of 0.20m. what is the average resistive force exertedby the block on the bullet ?
(A) 2.5 x 102 N (B) 25 N (C) 25 x 102 N (D) 2.5 x 104 N
53. A simple pendulum is released from Aas shown in figure. If 10 g and 100 cmrepresent the mass of the bob and lengthof the pendulum. what is the gain in K.E.at B ? ( g = 10 m/s2 )
(A) 0.5 J (B) 5 10-2 J(C) 5 J (D) 0.5 10-3 J
54. A rifle bullet loes th110 of its velocity in passing through a plank. The least
number of such planks required just to stop the bullent is
(A) 5 (B) 10 (C) 11 (D) 20
55. A sphere of mass m moving the velocity v enters a hanging bag of sand and stops.If the mass of the bag is M and it is raised by height h, then the velocity of thesphere was
(A) h29m
mm (B) h29
mM
(C) h29mM
m (D) h29
Mm
56. A particle is acted upon by a force Fwhich varies with position x as shownin figure. If the particle at x = 0 haskinetic energy of 20 J. Then thecalculate the kinetic energy of theparticle at x = 16 cm.
(A) 45 J (B) 30 J(C) 70 J (D) 135 J
600
B
A
10 •
5
0
-5
-10
•
•
•
•
F(N)
x(m)
90
57. A frictionless track 12345 endsin a circular loop of radius R. Abody slids down the track frompoint 1 Which is 6 cm. Maximumvalue of R for the body tosuccessfully complete the loop is
(A) 6 cm (B) 415 cm
(C) 125 cm (D) 5
12 cm
58. If the water falls from a dam into a turbine wheel 19.6m below, then the velocityof water at the turbine is ...... ( g = 9.8 m/s2)
(A) 9.8 m/s (B) 19.6 m/s (C) 39.2 m/s (D) 98.0 m/s
59. A bomb of 12 kg divedes in two parts whose ratio of masses is 1:4. If kineticenergy of smaller part is 288 J, then momentum of bigger part in kgm/sec will be
(A) 48 (B) 72 (C) 108 (D) Data is incomplete
60. An ice-cream has a marked value of 700kcal. How many kilo-watt-hour ofenergy will it deliver to the body as it is digested (J = 4.2 J/cal)
(A) 0.81 kwh (B) 0.90 kwh (C) 1.11 kwh (D) 0.71 kwh
61. A spherical ball of mass 15 kgstationary at the top of a hill of height82m. It slides down a smooth surfaceto the ground, then climbs up anotherhill of height 32m and finally slidesdown to horizontal base at a height of10 m above the ground. The velocityattained by the ball is
(A) 30 10 m/s (B) 10 30 m/s (C) 12 10 m/s (D) 10 12 m/s
62. A bomb of mass 10 kg explodes into 2 pieces of mass 4 kg and 6 kg. Thevelocity of mass 4 kg is 1.5 m/s, the K.E. of mass 6 kg is .......
(A) 3.84 J (B) 9.6 J (C) 3.00 J (D) 2.5 J
63. A bomb of mass 3.0 kg explodes in air into two pieces of masses 2.0 kg and 1.0kg. The smaller mass goes at a speed of 80m/s. The total energy imparted to thetwo fragments is
(A) 1.07 KJ (B) 2.14 KJ (C) 2.4 KJ (D) 4.8 KJ
64. The bob of simple pendulum (mass m and length l) dropped from a horizontalposition strike a block of the same mass elastically placed on a horizontalfrictionless table. The K.E. of the block will be
(A) 2 mgl (B) mgl /2 (C) mgl (D) zero
91
65. A gun fires a bullet of mass 40 g with a velocity of 50 m/s. Because of this thegun is pushed back with a velocity of 1 m/s. The mass of the gun is
(A) 1.5 kg (B) 3 kg (C) 2 kg (D) 2.5 kg
66. The decreases in the potential energy of a ball of mass 25 kg which falls froma height of 40 cm is
(A) 968 J (B) 100 J (C) 1980 J (D) 200 J
67. If a man increase his speed by 2 m/s, his K.E. is doubled, the original speed ofthe man is
(A) (2+2 2 ) m/s (B) (2 + 2 ) m/s (C) 4 m/s (D) (1+2 2 ) m/s
68. The potential energy of a conservative system is given by U(X) = (x2-5x) J.Then the equilibrium position is at....... (where x in m)
(A) x = 1.5 m (B) x = 2m (C) x = 2.5 m (D) x = 5 m
69. The potential energy of a particlevaries with distance x as shown in thegraph. The force acting on the particleis zero at
(A) C (B) B
(C) B and C (D) A and D
70. A nucleus at rest splits into two nuclear parts having same density and radii inthe ratio 1:2. Their velocites are in the ratio
(A) 2:1 (B) 4:1 (C) 6:1 (D) 8:1
71. A body of mass 1.5 kg slide down a curved trackwhich is quadrant of a circle of radias 0.75 meter.All the surfaces are frictionless. If the body startsfrom rest, its speed at the bottom of the track
is ..... ( g = 10 m/s2)
(A) 3. 87 m/s (B) 2 m/s
(C) 1.5 m/s (D) 0.387 m/s
72. A single conservative force F(x) acts on a 2.5 kg particle that moves along the x-axis. The potential energy U(x) is given by U(x) = (10 + (x - 4)2) where x is inmeter. At x = 6.0m the particle has kinetic energy of 20J. what is the mechanicalenergy of the system ?
(A) 34 J (B) 45 J (C) 48 J (D) 49 J
C
D
x
W( )x
0.75m
0.75
m
92
P O W E R73. A body initially at rest undergoes one dimensonal motion with constant
acceleration. The power delivered to it at time t is proportional to.....
(A) 21t (B) t (C) 2
3t (D) t2
74. An electric motor develops 5KW of power. How much time will it take to lifta water of mass 100 kg to a height of 20 m ? ( g = 10 m/s2 )
(A) 4 sec (B) 5 sec (C) 8 sec (D) 10 sec
75. Bansi does a given amount of work in 30 sec. Jaimeen does the same amountof work... in 15 sec. The ratio of the output power of Bansi to the Jaimeen is....
(A) 1:1 (B) 1:2 (C) 2:1 (D) 5:3
76. A rope-way trolly of mass 1200kg uniformly from rest to a velocity of 72 km/h in 6s. What is the average power of the engine during this period in watt ?(Neglect friction)
(A) 400 W (B) 40,000 W (C) 24000 W (D) 4000 W
77. A body of mass m is accelarated uniformly from rest to a speed v in time T. Theinstantaneous power delivered to the body in terms of time is given by.....
(A) t.T
mv2
2
(B) 22
2
t.T
mv(C) t.
T2mv2
(D) 22
t.T2
mv2
78. 1 kg apple gives 25 KJ energy to a monkey. How much height he can climb byusing this energy if his efficiency is 40%. (mass of monkey = 25 kg and g = 10m/s2)
(A) 20m (B) 4m (C) 30m (D) 40m
79. A force of ˆ ˆ ˆ2i 3j k N acts on a body for 5 second, produces a displacement
of ˆ ˆ ˆ3i 5j k . What was the power used ?
(A) 4W (B) 20 W (C) 21 W (D) 4.2 W
80. If the force F is applied on a body and it moves with a velocity v, the power willbe
(A) Fv (B) VF (C) 2
VF (D) 2Fv
81. From an automatic gun a man fires 240 bullet per minute with a speed of 360 km/h. If each weighs 20 g, the power of the gun is
(A) 400 W (B) 300 W (C) 150 W (D) 600 W
93
82. A body of mass M is moving with a uniform speed of 10 m/s on frictionlesssurface under the influence of two forces F1 and F2. The net power of the system
is 1 2F M F
(A) MFF10 21 (B) M)FF(10 21 (C) M)FF( 21 (D) Zero
83. A coolie 2.0m tall raises a load of 75 kg in 25 from the ground to his head andthen walks a distance to 40m in another 25. The power developled by the coolieis ( g = 10 m/s2)(A)0.25 kw (B) 0.50 kw (C) 0.75 kw (D) 1.00 kw
84. A body is moved along a straight line by a machine delivering a constantpower. The velocity gained by the body in time t is proportional to.....
(A) 43t (B) 2
3t (C) 41t (D) 2
1t
ELASTIC - INELASTIC (COLLISION)85. The coefficient of restitution e for a perfectly elastic collision is
(A) 1 (B) 0 (C) (D) -186. Two balls at same temperature collide. What is conserved
(A) Temperature (B) velocity (C) kinetic energy (D) momentum87. A particle of mass m moving with horizontal speed 6 m/s as shown in figure.
If m<<M then for one dimensional elastic collision, the speed of lighter particleafter collision will be
(A) 1 m/s in original direction.(B) 2 m/s opposite to the original direction.(C) 1 m/s opposite to the original direction.(D) 2 m/s in original direction.
88. A rubber ball is dropped from a height of 5 m on a planet where the accelerationdue to gravity is not known. On bouncing, it rises to 1.8 m. The ball losses itsvelocity on bouncing by a factor of
(A) 2516 (B) 25
9 (C) 53 (D) 5
2
89. Three objects A, B and C are kept in a straight line on a frictionless horizontalsurface. These have masses m, 2m and m respectivelly. The object A movestowards B with a speed 9 m/s and makes an elastic collision with it. Thereafter,B makes compelety inelastic collision with C. All motion occur on the samestraight line. Find final speed of the object C
94
(A) 3 m/s(B) 4 m/s(C) 5 m/s (D) 1 m/s
90. Two solid rubber balls P and Q having masses 200 g and 400 g respectively aremoving in opposite directions with velocity of P equal to 0.3 m/s. Aftercollision the two balls come to rest, then the velocity of Q is(A) 0.15 m/s (B) 1.5 m/s (C) -0.15 m/s (D) Zero
91. A sphere collides with another sphere of identical mass. After collision, the twosphere move. The collision is inelastic. Then the angle between the directionsof the two spheres is
(A) Different from 90 (B) 90 (C) O (D) 45
92. A ball is allowed to fall from a height 20m . If there is 30% loss of energy dueto impact, then after one impact ball will go up to(A) 18 m (B) 16 m (C) 12 m (D) 14 m
93. If a skater of weight 4 kg has intial speed 40 m/s and 2nd one of weight 6 kghas 6 m/s. After collision, they have speed (couple) 6 m/s. Then the loss in K.E.is.....(A) 48 J (B) zero (C) 96 J (D) None ofthese
94. A metal ball of mass 2 kg moving with a velocity of 36 km/h has a head oncollision with a stationary ball of mass 3 kg. If after the collision, the two ballsmove togather, the loss in kinetic energy due to collision is(A) 40 J (B) 60 J (C) 100 J (D) 140 J
95. A neutron having mass of 1.6710-27 kg and moving at 108 m/s collides with adeutron at rest and sticks to it. If the mass of the deutron is 3.3410-27 kg thenthe speed of the combination is(A) 3.33 107 m/s (B) 3105 m/s (C) 33.3107 m/s (D) 2.98105 m/s
96. A mass of 100g strikes the wall speed5 m/s at an angle as shown in figureand is rebounds with the same speed.If the contact time is 35 10 sec, whatis the force applied on the mass by thewall
(A) 3100 N to right
(B) 100 N to right
(C) 3100 N to left
(D) 100 N to left
95
97. Three identical sperical balls A, B and C are placed on a table as shown in thefigure along a straight line. B and C are at rest initially. The ball A hits B head onwith a speed of 10 m/s. Then after all collision A and B are brought to rest and Ctakes off with velocity of...... (elastic collision)
(A) 20 m/s (B) 2.5 m/s (C) 10 m/s (D) 7.5 m/s98. A ball dropped from a height of 4m rebounds to a height of 2.4m after hitting
the ground. Then the percentage of energy lost is(A) 40 (B) 50 (C) 30 (D) 600
99. A billiard ball moving with a speed of 8 m/s collides with an identical balloriginally at rest. If the first ball stops after collision, then the second ball willmove forward with a speed of..... (elastic collision)(A) 8 m/s (B) 4 m/s (C) 16 m/s (D) 1.0 m/s
100. A bullet of mass m moving with velocity v strikes a block of mass M at rest andgets embedded into it. The kinetic energy of the composite block will be
(A) )mm(Mmv
21 2
(B) M
)mm(mv21 2
(C) )Mm(mMV
21 2
(D) )Mm(
mmv21 2
101. A body of mass m1 is moving with a velocity v. It collides with anotherstationary body of mass m1. They get embeded. At the point of collision, thevelocity of the system(A) Increases(B) Decreases but does not become zero(C) Remains same(D) Become zero
102. Two small particles of equal masses startmoving in opposite directions from a point. Ain a horizontal circular orbit. Their tangentialvelocities are v and 2v, repectively, as shown inthe figure. Between collisions, the particlesmove with constant speeds. After making howmany elastic collisions, other than that at A,these two particles will again reach the pointA(A) 1 (B) 2(C) 3 (D) 4
96
103. Four identical balls are lined in a straight grove made on a horizontalfrictionless surface as shown. Two similar balls each moving with a velocity vcollide elastically with the row of 4 balls from left. What will happen
(A) One ball from the right rolls out with a speed 2v and the remaining balls willremain at rest.(B) Two balls from the right roll out speed v each and the remaining balls willremain statinary.
(C) All the four balls in the row will roll out with speed v4 each and the two
colliding balls will come to rest.(D) The colliding balls will come to rest and no ball rolls out from right.
(ASSERTION & REASON)
* Assertion and Reason are given in following questions. Each question havefour option. One of them is correct it.(1) If both assertion and reason and the reason is the correct explanation of theAssertion.(2) If bothe assertion and reason are true but reason is not the correctexplanation of the assertion.(3) If the assertion is true but reason is false.(4) If the assertion and reason both are fare.
104. Assertion : In the elastic collision between two bodies, the relative speed of thebodies after collision is equal to the relative speed before the collision.Reason : In the elastic collision the linear momentum of the system isconserved.(A)1 (B) 2 (C) 3 (D) 4
105. Assertion : When a gas is allowed to expand, work done by gas is positive.Reason : Force due to gaseous pressure and displacement (of position) on inthe same direction.(A) 1 (B) 2 (C) 3 (D) 4
97
106. Assertion : A light body and heavy body have same momemtum. Then they alsohave same kinetic energy.Reason : Kinetic energy does not depend on mass of the body.(A) 1 (B) 2 (C) 3 (D) 4
107. Assertion : Mountain roads rarely go straight up the slope.Reason : slope of mountains are large, therefore, more chances of vehicle toslip from roads.(A) 1 (B) 2 (C) 3 (D) 4
108. Assertion : The change in kinetic energy of a particle is equal to the work doneon it by the net force.Reason : Change in kinetic energy of particle is equal to the work done onlyin case of a system of one particle.(A) 1 (B) 2 (C) 3 (D) 4
109. Assertion : Work done in moving a body over a closed loop is zero for everyforce in nature.Reason : Work done does not depend on nature of force.(A) 1 (B) 2 (C) 3 (D) 4
110. Assertion : A weight lifter does no work in holding the weight up.Reason : Work done is zero because distance moved is zero.(A) 1 (B) 2 (C) 3 (D) 4
111. Assertion : stopping distance = KineticenergyStopping force
Reason : Work done in stopping a body is equal to K.E. of the body.(A) 1 (B) 2 (C) 3 (D) 4
112. Assertion : The mass equivalent of 1000 kwh energy is 40 microgram.
Reason : This follows from E = mc2 where 8C 3 10 m s
(A) 1 (B) 2 (C) 3 (D) 4113. Assertion : Work done by centripetal force is zero.
Reason : This is because entripetal force is always along the tangent.(A)1 (B) 2 (C) 3 (D) 4
114. Assertion : Two bodies of different masses have same momentum. Their kineticenergy are in the inverse ratio of their masses.
Reason : K.E. = 2mv21
(A) 1 (B) 2 (C) 3 (D) 4
98
115. Assertion : Linear momentum is conserved in both, elastic and inelasticcollisions.Reason : Total energy is conserved in all such collisins.(A) 1 (B) 2 (C) 3 (D) 4
116. Assertion : Both, a stretched spring and a compressed spring have potentialenergy.Reason : Work is done against the restoring force in each case.(A) 1 (B) 2 (C) 3 (D) 4
(COLUMN)117. A force F = kx (where k is positive constant) is acting on a particle. Match
column-I and column-II, regarding work done in displacing the particle.Column - I Column - ii(a) From x = -4 to x = -2 (P) Positive(b) From x = -2 to x = -4 (Q) zero(c) From x = -2 to x = +2 (R) negative(A) a - R, b - P, c - Q (B) a - P, b - Q, c - R(C) a - R, b - Q, c - P (D) a - Q, b - P, c - R
118. A body falls freely under the action of gravity from a height h above theground.Column - i Column - ii(a) P.E. = 2(K.E.) (P) constant at every point(b) P.E. = K.E. (Q) at height
3h
(c) P.E. = 21
(K.E.) (R) at height 3h2
(d) P.E.+ K.E. (S) at height 2h
(A) a - P, b - Q, c - R, d - S (B) a - Q, b - P, c - S, d - R(C) a - S, b - R, c - Q, d - P (D) a - R, b - S, c - Q, d - P
119. Two vehicles moving on a horizontal road are stopped by same retarding force.Column - I Column - ii(a) When they have same K.E. (P) faster body stop in larger distance(b) When they have different masses (Q) larger body stops in larger distance. but same velocity(c) When both have same momentum (R) heavier body stops in larger distance.(d) When both have same mass but (S) stopped in same distance.
different velocities
(A) a - P, b - Q, c - R, d - S (B) a - Q, b - P, c - S, d - R(C) a - S, b - R, c - Q, d - p (D) a - R, b - S, c - Q, d - P
99
KEY NOTE
1 B 26 A 51 A 76 B 101 B
2 B 27 B 52 C 77 A 102 B
3 D 28 A 53 B 78 D 103 B
4 D 29 B 54 A 79 A 104 D
5 A 30 C 55 A 80 A 105 A
6 D 31 D 56 B 81 A 106 D
7 C 32 B 57 D 82 D 107 A
8 C 33 A 58 B 83 C 108 B
9 D 34 B 59 A 84 D 109 D
10 D 35 B 60 A 85 A 110 A
11 A 36 B 61 C 86 D 111 A
12 B 37 C 62 C 87 A 112 A
13 D 38 C 63 D 88 D 113 C
14 B 39 B 64 C 89 B 114 B
15 D 40 A 65 C 90 C 115 B
16 A 41 A 66 B 91 A 116 A
17 C 42 A 67 A 92 D 117 A
18 C 43 A 68 C 93 D 118 D
19 B 44 C 69 C 94 B 119 C
20 D 45 A 70 D 95 A
21 D 46 D 71 A 96 C
22 B 47 A 72 A 97 C
23 B 48 A 73 B 98 A
24 A 49 D 74 A 99 A
25 B 50 A 75 B 100 D
100
HINT
1. W Fd cos
2. cosFdK = w
3. ;)d()Mg(dFW where M = m1 + m2
4. a g and 2 20v v 2 gd
2vod v 02 g
5. W = Area under curve of F x graph.
6. According 21W kx2
7. (2n 1)nh
8. 2 22 1
1W k x x2
9. According to work energy theorm W k
10. 3t2s
3
acceleration 2
2
d sa 4tdt
, work 0 0
F ds madsl l
W
11. W = F x = max = (m) (Area covered by curve of ax graph)12. work W = Area covered by curve of Fx graph13. Loss in P.E. of spring = gain in K.E. of ball
14. W = 2mgl2n
, where n is, fraction of length of the chain hanging from the table.
15. W = 22mgl
n ( n = 3 Given)
16. Tension in the cord, T = M (g- 2g
) = 2mg
. Work done by cord = -Td = - 2mgd
17. Fd cos = 21
mv2 - 21
mu2
18.F mgxK k
19.duFdx
101
20. 22mglw
n ( n = 4 given)
21. Gravitational force is a conservative force and work done against it is a point functioni.e. does not depends on the path.
22. 2 20v v 2ad , (v 0) and gh2v0
2o 2gh 2(g a)d a
23. Work done = gain in potential energy Area under curve = mgh24. Intial velocity at x = 0; v1 = 0
Final velocity at x = 2; v2= 325 2
Work done = 2 22 1
1 m ( v v )2
25. Intial velocity of particle v1= 20 m/sFinal velocity of particle v2= 0
2 22 1 2 1
1W k k k m (v v )2
26.
0
W FlsinU mgl(1 cos )
W U; ( 60 )
27. Work = Area under (F - d) graph28. Work = Area under (F - d) graph
29. W Fdy
= 2
2
a
Ay By C dy
3 2Ay By cy3 2
a
a
32Aa 2Ca3
30. Energy U x2 ( k constant)
102
31. 2
2
PKm2
PK ( m constant)
dk dp2k p
32. 2mv21k
mk2v
33. Energy lost due to air friction = 21
mv2 - mgh
34.2
1
2
1
mm
ppmP
35. mgR = 2mv21
36.m1E
m2pE
2
37. gh2V
38.2kxU
2
39. F. t = momentum, The resulatant fore is zero, therefore F. t = 0
40. Kinetic Energy (K.E) 2v and 22 tv ....
41. P.E (U) h and E = K + U = constant
42.2 2
21 mv 1 A V 1 A2 t 2 t 2
l v
43. Find the value of v
and than, K.E = 21 mv2
44. .dxkxU3
1
45. Spring constant 2kx21U,
xFK
46. Spring constant 2kx21U,
xFK
103
47. Hmax = 2 2 2 2o 0 0 0v Sin mv sin 1, U mg H max Ko
2g 2 2
48. According to equation 2
QQ2
pp vm21vm
21
p
Q
VV ,
Using impulse momentum, p pP
Q Q Q
m vFΔt =FΔT m v
49. 2mv21
area covered by curve of Fx , graph
50. K.E. is maximum than P.E. minimum.
so 0dxdu 0 OR
For x 4 .minU41)x(U
51. EE 1 and E44.1E2 . Than EP
52. F d = 21
m 2 22 1v v .
53. Loss of potential energy = Kinetic energy gained.54. Let the thickness of one planks be s and n planks are arranged just to stop bullet.
2 20 0
9v v 2as , put v v10
2
0v 100Now2as 19
20vn
2as
55. By conservation of linear momentum mv = (m+M) Vsys
By convervation of low of energy. 21
(m+M) v2 sys = (m+M) gh
Find Vsys and put it sin egn.(1)56. Work done = Area under FX graph with proper sign.
57. Condition for up direction loop (Parabola) h = 25
R
58. P.E. of water = K.E. at turbine. mgh = 2mv21
59. The bomb into two masses 1 2m + m =12kg and 1
2
m 1m 4
, K.E.
energy of smaller part = 21 1
1 m v2
104
60. 1 k cal = 103 calorie = 4200J = 6103.64200
kwh
6103.64200700calK700
kwh
61. From the conservation of energy,
mg 2mv21)hh( 21
62. 1 1 2 2m v +m v =0 and than K.E. = 22 2
1 m v2
63. ,vm 21 = ,vm 22
Total energy of system = 2 21 1 2 2
1 1m v m v2 2
64. The collision between bob and block is elastic P.E. K.E.
65. BBGG vmvm
66. mghU
67. Initial kinetic energy, E = 2mv21
Final kinetic energy, 2E = 2)2v(m21
68. For a conservation field. Force, dxduF ,
At equilibrium position, F = 0
69. dxduF it is clear that slope of U - x curve is zero at point B and C.
70. Given 1 2= 2
1
MM
=
3
2
1
RR
According to low conservation of linear momentum, 3
2
1
12
VV
71. Conservation of energy, mgh = 2mv21
72. P.E. at X = 6m is U = 10 + (6 - 4)2 = 14 J
Mechanical energy = K.E. + P.E. = 20 + 14 = 34 J
105
73. Power, P = Fv = mav
74. Power, P = tmgh
75. Power, P t1
76. m2P2E
77. P 2mE ; P E
78. P 2mE ; P E
79. tWP
td.FP
80. .............
81. Power of gun = TotalK.E.of fired bullet
time 21 mvn t2
82. Work done by forces = 0 ; ptW
83. KW75.02
)2)(10)(75(t
mghP
84. 21 mv kt (K,Cons tan t)2
, 1
22kv tm
, 1
22kdx t dt.m
Take integration.
85. ............86. ............
87.
21
221
21
211 mm
um2ummmmv where m1 = m and m2 = M; m << M.
m1 = 0, v1 = -u1+ 2u2
88. ;hh
vv
1
2
1
2 i.e. fractional loss in velocity 1
2
vv1
89.1 1
21 2
2m uv 6m / sm m
i.e. After elastic collision B strike to C with velocity of 6 m/s, Now for B and C
sys2m –– m 3m V6m/s Rest
(2m) (6) + 0 = (3m) (V sys)
90. p p Q Qm v m v 0
106
91. Angle will be 90o if collision is perfectly elastic.
92. mgh ' = 70% of mgh
93. Loss in K.E. = (initial K.E. - Final K.E.) of system 2 2 21 1 2 2 1 2
1 1 1m u m u (m m )v2 2 2
94. By low of conservation of momentum
210 = (2+3) v v 4m s
Loss in K.E. = 21
(2)(10)2 - 21
(5)(4)2
95. Momentum of neutron = Momentum of combination96. Force = Rate of change of momentum.
Intial momentum 1ˆ ˆp mvsin j mv cos j
Final momentum 2ˆ ˆp mvsin j mv cos j
tPF
97. In elastic head on collision velocity gets interchanged.98. U1 = mgh1 and U2= mgh2
% energy lost = 100U
UU
1
21
99. In elastic head on collision velocites gets interchanged.100. By conservation of momentum, mv + M(o) = (m+M) v
Velocity of composite block V = vMm
m
K.E. of composite block = 2v)mM(21
101. By conservation of momentum, v)mm()0(mVm 2121
102. Let initially particle x is moving in anticlock-wise direction and y in clockwisedirection.As the ratio of velocities of x and y
particles are Vx 1Vy 2
, therefore ratio of
their distance covered will be in the ratioof 2:1. It means they collide at B.so, after two collision these two particleswill again reach the point A.
103. Momentum and kinetic energy is conserved only in this case.
107
Unit - 5Rotational Motion
108
SUMMARY* Important Formula, Facts and Terms
1. Centre of mass of system of particles
n
1n n
nn
n21
nn2111cm
mrm
mmmrmrmrmR
for rigid body
nncm rmR.M
for Two body System
2211 rmrm
OR 1 2 1 2 2 1
1 12 2
r m r +r m +m+1= +1 OR =m mr r
1 2 1 22 11 1 2 1 22
r m +m rm rm = r = OR r =m m +m m +mr
2.1 2 n1 2 n
cmm v + m v + ......... + m vV =
M
1 2 .........cm nP M v p p p
similarly
Mamamama nn2211
cm
n21cm FFFaMF
3. Torque = T = Fr = I
sinrF
= product of force and perpendicular distance between point of rotation
and line of action.
Angular momentum = L r P I
| L | = rpsin = product of linear momentum and perpendicular distance between point ofrotation and line of action.
Moment of inertia = I = 2nn
222
211 rmrmrm
=
n
1n
2nnrm
109
4. Low of conservation of angular momentum
As L I
pd L dI Idt dt
d Ldt
when
= then L remains constant
Its geometrical representation in planetary motion
Let dtdA
is an areal velocity
Then m dA L dA L= OR =dt 2 dt 2m
5. Radius of gyration {K}
As I = 2nn
222
211 rmrmrm
If all particles have same mass then= m 2
n2
22
1 rrr
n
rrrnmI2
n2
22
1 Here (nm = M)
2mk
n)rrr(k
2n
22
21
6. Some relations between linear and rotational motion.v = rw
v w r
Here w = ddt
2
2d w ddt dt
r Ta a a
where Twv rra a
2 2 2 2 2 2r ta a a v r
2 2 2 2 2 4 2r r r
7. Equilibrium of a rigid body.
When 0FFFF n21 it is in linear equilibrium
When 21 0nP P P P
it is in rotational equilibrium8. Two theorm for moment of inertia
YXZ III Theorm of perpendicular axis.2
cm MdII Theorm of Parallel axis.
110
9. Rolling down of body on an inclined plane.
V 2 22 2
2 2 sin,1 1
gh aK K
R R
Condition for rolling without sliding
,KR1
tan
22s
For ring tan21
s (K = R)
disc
2RKtan
31
s
solid sphere
R
52Ktan
72
s
10. Rotational Kinetic Energy.
R.K.E = 2
2 2 22
1 1 &2 2
V VI MK I MKRR
I2L2
Total kinetic Energy = Rotational K.E. + Linear K.E.
2 2
2 2 22 2
1 1 1 12 2 2
V KMK MV MVR R
Now (1) 22
RK
E.LinearKE.K.R
(2) 2
2 22 2 2
2
.. 1
KR
KR
Rotational K E KTotal K E R K
Percentage rotational K.E. = %1001 2
2
22
RKR
K
111
(3)2
2
RK22
2
11
KRR
E.KTotalE.KnalTranslatio
Comparison between physical quantities of linear motion and rotational motion
Translational motion Rotational motion
Linear displacement, d Angular displacement,
Linear velocity, V Angular velocity, w
Linear acceleration, dtvda Angular acceleration,
dtwd
Mass, m Moment of inertia, I
Linear Momentum, vmP Angular momentum, wIL
Force, amF Torque,dtLd
Newton's Second Law of Motion, A result similar to newtown's Second Law,
dtPdF
dtLd
Translational kinetic energy K = 2mv21
Rotational kinetic energy K = 21 I2
m2p2
I2
L2
Work, W = dF Work, W = Power, P = Fv Power, P = wEquations of linear motion taking place Equations of rotational motion taking placewith constant linear acceleration with constant angular acceleration :
v = VO + at w = tWO
d = 2O
1v t+ at2
20 at
21tW
2ad = 220v -v 2a = 2
02 WW
Law of conservation of linear momentum Law of conservation of angular momentum
when 0F then P is constant Impulse when 0P
then L is constant Impulse
linear 12 PPtF rotational 2 1t L L
Value of V, a, t for some Rolling Bodies
Shape of Body velocity velocity acceleration Time 22
RK
Ring/Hollow cylinder gh 21
singl sing21
sing2
1
Disc/Solid cylinder gh34 2
1
singl34
sing3
2sing
32
1
Solid Sphere gh7
10 21
singl7
10
sing7
5sing
145
2
Shell/Hollow spher gh56 2
1
singl56
sing5
3sing
103
2
Moment of inertia an radius of gyration for some symmetric bodies
Body Axis Figure I K For rollingbody 2
2
RK
Thin rod of Passing through its 2ML121
32L
-
Length L centre and per pendicularto its length
Ring of Any diameter 2MR21
2R
-radius R
Ring of Passing through its 2MR R 1radius R centre and perpen-
dicular to its plane
Circular disc Passing through its 2MR21
2R
21
radius R centre and perpen-dicular to its plane of
Circular disc Any diameter 2MR41
2R -
of radius R
113
Body Axis Figure I K For rollingbody 2
2
RK
Hollow Geometrical 2MR R 1cylinder of axis of theradius R cylinder
Solid cylinder Geometrical axis 2MR21
2R
21
of radius R of the cylinder
Solid sphere Any diameter 2MR52 R
52
52
of radius R
Hollow Any diameter 2MR32 R
32 2
3sphereof radius R
MCQFor the answer of the following questions choose the correct alternative from among the given ones.1. The centre of mass of a systems of two particles is
(A) on the line joining them and midway between them(B) on the line joining them at a point whose distance from each particle is proportional
to the square of the mass of that particle.(C) on the line joining them at a point whose distance from each particle inversely
propotional to the mass of that particle.(D) On the line joining them at a point whose distance from each particle is proportional
to the mass of that particle.2. Particles of 1 gm, 1 gm, 2 gm, 2 gm are placed at the corners A, B, C, D, respectively
of a square of side 6 cm as shown in figure. Find the distance of centre of mass of thesystem from geometrical centre of square.{A} 1 cm
{B} 2 cm
{C} 3 cm
{D} 4 cm3. Three particles of the same mass lie in the (X, Y) plane, The (X, Y) coordinates of their
positions are (1, 1), (2, 2) and (3, 3) respectively. The (X,Y) coordinates of the centre ofmass are{A} (1, 2) {B} (2, 2) {C} (1.5, 2) {D} (2, 1.5)
114
4. Consider a two-particle system with the particles having masses 1M , and 2M . If the firstparticle is pushed towards the centre of mass through a distance d, by what distance shouldthe second particle be moved so as to keep the centre of mass at the same position?
{A} 21
1
MMdM
{B} 21
2
MMdM
{C} 2
1
MdM
{D} 1
2
MdM
5. Four particles A, B, C and D of masses m, 2m, 3m and 5mrespectively are placed at corners of a square of side x asshown in figure find the coordinate of centre of mass take Aat origine of x-y plane.
{A}
10x7,x2 {B}
7x10,x2
{C}
7x10,
2x
{D}
10x7,
2x
6. From a uniform circular disc of radius R, a circular disc of radius 6R and having centre at a
distance + 2R from the centre of the disc is removed. Determine the centre of mass of remaining
portion of the disc.
{A} 70R
{B} 70R
{C} 7R
{D} 7R
7. A circular plate of uniform thickness has a diameter of 56 cm. A circular portion of diameter42 cm. is removed from +ve x edge of the plate. Find the position of centre of mass of theremaining portion with respect to centre of mass of whole plate.{A} - 7 cm {B} + 9 cm {C} - 9 cm {D} + 7 cm
8. Two blocks of masses 10 kg an 4 kg are connected by a spring of negligible mass and placedon a frictionless horizontal surface. An impulse gives velocity of 14 m/s to the heavier block inthe direction of the lighter block. The velocity of the centre of mass is :{A} 30 m/s {B} 20 m/s {C} 10 m./s {D} 5 m/s
9. A particle performing uniform circular motion has angular momentum L., its angular frequency isdoubled and its K.E. halved, then the new angular momentum is :{A} ½ {B} ¼ {C} 2L {D} 4L
10. A circular disc of radius R is removed from a bigger disc of radius 2R. such that the circumferencesof the disc coincide. The centre of mass of the remaining portion is R from the centre of massof the bigger disc. The value of is.{A} ½ {B} 1/6 {C} ¼ {D} 1/3
11. Three point masses M1, M2 and M3 are located at the vertices of an equilateral triangle ofside 'a'. what is the moment of inertia of the system about an axis along the attitude of the trianglepassing through M1, ?
{A} 4aMM
221 {B} 4
aMM2
32 {C} 4aMM
231 {D} 4
aMMM2
321
115
12. A body of mass m is tied to one end of spring and whirled round in a horizontal planewith a anstant angular velocity. The elongation in the spring is one centimeter. If the angularvelocity is doubted, the elongation in the spring is 5 cm. The original length of spring is…{A} 16 cm {B} 15 cm {C} 14 cm {D} 13 cm
13. A cylinder of mass 5 kg and radius 30 cm, and free to rotate about its axis, receives an angularimpulse of 3 kg M2S-1 initially followed by a similar impulse after every 4 sec. what is the angularspeed of the cylinder 30 sec after initial imulse ? The cylinder is at rest initially.
{A} 1Srad7.106 {B} 1Srad7.206 {C} 107.6 rad S-1 {D} 207.6 rad S-1
14. Two circular loop A & B of radi ra and rb respectively are made from a uniform wire. The ratioof their moment of inertia about axes passing through their centres and perpendicular to their planes
is 8II
A
B then rarb
Ra is equal to…
{A} 2 {B} 4 {C} 6 {D} 815. If the earth were to suddenly contract so that its radius become half of it present radius, without
any change in its mass, the duration of the new day will be…{A} 6 hr {B} 12 hr {C} 18 hr {D} 30 hr
16. In HC1 molecule the separation between the nuclei of the two atoms is about m10A1A27.1 10 .
The approximate location of the centre of mass of the molecule is iA with respect of Hydrogenatom ( mass of CL is 35.5 times of mass of Hydrogen){A} 1 {B} 2.5 {C} 1.24 {D} 1.5
17. Two bodies of mass 1kg and 3 kg have position vector kj2i and (-3i-2j+k) respectivelythe center of mass of this system has a position vector……
{A} k2i2 {B} kji2 {C} kji2 {D} kji 18. Identify the correct statement for the rotational motion of a rigid body
{A} Individual particles of the body do not undergo accelerated motion{B} The center of mass of the body remains unchanged.{C} The center of mass of the body moves uniformly in a circular path{D} Individual particle and centre of mass of the body undergo an accelerated motion.
19. A car is moving at a speed of 72 km/hr the radius of its wheel is 0.25m. If the wheels arestopped in 20 rotations after applying breaks then angular retardation produced by the breaksis ……
{A} -25.5 2srad {B} -29.5 2s
rad {C} -33.5 2srad {D} -45.5 2s
rad
20. A wheel rotates with a constant acceleration of 2.0 2secrad If the wheel start from rest. Thenumber of revolution it makes in the first ten seconds will be approximately.{A} 8 {B} 16 {C} 24 {D} 32
21. Two discs of the same material and thickness have radii 0.2 m and 0.6 m their moment of inertiaabout their axes will be in the ratio{A} 1 : 81 {B} 1 : 27 {C} 1 : 9 {D} 1 : 3
116
22. A wheel of mass 10 kg has a moment of inertia of 160 kg 2m about its own axis. Theradius of gyration will be _______ m.{A} 10 {B} 8 {C} 6 {D} 4
23. One circular rig and one circular disc both are having the same mass and radius. The ratio oftheir moment of inertia about the axes passing through their centres and perpendicular to theirplanes, will be……{A} 1 : 1 {B} 2 : 1 {C} 1 : 2 {D} 4 : 1
24. One solid sphere A and another hollow sphere B are of the same mass and same outer radii.The moment of inertia about their diameters are respectively AI and BI such that…
{A} BA II {B} BA II {C} BA II {D} dBdA
II
B
A (radio of their densities)
25. A ring of mass M and radius r is melted and then molded in to a sphere then the moment ofinertia of the sphere will be…..{A} more than that of the ring {B} Less than that of the ring{C} Equal to that of the ring {D} None of these
26. A circular disc of radius R and thickness R/6 has moment of inertia I about an axis passing throughits centre and perpendicular to its plane. It is melted and recasted in to a solid sphere. The momentof inertia of the sphere about its diameter as axis of rotation is …
{A} I {B} 8I2
{C} 5I
{D} 10I
27. One quater sector is cut from a uniform circular disc of radius R. Thissector has mass M. It is made to rotate about a line perpendicular toits plane and passing through the centre of the original disc. Its momentof inertia about the axis of rotation is…
{A} 2MR21 {B} 2MR4
1
{C} 2MR81 {D} 2MR2
28. A thin wire of length L and uniform linear mass density is bentin to a circular loop with centre at O as shown in figure. The momentof inertia of the loop about the axis xx' is ….
{A} 2
2L
8
{B}
3
2L
16
{C}3
25 L
16
{D}
3
23 L
8
29. Two disc of same thickness but of different radii are made of two different materials such that
their masses are same. The densities of the materials are in the ratio 1:3. The moment of inertiaof these disc about the respective axes passing through their centres and perpendicular to theirplanes will be in the ratio.{A} 1 : 3 {B} 3 : 1 {C} 1 : 9 {D} 9 : 1
117
30. Let I be the moment of inertia of a uniform square plate about an axis AB that passesthrough its centre and is parallel to two of its sides CD is a line in the plane of the platethat passes through the centre of the plate and makes an angle of Q with AB. The momentof inertia of the plate about the axis CD is then equal to….
{A} I {B} 2sinI {C} 2cosI {D} 2cosI 2
31. A small disc of radius 2 cm is cut from a disc of radius 6 cm. If the distance between theircentres is 3.2 cm, what is the shift in the centre of mass of the disc…{A} -0.4 cm {B} -2.4 cm {C} -1.8 cm {D} 1.2 cm
32. A straight rod of length L has one of its ends at the origin and the other end at x=L If the massper unit length of rod is given by Ax where A is constant where is its center of mass.{A} L/3 {B} L/2 {C} 2L / 3 {D} 3L / 4
33. A uniform rod of length 2L is placed with one end in contact with horizontal and is then inclinedat an angle to the horizontal and allowed to fall without slipping at contact point. When it becomeshorizontal, its angular velocity will be…..
{A} w =L2
sing3 {B} w = sing3
L2{C} w =
Lsing6
{D} w = singL
34. A cubical block of side a is moving with velocity V on ahorizontal smooth plane as shown in figure. It hits a ridge atpoint O. The angular speed of the block after it hits O is ….
{A} 34
va {B} 3
2v
a {C} 32
va {D} zero
35. Consider a body as shown in figure, consisting of two identicalbulls, each of mass M connected by a light rigid rod. If animpulse J = MV is imparted to the body at one of its ends,what would be its angular velocity. What is V ?{A} V / L {B} 2V / L {C} V / 3L {D} V / 4L
36. A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocityw. Two objects each of mass m are attached gently to the opposite ends of a diameter of thering. The ring will now rotate with an angular velocity….
{A} ( 2 )
2M m
M m
{B} 2
MM m {C}
MM m
{D} 2M m
M
37. A smooth sphere A is moving on a frictionless horizontal plane with angular speed and centreof mass velocity v. It collides elastically and head on with an identical sphere B at rest. Neglectfriction everywhere. After the collision, their angular speeds are A and B respectively, Then{A} A < B {B} A = B {C} A = {D} = B
38. Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m andof negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniformangular speed. The point on the rod through which the axis should pass in order that the workrequired for rotation of the rod is minimum, is located at a distance of …..{A} 0.4 m from mass of 0.3 kg {B} 0.98 m from mass of 0.3 kg{C} 0.7 m from mass of 0.7 kg {D} 0.98 m from mass of 0.7 kg
118
39. In a bicycle the radius of rear wheel is twice the radius of front wheel. If Fr and rr arethe radius, vF and vr are speed of top most points of wheel respectively then...{A} vr = 2vF {B} vF = 2vr {C} vF = vr {D} vF > vr
40. From a circular disc of radius R and mass 9M, a small disc of radiusR/3 is removed from the disc. The moment of inertia of the remainingportion about an axis perpendicular to the plane of the disc and passingthrough O is….
{A} 2MR4 {B} 2MR940
{C} 2MR10 {D} 2MR9
37
41. A child is standing with folded hands at the centre of a platform rotating about its central axisthe kinetic energy of the system is K. The child now stretches his arms so that the moment ofinertia of the system doubles. The kinetic energy of the system now is…{A} 2 K {B} K/2 {C} K/4 {D} 4K
42. If the earth is treated as a sphere of radius R and mass M. Its angular momentum about theaxis of rotation with period T is…..
{A} 3MR
T
{B} T
MR 2 {C}
T5MR2 2
{D} T5
MR4 2
43. If the angular momentum of any rotating body increases by 200%, then the increase in its kineticenergy will be…..{A} 400% {B} 800% {C} 200% {D} 100%
44. The M.I. of a body about the given axis is 1.2 kgm2 initially the body is at rest. In order toproduce a rotational kinetic energy of 1500 J. an angular acceleration of 2secrad25 must beapplied about that axis for duration of ….{A} 4 sec {B} 2 sec {C} 8 sec {D} 10 sec
45. An automobile engine develops 100kw when rotating at a speed of 1800 r.p.m. what torque doesit deliver ?{A} 350 Nm {B} 440 Nm {C} 531 Nm {D} 628 Nm
46. The moment of inertia of two rotating bodies A and B are AI and BI . BA II and their angular
momentum are equal. If their K.E. be AK and BK respectively then….
{A} KA, KB {B} 1KAKB {C} 1KB
KA {D} 21
KAKB
47. The centre of mass of the disc undergoes S.H.M. with angular frequency equal to..
{A} mk
{B} mk2
{C} m3k2
{D} m3k4
48. Three rings, each of mass P and radius are arranged as shown in thefigure the moment of inertia of the arrangement about YY' axis will be.
{A} 2P27 {B} 2P7
2 {C} 2P52 {D} 2P2
5
119
49. If distance of the earth becomes three times that of the present distance from the sun thennumber of days in one year will be ….
{A} 365 3 {B} 365 27 {C} 33365 {D} 33365
50. A solid sphere and a solid cylinder having same mass and radius roll down the same incline theratio of their acceleration will be….
{A} 15 : 14 {B} 14 : 15 {C} 5 : 3 {D} 3 : 551. The ratio of angular momentum of the electron in the first allowed orbit to that in the second
allowed orbit of hydrogen atom is ……
{A} 2 {B} 21
{C} ½ {D} 2
52. A player caught a cricket ball of mass 150 gm moving at a rate of 20 m/s If the catching processis Comlitad in 0.1 sec the force of the flow exerted by the ball on the hand of theplayer ….. N{A} 3 {B} 30 {C} 150 {D} 300
53. Two disc one of the density 7.2 gm/cc and other of density 8.9 gm/cc are of the same massand thickness their moment of inertia are in the ratio of ……
{A} 9.82.7 {B} 2.79.8
1
{C} 2.79.8 {D} 2.79.8
54. Two identical hollow spheres of mass M and radius R are joinedtogether and the combination is rotated about an axis tangential toone sphere and perpendicular to the line connecting their centers.The moment of inertia of the combination is ________.
{A} 10 2MR {B} 2MR34 {C} 2MR3
32 {D} 2MR334
55. A rod of length L rotate about an axis passing through its centre and normal to its length withan angular velocity . If A is the cross-section and D is the density of material of rod. Findits rotational K.E.
{A} 312 AL D {B} 31
6 AL D {C} 3124 AL D {D} 31
12 AL D
56. Initial angular velocity of a circular disc of mass M is w1 Then two spheres of mass m are attachedgently two diametrically oppsite points on the edge of the disc what is the final angular velocityof the disc?
{A} 1wM
mM
{B} 1wM
m4M
{C} 1wm4M
M
{D} 1wm2M
M
57. A circular disc x of radius R is made from an iron plate of thickness t. and another disc Y ofradius 4R is made from an iron plate of thickness t/4 then the rotation between the moment of
inertia XI and yI is ___________
{A} xy I64I {B} xy I32I {C} xy I16I {D} xy II
58. A Pulley of radius 2 m is rotated about its axis by a force Nt5t20F 2 where t is in sec
applied tangentially. If the moment of inertia of the Pulley about its axis of rotation is 10 ,KgM 2
120
the number of rotations made by the pulley before its direction of motion is reversed is :{A} more than 3 but less then 6 {B}more than 6 but less then 9{C} more than 9 {D} Less then 3
59. Two spheres each of mass M and radius R/2 are connected witha mass less rod of length 2R as shown in figure. What will bemoment of inertia of the system about an axis passing throughcentre of one of the spheres and perpendicular to the rod?
{A} 2MR521
{B} 2MR52
{C} 2MR25
{D} 2MR215
60. Four particles each of mass 'm' are lying symmetrically on therim of the disc of mass M and radius R moment of inertiaof this system about an axis passing through one of the particlesand perpendicular to plane of disc is _____
{A} 2MR16 {B} 2
RM16M32
{C} 2
RM12M32
{D} Zero
61. A mass M is supported by a mass less string wound arounda uniform cylinder of mass M and radius R as in figure. Withwhat acceleration will the mass fall on release?{A} 2/3g {B} g/2{C} g {D} 4g/3
62. Calculate rotational K.E. of earth due to its rotation about its own axis.24
eM =6×10 kg Re=6400 Km
{A} 29102.6 Joule {B} 29106.2 Joule {C} 291062 Joule {D} 291026 Joule63. A cord is wound round the circumference of wheel of radius r. the axis of the wheel is horizontal
and moment of inertia about it is I A weight mg is attached to the end of the cord and fallsfrom the rest. After falling through the distance h. the angular velocity of the wheel will be….
{A} mrI
gh2
{B} 21
2mrImgh2
{C} 2
1
2mr2Imgh2
{D} gh2
64. If rotational K.E. is 50% of translational K.E. then the body is …..{A} Ring {B} solid cylinder {C} Hollow sphere {D} Solid sphere
65. A meter stick of mass 400 gm is pivoted at one end and displaced through an angle 600 theincrease in its P.E. is ______ J.{A} 2 {B} 3 {C} Zero {D} 1
66. Tow uniform rod of equal length but different masses are rigidlyjoined to form L shaped body which is then pivoted as shown.If in equilibrium the body is in the shown configuration ratioM/m will be…….{A} 2 {B} 3
{C} 2 {D} 3
121
67. A light rod carries three equal masses A, B and C as shown inthe figure the velocity of B in vertical position of rod if it is releasedfrom horizontal position as shown in the figure is …….
{A} g2 {B} 7g18
{C} 3g4
{D} 8g3
68. A gramophone record of mass M and radius R is rotating with angular speed W. If two piecesof wax each of mass M are kept on it at a distance of R/2 from the centre on opposite sidethen the new angular velocity will be…..
{A} 2
{B} m
M m
{C} M
M m
{D} M mM
69. A solid cylinder rolls down a smooth inclined plane 4.8m high without slipping what is itslinear speed at the bottom of the plane, if it starts rolling from the top of the plane?(take g = 10 m/S2){A} 4 m/S {B} 2 m/S {C} 10 m/S {D} 8 m/S
70. The M.I of a disc of mass M and radius R about an axis passing
through the centre O and perpendicular to the plane of disc is 2
MR 2
.
If one quarter of the disc is removed the new moment of inertiaof disc will be…..
{A} 3
MR 2
{B} 4
MR 2
{C} 2MR83
{D} 2MR23
71. The moment of inertia of a uniform rod about a perpendicular axis passing through one of itsends is I1. The same rod is bent in to a ring and its moment of inertia about a diameter is
I2, Then 2
1I
I is.
{A} 3
2{B}
34 2
{C} 3
8 2{D}
316 2
72. A molecule consist of two atoms each of mass 'm' and separated by a distance of 'd' If 'K' isthe average rotational K.E. of the molecule at particular temperature then its angular frequency is….
{A} mk
d2
{B} mk
2d {C} k
md2 {D} km
4d
73. A car is moving with a constant speed the wheels of the car make 120 rotations per minutethe breaks are applied and the car comes to rest in 8 sec how many rotation are completedby the wheels before the car is brought to rest.{A} 4 {B} 6 {C} 8 {D} 10
74. The angular momentum of a wheel changes from 2L to 5L in 3 seconds what is the magnitudesof torque acting on it?{A} L {B} L/2 {C} L/3 {D} L/5
122
75. A uniform disc of mass 500kg and radius 2 m is rotating at the rate of 600 r.p.m. what isthe torque required to rotate the disc in the opposite direction with the same angular speed ina time of 100 sec ?{A} Nm600 {B} Nm500 {C} Nm400 {D} Nm300
76. The moment of inertia of a meter scale of mass 0.6kg about an axis perpendicular to the scaleand passing through 30 cm position on the scale is given by (Breath of scale is negligible). ________{A} 0.104 2mkg {B} 0.208 2mkg {C} 0.074 2mkg {D} 0.148 2mkg
77. How much constant force should be applied tangential to equator of the earth to stop its rotationin one day ?
{A} N103.1 22 {B} N1026.8 28 {C} 231.3 10 N {D} None of these78. A constant torque of 1500 Nm turns a wheel of moment of inertia 300 kg 2m about an axis
passing through its centre the angular velocity of the wheel after 3 sec will be…….... rad/sec{A} 5 {B} 10 {C} 15 {D} 20
79. A disc of mass M and radius R is rolling with angularspeed w on a horizontal plane, as shown in figure. Themagnitude of angular momentum of the disc about theorigin O is ______
{A} 21 MR2 {B} 2MR {C} 23 MR2 {D} 22 MR
80. A mass m is moving with a constant velocity along the line parallel to the x-axis, away from theorigin. Its angular momentum with respect to the origin{A} Zero {B} remains constant {C} goes on increasing {D} goes on decreasing
81. A body is rolling down an incline plane. If the rotational K.E. of the body is 40% of its translationalK.E. then the body is ….{A} ring {B} Cylinder {C} solid sphere {D} hollow sphere
82. A spherical ball rolls on a table without slipping, then the fraction of its total energy associatedwith rotation is .{A} 2/5 {B} 3/5 {C} 2/7 {D} 3/7
83. A binary star consist of two stars A (2.2 Ms) and B(mass 11Ms) where Ms is the mass of sun.They are separated by distance d and are rotating about their centre of mass, which is stationary.The ratio of the total angular momentum of the binary star to the angular momentum of star B.about the centre of mass is _____.{A} 6 {B} ¼ {C} 12 {D} ½
84. A small object of uniform density rolls up a curved surface with initial
velocity 'u'. It reaches up to maximum height of 23
4v
g with respect
to initial position then the object is ____.{A} ring {B} solid sphere {C} disc {D} hollow sphere
85. A particle of mass m slides down on inclined plane and reaches the bottom with linearvelocity V. If the same mass is in the form of ring and rolls without slipping down the same inclinedplane. Its velocity will be______.
{A} V {B} 2V {C} 2V
{D} 2V
123
GRAPHICAL QUESTIONS.86. Moment of inertia of a sphere of mass M and radius R is I. keeping mass constant if
graph is plotted between I and R then its form would be.
{A} {B} {C} {D}
87. According to the theorem of parallel axis 2cm mdII the graph between dI will be
{A} {B} {C} {D}
88. The graphs between angular momentum L and angular velocity w will be.
{A} {B} {C} {D}
89. The graphs between loge L and loge P is ____ where L is angular momentum and P islinear momentum
{A} {B} {C} {D}
90. Let Er is the rotational kinetic energy and L is angular momentum then the graph betweenLEr elogandLoge can be
{A} {B} {C} {D}
124
MACHING COLUMN TYPE91. Match list I with list II and select the correct answer.
List - I List - II System Moment of inertia
(x) A ring about it axis2
MR)1(2
(y) A uniform circular disc about it axis 2MR52)2(
(z) A solid sphere about any diameter 2MR57)3(
(w) A solid sphere about any tangent 2MR)4( s
2MR59)5(
Select correct option
Option? X Y Z W
{A} 2 1 3 4
{B} 4 3 2 5
{C} 1 5 4 3
{D} 4 1 2 3
92. Match the shape of graph with given pair of physical quantities.Physical Quantities{X} Moment of inertiadistance {Z}Angular momentum(L) letargen parallel axis angular geloaty (w){Y} re Elog Lloge {W} Lloge Ploge
Shape of graph
{P} {Q} {R} {S}
Select Correct OptionOption? X Y Z W
{A} R S P Q{B} Q P S R{C} S Q R P{D} P R Q S
125
ASSERTION - REASONING TYPE92. In the following questions statement - 1 (Assertion) is followed by statement - 2 (Reason). Each
question has the following four choices out of which only one choice is correct.{A} Statement - 1 is correct (true), Statement - 2 is true and Statement- 2 is correct explanationfor Statement - 1{B} Statement -1 is true, statement -2 is true but statement-2 is not the correct explanation fourstatement -1.{C} Statement - 1 is true, statement-2 is false{D} Statement-2 is false, statement -2 is true
93. Statement -1 — The angular momentum of a particle moving in a circular orbit with a constantspeed remains conserved about any point on the circumference of the circle.Statement -2— If no net torque outs, the angular momentum of a system is conserved.
94. Statement -1— A sphere and a cylinder slide without rolling from rest from the top of an inclinedplane. They will reach the bottom with the same speed.Statement -2 — Bodies of all shapes, masses and sides slide down a plane with the sameacceleration.
95. Statement -1— Friction is necessary for a body to roll on surfaceStatement -2— Friction provides the necessary tangential force and torque.
96. Statement -1— A body tied to a string is moved in a circle with a uniform speed. If the stringsuddenly breaks the angular momentum of the body becomes zero.Statement -2 — The torque on the body equals to the rate of change of angular momentum.
97. Statement -1 — If there is no external torque on a body about its centre of mass, then the velocityof the centre of mass remains constant.Statement -2 — The Linear momentum of an isolated system remains constant.
98. Statement -1 —Two cylinder one hollow and other solid (wood) with the same mass and identicaldimensions are simultaneously allowed to roll without slipping down an inclined plane from thesame height. The hollow will reach the bottom of inclined plane first.Statement -2 — By the principle of conservation of energy, the total kinetic energies of both thecylinders are identical when they reach the bottom of the incline.
99. Statement -1 —A thin uniform rod AB of mass M and length L is hinged at one end A to thehorizontal floor initially it stands vertically. It is allowed to fall freely on the floor in the vertical
plane, The angular velocity of the rod when its ends B strikes the floor Lg3
Statement -2 — The angular momentum of the rod about the hinge remains constant through outits fall to the floor.
100 .Statement -1 —If the cylinder rolling with angular speed- w. suddenly breaks up in to two equalhalves of the same radius. The angular speed of each piece becomes 2w.Statement -2—If no external torque outs, the angular momentum of the system is conserved.
126
PASSAGE BASED QUESTIONSPassage - I
A Solid sphere of mass M and radius R is released from rest at the top of a frictionless inclined planeof length 'd' and inclination ?.
In case (a) it rolls down the plane without slipping and in case (b) it slides down the plane
101. The ratio of the acceleration of the sphere in case (a) to that in case (b) is[A] 1[B] 2/3 [C] 5/7 [D] 7/9
102. The ratio of the velocity of the spheres when it reaches the bottom of the plane in case (a) tothat in case (b) is
[A] 2 [B] 32
[C] 75
[D] 37
103. The ratio of time taken by the sphere to reach the bottom in case (a) to that in case (b) as
[A] 1 [B] 23
[C] 2 [D] 57
Passage - IIA uniform disc of mass M and radius R rolls without slipping down a plane inclined at an angle with the horizontal.
104. The acceleration of the centre of mass of the disc is
[A] sing [B] 3sing2
[C] 3sing
[D] 3cosg2
105. The frictional force on the disc is
[A] 3sinMg
[B] 3sinMg2
[C] sinMg [D] None
106. The magnitude of torque acting on the disc is
[A] MgR [B] sinMgR [C] 3sinMgR2
[D] 3sinMgR
107. If the disc is replaced by a ring of the same mass M and the same radius R, the ratio of thefrictional force on the ring to that on the disc will be[A] 3/2 [B] 2 [C] 2 [D] 1
Passage - IIIA solid cylinder of mass M and R is mounted on a frictionless horizontalaxle so that it can freely rotate about this axis. A string of negligible massis wrapped round the cylinder and a body of mass m is hung from thestring as shown in figure the mass is released from rest then___
108. The acceleration with which the mass falls is
[A] g [B] Mmg
[C]
mMmg
[D] m2MMng2
127
109. The tension in string is
[A] mg [B] mMMmg [C] m2M
Mmg2 [D] 2
MgM m
110. The angular speed of cylinder is proportional to hn, where h is the height through which massfalls, Then the value of n is___[A] zero [B] 1[C] ½ [D] 2
111. The moment of inertia of a uniform circular disc of mass M and radius R about any of its diameteris ¼ 2MR , what is the moment of inertia of the disc about an axis passing through its centreand normal to the disc?
[A] 2MR [B] 2MR21
[C] 2MR23
[D] 2MR2
112. A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The angularvelocity of the cylinder when it reaches the bottom of the plane will be.
[A] ghR2
[B] 2
ghR2
[C] 3
ghR2
[D] ghR21
113. A cylinder of mass m and radius r is rotating about its axis with constant speed v Its kineticenergy is _____
[A] 2mv2 [B] mv2 [C] 12 mv2 [D] mv2
114. A circular disc of mass m and radius r is rolling on a smooth horizontal surface with a constantspeed v. Its kinetic energy is _____
[A] 14 mv2 [B] 1
2 mv2 [C] 34 mv2 [D] mv2
115. A solid sphere is rotating about a diameter at an angular velocity w. if it cools so that its radiusreduces to 1/n of its original value. Its angular velocity becomes_____
[A] n [B] 2n
[C] n [D] n2
116. In above question (115)If the original rotational K.E. of the sphere is K, Its new value will be_____
[A] 2nK [B] 4n
K [C] Kn2 [D] Kn4
117. A solid sphere is rotating about a diameter due to increase in room temperature, its volume increasesby 5%, If no external torque acts. The angular speed of the sphere will.[A] increase by nearly 1/3 % [B] decrease by nearly 1/3 %[C] increase by nearly ½ % [D] decrease nearly by ½ %
118. A cylinder of mass M has length L that is 3 times its radius what is the ratio of its momentof inertia about its own axis and that about an axis passing through its centre and perpendicularto its axis?
[A] 1 [B] 31
[C] 3 [D] 23
128
119. A uniform rod of length L is suspended from one end such that it is free to rotate about anaxis passing through that end and perpendicular to the length, what maximum speed must be impartedto the lower end so that the rod completes one full revolution?
[A] gL2 [B] gL2 [C] gL6 [D] gL22120. The height of a solid cylinder is four times that of its radius. It is kept vertically at time t=o on
a belt which is moving in the horizontal direction with a velocity v = 2t45.2 where v in m/sand t is in second. If the cylinder does not slip, it will topple over a time t = ____[A] 1 second [B] 2 sec. [C] 3 sec. [D] 4 sec.
121. The moment of inertia of a thin rod of mass M and length L about an axis passing through thepoint at a distance L/4 from one of its ends and perpendicular to the rod is _____
[A] 48ML7 2
[B] 12
ML2
[C] 9
ML2
[D] 3
ML2
122. A thin uniform rod AB of mass M and length L is hinged at one end A to the horizontal floor.Initially it stands vertically. It is allowed to fall freely on the floor in the vertical plane. The angularvelocity of the rod when its end B strikes the floor is ____
[A] Lg
[B] Lg2
[C] Lg3
[D] Lg2
123. A circular disc of radius R is free to oscillate about an axis passing through a point on its rimand perpendicular to its plane. The disc is turned through an angle of 60? and released. Its angularvelocity when it reaches the equilibrium position will be__
[A] R3g
[B] R3g2
[C] Rg2
[D] Rg22
124. The moment of inertia of a hollow sphere of mass M and inner and outer radii R and 2R aboutthe axis passing through its centre and perpendicular to its plane is
[A] 2MR23
[B] 2MR3213
[C] 2MR3531
[D] 2MR3562
125. If a is aerial velocity of a planet of mass M its angular momentum is[A] M [B] 2 MA [C] MA2 [D] 2AM
126. A wheel having moment of inertia 2 kg 2M about its vertical axis, rotates at the rate of 60 rpmabout this axis. The torque which can stop the wheels rotation in one minute will be..
[A] Nm15
[B] Nm18
[C] mN152
[D] Nm12
127. A wheel is rotating at 900 rpm about its axis. When power is cut off it comes to rest in 1 minute,the angular retardation in rad / sec is ___
[A] 2
[B] 4
[C] 6
[D] 8
128. What is the moment of inertia of a solid sphere of density and radius R about its diameter?
[A] 5105176
R [B] 5176105
R [C] 2105176
R [D] 2176105
R
129
129. A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocityis zero. In the first two second it rotate through an angle 1, in the next 2 sec. it rotates through
an angle 2, find the ratio 2
1
= ____
[A] 1 [B] 2 [C] 3 [D] 4130. A gramophone record of mass M and radius R is rotating at an angular velocity w. A win of
mass M is gently placed on the record at a distance R/2. from its centre. The new angular velocityof the system is
[A] mM2wM2 [B] m2M
wM2 [C] [D] M
wm
KEYNOTEQ.No. Ans Q.No. Ans Q.No. Ans Q.No. Ans Q.No. Ans
1 C 31 A 61 A 91 D 121 A2 A 32 C 62 B 92 B 122 C3 B 33 A 63 B 93 D 123 B4 C 34 64 B 94 A 124 D5 D 35 A 65 D 95 A 125 B6 A 36 B 66 D 96 D 126 A7 C 37 C 67 D 97 D 127 A8 C 38 B 68 C 98 D 128 B9 B 39 C 69 D 99 C 129 C
10 D 40 A 70 C 100 A 130 A11 B 41 B 71 C 101 C12 B 42 D 72 A 102 C13 A 43 B 73 C 103 D14 A 44 B 74 A 104 B15 A 45 C 75 C 105 A16 C 46 B 76 C 106 D17 B 47 D 77 A 107 A18 B 48 A 78 C 108 D19 A 49 C 79 C 109 D20 B 50 A 80 110 C21 A 51 C 81 C 111 B22 D 52 B 82 C 112 C23 B 53 C 83 A 113 D24 C 54 D 84 C 114 C25 B 55 C 85 C 115 D26 C 56 D 86 D 116 C27 A 57 A 87 C 117 B28 D 58 A 88 A 118 A29 B 59 A 89 B 119 C30 A 60 B 90 B 120 A
B
A
130
Hints1. Let Rcm is at origin
2211cm rmrmRM
1 1 2 2O = m r + m r
1 1 2 2m r m r
1 2
2 1
r m = r m – ve sign ignore as distance
2. Here^ ^
Ar = 0 i + 6J
^ ^
Br = 6 i + 6J
^ ^
Cr = 6 i + 0J
^ ^
Dr = 0 i + 0J
M = 1 + 1 + 2 + 2 = 6 gm
DDCCBBAAcm rmrmrmrmRM
3. The x and y co.ordinates of centre of mass are
1 1 2 2 3 3
1 2 3
m x + m x + m xx = m + m + m as m1 = m2 = m3
1 2 31 (x + x + x) = 23
similtaraly for y = 2
(x, y) = (2, 2)
4. m1x1 = m2x2
and m1(x1 – d) = m2 (x2 – d’)
1 2 m d = m d ' 1
2
m d d ' = m
5. Here at origin A (0, 0), B (x, 0), C (x, x), D (0, x)
(m 0) + (2mx) + (3mx) + (4m 0) = m + 2m + 3m + 4mcmx
cm(m × 0) + (2m × 0) + (3mx) + (4m )y =
m + 2m + 3m + 4mx
r1 r2
m1 m2
131
6. Let mass per unit area of disc = m
Mass of disc = M = 2R m
Mass of removed disc = 2 2R R mM ' = m =
6 36
from figrue R00' = 2
RM 0 = M' + (M – M')x2
RM 'x = M' + Mx2
M' Rx = M – M' 2
7. Let mass per unit area of Plote = m
Mass of whole Plote = 256M = m
2
Mass of removed part = 2
142M = m2
Mass of remaining Portion M2 = M – M1C.M of whole disc R = O at originC.M of removed Plote = r1 = 28 – 21 = 7cmC.M of remaining Portion r2 = ?
i i 2 2M O = M r + M r
8. The Velocity of C.M. is given by
1 1 2 2
1 2
m + m = m + mcmv vV
9. 21 1 1E = = = 2 2 2
I I L
2E 2E' L = L' = '
10. Let m is the mass of unit area then mass of big disc = 2(2R) m = M
Let m is the mass of unit area then mass of small disc = R2m = 1MM = 4
Mass of remaining Portion = M2 = M – M1
23MM = 4
132
Let G be the C.M of remaining PortionM2(OG) = M1(OO’)
3M M ( R) = R4 4
1 = 3
11. The moment of inertia about AD = ?I = m1 (Perpendicular distance of m1 from AD)2
+ m2 (Perpendicular distance of m2 from AD)2
+ m3 (Perpendicular distance of m3 from AD)2
2 2
2 39 9 0 + m + m 2 2
2
2 3a (m + m ) 4
12. Let L is original length & K spring anstant thenm (L + x1) w1
2 = kx1 & M (L + x2) w22 = kx2
Taking ratio2
1 1 12
2 2
L + = L +
x xx x
given x1 = 1cm, x2 = 5cm and 2 = 1
13. Inital angular momentum = Li = Ii = I x 0 = 0angular mumentum after initial inpulse = 3kgm2s–1
angular mumentum after initial 4 sec = 3 + 3 = 6kgm2s–1
angular mumentum after initial 8 sec = 6 + 3 = 9kgm2s–1
angular mumentum after initial 28 sec = 24 kgm2s–1
angular mumentum after initial 30 sec = 24 kgm2s–1
I = 24 here 2
2 2MR 1I = = 5 (0.3) = 0.225kgm2 2
24 24 = = = 106.7I 0.225
rad s–1
14. IA = ma ra2, IB = mb rb
2
2
b bB
A a a
m rI = I m r
Let K is the mass of unit length of the wire then
a am = (2 r )k and b bm = (2 r )k
b b
a a
m r = m r
2 3
B b b b
A a a a
I m r r = 8 = = I m r r
b
a
r = 2r
133
15. Let M be the mass and R1 the initial radius of the earth 1 is the angular veloalty of the rotation
of the earth, the duration T1 of the day is 1 21 2
2 2T = and T =
According to law of conservation of angular momentumI11 = I12
16. m1 = 1 m2 = 35.5 r1 = 0^
2r = 1.27 i
1 1 2 2cm
1 2
m r + m rr = m + m
17. 1 1 2 2cm
1 2
m r + m rr = m + m
18. Theory [B] The centre of mass of lucky remains uncharged.
19. oG 72 1000/3600 = = 80rad/sec0.25o r
= O, = 2 n = 2 20 = 40 rad As 2 2o2 = w – w
20. 21 = wot + t = 100 rad2
21. 2 2 21 1I = MR = ( R t )R2 2
As t are same
4I R4
1 1
2 2
I R = I R
22. I = MK2 = 160
2 160 160 K = = = 16m 10
K = 4
23.2
2
I ring MR 2 = = 1I Disc 1MR2
2 : 1
24. 2 2A Solid
2I = I = MR = 0.4MR5
and 2 2B hollow
2I = I = MR = 0.66MR3 A B I < I
25. Iring = MR12 As Volume and Mass remain same
2Solid 2
2I = MR5 R2 <<< R1
134
26. Volume of disc = 3
2 11 1
RV = R t = 6
3312
R 4 = R6 3
Volume of Sphere = 32 2
4V = R3 3 3
1 2R = 8R 1 2 R = 2R
I1 = M.I of disc = 21
1I = MR2
I2 = M.I of sphere = 2
2 12
2 MR I MR = = 5 5 2 5
27. Mass of the ontire disc would be AM and its moment of inertia about the given axis would
be 212 (AM)R . For the given section the moment of inertia about the since axis be one qvarter
of this is 212 (AM)R .
28. Mass per unit length of the wire = Mass of L length M = LWhen it is lent in form of circularring
2 r = LL R = 2
Moment of inertia of ring about given axis = 23 MR2
29. M.I of disc = 2
21 1 M 1 M MR = M = 2 2 t 2 t
22
M M = R = R t t
As their mass & thickness are some1I
30. Let IZ be the M.I of square plote about the axis passing through the centre and perpendicularto the plane of square, hence according to Perfendicular axis theorm.IZ = IAB + IA'B' Also IZ = ICD + IC'D'
As axis are symmetric IAB = IA'B' = ZI2
And ICD = IC'D' = ZI2
So we can say that IAB = IA'B' = ICD = IC'D' = I
31. Let the radius of complete disc is a & that of small disc is b After small disc is cut from completedisc let the C.M. shift to O2 at distance x2 flem original centre O.The Position of new C.M. is givenly let 6 is mass percunitarea.
2
cm 2 2
–6 b 1X = 6 a – 6 b
135
32. Let the mass of an element of length dx of the rod located at a distance X away from left
and is M dxL . the x cordinate of the C.M. is given by..
Total mass of rod = 1 2
0
ALAx dx = 2
1
cm 20
1 1x = xdm = x (Ax dx)M AL
2
33. By Conservation of Energy P.E. of rod = Rotational K.E. M g 12 sind =
12 I2 =
12
22mL
3
3g sin = L
As here l = 2L
3g sin = 2L
34. Angular momentum of Block w.r.t. O before collision with
O = MV a2 on collision the block will rotate about the side
passing through O. Now its angular momentum = IwAcc. to law of conservation of angular momentum
MV a2
= I 2 2M M +
6 2a a
3 = 4vv
here I is the M.I of block about the axis perpendicular to the plane passing through O.
35. Given system of two purtides will rotate about its centre of mass.
initial angular momentum = LM2
v
Final angular momentum 2L 2I = 2M
2
By law of conservation of angular momentum
LMV2
= 2L2M
2
V = L
36. Initial angular momentum of ring L = I = MR2Final angular momentum of ring and particles = (MR2 + 2mR2) 'As No external forque so According to Law of conservation of angular momentum.MR2 = (MR2 + 2mR2) '
wM ' = (M + 2m)
136
37. As it is head-on elastic collision between two idential balls there fore they will exchange theirlinear vecocity is A comes to rest and B starts moving with linear velocity V.As there is no friction any where, forque on both the spheres about their centre of mass iszero and their angular velocities remains unchanged.Therefore A = and B = 0
38. I = 0.3x2 + 0.7 (1.4 – x)2
For minimum work moment of inertia of the system should be minimum is
dI = 0 = 03 2x – 0.7 2 (1.4 – x) = 0dx
x = 0.98 m [B] 0.98 m from mass of 0.3 kg39. Angular speed for both wheels are different but lincar speed for both same so VF = Vr40. M.I of complete disc about O point
I Total = 212 (gM) R
Radius of removed disc = R3
As 2M = R i.e. 2M R
Mass of removed portion = gM Mg
M.I of removed disc about it own axis = 2 21 R MRM =
2 3 18
M.I of removed disc about O. I removed = 22 2
2cm
MR 2R MRI + Mx = + M = 18 3 2
Itotal = I removed disc + I remaining part
41.2LE = = K
2I given 1K
I If L is constant when child stretches his arms the moment of inertia
of system get doubled so kinetic energy will becomes half i.e. K2
42. Moment of inertia of sphere 22I = MR5
about its axis
L = I
43.2LE =
2I2E L
44. Rotational K.E. = 21 I = 15002
According to w = wo + t
137
45. = 1800 rpm 2 1800 = 60 rad/sec
60
P =
46. As 2IK =
2I
2
AA
JK = 2I
and2
BB
JK = 2I
47. As M is the mass of disc, the force is producing angular accoration in the disc, therefore
24 4k– = –m( ) = 3 3mkx x
48. For Ring 1 and 2 I1 and I2 about ' 23YY = MR2
and for Ring 3. I3 about ' 21YY = MR2
M.I. of system 1 2 3I = I + I + I
49.2 3
2 22 3
1 1
T r = T r
50. 2 2
g sina = (1 + k / R )
1
5g sin a = 7
and 2
2g sina = 3
51. nhL = 2
52. Impulse = Change in momentum = F t
L F = t
53.
221 1
1 12
22 22 2
1 M RI R2 = = 1I RM R2
54. Apply theorm of parallel axis
55. Rotational 21K.E. = I2
56. If = 0 I1w1 = I2w2
57. M ass = Volume = M = R2t
M.I of x is 2x 1 1
1I = m R2
58. Here direction of Motion will be reversed when force F = 0 or 20 t – 5r2 = 0 or t = 4sec.
If is angular accellaration then forque = I = F.r OR 210 = (20t – 5t ) 2 OR
2 = 4t – t and dw = dt
Also d = tdt
138
2 2 3 d = .t dt = (4t – t ) t dt = (4t t ) Taking intigration3 44t t = –
3 4 If n rotations are completed in As then Putting t = 4
4 64 64 = 2 n = – 64 = 3 3
n = 3.4 which is 3 < 3.4 < 6
59.2 2
22 R 2 RI = M + M(2R) + M 5 2 5 2
221 MR5
60. According to the Parallel axis theorm M.I of disc about an axis passing through particle (3)
and perpendicular to plane of disc is 2 2 21 3 MR + MR = MR2 2
61. In falling through a height h.P.E. = K.E
2 2 2 2 21 1 1 1 1 = + = + 2 2 2 2 2
mgh mv I mv mr
According to V2 – v2 = 2ad taking v = 0 & d = h, V2 = 2ah so 3mgh = m 2ah4
62. 2 24 6 2 37 22 2I = MR = 6 10 (6.4 10 ) = 9.83 10 kgm5 5
Angularvelocity of earth 2 T
K.E. of rotation of the earth 21 I 2
63. We know 22
kr
2gh = 1 +
v 2 2 2 2
V 2gh 2mgh = = 2 r + k mr + mk
64. kr = 50% kT
65. Centre of mass of stick is at midpoint when it is displaced by 600 Its c.m. rises up to height
h from figure = – cos = (1 – cos )2 2 2l l lh
so increase in P.E. = mgh = (1 – cos )2lmg
o10.4 10 (1 – cos60 )2
= 1 Joule
139
66. Net forque about O should be zero
Hence o o1 1Mg sin 60 = Mg sin 302 2
67. Loss in PE = gain in agular kinetic energy
22 + mg + mg = 3 3 2l l lMg l Iw
Here I = Moment of inertia about fixed point2 2
2 22 14 M + m m = m3 3 9l l l l
From fig. 2 21 14 36g 2 36 892 = = As V r = 2 9 14l 3 14 7
lg lmgl mll
68. I = I''2 2 2MR MR R = + 2m '
2 2 4
M = (M + m) '
69. 2
2
2gh = k1 +
V
r
70. New Mass 3 M4 M.I of disc =
2MR2
New MI = ' 2 21 3 3 = = 2 4 8
I M R MR
[C]
71. For a rod of mass M and length L., the MI about a perpendicular axis passing through one
and is 2
1MLI =
3
when it is bent to form a ring, then L = 2 RL R = 2
The M.I of the ring about its diameter is 2 2 2
2 2 2
MR ML MLI = = = 2 4 2 8
72. The M.I of the molecule = 2 2d dM + m
2 2
2 2d mdI = 2m = 4 2
The Rotational K.E. of the moldule 21( ) = 2
K I
2K = I
140
73.2 120 = = 4 rad/sec
60o
Now = 0 + t
Total angle descrited in 8 second is
2o
1 = w t + t2
74.dL 5L – 2L 3L = = = = Ldt 3 3
75. For disc 2MR 500 4I = =
2 2
angular speed = 2 600 =
60
so angular momentum L = I
final angular momentum in opposite direction = 2kgm–1000 20
sec
So change in angular momentum = 2L = 2 1000 20 kgm /sec
dL 2 1000 20 = = = 400 N.m.dt 100
76. By Parallel axis therom solved problem.77. 1 = 2 rad/day and 2 = 0, t = 1 day
2 1w – w = t
Forque required to stop the earth = T = I = F.R.
I.F = R
78.d = I = Idt
79. The disc have two types of motion translational and rotational so there will be two types ofangularmotion there fore total angular momentum should be total of both
T RL = L + L80. L = Momentum x perpenlicular distance between
point of rotation and line of action= m.V.y all remain constantL = remaing constant
81.40R.K.E. = T.K.E
100
2 2 21 4 1 1 = = 2 10 2 5
I mv mv 2
2 22
1 V 2 mk = 2 5
mvr
141
82. Total energy 2 2 2 2 2 21 1 1 2 1 = + = + 2 2 2 5 2
E I mv mr mr
83. Here C.M. wrt A.
5
5 5
11M 5d d = 11M + 2.2M 6
2 2
5 55d d = 2.2 M w = 55M 6 36AL w
2 2
5 5d d = 11 M = 11M 6 36BL
84. At the highest point the whole energy is conserted to P.E. of the object.
P.E. = mgh,23vh =
4g
2 23v 3m PE = mg = 4 4
vg
Total K.E. of body = 2
2 2 22
1 1 1 1 + = + 2 2 2 2
IVmv Iw mvr
Now, K.E. = P.E.
85. Here in first case 21 = 2
mg mvh = 2ghv
In second case
12
22
KR
2gh' = 1 +
v
As for the ring K = R ' = ghv
86. 22I = MR5
2 I R
This relation shows that the graph between I and R will be paracola symatric to I axis
87. Graph should be paracola symmetric to I - axis, but it should not pass from origin becausethere is a constant value Icm plesent for d = 0
88. L = Iw L w If I constant so graph between L and w will be straight line with constantstop and passing from origin.
89. = L r P
e e e log L = log P + log rIf graph is drawn between loge L and loge P then it will be straight line which will not passthrough origin at loge P = 0 loge L = loge r
90.2
rLE = 2I
so Log Er = 2 log L – log (2I)
So the graph rlog E log L will be straightline with constant stop and when Log L = 0.Log Er = – log (2I)
142
93. The correct choice is [D] since the centrifetal force is radial. Forque is zero so L = constant.94. The correct choice is [A] If a body slides down an indined plane its acaleration is
a = g (sin – cos ) which depends only on g, and .
98.cm
2
g sina = I1 + mR
Icm of hollow cylinder is loss, so it will have more accderation and will take less time to reachbottom.
99. Loss in P.E. is equal to gain in rotational K.E. As the centre of mass of the rod falls through
the distance L2
22 2L ML 3gMg I =
2 2 3 L
100. If I is the M.I. of the complete cylinder, The M.I of each Riece becomes I2 since L = I,
the angular speed of each Piece becomes 2.
101. In case (a) accelaration down the plane is
1
2
g sin 5 = = g sinI 71 +
MR
a
So 1
2
a 5 = a 7
In ase (b) 2a = g sin102. using V2 = 2ad we can find.
21 1 1
22 22
v v 5 = = v 7v
aa
103. From 21d = at2
, we find that 1
2
d 7 = d 5
104. Mg sin – f = Ma
= I . As 21I = MR2
, a = R
and = fR
Hence 21 a 1fR = MR = MRa2 R 2
105.1f = Ma2
Mg sin f = 3
106.MgR sin fR =
3
143
107. For a ring I = M R2 Correct choice is
108. Mg – T = Ma, T is Tension in string forque on cylinder = TR = I
22
I 1 a 1 = = MR = MaR 2 R 2
2mga = (M + 2m) [D]
109.MMgT =
M + 2M [D]
110. From conservation of energy we have
2 21 1 = + 2 2
Mgh MV I
2 2 2 2 2 21 1 1 1 + (2 + M)
2 2 2 4MR MR m R
12
24mgh =
(M + 2m) R
As 12 h
111. According to perpendicular axis theorm
2x y
1I = I = MR4 (Considering two perpendicular diameter in x & y directions)
So 2 2 2x y c
1 1 1I I = I = MR + MR = MR4 4 2
112. For solid cylinder 21I = MR2
As 2 21 1 = + 2 2
Mgh MV I
113. K.E. in Rotation 21 2
Iw Here 21I mr2
and V = r
therefore 2
2 21 1 1K.E = MR = 2 2 r 4
v mr
114. Rotational 21K.E = I2 & Translational 21K.E =
2mv
2 21 1K.E = + 2 2
Mv I
144
115. A solid sphere 22M.I = 5
mr
As 2
21 1 1 1
2 2 = I = = 5 5
rIw mr mn
2112 = = n
n
116. 21K = 2
I and 21 1 1
1 = 2
K I
22 2 2 2
1 1 2
1 2 r 1 2 1K = m = mr (n )2 5 n 2 5 n
2 2 21
1K = I kn2
n
117. 34V = r3
4log V = log + 3 log r3
differentiating we have dv dr = 3v r
dr 1 dv 1 1 = = 0.5% = %r 3 v 3 6
No external forque so I = constant 22 mr5
= constant
2 r = constant 2 log r + log = log c
dr d 2 + = 0r
d dr 1 1 = –2 = –2 % = – %
dt 6 3
-ve sign for decrensing
118.2 2
21 2
1 L = , = M + 2 4 12
RI MR I
119. In one full revolution the incrase in P.E = MgL, where M is the mass of rod.2
2 21 1 MgL = = 2 2 3
MLI
120. The cylinder will topple when the forque mgr equals the forque h ma2
so hma = mgr2
2gr g a = = h 2
Now V = 2.45t2 and dV = dt
a
2da = [2.45(t) ] = 4.gtdt
145
121. Using I = I cm + Md2 Parallel axis theorm
L L Ld = – = 2 4 4 and
2
cmMLI = 12
2 2 2ML ML 7ML I = + = 12 16 48
122. Loss in P.E = gain in Rotational K.E. centre of Mass of rod is at L2 .
MgLP.E = 2
Gain in 2
2 21 1 1R.E = = 2 2 2 3
MLI
2 2MgL ML 3g = = 2 6 L
123. P.E. at 600 = Mgh (1 – cos )
Eqvillibrium at 21K.E = 2
I
For I According to Parallel axis theorm (as d = R)
2 2 2 2cm
1 3I = I + MR = MR + MR = MR2 2
Here 2 o1 = Mgh (1 – cos 60 )2
I
124. We can obtain hollow sphere as it solid sphere of radius R is removed from a solid sphere
of 2R mass of hollow sphere M = M1 – M2 It is density 31
4M = (2 )3
R and 32
4M = 3
R
228 = 3
M R
Moment of inertia of hollow sphere
2 21 2
2 2I = M (2R) – M R5 5 By substituting the values of M1 and M2
125. Areal Velocity 2RA =
T and 2T =
2 2
2
R R 2AA = = = 2 2 R
= IL
146
126. Here I = 2kgm2 n1 = 60rpm = 1rps
T = ? n2 = 0 t = 1mn = 60 sec
2 1I ( – )T = I. = t
127. Angular retardation 2 1 – = t
128. For sphere 2 3 22 2 4 = = 5 5 3
I MR R R
129. From 21 = wt + t2
21
1 = 0 + (2) = 22
21 2
1 + = 0 + (4) = 82
130. Angular momentum of Recoul L = I Where 21I = MR2
Let w' be the angular velocity after putting coin of mass m at distance R2 from centre the angular
momentum of system L' ( I + mr2) ' since T = 0 so L' = L 2 (I + mr ) ' I
2
2 2 2 2
2
12
12
I ' = = = I + mr + mr 2mr1 +
MRMR
MR
but Rr = 2
2 M ' = 2M + m
147
Unit - 6Gravitational
148
SUMMARY1. Gravitational force between two point masses is
1 22
Gm mF
r
2. Acceleration due to Gravity
(I) on the surface of earth = 22 9.81GM ms
R
(II) At a height h from surface of earth
12
2(1 )1
g hg gRh
R
if h << R
(III) At a depth d form the surfce of earth
1 /(1 )dg gR
g1 = g if d = R i. e. on the surface of earth(IV) Effect of rotation of earth at latitude
g1 = g – R2 cos2 - at the equator = 0g1 = g – R2 = minimum value- At the pole = 900
g1 = g – R2 = maximum value- At the equator effect of rotation of earth is maximum and value of g is minimum.- At the pole effect of rotation of earth is zero and value of g is maximum.
3. Field Strength Gravitational field strength at a point in gravirtational field is defined as,
FE
m
= gravitational force per unit mass
Due to point mass
2
GMEr
(towards the mass) 2r
1 E
Due to solid sphere
inside points 3i
GME rR
At r = 0, E = 0 at the center
At r = R, 2
GMER
i.e. on the surface
out side points Eo = 2
GM orr 2o r
1E
149
At r , E 0 Due to a sphericell shell -
inside points E=0
outside points E0 = 2
GMr
just outside surface 2
GMER
on the surface E – r graph is discontinuous on the axis of a ring
32 2 2( )
r
GMrER r
At r = 0, E = 0 i.e. at the center
If r >> R, 2GMEr
, i.e. ring behaves as a points mass
As 0 r E4. Gravitational potential :-
(i) Gravitational potential at a point in a gravitational field is defined as the negative of workdone in moving a unit mass from infinity to that point per unit mass, thus
pp
w wV
m
(ii) Due to point mass
GmVrm
0 0 v as r and v as r(iii) Due to solid sphere
inside points 2 2 – (1.5 0.5 )GMVi R rR
as r = R GMV
R i.e. on the surface
V - r graph is parabola for inside points
out side points GMvr
(iv) Due to sphercal shell
inside points GMViR
outside points GMVi
R
150
(v) on the axis of a ring
2 2
GMVrR r
at r = 0, GMV
R i.e. at center
5. Gravitational potential Energy(i) This is the work done by gravitational forces in arranging the system from infinite seprationin the present position(ii) Gravitational potential energy of two point massess is
1 2Gm mUr
(iii) To find the gravitational points energy of more than two points masses we have to makepairs of masses. Neighter of the pair should be repeated. For example in case of four pointmasses.
4 3 4 2 4 1 3 2 3 1 2 1
43 23 41 32 31 21
m m m m m m m m m m m mU G
r r r r r r
for n point masses total number of pairs will be
( 1)2
n n
(iv) If a point mass m is plaled on the surface of earth the potential energy here is Uo
oGMmU
R
and potential energy at height h is
hGMmU(R h)
the difference in potential energy would be
U = Uh – Uo = mgh
1 + h/rIf h << R, U = mgh
6. Relation between field strength
E and potential V(i) if V is a function of only one variable (Say r) then
dVEdr
slope of U r graph
(ii) If V is funtion at three coordinates variable; e x, y, and z then
v v vˆ ˆ ˆE i j kx y z
151
7. Escape velocity(i) From the surface of earth
2
22 GM GMVe gR as gR R
=11.2 km / sec(ii) Escape velocity does not depend upon the angle which particle is projected form the surfaceand the mass of body
8. Motion of satellites
(i) orbital speed GMVor
(ii) time period 3/ 2 2 32T r T rGM
(iii) Kinetic energy 2
GMmK
r
(iv) Potential energy GMmUr
...
(v) Total Mechanical energy. GMmEr
9. Kepler’s laws- First law :Each planet moves in an elliptical orbit with the sun at one focus of ellipse- Second law :The radius vectors drawn form the sun to a planet, sweeps out equal area in equal time intervali.e. areal Velocity is constant.this law is derived from the law of conservation of angular momentum
2dA Ldt m
= constant here L is the angular momentum and m is mass of planet
- Third law2 3T r
where r is semi-major axis of elliptical path- The gravitational force acting between two bodies is always attractive. It is independent of medium
between bodies. It holds good over a wide range of distance. It is an action and reaction pair.It is conservative force. It is a central force and obey inverse square law as 21/F r
- The value of G is never zero any where but the value of g is zero at the center of earth.- the acceleration due to gravity is independent of mass, shape, size etc of falling body.- the rate of decrease of the acceteration due to gravity with height is twice as compared to that
with depth.- It the rate of rotation of earth increases the value of acceleration due to gravity decreases at
all points on the surface of earth except at poles.
152
- If the radius of planet decreasees by n% keeping its mas unchanged, the accelreotion due togravity on its surface increases by 2n%.
- If the mass of a planet increases by n% keeping its radius unchanged the acceleration due togravity on its surface increases by n%.
- The value of g at a location gives the value of intensity of gravitational field at the location.- The orbital velocity of a satellite is independent of mass of the satellite but depends upon the
mass and radius of planet around which the rotation is taking place- The value of orbital velocity for a satellite near the surface of earth is 7.92 kms–1.- The direction of orbital velocity of satellite at an instant is along the tangent to the orbital path
at that instant.- The work done by a satellite in a complete orbit is zero.- For a satellite orbiting close to the surface of earth (h << R), the time period of revolution
is 2 / 84.6R g minutes and angular velocity. 2 / 0.001237 / secg R radT
- A geostationary satellite revolves around the earth from west to east. Its period of revolutonis one day i.e. 24 hours. The orbital velocity of geostationary satellite is 3.08 kms–1. Its heightabove the surface of earth is about 36000 km. The relative angular velocity of geostationarysatellite w.r.t earth is zero.
- When a satelilte is orbiting in its orbit, no energy is required to keep it in its orbit.- When the total energy of a satellite is negative, it will be moving in either a circular or an elliptical
orbit.- When the total energy of a satellite is zero, it will escape away from its orbit and its path becomes
parabolic.- When the height of satellites is increased its potential energy will increase and K.E. will decrease.- When the velocity of satellite is increased, its total energy will increase and it will start orbiting
in a circular path of larger radius.- For a satellite orbiting in a circular orbit, the value of potenial energy is always greater than
its K.E.- If the velocity of a satellite orbiting the earth is increased by 41.4% or its K.E. is doubled,
then it will escape away from the gravitational field of earth- Escape velocity =. 2 orbital velocity- If the gravitational force is inversely proportional to the nth power of distance r, then the orbital
velocity of a satellite
/ 20
nV r and time period is ( 2)
2n
T r
- When a body is projected horizontally with velocity v, from any height from the surface of earth,
then the following possibilities are there.(i) If v < v0, the body fails to revolve around the earth and finally falls to the surface of earth.(ii) If v = v0, the body will revolve around the earth in circular orbit.(iii) If v < ve the body will revolve around the earth in elliptical orbit.(iv) If v = ve, the body will escape from the gravitational field of earth.(v) If v > ve the body will escape, following a hyperbolic path.
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- If earth suddently contracts to1 thn of its present size without any change in mass, the duration
of the day will be nearly 24/n2 hours.- Force function F(r) is related with potential energy function U(r) by a relaion
dUFdr
- A given planet will have atmosphere if the root mean square velocity of molecules in its atmosphere
(i.e. Vrms =3RTM
) is smaller than escape velocity for that planet.
- In the weightlessness state, the bodies donot have weight but they do possess inertia on accountof their mass. the bodies floating inside the space craft may collide with each other and crash.
- If a body is released from a height aqual to n times the radius of earth, then its striking velocityon the surface of earth is
21
ngRn
- If polar ice caps melt then moment of inertia, Angular velocity will decrease and period of rotaitonof earth increase.
- The, line joining the places on earth having same value of g are called isogams.- Gravity meter and Etvos gravity balance are used to measure changes in accelaration due to
gravity.
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Newton’s Law of GravitationFor the answer of the following questions choose the correct alternative from among thegiven ones.1. Two identical solid copper spheres of radius R are placed in contact with each other. The
gravitational force between them is proportional to(A) R2 (B) R-2 (C) R-4 (D) R4
2. The gravitational force Fg between two objects does not depend on(A) sum of the masses (B) product of masses(C) Gravitational constant (D) Distance between the masses
3. The atmosphere is held to the earth by(A) clouds (B) Gravity (C) Winds (D) None of the above
4. Two sphere of mass m1 and m2 are situated in air and the gravitational force between themis F. The space around the masses is now filled with liquid of specific gravity 3. The gravitationalforce will now be(A) F (B) 3F (C) F/3 (D) F/9.
5. A satellite of the earth is revolveing in a circular orbit with a uniform speed v. If the gravitationalforce suddennly disappears, the satellite will(A) Contineue to move with velocity v along the original orbit.(B) Move with a Velocity v, tangentially to the original orbit.(C) Fall down with increasing velocity.(D) Ultimately come to rest somewhere on the original orbit.
6. Correct form of gravitational law is
(A) 1 22
Gm m = –Fr (B) 1 2
2
Gm m = –Fr
(C) 1 2
3
Gm m ˆ = –F rr
(D) 1 2
3
Gm m = –F rr
7. Mass M is divided into two parts xM and (1 - x) M. For a given separation, the value ofx for which the gravitational force between the two pieces becomes maximum is(A) 1 (B) 2 (D) 1/2 (D) 4/5
8. The earth (mass = 6 x 1024 kg) revolves around the sun with angular velocity 2 10–7 rad/sec in a circular orbit of radius 1.5 x 108 km. The force exerted by the sun on the earth is= ................... N(A) 18 1025 (b) zero (C) 27 1039 (D) 36 1021
9. Two particle of equal mass go round a circle of radius r. Under the action of their mutualgravitational force.The speed of each particle is =..................
(A)1 12r Gm
(B) 2Gm
r (C)
12
Gmr
(D)4Gm
r
10. The distance of the moon and earth is D the mass of earth is 81 times the mass of moon.At what distance from the center of the earth, the gravitational force will be zero(A) D/2 (B) 12D/3 (C) 4D/3 (D) 9D/10
11. One can easily “ Weight the earth ” by calculating the mass of earth using the formula(in usual notation)
(A) RegG (B) g/G Re2 (C)
2ReGg (D)
3ReGg
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12. Three equal masses of m kg each are plced the vertices of an equilateral triangle PQR and
a mass of 2m kg is placed at the centroid 0 of the triangle which is at a distance of 2 mfrom each of vertices of triangle. The force in newton. acting on the mass 2m is = ............(A) 2 (B) 1 (C) 2 (D) zero
13. Which of the follwing statement about the gravitational constant is true(A) It is a force(B) It has no unit(C) It has same value in all system of unit(D) It depends on the value of the masses.
14. Two point masses A and B having masses in the ratio 4 : 3 are seprated by a distance oflm. When another point mass c of mass M is placed in between A and B the forces A andC is 1/3rd of the force between band C, Then the distance C form A is = ............... m(A) 23 (B) 1/3 (C) 1/4 (D) 2/7
15. The gravitational force between two point masses m1 ans m2 at separation r is given by
1 22
m mF Gr
The constant k ...............
(A) Depends on system of units only.(B) Depends on medium between masses only.(C) Depends on both (a) and (b)(D) is independent of both (a) and (b)
Acceleration Due to Gravity16. As we go from the equator to the poles, the value of g ...............
(A) Remains constant (B) Decreases (C) Increases (D) Decreases upro latitude of 45o
17. If R is the radius of the earth and g the acceleration due to gravity on the earth’s surface,the mean density of the earth is = ...............
(A) 4 3/G gR (B) 3 4/R gG (C) 3 4/g RG (D) 12/RG g18. The radius of the earth is 6400 km and g=10ms-2. In order that a body of 5 kg weights zero
at the equator, the angular speed of the earth is = ............... rad/sec(A) 1/80 (B) 1/400 (C) 1/800 (D) 1/600
19. The time period of a simple pendulum on a freely moving artificial satellite is ............... sec(A) 0 (B) 2 (C) 3 (D) Infinite
20. A spherical planet far out in space has mas Mo and diameter Do. A particle of m falling nearthe surface of this planet will experience an acceleration due to gravity which is equal to
(A) 20 /GM Do (B) 24 /mGMo Do (C) 24 /GMo Do (D) 2/GmMo Do
21. A body weights 700 g wt on the surface of earth How much it weight on the surface of planetwhose mass is 1/7 and radius is half that of the earth(A) 200g wt (B) 1400g wt (C) 50 g wt (D) 300g wt.
22. The value of g on he earth surface is 980 cm/sec2. Its value at a height of 64 km from theearth surface is ............... cm5–2
(a) 960.40 (B) 984.90 (C) 982.45 (D) 977.55
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23. If earth rotates faster than its present speed the weight of an object will.(A) increases at the equator but remain unchanged of the poles.(B) Decreases at the equator but remain unchanged at poles.(C) Remain unchanged at the equator but decreases at poles.(D) Remain unchanged at the equator but increases at the poles.
24. The moon’s radius is 1/4 that of earth and its mass is 1/80 times that of the earth. If g representsthe acceleration due to gravity on the surface of earth, that on the surface of the moon is..............(A) g/4 (B) g/5 (c) g/6 (D) g/8
25. The depth of at which the value of acceleration due to gravity becomes 1/n the time the valueof at the surface is (R = radius of earth)
(A) R/n (B) ( 1)n
nR
(C) R/n2 (D) 1
nRn
26. If the density of smalll planet is that of the same as that of the earth while the radius of theplanet is 0.2 times that of lthe earth, the gravitational acceleration on the surface of the planetis ...............(A) 0.2g (B) 0.4g (c) 2g (D) 4g
27. If mass of a body is M on the earth surface, than the mass of the same body on the moonsurfae is(A) M/6 (B) 56 (C) M (D) None of these
28. An object weights 72 N on the earth. Its weight at a height R/2 from earth is = .............. N(A) 32 (B) 56 (C) 72 (D) zero
29. If the radius of earth is R then height ‘h’ at which value of ‘g’ becomes one - fourth is(A) R/4 (B) 3R/4 (C) R (D) R/8
30. If the mass of earth is 80 times of that of a planet and diameter is double that of planet and‘g’ on the earth is 9.8 ms-2 , then the value of ‘g’ on that planet is = ............... ms-2
(A) 4.9 (B) 0.98 (c) 0.49 (D) 4931. Assuming earth to be a sphere of a uniform density, what is value of gravitational acceleration
in mine 100 km below the earth surface = ............... ms-2
(A) 9.66 (B) 7.64 (C) 5.00 (D) 3.132. Let g be the acceleration due to gravity at earth’s surface and k be the rotational K.E. of earth
suppose the earth’s radius decreases by 2% keeping alt other quantities same then(A) g decreases by 2% and K decreases by 4 %(B) g decreases by 4% and K increases by 2%(C) g increases by 4% and K increases by 4%(D) g decreases by 4% and K increases by 4%
33. A body weight 500 N on the surface of the earth. How much would it weight half way belowthe surface of earth(A) 125N (B) 1250N (C) 500N (D) 1000N
34. The radii of two planets are respectively R1 and R2 and their densities are respectively 1 and2 the ratio of the accelerations due to gravity at their surface is ...............
(A) 1 21 2 2 2
1 2
: .g gR R
(B) 1 2 1 2 1 2: :g g R R
(C) 1 2 1 2 2 1: :g g R R (D) 1 2 1 1 2 2: :g g R R
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35. At what height over the earth’s pole, the free fall acceleration decreases by one percent= .................. km (Re = 6400 km).(A) 32 (B) 80 (C) 1.253 (D) 64
36. Weight of a body is maximum at(A) moon (B) poles of earth (C) Equator of earth (D) Center of earth
37. At what distance from the center of earth, the value of aceeleration due to gravity g will behalf that of the surfaces (R = Radius of earth)(A) 2R (B) R (C) 1.414 R (D) o.414 R
38. The acceleration due to gravity near the surface of a planet of radius R and density d is proportionalto(A) d/R2 (B) dR2 (C) dR (D) d/R
39. The acceleration due to gravity is g at a point distance r from the center of earth R. if r < Rthen(A) g r (B) g r2 (c) g r-2 (D) g r-1
40. The density of a newly discoverd planet is twice that of earth. The acceleration due to gravityat the surface of the planet is equal to that at the surface of earth. If the radius of the earthis R, the radius of planet would be .....................(A) 2R (B) 4R (C) 1/4 R (D) .......
41. Density of the earth is doubled keeping its radius constant then acceleration, due to gravity willbe .................. ms-2(g = 9.8 ms2)(A) 19.6 (B) 9.8 (C) 4.9 (D) 2.45
42. Weight of body of mass m decreases by 1% when it is raised to height h above the earth’ssurface. If the body is taken to a depth h in a mine. change in its weight is(A) 2% decreases (B) 0.5% decreases(C) 1% increases (D) 0.5% increases
43. If density of earth increased 4 times and its radius becomes half of then out weight will be...(A) Four times it present calue (B) doubled(C) Remain same (D) halved
44. A man can jump to a height of 1.5 m on a planet A what is the height ne may be able tojump on another planet whose density and radius are respectively one- quater and one- thirdthat of planet A(A) 1.5 m (B) 15 m (C) 18 m (D) 28 m
45. If the value of ‘g’ acceleration due to gravity, at earth surface fis 10ms–2. its value in ms–2
at the center of earth, which is assumed to be a sphere of Radius ‘R’ meter and uniform density is(A) 5 (B) 10/R (C) 10/2R (D) zero
46. A research satellite of mass 200 kg. circles the earth in an orbit of avrage radius 3R/2 whereR is radius of earth. Assuming the gravitational pull 10 N, the pull on the satellite will be =..........N(A) 880 (B) 889 (C) 890 (D) 892
47. Acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth. If the
ratio of densities of earth e and moon m is 5 3e
m
then radius of moon Rm in terms
of Re will be ...............
(A) 5 Re18 (B)
1 Re6 (C)
3 Re16 (D)
1 Re2 3
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48. The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R fromthe surface of the earth is = ...............(g = acceleration due to gravity at the surface of earth)(A) g/9 (B) g/3 (C) g/4 (D) 9
49. The height at which the weight of a body becomes 1/16 th its weight on the surface of(radius R) is(A) 3R (B) 4R (C) 5R (D) 15R
50. A spherical planet has a mass Mp and diameter Dp A particle of mass m falling freely nearthe surface of this planet will experience an acceleration due to gravity, equal to
(A) 2
4 GMpDp (B) 2
GMpmDp (C) 2
GMpDp (D) 2
4GMpmDp
51. Assuming the earth to have a constant density, point out which of following curves show thevariation acceleration due to gravity from center of earth to points far away from the surfaceof earth ...............
(D) None of these
Gravitational potential, Energy and Escape Velocity52. In a gravitational field, at a point where the gravitational potential is zero
(A) The gravitational field is necessarily zero(B) The gravitational field is not necessarily zero(C) Nothing can be said definetely, about the gravitational field(D) None of these
53. The mass of the earth is 6.001024 kg and that of the moon is 7.401022 kg. The constantof gravitation G = 6.6710-11 Nm2 kg–2. The potential energy of the system is -7.791028
Joules. the mean distance between the earth and moon is = ............. meter.(A) 3.80108 (B) 3.37108 (C) 7.60108 (D) 1.90102
54. The masses and radii of earth and moon are M1, R1 and M2, R2 respectively. Their centresare d distance of apart. The minimum velocity with which a particle of mass m should be projectedfrom a point midway between their centres so that it esacapes to infinity is...............
(A) 1 22 ( )G M Md
(B) 1 222 ( )G M Md
(C) 1 22 ( )Gm M Md
(D) 1 2
1 2
( )2
( )Gm M M
d R R
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55. A rocket is launched with velocity 10 kms-1. If radius of earth is R then maximum height attainedby it will be = ..............(A) 2R (B) 3R (C) 4R (D) 5R
56. What is the intensity of gravitational field at the center of spherical shell
(A) 2
Gmr (B) g (C) zero (D) None of these
57. Escape velocity of a body of 1 kg. on a planet is 100 ms-1. Gravitational potential energy ofthe body at the planet is =............... J(A) -5000 (B) -1000 (C) -2400 (D) 5000
58. A body of mass m kg starts falling from a point 2R above the earth’s surface. Its K.E. whenit has fallen to a point ‘R’ above the Earth’s surface = .....................[R - Radius of Earth, M-mass of Earth G-Gravitational constant]
(A) 12
GMmR (B)
16
GMmR (C)
23
GMmR (D)
13
GMmR
59. The Gravitational P.E. of a body of mas m at the earth’s surface is -mgRe. Its gravitationalpotential energy at a height Re from the earth’s surface will be = ............ here(Re is the radius of the earth)
(A) –2 mgRe (B) 2 mgRe (C) 1 Re2
mg (D)1 Re2
mg
60. A body is projected vertically upwards from the surface of a planet of radius R with a velocityequal to half the escape velocity for that planet. The maximum height attained by the body is..........(A) R/3 (B) R/2 (C) R/4 (D) R/5
61. Energy required to move a body of mass m from an orbit of radius 2R to 3R is ................
(A) 2
GMm 12R (B) 23
G mR
(C) 8G m
R
(D) 6G m
R
62. Radius of orbit of satellite of earth is R. Its K.E. is proportional to
(A) 1R (B)
1R (C) R (D) 3/2
1R
63. A particle falls towards earth from infinity. It’s velocity reaching the earth would be.............
(A) infinity (B) 2gR (C) 2 gR (D) zero64. The escape velocity of a sphere of mass m from earth having mass M and Radius R is given by
(A) 2GM
R(B) 2 GM
R(C)
2GMmR
(D) GM
R65. The escape velocity for a rocket from earth is 11.2 kms–1 value on a planet where acceleration
due to gravity is double that on earth and diameter of the planet is twice that of earth will be= ........... kms–1
(A) 11.2 (B) 22.4 (C) 5.6 (D) 53.666. The escape velocity from the earth is about 11 kms–1. The escape velocity from a planet having
twice the radius and the same mean density as the earth is =............ kms–1.(A) 22 (B) 11 (C) 5.5 (C)15.5
160
67. If g is the acceleration due to gravity at the earth’s surface and r is the radius of the earth,the escape velocity for the body to escape out of earth’s gravitational field is.................
(A) gr (B) 2gr (C) g/r (D) r/g
68. The escape velocity of a projectile from the earth is approximately(A) 11.2 kms–1 (B) 112 kms–1 (C) 11.2 ms–1 (D) 1120 kms–1
69. The escape velocity of a particle of mass m varies as...................
(A) m2 (B) m (C) m0 (D) m-1
70. The escape velocity of an object from the earth depends upon the mass of earth (M), its meandensity (), its radius (R) and gravitational constant (G), thus the formula for escape veloctiy is
(A) 83
R G (B) 83
M GR (C) 2GMR (D) 2
2
GMR
71. Two small and heavy sphere, each of mass M, are placed distance r apart on a horizontal surfacethe gravitational potential at a mid point on the line joining the center of spheres is
(A) zero (B) GM
r (C)
2GMr
(D) 4
GMr
72. The escape velocity of a body from earth’s surface is Ve. The escape velocity of the samebody from a height equal to 7 R from earth’s surface will be
(A) 2Ve
(B) 2Ve
(C) 2 2Ve
(D) 4Ve
73. An artificial satellite is revolving round the earth in a circular orbit. its velocity is half the escapevelocity. Its height from the earth surface is = ............... km
(A) 6400 (B) 12800 (C) 3200 (D) 1600
74. The escape velocity of a planet having mass 6 times and radius 2 times as that of earth is..........
(A) 3 Ve (B) 3 Ve (C) 2 Ve (D) 2Ve
75. There are two planets, the ratio of radius of two planets is k but the acceleration due to gravityof both planets are g what will be the ratio of their escape velocity.
(A) 1/ 2kg (B) 1/ 2kg (C) 2kg (D) 2kg
76. The escape velocity of a body on the surface of the earth is 11.2 km/sec. If the mass of theearth is increases to twice its present value and the radius of the earth becomes half, the escapevelocity becomes = ............... kms–1
(A) 5.6 (B) 11.2 (C) 22.4 (D) 494.8
77. Given mass of the moon is 181 of the mass of the earth and corresponding radius is
14 of
the earth, If escape velocity on the earth surface is 11.2 kms–1 the value of same on the surfaceof moon is = ....... kms–1.(A) 0.14 (B) 0.5 (C) 12.5 (D) 5
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78. 3 particle each of mass m are kept at vertices of an equilateral triangle of side L. The gravitationalfield at center due to these particies is ................
(A) zero (B) 2
3GML (C) 2
9GmL (D) 2
123
GmL
79. Escape velocity on the surface of earth is 11.2 kms–1 Escape velocity from a planet whose massesthe same as that of earth and radius 1/4 that of earth is = ....... kms–1
(A) 2.8 (B) 15.6 (C) 22.4 (D) 44.880. The velocity with which a projectile must be fired so that it escapes earth’s gravitational does
not depend on ..................(A) mass of earth(B) Mass of the projectile(C) Radius of the projectiles’s orbit(D) Gravitational constant
81. The escape velocity for a body projected vertically upwards from the surface of earth is 11kms-1. If the body is projected at an angle of 450 with the vertical, the escape velocity willbe ............kms-1.
(A) 112 (B) 11 2 (C) 22 (D) 11
82. The acceleration due to gravity on a planet is same as that on earth and its radius is four timesthat of earth. What will be the value of escape velocity on that planet if it is Ve on the earth(A) Ve (B) 2 Ve (C) 4 Ve (D) Ve/2
83. A particle of mass 10g is kept on the surface of a uniform sphere of mass 100 kg and radiius10 cm. Find the work to be done aginst the gravitational force between them to take the particleis away from the sphere (G = 6.67 10-11 SI unit)(A) 6.6710-9 J (B) 6.6710-10 J (C) 13.3410-10 J (D) 3.3310-10 J
84. A particle of mass M is situated at the center of a spherical shell of same mass and radiusa the magnitude of gravitational potential at a point situated at a / 2 distance from the centerwill be
(A) 4GM
a (B) GM
a (C) 2GM
a (D) 3GM
a
85. The mass and radius of the sun are 1.99 x 1030 kg and R = 6.96 x 108 m. The escapevelocity of rocket from the sun is =.......... km/sec(A) 11.2 (B) 12.38 (C) 59.5 (D) 618
86. The mass of a space ship is 1000 kg. It is to be lauched from earth’s surface out into freespace the value of g and R (radius of earth) are 10ms-2 and 6400 km respectively the requiredenergy for this work will be = ................. J(A) 6.41011 (B) 6.4108 (C) 6.4109 (D) 6.41010
162
87. The diagram showing th variation of gravitational potential of earth with distance from the centerof earth is
(A) (B)
(C) (D)
88. A shpere of mass M and Radius R2 has a concentric cavity of Radius R1as shown in figure.The force F exerted by the shpere on a particle of mass m located at a distance r from thecenter of shhere varies as )r(O
(A) (B)
(C) (D)
163
89. Which one of the following graphs represents correctly the variation of the gravitational fieldwith the distance (r) from the center of spherical shell of mass M and radius a
(A) (B)
(C) (D)
90. Which of the following graphs represents the motion of a planet moving about the sun.
(A) (B)
(C) (D)
91. The curves for P.E. (U) K.E. (Ek) of two particle system ae shown in figure . At what pointssystemwill be bound.(A) Only at point D(B) Only at point A(C) At point D and A(D) At points A, B and C
Ek
164
92. The correct graph representing the variation of total energy (E) kinetic energy (K) and potentialenergy (U) of a satellite with its distance from the centre of earth is...........
(A) (B)
(C) (D)
93. A shell of mass M and radius R has a point mass m placed at a distance r from its center.The gravitational potential energy U(r) –v will be
(A) (B)
(C) (D)
Motion of satellite94. If Ve and Vo are represent the escape velocity and orbital velocity of satelllite correspoinding
to a circular orbit of radius r, then
(A) Ve = Vo (B) 2 Vo Ve (C) / 2Ve Vo (D)ve and vo are not related
Ener
gyEn
ergy
Ener
gyEn
ergy
165
95. If r represents the radius of the orbit of a saltellite of mass m moving around a planet of masM, the velocity of the satellite is given by
(A) 2 gMr
(B) 2 GMmr
(C) GMr
(D) 2 GMr
96. Two satellites of mass m1 and m2 (m1 > m2 ) are revolving round the earth in cirular orbitsof r1 and r2 ( r1 > r2) respectively. Which of the following statement is true regarding theirspeeds V1 and V2
(A) V1 = V2 (B) V1 < V2 (C) V1 > V2 (D) 1 2
1 2
V Vr r
97. A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance fromthe centere of earth in new orbit is two times of the earlier orbit. The time period in secondorbit is .........hours.
(a) 4.8 (B) 48 2 (C) 24 (D) 24 2
98. As astronaut orbiting the earth in a circular orbit 120 km above the surface of earth,gently dropsa spoon out of space-ship. The spoon will(A) Fall vertically down to the earth(B) move towards the moon(C) Will move along with sapace - ship(D) Will move in an irregualr way then fall down to earth
99. The period of a satellite in cirular orbit around a planet is independent of(A) the mass of the planet (B) the radius of the planet(C) mass of the satellite (D) all the three parameters (A), (B) and (C)
100. Two satellites A and B go round a planet p in circular orbits having radii 4 R and R respectivelyif the speed of the satellite A is 3V, the speed if satellite B will be(A) 12V (B) 6V (C) 4/3 V (D) 3/2 V
101 A small satellite is revolveing near earth’s surface. Its orbital velocity will be nearly= .......... kms–1.
(A) 8 (B) 4 (C) 6 (D) 11.2102 A satellite revolves around the earth in an elliptical orbit. Its speed
(A) is the same at all points in the orbit(B) is greatest when it is closest to the earth(C) is greatest when it is farthest to the earth(D) goes on increasing or decresing continuously depending upon the mass of the satellite
103. If the height of a satellite from the earth is negligible in comparison of the radius of the earthR, the orbital velocity of the stellite is ............
(A) gR (B) gR/2 (C) /g R (D) gR
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104 A satellite is moving around the earth with speed v in a circular orbit of radius r. If the orbitradius is decreasd by 1% its speed will(A) increase by 1% (B) increase by 0.5%(C) decrrease by 1% (C) Decrrease by 0.5%
105. orbital velocity of an artifiicial satellite does not depend upon(A) mass of earth (B) mass of satellite(C) radius of earth (D) acceleration due to gravity
106. orbital velocity of eath’s satellite near the surface is 7 kms–1. when the radius of orbit is 4 timesthat of earth’s radius, then orbital velocity in that orbit is =........kms-1
(A) 3.5 (B) 17 (C) 14 (D) 35107. Two identical satellites are at R and 7R away from each surface, the wrong statement is
(R - Radius of earth)(A) ratio of total energy will be 4(B) ratio of kinetic energes will be 4(C) ratio of potential energies will be 4(D) ratio of total energy will be 4 but ratio of potential and kinetic energies will be 2
108. Which one of follwoing sttements regarding artificial satellite of earth is incorrect(A) The orbital velocity depends on the mass of the satellite(B) A minimum velocity of 8kms-1 is required by a satellite to orbit quite close to the earth.(C) The period of revolution is large if the radius of its orbit is large(D) The height of geostationary satellite is about 36000 km from earth
109. The weight of an astronuat, in an artificial satellite revolving around the earth is(A) zero(B) Equal to that on the earth(C) more than that on earth(D) less than that on the earth
110. The distance of a geo-stationary satellite from the center of the earth (Radius R= 6400km)is nearest to(A) 5R (B) 7R (C) 10R (D) 18R
111. A geo-stationary satellite is orbiting the earth of a height of 6R above the surface of earth, Rbeing the radius of earth. The time period of another satellite at a height of 2.5 R from thesurface fo earth is = ...............hr(A) 6 (B) 6 2 (C) 10 (D) 6 / 2
112. If the gravitational force between two objects were proportional to 1R (and not as 2
1R )where
R is separation between them, then a particle in circular orbit under such a force would haveits orbital speed v proportional to
(A) 2
1R (B) R0 (C) R1 (D)
1R
113. A satellite moves around the earth in a circular orbit of radius r with speed v, If mass of thesatellite is M , its total energy is
(A) 212
MV (B) 212
MV (C) 232
MV (D) MV2
167
114. A satellite with K.E. Ek is revolving round the earth in a circular orbit. How much more K.E.should be given to it so that it may just escape into outerspace ?
(A) Ek (B) 2 Ek (C) 1 E2 k (D) 3 Ek
115. Potential energy of a satellite having mass m and rotating at a height of 6.4 x 106 m fromthe surface of earth(A) -0.5 mg Re (B) -mg Re (C) -2mgRe (D) 4 mgRe
116. When a satellite going round the earth in a circular obrit of radius r and speed v loses someof its energy, then r and v changes as(A) r and v bothe will increase(B) r and v both will decease(C) r will decrease and v will increase(D) r will increase and v will decrease
117. The time period of a satellite of earth is 5 hours If the sepration between the earth and thesatellite is increased to four times the previous value, the new time period will become ...... hours(A) 10 (B) 120 (C) 40 (D) 80
118. A person sitting in a chair in a satellite feels weightless because(A) the earth does not attract the objects in a satellite(B) the normal force by the chair on the person balances the earth’s attraction(C) the normal force is zero(D) the person in satellite is not accelerated
119. Two satellites A and B go round a planet in cirular orbits having radii 4R and R respectivelyIf the speed of satellite A is 3v, then speed of satellite B is(A) 3v/2 (B) 4v/2 (C) 6v (D) 12v
120. A satellite moves in a circle around the earth, the radius of this circlr is equal to one half ofthe radius of the moon’s orbit the satellite completes one revolution in .........lunar month(A) 1/2 (B) 2/3 (C) 2–3/2 (D) 123/2
121. The additional K.E. to be provided to a satellite of mass m revolving around a planet of massM, to transfer it from a circular orbit of radius R1 to another radius R2 (R2 > R1) is
(A) 2 21 2
1 1
G M m
R R(B)
1 2
1 1
G M m
R R
(C) 1 2
1 12
G M mR R
(D) 1 2
1 1 12
G M m
R R
122. Rockets are launched in eastward direction to take advantage of(A) the clear sky on eastem side(B) the thiner atmosphere on this side(C) earth’s rotation(D) earth’s tilt
168
123. A satellite of mass m is orbiting close to the surface of the earth (Radius R = 6400 km) hasa K.E. K. The corresponding K.E. of satellite to escape from the earth’s gravitational field is(A) k (B) 2 k (C) mg R (D) m k
124. A planet moving along an elliptical orbit is closest to the sun at a distance r1 and farthest awayat a distance of r2. If v1 and v2 are the liner velocities at these points respectively, then the
ratio 1
2
vv is ............
(A) 1
2
rr (B)
2
1
2
rr
(C)2
1
rr (D)
2
1
2
rr
125. The time period T of the moon of planet Mars(Mm) is related to its orbital radius R as(G = Gravitational constant)
(A) 2
2 4
RTGMm (B)
2 22 4π GRT
Mm
(C) 2
2 2
R GTMm (d) T2 = 4 Mm GR2
126. A geostationary satellite is orbiting the earth at a height of 5 R above that of surface of theearth. R being the radius of the earth. The time period of another satellite in hours at a heightof 2R from the surface of earth is ............ hr
(A) 5 (B) 10 (C) 6 2 (D) 6 / 2127. The figure shows ellipticall orbit of a planet m about the sun s. the shaded area SCD is twice
the saded area SAB. If t1 is the time for the planet to move from C and D and t2 is the timeto move from A to B then
(A) t1 > t2 (B) t1 > 4t2 (C) t1 = 2t2 (D) t1 = t2128. The period of a satellite in a circular orbit of radius R is T. the period of another satellite
in a circular orbit of radius 4R is(A) 4T (B) T/4 (C) 8T (D) T/8
129. If the earth is at one- fourth of its present distance from the sun the duration of year will be(A) half the pesent Year(B) one-eight the present year(C) one-fourth the present year(D) one-sixth the present year
130. The orbital speed of jupiter is(A) greater than the orbital speed of earth (B) less than the orbital speed of earth(C) equal to the orbital speed of earth (D) zero
169
131. Kepler’s second law regarding constancy of aerial velocity of a palnet is consequence of thelaw of conservation of(A) energy (B) angular Momentum(C) linear momentum (D) None of these
132. The largest and shortest distance of the earth from the sun are r1 and r2 its distance fromthe sun when it is at the perpendicular to the major axis of the orbit drawn from the sun
(A) 1 2
4r r
(B) 1 2
1 2
r rr r (C)
1 2
1 2
2r rr r (D) 1 2
3r r
133. According to keplar, the period of revolution of a planet (T) and its mean distance from thesun (r) are related by the equation
(A) 3 3 tanT r cons t (B) 2 3 tanT r cons t
(C) 3 tanTr cons t (D) 2 tanT r cons t
134. A satellite of mass m is circulating around the earth with constant angular velocity. If radius ofthe orbit is Ro and mass of earth M , the angular momentum about the center of earth is
(A) m GMRo (B) M GMRo
(C) GMmRo
(D) GMMRo
135. The earth E moves in an elliptical orbit with the sun s at one of the foci as shown in figure.Its speed of motion will be maximum at a point ..........
(A) C (B) A (C) B (D) D136. The period of revolution of planet A around the sun is 8 times that of B. The distance of A
from the sun is how many times greater than that of B from the sun.(A) 2 (B) 3 (C) 4 (D) 5
137. The earth revolves round the sun in one year. If distace between then becomes double the newperiod will be .......... years.
(A) 0.5 (B) 2 2 (C) 4 (D) 8138. The maximum and minimum distance of a comet from the sun are 8 x 1012m and 1.6 x 1012
m. If its velocity when nearest to the sun is 60 ms-1, What will be its velocity in ms-1 whenit is farthest ?(A) 6 (B) 12 (C) 60 (D) 112
139 The period of moon’s rotation around th earth is nearly 29 days. If moon’s mass were 2 foldits present value and all other things remained unchanged the period of moons’s rotation wouldbe nearly .....days
(A) 29 2 (B) 29 / 2 (C) 29 2 (D) 29
170
140. If the velocity of planet is given by a b cG M R then
(A) a =1/3 b = 1/3 c = -1/3(B) a =1/2 b = 1/2 c = -1/2(C) a =1/2 b = -1/2 c = 1/2(D) a =1/2 b = -1/2 c = -1/2
141. The radius of orbit of a planet is two times that of earth. The time period of planet is.........years.(A) 4.2 (B) 2.8 (C) 5.6 (D) 8.4
142. If r denotes the distance between the sun and the earth, then the angular momentum of theearth around the sun is proportional to(A) r3/2 (B) r (C) r1/1 (D) r2
143. What does not change in the field of central force(A) potential energy (B) Kinetic energy(C) linear momentum (D) Angular momentum
144. A thin uniform annular disc (See figure) of mass M has outerradius 4R and inner radius 3R. The work required to take aunit mass from point P on its axix to infinity is..............
(A) 2 (4 2 5)
7GM
R (B)
2 (4 2 5)7GM
R
(C) 4GM
R (D) 1)2(5R
2GM
145. Suppose the gravitational force varies inversely as the nth power of distance the time periodof planet in circular orbit of radius R around the sun will be proportional to
(A) 1
2n
R
(B)
12
n
R (C) Rn (D) 2
2
n
R
146. If the radius of the earth were to shrink by 1% its mass remaing the same, the accelration dueto gravity on the earth’s surface would(A) decrease by 2% (B) remain Unchanged(C) increase by 2% (D) increases by 1%
147. A body of mass m is taken from earth surface to the height h equal to radius of earth, theincrease in potential energy will be
(A) mg R (B) 12
mgR (C) 2 mgR (D)14
mgR
148. An artificial satellite moving in a circular orbit around earth has a total (kinetic + potential lenergy)Eo, its potential energy is(A) -Eo (B) 1.5 Eo (C) 2 Eo (D) Eo
171
149. Two bodies of masses m1 and m2 are initially at rest at infinite distance apart. They are then
allowed to move towards each other under mutual agravitational attraction Their relative velocityof apporach at sepration distance r between them is
(A) 12
1 22 ( )
G m mr
(B) 12
1 22 ( )
G m mr
(C)
12
1 22
rG m m
(D) 12
1 22 r
G m m
150. A geostationary satellite orbits around the earth in a circular orbit of radus 3600 km the timeperiod of a satellite orbiting a few hundred kilometers above the earth’s surface (Rearth = 6400km) will approximately be = ...... h(A) 1/2 (B) 1 (C) 2 (D) 4
KEY NOTE1 . C 25. B 49. C 73. A 97. B 121 D 145. A2 . A 26. A 50. A 74. A 98. C 122. C 146. C3. B 27. C 51. C 75. A 99. C 123 B 147. B4. A 28. A 52. A 76. C 100. B 124. C 148. C5. B 29. C 53. A 77. C 101. A 125. A 149. B6. C 30. C 54. A 78. A 102. B 126. C 150. C7. A 31. A 55. C 79. C 103. D 127. C8. D 32. C 56. C 80. B 104. B 128. C9. C 33. B 57. A 81. D 105. B 129. B10. D 34. D 58. B 82. B 106. A 130. B11. B 35. A 59. D 83. B 107. D 131. B12. D 26. B 60. A 84. D 108. A 132. C13. A 27. C 61. D 85. D 109. A 133. B14. A 38. C 62. A 86. D 110. B 134. A15. A 39. A 63. B 87. C 111. D 135. B16. C 40. D 64. A 88. B 112. B 136. C17. C 41. A 65.B 89. D 113. A 137. B18. C 42. B 66. A 90. C 114. A 138. A19. D 43. B 67. B 91. D 115. A 139. D20. C 44. C 68. C 92. C 116. C 140. B21. B 45. D 69. C 93. C 117. D 141. C22. A 46. B 70. A 94. B 118. C 142. C23. B 47. A 71. D 95. D 119. C 143. D24. B 48. A 72. C 96. B 120. C 144.A
1 . C 25. B 49. C 73. A 97. B 121 D 145. A 2 . A 26. A 50. A 74. A 98. C 122. C 146. C3. B 27. C 51. C 75. A 99. C 123 B 147. B4. A 28. A 52. A 76. C 100. B 124. C 148. C5. B 29. C 53. A 77. C 101. A 125. A 149. B6. C 30. C 54. A 78. A 102. B 126. C 150. C7. A 31. A 55. C 79. C 103. D 127. C8. D 32. C 56. C 80. B 104. B 128. C9. C 33. B 57. A 81. D 105. B 129. B10. D 34. D 58. B 82. B 106. A 130. B11. B 35. A 59. D 83. B 107. D 131. B12. D 26. B 60. A 84. D 108. A 132. C 13. A 27. C 61. D 85. D 109. A 133. B14. A 38. C 62. A 86. D 110. B 134. A15. A 39. A 63. B 87. C 111. D 135. B16. C 40. D 64. A 88. B 112. B 136. C17. C 41. A 65.B 89. D 113. A 137. B18. C 42. B 66. A 90. C 114. A 138. A19. D 43. B 67. B 91. D 115. A 139. D20. C 44. C 68. C 92. C 116. C 140. B21. B 45. D 69. C 93. C 117. D 141. C22. A 46. B 70. A 94. B 118. C 142. C23. B 47. A 71. D 95. D 119. C 143. D24. B 48. A 72. C 96. B 120. C 144.A
172
HINT
(1) (c) 3. 2
2 2 42 2
4( )( )( ) 43(2 ) 4 3
G RG m mF R
R R
(4) (A) Gravitational force does not depend upon the medium(5) (B) Due to inertia to direction(6) (D) the force exerted by sun on the earth
F = m2R 24 7 2 11(6 10 ) (2 10 ) (1.5 10 )
2136 10 N
(7) 2( 1) ( 1)F xm x m m x x
for maximum force 0dFdx
2 22 0 1/ 2 dF m m x xdx
(9) (C) cenripetal force provided by the gravitational force of attraction between two particites2
2
( )( )(2 )
m G m m
r r
12
Gmr
(10) (D)
For will be zero at the point of zero intensity
1
1 2
81 91081
m mx D Dm m m m
11. B 2Re eGM mmg where Me and Re is the mass and radius of the earth respectivel
2Re egMG
12. (D) Here FOA = FOB = FOC = 2
( ) (2 )G m mr
OA OB OCF F F F
(D - X)
173
=
= 014. (A) let a point mass C is placed at a distance of x m from the point mass A as show in figure
43a
b
mherem
2
( )( ) .........( )G m mAFAC i
x
2
( )( ) .........( )(1 )
G m mBFBC iix
According to given problem 14
FAC FBC
with the help of equ (i) and (ii)
2 2
( )( ) 1 ( )( )3 (1 )
G m mA G m mBx x
2 2 2
2 2 2
4 43(1 ) 3 3(1 ) (1 )
A
B
m x x xm x x x
2 2 21
x x xx
23 2 3x x m
17. (c) 32
43
GMg and M RR
32
4 3.3 4
G gg RR RG
18. (c) for condition of weight lessness at equator
3
10 1 sec6400 10 800
gR
rad
19. (D) Time peripd of simple pendylym 2 'T l g
In artificial satellite g’ = 0 T
mA m mB(1–x)C
x
174
20. (C) 2 2 2
4( 2)
GM GMo GMogR Do Do
21. (B) we know that 2
GMgR
on the planet gp 2
7 4 74
GM gR
Hence the weight on the planet = 700 47 = 400 gm t
22.22
2' 6400 ' 960.400( ) 6400 64
g R g msg R h
23. 1 2 2 g g R Cos Rotation of the earth results in the decreased weight apparently.this decrease in weight is notfelt at the poles as the angle of latitude is 90’
24. (B) using 2
GMgR
we get gm = g/5
25. (B) 1 d ( 1)g g (1 )R
nd Rn
26. (A) 4g πGR and g' 4/3πGR'3
g' R' 0.2 g' 0.2gg R
27. (C) mass does not vary from place to place
28. (a) 22
2
4'9
R
R Rg g g gR h R
4W' 4/9 w (72) 32 N9
29. (c) RhsolvingbyhR
Rgg
4/9'
2
30. (c) 2
2 2p e
Mp Re 1 9.8g g 9.8 (2) 0.49msMe Rp 80 20
31. (A) 2d 100g' g 1 9.8 1 9.66 ms
R 6400
175
32. (C) 2RGMg and k =
2L2I
If mass of the earth and its angular momentum ramains constant then 2
1gR
and 2
1kR
i.e. if radius of earth decreases by 2% then g and k both increase by 4%33. (B) Weight on surface of earth, mg = 500N and weight below the surface of earth at
d = R2 , mg1 = d mgmg 1 mg 1 1/2 250N
R 2
34. (D) 1 1 1
2 2 2
g4 Rg π GR 3 g R
35. (A) 2rGMαg g
1αrorr1αg 2
If g decreases by one percent then r should be increase by 1/2 % i.e. 1 6400 32
2 100R km
37. (C) RhRhR
RhR
Rgg 22
1'2
h 2 1 R 0.414 R
Hence distance form center = R + 0.414R = 1.414R
38. (C) 4g π GR g α d R ( d given in the problem)3
39. (A) insided the earth π34g' Gr rg '
40. (D) 4g π GR3
1(1)
Re 2
p e
e p
gRpg
Re
22RRp
41. (A) g
42. (B) For height 100424100%
gΔg
%
For depth 0.51/2Rh
Rd100%
gΔg
43. (B) g R
176
44. (C) 2
B A
A B
V 1 H gH Hα2g g H g
Now AB
gg12
as g R
B AB A
A B
H g 12 H 12 H 12 1.5 18 mH g
46. (B) 2
2R Rg' g g 4/9.g(g 10 ms )RR h 3 2
47. (A) e e
m m
g4 Reg π GR g α R .3 g Rm
RmRe.
35
16
Re185Rm
48. (A) ' /92 2
R Rg g g gR R R R
49. (C) 2)/1('
Rhgg
2)/1(16 Rhgg
1612
Rh
41 Rh
3Rh
Rh 3
50. (A) Gravitational attraction fore on particle B
2)2/(DpGMpmFg
Acceleration of paritcle due to particle due to gravity 24DpGMp
mFga
177
51. (C) 2
1g α r it(r r) and g α (it r R)r
52. (A) dxdvI
53. (A) rGMmU
r
24221128 1061071067.61079.7
m103.8r 8
54. (A) 2/2/21
dGM
dGMV
1 22GMNow P.E mv (m m )
d (m mass of particie)
K.E = P.E
21 2
1 2 ( )2
GMm m md
1 2( )2 G M Md
55. (C) If the body is projected with velocity Ve)(υυ then height up to where it rises
2 2
2
4 ( )11.2110
R Rh R approxVeV
57. (A) 2GM GMmVe P.E. U 5000 JR R
RGM
100
5000R
GM
58. (B) P.E.
GMm GMmUr R h
Uinitial = – 3GMm
R and Ufinal = – 2GMm
R
loss of P.E. = gain in K.E = 2GMm
R – 3GMm
R = 6GMm
R
178
59. (D) 2 1mgh mgRe mgReΔU U U Re1 h/Re 21
Re
2mg ReU ( mg Re)
2 mgRe
21U2
60. (A) If body is projected with velocity Ve)(υυ then height up to which it will rise
R/314
R
1Ve/2Ve
Rhve/2ubut 2
61. (D) change in potential energy in displacing a body from r1and r2 is given by
1 2
1 1 1 12 3 6
GMmU GMm GMmr r R R R
62. (A) K.E = 2GMm
R R1αK.E.
63. (B) this should be equal to escape velocity i.e. = 2gR64. (A) A Escape velocity does not depend on the mass of the projectile
65 (B) p
e
gVp Rp. 2 2 2Ve g Re
122.4Kms11.222VeVe
66. (A) 83
2GMVe R GR
Ve α R if constant
since the planet is having double radius incomparision to earth therefore the escape velocitybecomes twice i.e. 22 kms-1
69. (C) Because it does not depend on the mass of projetile
71. (D)
Gravitational potential of A at 0 rGM
rGM 2
2/
of B at 0 rGM
rGM 2
2/
Total potential at 0 4GM
r
179
72. (C) r1αVe where r is the position of body from the surface
1 2 12
2 1
V r R 7R V2 2 VV r R 2 2
73. (A) R
2GM21
hRGMV
h)2(R4R
h R 6400 km
74. (A) 3216
RpRe
MeMp
VeVp
Ve3Ve
75. (A)1/2(kg)gk
R2R1.
g2g1
V2V12gRυ
76. (C) RMαVe
R2GMVe
If M becomes double and R becomes half then escape velocity; becomes two times
77. (C) on earth Ve = 12GM 11.2 kmsR
on moon 12GM 4 2 2GM 2 11.2Vm . 2.5 kmsR R 9 R 9
78. (A) Due to three particles net intensity at the center
A B CI I I I 0
because out of these three intensities ARE equal in magnitudeand between each other is 120’
79. (C) R1αVe if R becomes 4
1 then Ve will be 2 times
80. (D) Escape velocity does not depends upon the angle of projection
82. (B) p
e
gVp Rp2gR . 1 4 2Ve g Re
Ve = 2Ve83. (B) potential energy of system of two mass
J106.671010
1010100106.67R
GMmU 102
311
so, the amount of work done to take the particle up to infinte will be 6.6710-10 J
180
84. (D) Vp = Vsphere + Vpartical aGM
aGM
aGM 3
2/
85. (D) Ve = R
GM2=
11 3012 6.67 10 1.99 10 618 kms
6.98 108
86. (D)2GMm GMm mW 0- - gR mgR
R R R
3106400101000 91064
J106.4 10
87. (C) GM GM GMVin Vsurface VoutR R r
88. (B) 1F 0 When 0 r R
beccause intensity is zero inside Caviy
Fincrease When 21 RrR
2RrwhenR1Fα
89. (D) Intensity will zero inside the spherical shell
I = upto r = a and I 2
1r when r > a
90. (C) kepler’s law T2 α r3
91. (D) System will be bound at points where total energy is negative. In the given curve at pointA, B, and C the P.E. is more than K.E.
92. (C) 2rGMmEand
2rGMmK
rGMm-U
For satellite U, K, and E varies with r and also U and E remains negativw where K reaminalways positive
93. (C) Gravitational P.E. = m x gravitational potential U = mv so the graph of U will be sameas that of V for a spherical shell
94. (B) 0 0Ve 2gR and V gR Ve 2 V
96. (B) 2121 VVthenrritr
GMυ
181
97. (B) 3
2T α r if r becomes double then time period will become (2)3/2 times so new time period
will be 24 x 2 2 hr i.e. = 48 298. (C) The velocity of the spoon will be equal to the orbital velocity when dropped out of the
space ship
100. (B) A B
B A
GM V R R 1υR V R 4R 2
AB
B
V 3V V 6VV VA
104. (B) r1αυ
% increase in speed = 1 %2 decrease in radius
= 12 (1%) = 0.5 %
105. (B)r
GMυ
106. (A) r1αυ it orbital radius becomes 4 times then orbital velocity will becomes halt
107. (D) orbital rudius of satellites r1 = R+R = 2Rr2 = R+7R = 8R
22
11 r
GMmUandr
GMmU
1 21 2
GMm GMmK and K2r 2r
4EE
KK
UU
2
1
2
1
2
1
110. (B) 6R from the surface of earth and 7R from the center111. (D) Distance of satellite from the center are 7R and 3.5R respectively
2
3/2 3/22 2
1 1
T r 3.5RT 24 6 24 hrT r 7R
112. (B Gravitational force provides the required centripetal force for orbiting the satellite
RK
Rmυ2
becuse R1αF
oυ α R
182
113. (A) Total energy = -K.E. = 21
2m
114. (A) B.E. = -K.E.And it this amount of energy(Ek) given to satellite it will escape into outer space
115. (A) P.E. = GMmr
= –Re GMm
h = –
2 eGMm
R = –
2Re2 Re
g m = –
12 mg Re = 0.5 mg Re
116. (C) B. E. = GMm
r if B.E. decreases the r also decreases and V increases as 1r
117. (D) 3/2
3/222 1 1 1
1
RT T T (4) 8T 40 hrR
118. (C) 4 2 B A
A B
V r RV r R
120. time period of revolution of moon around the earth = 1 lunar month3/2 3/2
2/3e ee
m m
T r 1 T 2T r 2
lunar month
121 (D) 21 2R
GMmK.E.2R
GMm
21
112
.RR
GMmEK
122. Because Earth rotation from west to east direction
123. (B) 2GMmk
R
124. (C) v1 r1 = v2r2 (angular momentum is constant)
125. (A) Time period 2/322
GMR
RGMm
RT
GMmRT
322 4
126. (C) 2 3 3
1 12 3 3
2 2
T R (6R ) 8(3R )T R
22
24 24 728
T
2 6 2T
183
127. (C) ΔA L Δ A α Δ tΔt 2m
SCD 11 2
SAS 2
ΔA t 2A t 2tΔA t A
128. (C)
3 32 21 1
2 2
T R RT R 4 R
2 18T T
129. (B) 32 rαT since T81T
41
TT 1
321
130. (B) orbital radius of Jupiter > orbital radius of Earth J e
e j
V rV r
As rj < re there fore Vj < Vee
131. (B) 2LdA
dt m = constant
132. (C) The earth moves around the sun in elliptical path, so by using the properties of ellipse
e)(1r1 a and r2 = (1-e) r2, a = 1 2
2r r
2221 )1( aerr
where a= semi major axisb= semi minor axise= eccentircity
Now required distance = sem latysrectum = b2/9
21
21
21
212
2
rrr2r
)/2r(rrr
a)e(1a
133. (B) T2/ r3 = constant T2 r-3 = constant134. (A) Angular momentum = Mass orbital velocity x Radius
00
RR
GMm
0m GMR
135. (B) speed at the earth will be maximum when its distance from the sun is minimum becausem r = constant
136 (C) 3 3
2 2 3A A A 2A B B
B B B
T r 48 r 4 8 4rT r 4
137. (B) years22T22(2)rr
TT
22
323
1
2
1
2
184
138. (A) By conservation of angular momentum (mr) = constantVmin X rmax = Vmax X rmin
121
12
60 1.6 10 60Vmin 12 ms8 10 5
139. (D) Time period does not depends upon the mass of satellite
140. (B) 1 1 1
2 2 2GMυ G M RR
141. (B) yar2.821RRTT 2
32
3
1
212
142. (C) Angular momentum of earth around the sunL = MEVo r
2SE E S
GMM .r M .GM .rr
ME = mass of earthMS = mass of Sunr = Distance between sun and the earth
rαL143. (D) For central force toraqe is zero
dL 0 L Constantdt
144. (A) Wext FUU
Gdm0 (1)
x
2 2 2
M 2π r drGπ 7R 16R r
2 2 2
2 GM rdr7R 16 R r
2 2
GM π d2 GM (2)7R a 7R
4R
3R22
2 r16RR7GM2
5RR247RGM2
5247RGM2
185
145. (A) 2
2 2 n 1n 2 n
1 4π 1m Rα m R α T α RR T R
n+12T α R
146. (C) 2RGMg is mass ramains constant then
2
1R
147. (B) mgR21
Rh1
mghΔU
148. (C) P.E. = 2. total energy = 2E0
Because we know U = – GMM
r and Eo = – 2GMM
r149. (B) let velocities of these masses at r distance from each other be v1 and v2 respectively
By conservation of momentum 1 1 2 2 1 1 2 2m V m V 0 m V m V ____________ 1
By conservation of energychange in P.E. = change in K.E.
222
211
21 Vm21Vm
21
rmGm
ii_______r
mGm2m
VmmVm 21
2
222
1
211
on solving (i) and (ii)
21
2
122mmr
Gm
and 21
2
2 mmr2Gm
υ 1
1 21 2
2G m mV app V V
r
150. (C)
3 32 22 2
21 1
T r 6400T 24 2 hoursT r 3600
186
Assertion - Reason Type QuestionsDirection (Read the following uestions and choose)(A) If both Assertion and Reason are true and the Reason is correct explanation of assertion(B) If both Assertion and Reason are true, but reason is not correcte explanation of the Assertion(C) If Assertion is true, but the Reason is false(D) If Assertion is faluse, but the Reason is true
1. Assertion : The value of acc. due to gravity (g) does not depend upon mass of the body
Reason : This follows from 2
GMgR
, where M is mass of planet (earth) and R is radius of
planet (earth)(a) A (b) B (c) C (d) D
2. Assertion : Unit of gravitational field intensity is N/kg or ms-2
Reason : Gravitational field intensity
22
mskg
kg.m/seckgN
massForce
(a) A (b) B (c) C (d) D3. Assertion : The time period of a geostationary satellite is 24 hours
Reason : Such a satellite must have the same time period as the time taken by the earth tocomplete one revolution about its axis
(a) A (b) B (c) C (d) D4. Assertion : Even when orbit of a satellite is elliptical, its plane of rotaiion passes through the
center of earthReason : This is in accordance with the principle of conservation of angular momentum(a) A (b) B (c) C (d) D
5. Assertion : The time Period of pendulum, on a satlellite orbiting the earth is infinity
Reason : Time period of a pendulum is inversely proportional to g(a) A (b) B (c) C (d) D
6. Assertion : The escape velocity on the surface of a planet of the same mass but 14 times the
radius of earth is 5.6 kms-1
Reason : The escape velocity Ve = 2gR(a) A (b) B (c) C (d) D
7. Assertion : The comet does not obey kepler’s law of planetaty motionReason : The comet does not have ellitical orbit(a) A (b) B (c) C (d) D
8. Assertion : The square of the period of revolution of a planet is proportional to the cube ofits distance from the sun.
Reason : Sun’s gravitational field is inversely proportional to the square of its distance fromthe planet
(a) A (b) B (c) C (d) D9. Assertion : Space ship while entering the earth’s atmoshere is likely to catch fire
187
Reason : Temperature of upper atomosphere is very high(a) A (b) B (c) C (d) D
10 Assertion : The earth is slowing down and as a result the moon is coming nearer to itReason : The angular momentum of earth - moon system is not conserved(a) A (b) B (c) C (d) D
188
Assertion - Reason Type question :1. A Both the asseration and reason are true and the latter is correct expalnation of the former.2. A Both the asseration and reason are true and the latter is correct expalnation of the former.3. A As the satellite is to be stationary over a particular place, its time period of revolution = 24
hours= time peroid of revolution of earth about its axis.
4. A g2πT on a satellite, there is a weightless -ness, so g= 0 hence T
Thus both the asseration and reason are correct5. A As no torque is acting on the planet, its angular momentum must stay constant in a magnitude
as well as direction there for planet of rotation must pass throught the center of earth
6. DR1αVei.e.
RGM2Ve
1RVe' = Ve 2 Ve 2 11.2 22.4 kmsR4
thus assertion is wrong but Reason is correct7. A Both the asseration and reason are correct and the Reason is correct explanation of Asseration8. A A Both the asseration and reason are correct and the Reason is correct explanation of Asseration9. C Here Asseration is correct but Reason is wrong because the space ship while entering the earth’s
atmosphere may catch fire due to atmoshperic air friction10. D Here Both the asseration and reason are wrong because the angular momentum of earth
moon system is conserved in the absence of extermal touque.
189
Comperehensions Type Questions(1) If a smooth tunnel is dug across a diameter of earth and a particle is related from the surface
of earth, the particle oscillates simple harmonically along it.(1) Time period of the particle is not equal to
(A) gR2π (B) 2
3R
GM2π
(C) 84.0 min (D) Nome of these(2) Maximum speed of the these
(A) RGM2
(B) R
GM(C)
R 2GM3
(D) R 2
GM
2. When a paricle is projected from the surface of earth, it mechanincal energy and angular momentumabout center of earth at all time is constant(i) A particle of mass m is projected from the surface of earth with velocity vo at angle
with horizontal suppose h be the maximum height of particle from surface of earth and vits speed at that point them v is(A) v0coso (B) >v0coso(C) <v0coso (D) zero
(ii) Maximum height h of the paritcle is
(A) 2gθsinVo 22
(C) 2gθsinVo 22
(B) 2gθsinVo 22
(D) can be greater than or less than 2gθsinVo 22
3. A solid sphere of mass M and radius R is surrounding by a sphericalshell of same mass M and raius 2R as shown. A small particleof mass m is relased from rest from a height (h <<R) aobove theshell there is a hole in the shell.(i) in what time will it enter the hole at A
(A) GM
hR2 2
(C) GMhR 2
(B) GM
2hR 2
(D) None of these
(ii) what time will it take to move from A to B ?
(A) 2R
GMh (C)
2RGMh
(B) 2R
GMh (D) None of these
190
(iii) with what apporximate speed will it colide at B ?
(A) 2 GM
R(C)
2RGM
(B) 2RGM3
(D) R
GM
4. A planet is revolving round the sun in elliptical orbit. Velocity at perigee position (nearest) isv1| and at apogee position (farthest) is v2 Both these velocities are perpendicular to the joingingcenter of sun and planet r is the minimum distance and r2 the maximum distance.(i) when the planet is at perigee position, it wants to revolve in a circular orbit by itself. For
this value of G(A) Should increase (B) Should decrease(C) data is in sufficeint (D) will not depend on the value of G
(ii) At apogee position suppose speed of planer is slightly decreased from v2, then what willhappen to minimum distance r1 in the subsequent motion(A) r1 and r2 bothe will dicreases(B) r1 and r2 bothe will increases(C) r2 will remain as it is while r1 will increase(D) r2 will remain as it is while r1 will decrease
5. Garvitational potential at any point inside a spherical shall is uniform
and is given by GM
R where M is the mass of shell and R its
radius. At the center solid sphere, potential is 3
2
GMR
(i) There is a concentric hole of radius R in a solid sphere of radius 2R Mass of the remainingportion is M What is the gravitation potential at center?
(A) 7R3GM
(B) 5GM7R
(C) 14GM7
(D) R 14GM9
191
Solution
(1) (i) haveweRGMgputting
gR2πT 2
322πT R
Gm
(ii)maximum speed is at centre from conservation mechanicall energy(from cerface to center)increase in K.E. = decrease in P.E.
2i f i f
1 m U U m(V - V )2
n
i fυ 2(v v )
RGM
12GM3
RGM2
(2) (i) From conservation of angular momentum at A and B
mVocosθ m R h
0R V cosθ
R h
oV cosθ
(ii) From consercation of mechanical energy, Decrease in K.E. = increase in P.E.
R
h1mghVVm
21 22
0
θcosVVherebutR
h1gh2VV 0
220
xVVletsoθsinVVV 220
220
220
θsinVxhere 220
Rh1
2ghx
Rx2g
xh
i.e. h >x
2g i.e. h >2 2
O sin2g
192
(3) (i) Acceleration due to gravity near the surface of shell can be assumed to be uiform
22 2RGM
(2R)G(2M)g
From 21h2
gt
t = GMhR
gh 2
22
(ii) 2 22 22AGM GMhV gh h
R R
From A to B field due to shell is zero, but field due to sphere is not-zero
hence tAB
2
A
R RV GMh
(iii) KA=0 polential between A and B due to shell isFrom energy conservationKA + UA = KB + UBKB = (UA = UB) = m(VA – VB)
12 mVB
2 = m(VA – VVB)
B A BV 2(V V ) = GM
R(4) At perigee position v1 > v0 where v0 is the orbital velocity for circular motion
0GMV G
r
so value of G should incease , so that v0 will increase for this position and whichwill become equal to v1(ii) path will becomemore elliptical , keeping r2 constant and r1 to decrease
(5) Density of given material
3 3 343
M 3Mπ[(2R) R ] 28πR
Vwhole = Vhole + VremaingVremaing = Vwhole – Vhole
1 23 GM GM2 2R R
193
here 31
4 8M π (2R) M3 7
32
4 MM ( ) R3 7
14R9GMVremaining
Match the Column(1) on the surface of earth acceleration due to gravity is g and gravitational potential is V match
the followingTable - 1 Table -2
(A) At height h = R value of g (P) decreases by a factor 14
(B) At height h = R/2 value of g (Q) decreases by a factor 12
(C) At height h = R value of v (R) increases by a factor 11/8(D) At depth h = R/2 value of V (S) increases by a factor 2
(T) None(2) Density of planet is two times the density of earth Radius of this planet is half (As compared
to earth)Match the following
Table-1 Table-2(A) Acceleration due to gravity on (P) Half
this planet’s surface(B) Gravitational potential (Q) same
on the surface(C) Gravitational potential (R) Two times
at centre(D) Gravitational field strength (S) four times
at centre(3) let V and E denote the gravitational potential and gravitational field at a point. Then the match
the follwingTable-1 Table-2
(A) E = 0, V = 0 (P) At center of Spherical shell(B) E 0, V= o (Q) At centre of solid sphere(C) V 0, E -0 (R) At centre of circular ring(D) V 0, E 0 (S) At centre of two point masses of equal magnitude
(T) None
194
(4) Match the followingTable-1 Table -2
(A) time period of an earth satellite in circular orbit (P) Independent of mas of satellite(B) Orbital velocity of satellite (Q) independent of radius of orbit(C) Mechnical energy of stellite (R) independent of mass of earth
(S) none(5) Match the following (for a satellite in circular orbit)
Table-1 Table-2
(A) kinetic energy (P) 2GMm
r
(B) potential energy (Q)GM
r
(C) Total energy (R)GMm
r
(D) orbital velocity (S) 2GMm
r
solution :(1) A-p, B-Q, C-s D-t(2) A-Q, B-p, C-p D-o(3) A-t, B-t, C-p,q,r,s, D-t(4) A-q, B-t, C-r, D-s(5) A-s, s-B, c-p, D-q
• • •
195
Unit - 7Proporties ofLiquid a Solid
196
SUMMARY
Surface tension of a liquid is measured by the forceacting per unit length on either side of an imaginary linedrawn on the free surface of liquid, the direction of thisforce being perpendicular to the line and tangential tothe free surface of liquid. So if F is the force acting onone side of imaginary line of length l. then T = Fl.
1. It depends only on the nature of liquid and is independent of the area of surface or length of lineconsidered.
2. It is a scalar as it has a unique direction which is not to be specified.
3. Dimension 201 TLM
4. Unit : SImN
5. It is a molecular phenomenon and its root cause is the electromagnetic forces.
Force Due to Surface TensionIf a body of weight W is placed on the liquid surface, whose surface tension is T, and if F is theminimum force required to pull it away from the water then value of F for different bodies can becalculated by the following table.
Body Force
1. Needle (length l) wT2F
2. Hollow disc (inner radius r1 Outer radius r2) wTrr2F 21
3. Thin ring (radius r) wrT4F
4. Circular plate (or disc) (radius r) wrT2F
5. Square frame (side l) wT8F
6. Square plate wT4F Examples of Surface Tension :
1. When mercury is split on a clean glass plate, it formsglobules. Tiny globules are spherical on the account ofsurface tension because force of gravity is negligible. Thebigger globules because force of gravity is negiligible. Thebigger globules get plattened from the middle but haveround shape near the edges.
197
2. When a molten metal is poured into water from asuitable height, the falling stream of metal breaks upand the detached portion of the liquid in small quantityacquire the spherical shape.
3. Rain drops are spherical in shape because each drop tends to acquire minimum surface area due tosurface tension, and for a given volume, the surface area of sphere is minimum.
4. Oil drop spreads on cold water. Whereas it may remain as a drop on hot water. This is due to thefact that the surface tension of oil is less than that of cold water and is more than that of hot water.
5. If a small irregular piece of camphor is floated on the surface of pure water, it does not remain steadybut dances about on the surface. This is because, irregular shaped camphor dissolves unequally anddecreases the surface tension of the water locally. The unbalanced forces make it to move haphazardlyin different directions.
6. Take a frame of wire and dip it in soap solutionand take it out, a soap film will be formed in theframe. Place a loop of wet thread gently on thefilm. It will remain in the form, we place it on thefilm according to figure.Now, piercing the film with a pin at any pointinside the loop, It immediately takes the circularfrom as shown in figure.
7. When a greased iron needle is placed genetly on the surface ofwater at rest, so that it does not prick the water surface, the needlefloats on the surface of water despite it being heavier because theweight of needle is balanced by the vertical components of the forcesof surface tension. If the water surface is pricked by one end of theneedle, the needle sinks down.
8. Hair of shaving brush / painting brush when dipped inwater spread out, but as soon as it is taken out, its hairstick together.
Factors affecting surface tension1. Temperature :
The surface tension of liquid decreases with rise of temperature. The surface tension of liquid is zeroat its boiling point and it vanishes at critical temperature. At critical temperature, intermolecularforces for liquid and gases becomes equal and liquid can expand without any restriction. For smalltemperature differences, the variation in surface tenstion with temperature is linear and is given bythe relation.
198
t1TT 0t
Where Tt and TO are the surface tensions at tOc and 0OC respectively and is the temperaturecoefficient of surface tension.Examples :(i) Hot soup tastes better than the cold soup.(ii) Machinery parts get jammed in winter.
2. Impurities :The presence of impurities either on the liquid surface or dissolved in it, considerably affect thesurface tension, depending upon the degree of contamination. A highly soluble substance like sodiumchloride when dissolved in water, increases the surface tension of water. But the sparingly solublesubstances like phenol when dissolved in water, decreases the surface tension of water.Applications of Surface Tension
1. The oil and grease spots on cloths cannot be removed be pure water. On the other hand, whendetergents (like soap) are added in water, the surface tension of water decreases. As a result of this,wetting power of soap solution increases. Also the force of adhesion between soap solution and oilor grease on the clothes increases. Thus, Oil, grease and dirt particles get mixed with soap solutioneasily. Hence, clothes are washed easily.
2. Surface tension of all lubricating oils and paints is kept low so that they spread over a large area.3. A rough sea can be calmed by pouring oil on its surface.4. In soldering, addition of 'flux' reduces the surface tension of molten tin, hence, it spreads.
Molecular Theory of Surface TensionThe maximum distance upto which the force of attraction betweentwo molecules is appriciable is called molecular range m10 9 .A sphere with a molecule as centre and radius equal to molecularrange is called the sphere of influence. The liquid enclosed betweenfree surface (PQ) of the liquid and an imaginary plane (RS) at adistance r (equal to molecular range) from the free surface of theliquid form a liquid film.To understand the concept of tension acting on the free surface of a liquid, let us consider four liquidmolecules like A, B, C and D. Their sphere of influence are shown in the figure.
1. Molecule A is well within the liquid, so it is attracted equally in all directions. Hence the net force onthis molecule is zero and it moves freely inside the liquid.
2 Molecule B is little below the free surface of the liquid and it is also attracted equally in all directions.Hence the resultant force acting on it is also zero.
3. Molecule C is just below the upper surface of the liquid film and the part of its sphere of influence isoutside the free liquid surface. So the number of molecules in the upper half (attracting the moleculesupward) is less than the number of molecule in the lower half (attracting the molecule downward).Thus the molecule C experiences a net downward force.
4. Molecule D is just on the free surface of the liquid. The upper half of the sphere of influence has noliquid molecule. Hence the molecule D experiences a maximum downward force.
199
Thus all molecules lying on surface film experiences a net downward force. Therefore, free surfaceof the liquid behaves like a stretched membrane.Surface Energy :The potential energy of surface molecules per unit area of the surface is called surface energy.
Unit : 2mJ
Dimension : 2ML
Suppose that the sliding wire LM is moved through asmall distance x, so as to take me position . .ML Inthis process, area of the film increases by x2 (onthe two sides) and to do so, the work done is given by,
ATx2Tx2TfxW
AWT
i.e. Surface tension may be defined as the amount of work done in increasing the area of the liquidsurface by unity against the force of surface tension at constant temperature.Work done in Blowing a liquid drop or soap-bubble
1. If the intial radius of liquid drop is r1 and the final radius of liquid drop is r2 then,
AreainincrementTW
21
22 rr4T
2. In case of soap-bubble,
21
22 rr8TW (Bubble has two free surfaces)
Splitting of Bigger Drop :When a drop of radius R splits into n smaller drops, (each of radius r) then surface area of liquidincreases. Hence the work is to be done against surface tension.Since the volume of liquid remains constant therefore
33 r34nR
34
33 nrR
Work done AT 22 R4rn4T
Formation of Bigger drop :Amount of surface energy released = Initial surfaceenergy - final surface energy
TR4nTr4E 22
Excess Pressure :Due to the property of surface tension a drop or bubble tends to contract and so compresses thematter enclosed. This in turn increases the internal pressure which prevents further contraction and
200
equilibrium is achieved. So in equilibrium the pressure inside a bubble or drop is greater than outsideand the difference of pressure between two sides of the liquid surface is called excess pressure.Excess pressure in different cases is given in the following table :Plane surface Concave surface
P = 0 P = 2TR
Convex surface Drop
P = 2TR P =
2TR
Bubble in air Bubble in liquid
P = 4TR P =
2TR
Cylindrical liquid surface Liquid film of unequal radii
RIP
21 R1
R1T2P
Shape of liquid meniscus :When a capillary tube is dipped in a liquid, the liquid surface becomes curved near the point ofcontact. This curved surface is due to the resultant of two force i.e. the force of cohesion and theforce of adhesion. The curved surface of the liquid is called meniscus of the liquid.If liquid molecule A is in the contact with solid. (i.e. wall of capillary tube) then forces acting onmolecule A are(i) Force of adhesion Fa
(ii) Force of cohesion FC
(iii) Resultant force FN makes an angle with Fa.
Ca
Co
Ca
oC
FF2F
135cosFF135sinFtan
Example :Pure water in silver coated capillary tube.
201
Angle of contact :Angle of contact between a liquid and a solid is defined as the angle enclosed between the tangentsto the liquid surface and the solid surface inside the liquid, both the tangents being drawn at the pointof contact of the liquid with the solid.Capillarity :If a tube of very narrow bore is dipped in a liquid, it is found that the liquid in the capillary eitherascends or descends relative to the surrounding liquid. This phenomenon is called capillarity.The root cause of capillarity is the difference in pressure on two sides of (concave and convex)curved surface of liquid.Examples :(i) Ink rises in the fine pores of bloting paper leaving the paper dry.(ii) A towel soaks water.(iii) Oil rises in the long narrow spaces between the threads of wick.
Ascent Formula :
2ThρgR
rhρg RhρgT2cosθ 2
Useful Facts and Formulae :(1) Formation of double bubble :
Radius of the interface 1...rr
rrr12
21
and 2...r1
r1T4P
21
(2) Formation of a single bubbleUnder isothermal condition two soap bubble of radii "a"and "b" coalesce to from a single bubble of radius C.
Now CT4PP,
bT4PP,
aT4PP 0C0b0a
3a a
34V
Now as mass is conserved,
3c c
34V
cba
3c c
34V
c
cc
b
bb
a
aa
RTVP
RTVP
RTVP
202
Temp is constant,
cba TTT
3
003
0 c34
CT4P
bT4Pa
34
aT4P
222
3330
cba4bacPT
Hooke's Law and Modulus of Elasticity :According to this Law, within the elastic limit, stress is proportional to the strain.i.e. stress strain
StressStrain
= E = constant
The constant E is called the modulus of elasticity.There are three modulii of elasticity namely (i) Young'smodulus (y), (ii) Bulk modulus (B), and modulus of rigidity(), responding to three types of the strain.
(1) Young's modulus (Y)It is defined as the ratio of normal stress to longitudinal strain within limit of proportionality
AFL
L
AF
strainallongitudinstressNormalY
2rmgLY
Force constant of wire,
LYAFK
Bulk Modulus :When a solid or fluid (liquid or gas) is subjected to a uniform pressure all over the surface, such thatthe shape remains the same, then there is a change in volume.Then the ratio of normal stress to the volumetric strain within the elastic limits is called as Bulkmodulus. This is denoted by K.
strainVolumetricstressNormalK
vpV
Vv
AF
203
Modulus of Rigidity :Within limits of proportionality, the ratio of tangential stress to the shearing strain is called modulus ofrigidity of the material of the body and is denoted by ,
strainShearingstressTangential
In this case the shape of a body changes but its volume remains unchanged.
AFA
F
Pascal's Law :It states that if gravity effect is neglected, the pressure at every point of liquid in equilibrium of rest issame.Working of hydraulic lift, hydraulic press and hydroulic breaks :It is used to lift the heavy loads. If a small force f isapplied on pistion of C then the pressure exerted onthe liquid
afP
[a = Area of cross section of the pistion in C]This pressure is transmitted equally to piston of cylinder D.
AfA
afPAF As A >> a, F << f
Thermal Expansion :When matter is heated without any change in its state it usually expands. According to atomic theoryof matter, a symmetry in potential energy curve is responsible for thermal expansion. As energy ofatoms increases, hence the average distance between the atoms increases. So the matter as a wholeexpands.
(1) Thermal expansion is minimum in case of solids but maximum in case of gases because intermolecularforce is maximum in solids but minimum in gases.
(2) Solids can expand in one dimension (linear expansion) two dimensions (superficial expansion) andthree dimensions (volume expansion) While liquids and gases usually suffer change in volume only.Heat :
(1) The form of energy which is exchanged among various bodies or system on account of temperaturedifference is defined as heat.
(2) We can change the temperature of a body by giving heat (temperature rises) or by removing heat(temperature falls) from body.
(3) Heat is a scalar quantity. It's units are joule, erg, cal, kcal etc.(4) 1 kcal = 1000 cal = 4186 J and 1 cal = 4.18 J
204
Specific Heat :(1) The amount of heat energy required to raise the temperature of unit mass of a body through (or K)
is called specific heat of the material of the body.
If Q heat changes the temperature of mass m by then specific heat
mC
122 TLUnit (2) Molar specific Heat :
Molar specific heat of a substance is defined as the amount of heat required to raise the temperatureof one gram mole of the substance through a unit degree, it is represented by C
Q1QmM
MMQC
Unit : 1221 TLM
(3) Latent Heat :The amount of heat required to change the state of the mass m of the substance while its temperatureremaining constant is written as : Q = mL, where L is the latent heat Latent heat is also called as
Heat of Transformation, It's unit is gmcal
or kgJ
and dimension : 22 TL
Elastic behaviour :The propertory of matter by virture of which a body
tends to regain its original shape and size after the removal ofdeforming force is called elasticity.
In solids, atoms and molecules are arranged in such away that each molecule is acted upon by the forces due toheighbouring molecules. These forces are known asintermolecular forces.
For simplicity, the two molecules in their equilibrium positions (at inter-molecular distancer = r0) are shown by connecting them with a spring.
In fact, the spring connecting the two molecules represents the inter-molecular force betweenthem on applying the deforming forces, the molecules either come closer or go far apart from eachother and restoring forces are developed. When the deforming force is removed, these restoringforces bring the molecules of the solid to their respective equilibrium position (r = r0) and hence thebody regains its original form.Thermal Capacity : It is defined as the amount of heat required to raise the remperature of the whole body (mass m)through 10C or 1K.
Thermal capacity
Qcmc
The value of thermal capacity of a body depends upon the nature of the body and its mass.
Dimensions 1221 TLM Units : kJ,
CCalo
205
Principle of Calorimetry :Calorimetry means 'measuring heat'.When two bodies (one being solid and other liquid or both being liquid) at different temperatures aremixed, heat will be transfered from body at higher temperature to a body at lowe temperature tillboth acquire same temeprature. The body at higher temperature releases heat while body at lowertemeprature absorbs it, so that
Heat lost = Heat gainedi.e., Principle of calorimetry represents the law of conservation of heat energy.Thermometry :A branch of science which deals with the measurement of temperature of a substance is known asthermometry.Phase change :
We use the term phase to describe a specific state of matter, such as solid, liquid or gas. Atranstion from one phase to another is called a phase change.
For any given pressure a phase change takes place at a definite temperature, usuallyaccompanied by absorption or emission of heat and a change of volume and density.
In phase change 10ce at 00C melts into water at 00C. Water at 1000C boils to form steam at1000C.Streamline flow :
Stream line flow of a liquid is that flow in which each element of the liquid passing through apoint travels along the same path and with the same velocity as the preceding element passes throughthat point.
A streamline may be defined as the path, straight orcurved, the tangent to which at any point gives the directionof the flow of liquid at that point.
The two streamlines cannot cross each other andthe greater is the crowding of streamlines at a place, thegreater is the velocity of liquid particles at that place.Laminar Flow :
If a liquid is flowing over a horizontal surface with a steady flow and moves in the form oflayers of different velocities which do not mix with each other, then the flow of liquid is called laminarflow.
In this flow, the velocity of liquid flow is always less than the critical velocity of the liquid. Thelaminar flow is generally used synonymously with streamlined flow.Turbulent flow :
When a liquid, moves with a velocity greater thanits critical velocity, the motion of the particles of liquidbecome disordered or irregular. Such a flow is called aturbulent flow.
In a turbulent flow, the path and the velocity of the particles of the liquid change continuouslyand haphazardly with time from point to point. In a turbulent flow, most of the external energy
206
maintaining the flow is spent in producint eddies in the liquid and only a small fraction of energy isavailable for forward flow.Critical velocity :
The critical velocity is that velocity of liquid flow upto which its flow is streamlined and abovewhich its flow becomes turbulent.Reynold's number :
Reynold's number is a pure number whichdetermines the nature of flow of liquid through apipe.
areaunitperforceViscousareaunitperforceInprtialNR
3AVdtdmNow
Inertial force per unit area AdtdmV
Adtd
2VAAVV
Viscous force per unit area rnv
AF
VrNR
If ,2000N0 R If 3000N2000 R If 3000NR
laminar or stream line Unstable flow definitely turbulent flowEquation of continuity :
The equation of continuity is derived from theprinciple of conservation of mass.
A non-viscous liquid in streamline flow passesthrough a tube AB of varying cross section. Let the crosssectional area of the pipe at points A and B be a1 and a2respectively.
Mass of the liquid entering per second at A = Massof the liquid leaving per second at B.
222111 vava 2211 vava
If the liquid is incompressible av = constantwhich is the equation of continuity.
207
MCQFor the answer of the following questions choose the correct alternative from amongthe given ones.1. Two wires are made of the same material and have the same volume. However, wire 1 has cross-
sectional area A and wire 2 has cross-sectional Area 3A. If the length of wire 1 increases by onappling force F. How much force is needed to stretch wire 2 by same amount.(A) F (B) 4F (C) 6F (D) 9F
2. The increases in length in l of a wire of length L by longitudinal stress. Then the stress is propotionalto..............
(A) L
(B) L (C) (D) L2
3. The dimensions of four wires of the same material are given below, in which wire the increase inlength will be maximum when the same strain is applied.(A) Length 100 cm, Diameter 1 mm (B) Length 200 cm, Diameter 2 mm(C) Length 300 cm, Diameter 3 mm (D) Length 50 cm, Diameter 0.5 mm
4. The young's modulus of a wire of length L and radius r is Y 2mN
. If the length and radius are
reduced to 2L
and 2r
. Then what will be its young's modulus ?
(A) 2Y
(B) Y (C) 2Y (D) 4Y
5. A beam of metal supported at the two ends is loaded at the centre. The depression at the centre ispropotional to..............
(A) Y2 (B) Y (C) Y1
(D) 2Y1
6. A wire is loaded by 6 kg at its one end, the increase in length is 12 mm. If the radius of the wire isdoubled and all other magnitudes are unchanged, then increase in length will be.............(A) 6 mm (B) 3 mm (C) 24 mm (D) 48 mm
7. On increasing the length by 0.5 mm in a steel wire of length 2 m and area of cross-section 2 mm2 theforce required is................
[Y for steel mN102.2 11 ]
(A) N101.1 5 (B) N101.1 4 (C) N101.1 3 (D) N101.1 2
8. A stress of 28 Nm1018.3 is applied to steel rod of length 1 m along its length. Its young's
modulus is 211
mN102 . Then what is the elongation produced in the rod in mm ?
(A) 3.18 (B) 6.36 (C) 5.18 (D) 1.59
208
9. Two springs P & Q of force constant KP & KQ PQ
KK2
are stretched by applying force equal
magnitude. If the energy stored in Q is E. Then what is the energy stored in P ?
(A) E (B) 2E (C) 2E
(D) 4E
10. A force F is needed to break a copper wire having radius R, The force needed to break a copperwire of radius 2R will be........
(A) 2F
(B) 2F (C) 4F (D) 4F
11. A rubber cord 10m long is suspended vertically. How much does it stretch under its own weight.
( Density of rubber is 22
83 s
m10g,mN105Y,
mkg1500 )
(A) m1015 4 (B) m105.7 4 (C) m1012 4 (D) m1025 4
12. The young's modulus of the material of a wire is equal to the ...........(A) stress required to increase its length four times(B) strtess required to prdouce unit strain(C) strain produced in it (D) stress acting on it
13. If x, longitudinal strain is produced in a wire of young's modulus y then energy stroed in the materialof the wire per unit volume is..........
(A) yx2 (B) 2yx2 (C) xy21 2 (D) 2yx
21
14. A steel wire of cross-sectional area 26 m103 can with stand a maximum strain of 10-3 Young's
modulus of steel is .mN102 2
11 The maximum mass the wire can hold is.............
2s
m10g
(A) 40 kg (B) 60 kg (C) 80 kg (D) 100 kg
15. The young's modulus of a rubber string 8 cm long and density 3mkg5.1 is .
mN105 2
8 What will
be the length increase due to its own weight ?
(A) m106.9 5 (B) m106.9 11
(C) m106.9 3 (D) m6.9
16. A and B are two wires. The radius of A is twice that of B. They are stretched by the same load. Thenwhat is the stress on B ?(A) Equal to that on A (B) Four times that on A(C) Two times that on A (D) Half that on A
17. If the length of wire is reduced to half then it can hold the .............load.(A) Half (B) Same (C) Double (D) One fourth
209
18. There are two wires of same material and same length. While the diameter of second wire is 2 times,the diameter of first wire. Then what will be the ratio of extension produced in the wire by applyingsame load ?(A) 1 : 1 (B) 2 : 1 (C) 1 : 2 (D) 4 : 1
19. When the length of a wire having cross-section area 26 m10 is stretched by 0.1% then tension init is 100 N. Young's modulus of material of wire is...........
(A) 212
mN10 (B) 2
2
mN10 (C) 2
10
mN10 (D) 2
11
mN10
20. Two wires of equal lengths are made of the same material wire A has a diameter that is twice as thatof wire B. If identical weights are suspended from the ends of these wires the increase in lengthis............(A) Four times for wire A as for wire B. (B) Twice for wire A as for wire B.(C) Half for wire A as for wire B. (D) One-fourth for wire A as for wire B.
21. Steel and copper wires of same length are stretched by the same weight one after the other. Young'smodulus of steel the ratio increase in length ?
(A) 52
(B) 53
(C) 45
(D) 25
22. An area of a cross-section of rubber string is 2 cm3. Its length is doubled when stretched with alinear force of 5102 dynes. What will be young's modulus of the rubber in dynes ?
(A) 5104 (B) 5101 (C) 5102 (D) 4101
23. A substance breaks down by a stress of If the density of the material of the wire is then the length ofwire of the substance which will break under its own weight when suspended vertically is.............(A) 66.6 m (B) 60.0 m (C) 33.3 m (D) 30.0 m
24. The temperature of a wire of length 1 meter and area of cross-sectional section 1 cm2 is increasedfrom to If the rod is not allowed to increase in length. What will be the force required ?
211o5
mN10Y,C/10
(A) 103 N (B) 104 N (C) 105 N (D) 109 N25. If longitudinal strain for a wire is 0.03 and its poisson's ratio is 0.5, then what is its lateral strain ?
(A) 0.003 (B) 0.0075 (C) 0.015 (D) 0.4
26. An aluminium rod (Young's modulus 29
mN107 ) has a breaking strain of 0.2 % what is the
minimum cross-sectional area of the rod in order to support a load of 104 Newtons ?
(A) 22 m101 (B) 23 m104.1
(C) 23 m105.3 (D) 24 m101.7
210
27. Two wires of copper having the length in the ratio 4 : 1 and their are as 1 : 4 are stretched by thesame force. What will be the ratio of longitudinal strain in the two wires ?(A) 1 : 16 (B) 16 : 1 (C) 1 : 64 (D) 64 : 1
28. A wire elongates by 1 mm when a load W is hanged from it. If the wire goes ever a pulley and twoweight W each are hang at the two ends. What will be the elongation of the wire ? (in mm)
(A) 2 (B) zero (C) 2
(D)
29. 200 kg weight hanged at a free edge of vertical wire of length 600.5 cm when removed the weight,length reduced by 0.5 cm and it gets original status then what is the young moduluse of wire ?(A) 2.35 1012 N/m2 (B) 1.35 1010 N/m2
(C) 13.5 1011 N/m2 (D) 23.5 109 N/m2
30. The ratio of diameter of two wires of same material is n : 1 the length of wires are 4 m each. Onapplying the same load. What will be the increase in length of their wire ?(A) n2 times (B) n times (C) 2n times (D) None of the above
31. Longitudinal stress of 2mmN1 is applied on a wire what is the % increase in length ?
2
4
mN10Y
(A) 0.002 (B) 0.001 (C) 0.003 (D) 0.0132. A steel wire is stretched with a definate load of If the young's modulus of the wire is Y. For decreasing
the value of Y............(A) Radius is to be decreased (B) Radius is to be increased(C) Length is to be increased (D) None of the above
33. The area of cross-section of a steel wire is
2
11
mN102Y is 0.1 cm2, what will be the force
required to double its length ?
(A) N102 12 (B) N102 11 (C) N102 10 (D) N102 6
34. Two wires A & B are of same materials. Their lengths in the ratio 1 : 2 and diameters are in the ratio2 : 1 when stretched by force BA FandF respectively, they get equal increase in lengths Then the
ratio FBFA
should be...........
(A) 1 : 2 (B) 1 : 1 (C) 2 : 1 (D) 8 : 1
35. The mean distance between the atoms of iron is 10103 and interatomic force constant for iron is
mN7 . What is the young's modulus of elasticity for iron ?
(A) 25
mN1033.2 (B) 2
10
mN103.23
(C) 210
mN10233 (D) 2
10
mN1033.2
211
36. A force of 200 N is applied at one end of a wire of length 2 m and having area of cross-section22 cm10 , the other end of the wire is rigidly fixed. If of linear expansion of the wire
C/108 o6 and young's modulus 211
mN102.2Y and its temperature is increased by
50C then the increase in the tension of the wire will be..........(A) 4.2 N (B) 4.4 N (C) 2.4 N (D) 8.8 N
37. A uniform plank of young's modulus Y is moved over a smooth horizontal surface by a constanthorizontal force F, The area of cross-section of the plank is A. What is the compressive strain on itsplank in the direction by the force ?
(A) AYF
(B) AYF2
(C)
AYF
21
(D) AYF3
38. The length of a wire is 1.0 m and the area of cross-section is 22 cm100.1 . If the work done forincrease in length by 0.2 cm is 0.4 joule. Then what is the young's modulus ? Of material of the wire?
(A) 210
mN100.2 (B) 2
10
mN100.4
(C) 211
mN100.2 (D) 2
11
mN100.4
39. A rubber cord catapult has cross-sectional area 25 mm2 and initial length of cord is 10 cm. It isstretched to 5 cm and then released to project a missile of mass 5 gm. Taking
28
rubber mN105Y velocity of projected missile is............
(A) sm20 (B) 1s
m100 (C) 1sm250 (D) 1s
m200
40. A wire of cross-section 4 mm2 is stretched by 0.1 mm by a certain weight. How far (length) will bewire of same material and length but of area 8 mm2 stretched under the action of same force.(A) 0.05 mm (B) 0.10 mm (C) 0.15 mm (D) 0.20 mm
41. According to Hooke's law of elasticity stress is increased the ratio of stress toi strain.........(A) increases (B) Decreases(C) becomes zero (D) Remains constant
42. Young's modulus of rubber is 24
mN10 and area of cross section is 2 cm2 if force of 5102 dyne is
applied along its length then its initial length L becomes .............(A) 3L (B) 4L(C) 2L (D) None of the above
212
43. A copper wire of length 4 m and area of cross-section 1.2 cm2 is stretched with a force of 3108.4 N.
If young's modulus for copper is 211
mN102.1 What will be the length increase of the wire ?
(A) 1.33 mm (B) 1.33 cm (C) 2.66 mm (D) 2.66 cm
44. If the interatomic spacing in a steel wire is &A3 o .mN1020Y 2
10Steel the force constant =
.............
(A) AN106 2 (B) A
N106 9 (C) AN106 5 (D) A
N106 5
45. A wire of length 2 m is made from 10cm3 of copper. A force F is applied so that its length increasesby 2mm. Another wire of length 8 m is made from the same volume of copper. If the force F isapplied to it, its length will increase by............(A) 0.8 cm (B) 1.6 cm (C) 2.4 cm (D) 3.2 cm
46. A wire of length L and radius r is rigidly fixed at one end on stretching the other end of the wire aforce F the increase in its lengths is L. If another wire of same material but of length 2L and radius 2ris stretched with a force of 2F, the increase in its length will be ..........
(A) (B) 2 (C) 2
(D) 4
47. In steel the young's modulus and the strain at the breaking point are 211
mN102 and 0.15
respectively the stress at the breaking point for steel is therefore ...........
(A) 211
mN1033.1 (B) 2
12
mN1033.1
(C) 213
mN105.7 (D) 2
10
mN103
48. Which of the following statement is correct(A) Hooke's law is applicable only within elastic limit.(B) The adiabatic and isothermal elastic constants of a gas area equal.(C) Young's modulus is dimensionsless.(D) Stress multiplied by strain is equal to stored energy.
49. The force required to stretch a steel wire of 2cm1 cross-section to 1.1 times its length would be
2
11
mN102Y
(A) N102 6 (B) N102 3 (C) N102 6 (D) N102 7
213
50. Which one of the following quantities does not have unit of force per unit area..........(A) stress (B) strain(C) Young's modulus of elasticity (D) pressure
51. A copper wire and a steel wire of same diameter and length are connected end to end a force isapplied, which stretches their combined length by 1 cm, the two wires will have.......(A) different stresses and strains (B) the same stress and strain(C) the same strain but different stresses (D) the same stress but different strains
52. A steel ring of radius r and cross-section area 'A' is fitted on to a wooden disc of radius R(R > r ) If young's modulus be E then what is force with which the steel ring is expanded ?
(A) rRAE (B)
rrRAE (C)
E R rA R
(D) AREr
53. A wire of diameter 1 mm breaks under a tension of 100 N. Another wire of same material as that ofthe first one, but of diameter 2 mm breaks under a tension of.........(A) 500 N (B) 1000 N (C) 10,000 N (D) 4000 N
54. A fixed volume of iron is drawn into a wire of length L. The extension x produced in this wire be aconstant force F is propotional to..........
(A) 2L1
(B) L1
(C) L2 (D) L
55. On applying a stress of 28
mN1020 the length of a perfect elastic wire is doubled. What will be its
Young's modulus ?
(A) 28
mN1040 (B) 2
8
mN1020
(C) 28
mN1010 (D) 2
8
mN105
56. To keep constant time, watches are fitted with balance wheel made of........(A) invar (B) stainless steel (C) Tungsten (D) platinum
57. A wire is stretched by 0.01 m by a certain force F. Another wire of same material whose diameterand length are double to the original wire is stretched by the same force ? Then what will be itselongation ?(A) 0.005 m (B) 0.01 m (C) 0.02 m (D) 0.002 m
58. The Coefficient of linear expansion of brass & steel are 21 & If we take a brass rod of length 1& steel rod of length 2 at C0o , their difference in length 12 will remain the same at atemperature if.................
(A) 1221 (B) 1212
(C) 2221
22 (D) 2211
214
59. A rod is fixed between two points at C20 o The Coefficient of linear expansion of material of rad is
C/101.1 o5 and Young's modulus is 211
mN102.1 . Find the stress developed in the rod if
temperature of rod becomes C10 o .
(A) 27
mN1032.1 (B) 2
15
mN1010.1 (C) 2
8
mN1032.1 (D) 2
6
mN1010.1
60. How much force is required to produce an increase of 0.2% in the length of a bross wire of diameter
0.6 mm (Young's modulus for brass 211
mN109.0 ).
(A) Nearly 17 N (B) Nearly 34 N (C) Nearly 51 N (D) Nearly 68 N
61. A 5m long aluminium wire
2
10
mN107Y of diameter 3mm supports a 40 kg mass. In order
to have the same elongation in a copper wire 210
mN1012Y of the same length under the same
weight, the diameter should now be in mm............(A) 1.75 (B) 1.5 (C) 2.5 (D) 5.0
62. Two similar wires under the same load yield elongation of 0.1 mm and 0.05 mm respectively. If thearea of Cross - section of the first wire is 4mm2. Then what is the area of cross - section of thesecond wire ?
(A) 2mm6 (B) 2mm8 (C) 2mm10 (D) 2mm12
63. An iron rod of length 2m and cross-section area of 2mm50 stretched by 0.5 mm, when a mass of250 kg is hung from its lower end. What is young's modulus of the iron rod ?
(A) 210
mN106.19 (B) 2
15
mN106.19
(C) 218
mN106.19 (D) 2
20
mN106.19
64. A load W produces an extension of 1mm in a thread of radius r. Now if the load is made 4 w andradius is made 2r all other things remaining same the extension will becomes..............(A) 4 mm (B) 16 mm (C) 1 mm (D) 0.25 mm
65. A steel wire of 1m long and 2mm1 cross sectional area is hung from rigid end when weight of 1 kg
is hung from it then change in length will be............
2
11
mN102Y
(A) 0.5 mm (B) 0.25 mm (C) 0.05 mm (D) 5 mm
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66. Calculate the work done, if wire is loaded by 'M g' weight and the increase in length is 'l' ?
(A) mgl (B) Zero (C)mg
(D) 2mgl
67. Two wires of same diameter of the same material having the length and 2 . If the force F isapplied on each, what will be the ratio of the work done in the two wires ?(A) 1 : 2 (B) 1 : 4 (C) 2 : 1 (D) 1 : 1
68. A 5 meter long wire is fixed to the ceiling. A weight of 10 kg is hung at the lower end and is 1 meterabove the four. The wire was elongated by 1 mm. What is the stored in the wire due to stretching ?(A) Zero (B) 0.05 Joule(C) 100 Joule (D) 500 Joule
69. If the force constant of a wire is k. What is the work done in increasing the length of the wire by ?
(A) 2k
(B) k (C)2
k 2(D) 2k
70. Wire A and B are made from the same material. A has twice the diameter and three times the lengthof B. If the elastic limits are not reached when each is stretched by the same tension, what is the ratioof energy stored in A to that in B ?(A) 2 : 3 (B) 3 : 4 (C) 3 : 2 (D) 6 : 1
71. A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to thelower and. The weight stretches the wire by 1 mm. Then what is the elastic energy stored in the wire?(A) 0.1 J (B) 0.2 J (C) 10 J (D) 20 J
72. A brass rod of cross sectional area 1 cm2 and length 0.2 m is compressed length wise by a weight of
5 kg. If young's modulus of elasticity of brass is 211
mN101 and 2s
m10g Then what will be
increase in the energy of rod ?
(A) J10 5 (B) J105.2 5 (C) J105 5 (D) J105.2 4
73. Young's modulus of the material of a wire is Y. On pulling the wire by a force F the increase in itslength is x, what is the potential energy of the stretched wire ?
(A) Fx21
(B) Yx21
(C) 2Fx21
(D) None of these
74. The work per unit volume to stretch the length by 1% of a wire with cross - sectional area 1 mm2 will
be............
2
11
mN109Y
(A) J109 11 (B) J105.4 7 (C) J109 7 (D) J105.4 11
216
75. A wire of length 50 cm and cross - sectional area of 1 mm2 is extended by 1mm what will be the
required work ? 211 Nm102Y
(A) J106 2 (B) J102 2 (C) J104 2 (D) J101 2
76. If a spring extends by x cm loading then what is the energy stored by the spring ? (If T is tension inthe spring & K is spring constant)
(A)x2
T2
(B)k2
T2
(C) 2Tx2
(D)kT2 2
77. On stretching a wire what is the elastic energy stored per unit volume ?
(A)F
2AL (B) L2FA
(C) A2FL
(D) 2FL
78. When a force is applied on a wire of uniform cross-sectional area 26 m103 and length 4m, the
increase in length is 1 mm. what will be energy stored in it ?
2
11
mN102Y
(A) 62. 50 J (B) 0.177 J (C) 0.075 J (D) 0.150 J79. k is the force constant of a spring what will be the work done in increasing its extension form
1 to 2 be ?
(A) 1 2k (B) 122k (C) 2
12
2k (D) 21
222
k
80. When a 4 kg mass is hung vertically on a light spring that obeys Hook's law, the spring stretches by2 cm what will be the work required to be done by an external agent in stretching this spring by5 cm ?(A) 4.900 Joule (B) 2.450 Joule (C) 0.495 Joule (D) 0.245 Joule
81. The isothermal elasticity of a gas is equal to....................(A) Density (B) Volume (C) Pressure (D) Specific heat
82. If the volume of a block of aluminium is decreased, by the pressure (stress) on its surface is increased
by.................(Bulk modulus of 210 Nm105.7A )
(A) 210
mN105.7 (B) 2
8
mN105.7 (C) 2
6
mN105.7 (D) 2
4
mN105.7
83. The specific heat at constant pressure and at constant volume for an ideal gas are vp CandC and
isothermal elasticities are EandE respectively. What is the ratio of EandE .
(A)p
v
CC
(B)v
p
CC
(C) vp CC (D)vp CC
1
217
84. What is the ratio of the adiabatic to isothermal elasticities of a triatomic gas ?
(A) 43
(B) 34
(C) 1 (D) 35
85. To what depth below the surface of sea should a rubber ball be taken as to decrease its volume by
0.1% (Take : density of sea water 3mkg1000 Bulk modulus of rubber 2
8
mN109 ,acceleration
due to gravity 2sm10 )
(A) 9 m (B) 18 m (C) 180 m (D) 90 m
86. The compressibility of water 5104 per unit atmospheric pressure. The decrease in volume of100 cubic centimeter of water under a pressure of 100 atmosphere will be................
(A) CC104 5 (B) CC104 5 (C) 0.025 CC (D) 0.004 CC
87. If a rubber ball is taken at the depth of 200m in a pool, Its volume decreases by 0.1% . If the density
of the water is 233
sm10g&
mkg101 . Then waht will be the volume elasticity in ?
mN
2
(A) 810 (B) 8102 (C) 910 (D) 9102
88. For a constant hydraulic stress on an object, the fractional change in the object volume
and
its bulk modulus (B) are related as...............(A) 0.01 (B) 0.06 (C) 0.02 (D) 0.03
89. When a pressure of 100 atmosphere is applied on a spherical ball then its volume reduces to 0.01%
What is the bulk moduls of the material of the rubber in 2cmdyne .
(A) 121010 (B) 12101 (C) 1210100 (D) 121020
90. The pressure applied from all directions on a cube is p. How much its temperature should be raisedto maintain the orginal volume ? The volume elasticity. of the cube is B and the coefficient of volumeexpansion is .
(A) P
(B) P
(C)p
(D) p
91. A uniform cube is subjected to volume compression. If each side is decreased by 1% Then what isbulk strain ?(A) 0.01 (B) 0.06 (C) 0.02 (D) 0.03
218
92. The ratio of two specific heats of gas v
p
CC
for Argon is 1.6 and for hydrogen is 1.4. Adiabatic
elasticity of Argon at pressure p is E. Adiabatic elasticity of hydrogen will also be equal to E at thepressure.
(A) p (B) P87
(C) P78
(D) 1.4 P
93. What is the isothermal bulk modulus of a gas at atmospheric pressure ?(A) 1 mm of Hg (B) 13.6 mm of Hg
(C) 25
mN10013.1 (D) 2
5
mN10026.2
94. The bulk modulus of an ideal gas at constant temperature.........(A) is equal to its volume V (B) is equal to P/2(C) is equal to its pressure P (D) cannot be determined
95. A material has poisson's ratio 0.50. If uniform rod of it suffers a longitudinal strain of 3102 . Thenwhat is percentage change in volume ?(A) 0.6 (B) 0.4 (C) 0.2 (D) 0
96. There is no change in the volume of a wire due to change in its length on stretching. What is thepossion's ratio of the material of the wire....(A) +0.5 (B) -0.50 (C) 0.25 (D) -0.25
97. Which statement is true for a metal......
(A) Y (B) Y (C) Y (D) 1Yη
98. Which of the following relation is true
(A) 61kY3 (B) nygnyk
(C) Ynk66 (D) nnY5.06
99. Two wires A & B of same length and of the same material have the respective radius r, & r2 their oneend is fixed with a rigid support and at the other end equal twisting couple is applied. Then what willwe be the ratio of the angle of twist at the end of A and the angle of twist at the end of B.
(A) 22
21
rr
(B) 21
22
rr
(C) 41
42
rr
(D) 42
41
rr
100. The poisson's ratio can not have the value...........(A) 0.7 (B) 0.2 (C) 0.1 (D) 0.5
219
101. The value of poisson's ratio lies between........
(A) 21to1 (B) 2
1to43
(C) 1to21
(D) 1 to 2
102. If the young's modulus of the material is 3 times its modulus of rigidity. Then what will be its volumeelasticity ?
(A) zero (B) infinity (C) 210
mN102 (D) 2
10
mN103
103. For a given material the Young's modulus is 2.4 times that of rigidity modulus. What is its poisson'sratio ?(A) 2.4 (B) 1.2 (C) 0.4 (D) 0.2
104. The lower surface of a cube is fixed. On its upper surface is applied at an angle of 300 from itssurface. What will be the change of the type ?(A) shape (B) size (C) none (D) shape & size
105. The upper end of a wire of radius 4 mm and length 100 cm is clamped and its other end is twistedthrough an angle of 300. Then what is the angle of shear ?(A) 120 (B) 0.120 (C) 1.20 (D) 0.0120
106. Mark the wrong statement.(A) Sliding of moleculawr layer is much easier than compression or expansion.(B) Receiprocal of bulk modulus is called compressibility.(C) Twist is difficult in big rod as compared to small rod.(D) Which is more strong out of hollow and solid cylinder having equal length and mass ?
107. A 2m long rod of radius 1 cm which is fixed from one end is gien a twist of 0.8 radians. What will bethe shear strain developed ?(A) 0.002 (B) 0.004 (C) 0.008 (D) 0.016
108. Shearing stress causes change in(A) length (B) breadth (C) shape (D) volume
109. What is the relationship between Young's modulus Y, Bulk modulus k and modulus of rigidity ?
(A) k3k9Y
(B) k3yky9Y
(C) k3
k9Y
(D) k9k3Y
110. What is the possible value of posson's ratio ?(A) 1 (B) 0.9 (C) 0.8 (D) 0.4
111. The graph shown was obtained from experimentalmeasurements of the period of oscillations T for differentmasses M placed in the scale pan on the lower end of thepassing through the origin is that the......(A) Spring did not obey Hooke's law(B) Amplitude of the oscillations was large(C) Clock used needed regulating(D) Mass of the pan was neglected
220
112. A graph is shown between stress and strainfor metals. The part in which Hooke's lawholds good is(A) OA (B) AB(C) BC (D) CD
113. The strain-stress curves of three wires of different materialsare shown in the figure. P, Q and R the elastic limits of thewires the figure shows that(A) Elasticity of wire P is maximum.(B) Elasticity of wire Q is maximum.(C) Tensile strength of R is maximum(D) none of above
114. The diagram shows a force extension graph for a Rubberband consider the following statements.(I) It will be easier to compress these
rubber than expand it.(II) Rubber does not return to its original
length after it is stretched.(III) The rubber band will get heated if it
is stretched and relased.Which of these can be deduced from the graph.(A) III only (B) II and III (C) I and III (D) I only
115. The stress versus strain graph for wires of two material A& B are as shown in the figure. If YA & YB are the young'smodulus of the materials then(A) YB = 2YA
(B) YA = YB
(C) YB = 3YA
(D) YA = 3YB
116. The load versus elongation graph for four wires of the samematerial is shown in the figure. The thickest wire isrepresented by the line.(A) OD (B) OC(C) OB (D) OA
221
117. The adjacent graph shows the extension of a wire of length1m suspended from the top of a roof at one end with theload W connected to the other end. If the cross sectionalarea of the wire is 10-6 m2, calculate the young's modulusof the material of the wire.
(A) 211
mN102 (B) 2
11
mN102
(C) 212
mN103 (D) 2
12
mN102
118. The graph is drawn between the applied force F and thestrain (x) for a thin uniform wire the wire behaves as aliquid in the part.(A) ab (B) bc(C) cd (D) oa
119. The graph shows the behaviour of a length of wire in theregion for which the substance obey's Hooke's law. P &Q represent.(A) p = applied force, Q = extension(B) p = extension, Q = applied force(C) p = extension, Q = stored elastic energy(D) p = stored elastic energy, Q = extension
120. The potential energy U between two nmolecules as afunction of the distance x between them has been shownin the figure. The two molecules are.(A) Attracted when x lies between A & B
and are repelled when x lies between B & C.(B) Attracted when x lies between B and
C and are repelled when x lies between A and B.(C) Attracted when they reach B.(D) Repelled when they reach B.
121. The value of force constant between the applied elasticforce F and displacement will be
(A) 3 (B) 31
(C) 21
(D)23
222
122. The diagram shows stress v/s strain curvefor the materials A and B. From the curveswe infer that(A) A is brittle but B is ductile(B) A is ductile and B is brittle(C) Both A & B are ductile(D) Both A & B are brittle
123. Which are of the following is the young'smodulus (in N/m2) for the wire having thestress strain curve in the figure.
(A) 111024
(B) 11100.8
(C) 111010
(D) 11100.2
124. The diagram shows the change x in thelength of a thin uniform wire caused by theapplication of stress F at two differenttemperature T1 & T2. The variation
(A) 21 TT
(B) 21 TT
(C) 21 TT
(D) None of these
125. The point of maximum and minimumattraction in the curve between potentialenergy (U) and distawnce (r) of a diatomicmolecules are respectively.(A) S and R (B) T and S(C) R and S (D) S and T
Assertion and Reason :Read the assertion and reason carefully to mark the correct option out of the option given below(A) If both assertion and reason are true and reason is the correct explanation of the assertion.(B) If both assertion and reason are true but reason is not the correct explanation of the assertion.(C) If assertion is true but reason is false.(D) If assertion and reason both are false.
223
126. Assertion : The stretching of a coil is determined by its shear modulus.Reason : Shear modulus change only shape of a body keeping its dimensions unchanged.(A) a (B) b (C) c (D) d
127. Assertion : The bridges are declared unsafe after a long use.Reason : Elastic strength of bridges decrease with(A) a (B) b (C) c (D) d
128. Assertion : Two identical solid balls, one of ivory and the other of wet clay are dropped from thesame height on the floor. Both the balls will rise to same height after.Reason : Ivory and wet-clay have same elasticity.(A) a (B) b (C) c (D) d
129. Assertion : Young's modulus for a perfectly plastic body is zero.Reason : For a perfectly plastic body is zero.(A) a (B) b (C) c (D) d
130. Assertion : Identical spring of steel and copper are equally stretched more will be done on the steelspring.Reason : Steel is more elastic than copper.(A) a (B) b (C) c (D) d
131. A wire is stretched to double the length which of the following is false in this context ?(A) Its volume increased (B) Its longitudinal strain is I(C) Stress = Young's modlus (D) Stress = 2x Young's modulus
132. Which is the dimensional formula for modulus of rigidity ?
(A) 211 TLM (B) 211 TLM (C) 121 TLM (D) 221 TLM
133. When more than 20 kg mass is tied to the end of wire it breaks what is maximum mass that can betied to the end of a wire of same material with half the radius ?(A) 20 kg (B) 5 kg (C) 80 kg (D) 160 kg
134. When 100 N tensile force is applied to a rod of 10-6 m2 cross-sectional area, its length increases by1% so young's modulus of material is..........(A) 1012 Pa (B) 1011 Pa (C) 1010 Pa (D) 102 Pa
135. A composite wire is made by joining ends of two wires of equal dimensions, one of copper and theother of steel. When a weight is sttached to its end the ratio of increase in their length is .........
CopperSteel Y720Y
(A) 20.7 (B) 10.7 (C) 7:20 (D) 1:7136. A rubber ball when taken to the bottom of a 100 m deep take decrease in volume by 1% Hence, the
bulk modulus of rubber is............
2s
m10g
(A) 106 Pa (B) 108 Pa (C) 107 Pa (D) 109 Pa
224
137. Young's modulus of a rigid body is(A) 0 (B) 1 (C) (D) 0.5
138. Pressure on an object increases from .Pa10165.1toPa1001.1 55 He volume decrease by10% at constant temperature. Bulk modulus of material is........
(A) Pa1055.1 5 (B) Pa102.51 5
(C) Pa104.102 5 (D) Pa108.204 5
139. Cross-sectional area oif wire of length L is A. Young's modulus of material is Y. If this wire acts as aspring what is the value of force constant ?
(A) LYA
(B) L2YA
(C) LYA2
(D) AYL
Surface Tension140. Writing on black board with a pieace of chalk is possible by the property of
(A) Adhesive force (B) Cohesive force (C) Surface force (D) Viscosity
141. When there is no external force, the shape of liquid drop is determined by
(A) Surface tension of liquid (B) Density of Liquid
(C) Viscosity of liquid (D) Tempreture of air only
142. Soap helps in cleaning because
(A) chemicals of soap change
(B) It increase the surface tension of the soluiton.
(C) It absorbs the dirt.
(D) It lowers the surface tension of the solution
143. A beaker of radius 15 cm is filled with liquid of surface tension 0.075 N/m. Force across an imaginarydiameter on the surface ofliquid is
(A) 0.075 N (B) N105.1 2
(C) 0.225 N (D) N1025.2 2
144. A square frame of side L is dipped in a liquid on taking out a membrance is fomed if the surfacetension of the liquid is T, the force acting on the frame will be.
(A) 2 TL (B) 4 TL (C) 8 TL (D) 10 TL
145. The force required to separate two glass plates of area 10-2 m2 with a film of water 0.05 mm thick
between them is (surface tension of water is mN1070 3 )
(A) 28 N (B) 14 N (C) 50 N (D) 38 N
225
146. Surface tension of a liquid is found to be infuenced by
(A) It increases with the increase of temprature.
(B) Nature of the liquid in contact.
(C) Presense of soap that increaase it.
(D) Its variation with the concentration of the liquid.
147. A thm metal disc of radius r floats on water surface B and bends the surface down wards along theperimeter making an angle Q with vertical edge of the disc. If the disc displaces a weight of water Wand surface tension of water is T, then the weight at metal dis
(A) wrT2 (B) wcosrT2
(C) wcosrT2 (D) cosrT2w148. A thin liquid film formed between a u-shaped wire and
a light slider supports a weight of N105.1 2 (seefigure). The length of the slider is 30 cm and its weightnegligible. The surface tension of the liquid film is.
(A) 1mN0125.0 (B) 1mN1.0
(C) 1mN05.0 (D) 1mN025.0
Surface Energy149. Radius of a soap bubble is 'r', surface tension of soap solution is T. Then without incresing the
temprature how much energy will be needed to double its radius.
(A) Tr4 2 (B) Tr2 2
(C) Tr12 2 (D) Tr24 2
150. The amount of work done in blowing a soap bubble such that its diameter increases from d to D is(T = Surface tension of solution)
(A) TdD4 22 (B) TdD8 22
(C) TdD 22 (D) TdD2 22
151. A soap bubble of radius r is blown up to form a bubble of radius 2r under isothermal conditions if theT is the surface tension of soap solution the energy spent in the slowing is.
(A) 2Tr3 (B) 2Tr6 (C) 2Tr12 (D) 2Tr24152. The surface tension of a liquid is 5 N/m. If a thin film of the area 0.02 m2 is formed on a loop, then
its surface energy will be
(A) J105 2 (B) J105.2 2 (C) J102 1 (D) J105 1
226
153. A frame made of a metalic wire enclosing O surface area A is covered with a soap film. If the areaof the frame metalic wire is reduced by 50% the energy of the soap film will be changed by(A) 100% (B) 75% (C) 50% (D) 25%
154. Two small drops mercury, each of radius R, coaless the form a single large drop. The ratio of thetotal surface energies before and after the change is.
(A) 21
2:1 (B) 1:221
(C) 2 : 1 (D) 1 : 2
155. The work done is blowing a soap bubble of 10 cm radius is
(surface tension od the soap solution is m/N100
3)
(A) Joule1036.75 4 (B) Joule1068.37 4
(C) Joule1072.150 4 (D) Joule36.75
156. The work done increasing the size of a soap film drom cm11cm10 is Joule103 4 . Thesurface tension of the film is
(A) mN105.1 2 (B) m
N100.3 2 (C) mN100.6 2 (D) m
N100.11 2
157. A big drop of radius R is formed by 1000 small droplets of coater then the radius of small drop is
(A) 2R
(B) 5R
(C) 6R
(D) 10R
158. 8000 identioal water drops are combined to form a bigdrop. Then the ration of the final surfaceenergy to the intilial surface energy of all the drops together is(A) 1 : 10 (B) 1 : 15 (C) 1 : 20 (D) 1 : 25
159. The relation between surface tension T. Surface area A and surface energy E is given by.
(A) AET (B) EAT (C) A
TE (D) EAT
Angle of Contact160. If a coater drop is kept between two glasses plates then its shape is
(A) (B) (C)
(D) None of these
227
161. A liquid wets a solid completely. The menisions of the liquid in a surfficently long tube is(A) Flat (B) Concave (C) Convex (D) Cylindrical
Pressure Difference162. When two soap bubbles of radius r1 and r2 (r2 > r1) coalesce, the radius of curvature of common
surface is...........
(A) 12 rr (B)21
12
rrrr
(C)12
21
rrrr (D) 12 rr
163. The excess of pressure inside a soap bubble than that of the other pressure is
(A) rT2
(B) rT4
(C) r2T
(D) rT
164. The radill of two soap bubbles are r1 and r2. In isothermal conditions two meet together is vacumThen the radius of the resultant bubble is given by
(A) 2/rrR 21 (B) 3211 rrrrR
(C) 22
21
2 rrR (D) 21 rrR
165. A spherical drop of coater has radius 1 mm if surface tension of contex is m/N1070 3 differenceof pressures between inside and outside of the spherical drop is
(A) 2mN35 (B) 2m
N70 (C) 2mN140 (D) zero
166. In capilley pressure below the curved surface at water will be(A) Equal to atomospheric (B) Equal to upper side pressure(C) More than upper side pressure (D) Lesser than upper side pressure
167. Two bubbles A and B (A>B) are joined through a narrow tube than(A) The size of a will increase(B) The size of B will increase(C) The size of B will increase untill thenpressure equals(D) None of these
168. If the excess pressure inside a soap bubble is balanced by oil column of height 2 mm then the surfacetension of soap solution will be.Cr = 1 and density d = 0.8 gm/cc)
(A) mN9.3 (B) m
N109.3 2 (C) mN109.3 3 (D) m
N109.3 1
Capillarity169. A capillary tube at radius R is immersed in water and water rises in it to a height H. Mass of water
in the capillary tube is M. If the radius of the tube is doubled. Mass of water that will rise in thecapillary tube will now be(A) M (B) 2M (C) (D) 4M
228
170. A vesel whose bottom has round holes with diametre of 0.1 mm is filled with water. The maximumheight to which the water can be filled without leakage is
(S.T. of water = cmdyne75
2sm1000g )
(A) 100 cm (B) 75 cm (C) 50 cm (D) 30 cm171. The correct relation is
(A) hdgcosT2r
(B)
cosT2
hdgr (C)
cos
dghT2r (D) hdg2cosTr
172. In a capillary tube water rises by 1.2 mm. The height of water that will rise in another capillary tubehaving half the radius of the first is(A) 1.2 mm (B) 2.4 mm (C) 0.6 mm (D) 0.4 mm
173. Water rises in a vertical capillary tube upto a beight of 2.0 cm. If tube is inclined at an angle of 600
with the vertical then the what length the water will rise in the tube.
(A) 2.0 cm (B) 4.0 cm (C) cm3
4(D) cm22
174. The lower end of a glass capillary tube is dipped in water rises to a height of 8 cm the tube is thenbroken at a height of 6 cm. The height of water column and engled as contact will be
(A) 1 36 cm . sin4
(B) 1 46 cm . sin5
(C) 43cos.cm6 1 (D) 1 16 cm . sin
2
175. A large number of water drops each of rdius r combine to have a drop of radius R. If the surfacetension is T and the mechanical equllvalent at heat is J then the rise in tempreature will be
(A) rJT2
(B) RJT3
(C)
R1
r1
JJ3
(D)
R1
r1
JT2
Caraphical Question176. The correct curve between the height or depression h of liquid in a capillary tube and its radius is
229
177. A soap bubble is blown with the help of a mechanical pump at the mouth-of a tube the pumpproduces a certion increase per minit in the volume of the bubble irrespective of its internal pressurethe graph between the pressure inside the soap bubble and time t will be
178. Which graph present the variation of surface tension with temperature over small temperature rangesfor coater.
Comprehension type questionPassage - I
When liquid medicine of desting S is to be put in the eye. It is done with the help of a dropperas the bulb on the top of the dropper is pressed a drop forms at the opening od the dropper we wishto estimate the size of the drop. We dirst assume that the drop formed at the opening is sphericalbecause the requires a minimum increase in its surface energy. To determine the size we calculate thenet vertical force due to surface tension T when the radius od the drop is R. When this forcebecomes smaller than the weight of the drop the drop gets detched from the dropper.
179. If the radius od the opening od the dropper is r ; the vertical force due to the surface tension on thedrop of radius R. (cassuming r << R) is
(A) Tr2 (B) TR2 (C)R
Tr2 2(D)
rTR2 2
180. If 2334 ms10mkg10,m105 T = 0.11 N m-1 the radius of the drop when itdetaches from the dropper is approximately
(A) m104.1 3 (B) m103.3 3 (C) m100.2 3 (D) m101.4 3
230
181. After the drop detaches its surface energy is
(A) J104.1 6 (B) J107.2 6 (C) J104.5 6 (D) J101.8 6
Assertion & Reason type questionsRead the assertion and reason carefully and mark the correct option given below.(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.(b) If both assertion and reason are true but reason is not the correct explanation of the assertion.(c) If assertion is true but reason is false.(d) If the assertion and reason both are false.
182. Assertion : It is better to wash the clothes in cold soap solution.Reason : The surface tension of cold solution is more then the surface tension of hot solution.(A) a (B) b (C) c (D) d
183. Assertion : When height of a tube is less then liquid rise in the capillary tube the liquid does notovertow.Reason : Product of radius of meniscus and height of liquid incapilling tube always remains constant.(A) a (B) b (C) c (D) d
184. Assertion : The impurities always decrease the surface tension of a liquid.Reason : The change in surface tension of the liquid depends upon the degree of cont amination ofthe impurity.(A) a (B) b (C) c (D) d
185. Assertion : The concept of surface tension is help only for liquids.Reason : Surface tension does not hold for gases.(A) a (B) b (C) c (D) d
186. Assertion : The water rises higher in a capillary tube of small diametre than in the capillary tube oflarge diamtre.Reason : Height through which liquid rises in a capillay tube is inversely proportional to the diameterof the capillary tube.(A) a (B) b (C) c (D) d
187. Assertion : Tiny drops of liquid resist deforming forces better than bigger drops.Reason : Excess pressure inside a drop is directly proportional to surface tension.(A) a (B) b (C) c (D) d
Pressure and Density188. When a large bubble rises drom the bottom of a lake to the surface. Its radius double S. It atmospheric
pressure in euqal to that of colurnn of colurnn of water height H then the depth of Lake is(A) H (B) 2H (C) 7H (D) 8H
231
189. A trangular lamind of area A and, height h is immersed in a liquid of density S in a vertical plane withits base on the surface of the liquid. The thrust on lamina is
(A) ghA21
(B) ghA31
(C) ghA61
(D) ghA32
190. The density S of coater of bulk modulus B at a depth y in the ocean is related to the density atsurface so by the relation.
(A)
Bgy1 0
0 (B)
Bgy1 0
0
(C)
Bgyh1 0
0 (D)
gy
B10
0
191. By sucking through a straw, a student can reduce the pressure in his lungs to 750 mm of Hg (density
3cmgm6.13 ) using the straw, he can drink water from a glass up to a maximum depth of
(A) 10 cm (B) 75 cm (C) 13.6 cm (D) 1.36 cm192. The pressure on a swimmer 20 m below the surface of coater at sea level is
(A) 1.0 atm (B) 2.0 atm (C) 2.5 atm (D) 3.0 atm
Pascal's Law and Archimedes Principal193. A spherical solid ball of volume V is made of a material of density S. It is falling through a liquid of
density S2 (S2 < S1). Assume that the liquid opplies a viscous force on the ball that is proportional tosquare of its speed V. i.e. Fviscous = - KV2 (K > 0) The terminal speed of the ball is
(A)vg
K (B)
vgK
(C) 1 2gvK
(D) 1 2gv
K
194. The fraction of floating object of volume VO and density do above the surface of a Liquid as densityd will be
(A) dd0 (B)
0
0
dddd (C) d
dd 0(D)
0
0
dddd
195. A body floats in water with one-thired od its volume above the surface of water. It is placed in oil itfloats with half of : Its volume above the surface of the oil. The specific gravity od the oil is.
(A) 35
(B) 34
(C) 23
(D) 1
196. If there were no gravity which of the following will not be there for a fluid.(A) Viscosity (B) Surface tension(C) pressure (D) Archime des's upward thrust
232
197. A piece of solid weighs 120 g in air, 80 g in water and 60 g in liquid the relative density of the solidand that of the solid and that of the liquid are respectively.
(A) 3, 2 (B) 43,2 (C) 2,
43
(D) 23,3
198. Ice pieces are floating in a beaker A containing watre and also in a beakre B containing miscibleliquid of specific gravity 1.2 Ice melts the level of(A) water increases in A (B) water decreases in A(C) Liquid in B decrease B (D) Liquid in B increase
Fluid Flow199. An engine pumps water continuously through a hose water leares the hose with a velocity V and m
is the mass per unit length of the watre Jet what is the rate at which kinetic energy is imperted towater.
(A)3mv
21
(B) 3mv (C) 2mv21
(D) 2vmv21
200. The height of the dam in an hydro electric power station is 10 m. In order to generate 1 MW ofelectric power, the mass of water (in kg) that must full per second on the brades of turbine is(A) 106 (B) 105 (C) 103 (D) 104
201. Eight drops of a liquid of density 3 and each of radius a are fallingt through air with a constantvelocity 3.75 cm S1 when the eight drops coalesce to form a single drop the terminal velocity of thenew drop will be
(A) 12 ms1015 (B) s/m104.2 2 (C) 12 ms1075.0 (D) s/m1025 2
202. A cylinder of height 2.0 m is completely filled with water. The velocity of efflux of water cim m/sthrough a small hole on the side will of the cylinder mear its bottom is(A) 10 (B) 20 (C) 25.5 (D) 5
203. There is a hole in the bottom of tank having water. If total pressure at bottom 3 atm (1 atm ) then thevelocity of water flowig drom hole is
(A) sm400 (B) s
m60 (C) sm600 (D) None of these
204. Two drops of the same radius are falling through air with a steady velocity for 5 cm per sec. If thetwo drops coakesce the terminal velocity would be(A) 10 cm per sec (B) 2.5 cm per sec
(C) secpercm45 31
(D) secpercm25
205. An application of Bernouli's equation for liquid flow is found in(A) Dynamic lift of an aeroplane (B) Viscocity metere(C) Capillary rise (D) tly dulic press
233
206. An L-Sp aped tube with a small office is held in a waterstream as shwon in fig. The upper end of the tube is10.6 cm above the surface of water. What will be theheight of the set of water coming from the office velocityof water stream is 2.45 m/s.(A) zero (B) 20.0 cm(C) 10.6 cm (D) 40 cm
207. A tank is filled with water up to a height H. Water isallowed to come out of a hole P in one of the walls at adepth D below the surface of water express thehorizontal distunce x in terms of H and D.
(A) DHDx (B)
2
DHDx
(C) 2 DHDx (D) DHD4x
208. An incomepressible fluid flows steadily through a cylinderical pipe which has radius 2r at point A andradius r at B further along the flow direction. It the velocity at point A is V, its velocity at point B.
(A) 2v (B) v (C) 2v
(D) 4v
209. If the terminal speed of a sphere of gold
3m
kg5.19density is 2.0 m/s in a viscous liquid
3m
kg5.1density find the terminal speed of a sphere of siher
3m
kg5.10density of the
same size in the same liquid.
(A) sm133.0 (B) s
m2.0 (C) sm1.0 (D) s
m4.0
210. Two solid spheres of same metal but of mass M and 8M full simutineously on a viscous liquid andtheir terminal velocity are V and 'nv' then value of 'n' is(A) 16 (B) 8 (C) 4 (D) 2
211. Two metal Spheres are falling through a liquid of density 33
mkg102 with the same uniform speed
the meterical density at sphere 1 and sphere 2 are 33
mkg108 and 3
3
mkg1011 respectively. The
ration of their radii is.
(A) 811
(B)811
(C) 23
(D)23
234
212. Water is flowing continuously dram a temp having an internal diameter m108 3 . The water velocity
as it leaves the tap is 0.4 m/s. The diameter of the water stream at a distance m102 1 below thetap is close to
(A) m100.5 3 (B) m105.7 3 (C) m106.9 3 (D) m106.3 3
213. A large open tank has two holes in the wall one is a square hole of side L ata depth y froam the topand the other is a circular hole of radius R at a depth ay from the top. When the tank is completelyfilled with water the quantities of water following out per second from both the holes are the somethen R is equal to
(A) L2 (B) 2L
(C) L (D)2
L
214. A block of ice floats on a liquid of density 1.2 in a beaker then level of liquid when ice completelymelt.(A) Remains same (B) Rises (C) Lowers (D) (A) (B) or (C)
Assertion & Reason type questionsRead the assertion and reason carefully to mark the correct option out of the options given below(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.(b) If both assertion and reason are true but reason is not the correct explanation of the assertion.(c) If assertion is true but reason is false.(d) IF the assertion and reason are false.(e) If assertion is flase but reason is true.
215. Assertion : The blood pressure in humans is greater at the feet than at the brain.Reason : Pressure of liquid at any point is proportional to height clensity of liquid.(A) a (B) b (C) c (D) d
216. Assertion : To empty an oil tank two holes so it will made.Reason : Oil will come out of two holes so it will be emptied faster.(A) a (B) b (C) c (D) d
217. Assertion : A bubble comes from the bottom of a lake to the top.Reason : Its radius increases.(A) a (B) b (C) c (D) d
Ordinary ThinkingThermometary
218. Oxygen boils at 1830C. This temperature is approximately.(A) 2150 F (B) -2970 F (C) 3290 F (D) 3610 F
235
219. The resistance of a rasistance thermometer has values 2.71 and 3.70 ohm at 100C and 1000C. Thetemprature at which the resistance is 3.26 ohm is(A) 400 C (B) 500 C (C) 600 C (D) 700 C
220. Maximum density of H2O is at the temprature.(A) 320 F (B) 39.20 F (C) 420 F (D) 40 F
221. At what temprature the centigrade (celsius) and Fahrenheit readings at the same.(A) -400 (B) +400 C (C) 36.60 (D) -370 C
222. Mercury thermometers can be used to measure tempratures up to(A) 1000 C (B) 2120 C (C) 3600 C (D) 5000 C
223. If temperature of an object is 1400 F then its temperature in centigratde is(A) 1050 C (B) 320 C (C) 1400 C (D) 600 C
224. When the room temprature becomes equal to the dew point the relative humidity of the room is(A) 100 % (B) 0 % (C) 70 % (D) 85 %
225. If the length of a cylinder on heating increases by 2% the area of its base will increase by.(A) 0.5 % (B) 2 % (C) 1 % (D) 4 %
226. Density of substance at 00C is 10 gm/cc and at 1000Cits density is 9.7 gm/CC. The coefficient oflinear expansion of the substance will be(A) 102 (B) 10-2 (C) 10-3 (D) 10-4
227. A beaker is completely filled with water at 40C It will overflow if(A) Heated above 40C(B) Cooled below 40C(C) Both heated and cooled above and below 40C respectively(D) None of these
228. An iron bar of length 10m is heated from 00C to 1000C. If the coefficient of linear thermal expansion
of iron is C1010 6
the increase in the length of bar is
(A) 0.5 cm (B) 1.0 cm (C) 1.5 cm (D) 2.0 cm
Calorimetry229. Melting point of ice.....
(A) Increases with increasing pressure (B) Decreases with increasing pressure(C) Is independent of pressure (D) is proportional of pressure
230. Amount of heat required to raise the temprature of a body through 1k is called it is(A) Water equivalent (B) Thermal capacity(C) Entropy (D) Specific heat
231. A vessel contains 110 g of water the heat capacity of the vessel is equal to 10 g of water The initicaltemprature of water in vessel is 100C If 220 g of hot water at 700C is poured in the vessel the Finaltemperature meglecting radiation loss will be(A) 700 C (B) 800 C (C) 600 C (D) 500 C
236
232. In a water fall the water falls from a height of 100 cm. If the entire K.E. of water is converted in toheat the rise in temperature of water will be(A) 0.230 C (B) 0.460 C (C) 2.30 C (D) 0.0230 C
233. The temprature at which the vapour pressure of a liquid becomes equals of the external pressure isits.(A) Melting point (B) sublimation point(C) Critical temprature (D) Boiling point
234. 10 g of ice at 00C is mixed with 100 g of water at 500 C what is the resultant temprature of mixture.(A) 31.20 C (B) 32.80 C (C) 36.70 C (D) 38.20 C
Critical Thinking235. Two metal strips that constituate a thermostant must necessarily in their
(A) Mass (B) Length(C) Resistivity (D) Coefficient of liner expansion
236. 2 kg of Ice at -200 C is mixed with 5 kg of water at 200 C in an insulating vessel having a megligibleheat capacity calculate the final mass of water remaining in the container. It is given that the specificheats of water and ice care 1 keal/kg per0C and 0.5 Keal/kg 10C while the latent heat of fusion ofice is 80 kcoil/kg.(A) 7 kg (B) 6 kg (C) 4 kg (D) 2 kg
237. A lead bullet at 270 C just melts when stopped by an obstancle Assuming that 25% of heat isobsorbed by the obstacle then the velocity of the bullet at the time of striking.[M.P. of lead = 3270 C, specific heat of lead = 0.03 cal/x latent heat of fusion of lead = 6 cal/g andJ = 4.2 Jute / cal](A) 410 m/s (B) 1230 m/s (C) 307.5 m/s (D) None of these
238. An electric kettle takes 4A current at 220V How much time will it take to boil 1 kg of water fromtemprature 20C ? The temprature of boiling water is 100 C.(A) 12.6 min (B) 4.2 min(C) 6.3 min (D) 8.4 min
Graphical options239. Ablock of ice at -100C is slowly heated and covered to steam at 1000 C which of the following
curves represents the phenomenon qualitatively.
237
240. The variation of density of water with temprature is represented by the
Assertion & ReasonRead the assertion and reason carefully to mark the correct option out of the option given below.(a) If both asseration and reason are true and the reason is the correct explanation of the reason.(b) If both assertion and reason are true but reason is not the correct explanation of the assertion.(c) If assertion is true but reason is false.(d) If the assertion and reason both are false.(e) If assertion is false but reason is true.
241. Assertion : The melting point of the ice decreases with increases pressureReason : Ice contracts on melting(A) a (B) b (C) c (D) d
242. Assertion : Fahrenheit is the smallest unit measuring temprature.Reason : Fahrenheit was the first temprature scale used for measuring temprature.(A) a (B) b (C) c (D) d
243. Assertion : Melting of solid causes no change in internal energy.Reason :Latent heat is the heat required melt a unit mass of solid.(A) a (B) b (C) c (D) d
244. Assertion : Specific heat capacity is the cause of formation of land and sea breeze.Reason : The specific heat of water is more then land.(A) a (B) b (C) c (D) d
245. Assertion : The moleculs of 00C ice and 00C water will have same potential energy.Reason : Potential energy depends only on temprature of the system.(A) a (B) b (C) c (D) d
246. Assertion : A beaker is completely filled with water at 40C. It will overflow both where heated orcooled.Reason : There is expansion of water below and above 40C.(A) a (B) b (C) c (D) d
238
Answers Key1 D 42 C 83 B 124 A 165 C 206 B2 B 43 A 84 B 125 D 166 D 207 C3 D 44 B 85 D 126 A 167 A 208 D4 B 45 D 86 A 127 A 168 B 209 B5 C 46 A 87 D 128 D 169 B 210 C6 B 47 D 88 B 129 A 170 D 211 D7 D 48 A 89 C 130 A 171 A 212 D8 D 49 A 90 A 131 D 172 B 213 B9 C 50 B 91 D 132 B 173 B 214 B
10 C 51 D 92 B 133 B 174 C 215 A11 A 52 B 93 C 134 C 175 C 216 C12 B 53 D 94 C 135 A 176 B 217 A13 D 54 C 95 B 136 B 177 A 218 B14 B 55 B 96 A 137 C 178 B 219 B15 B 56 A 97 C 138 A 179 C 220 B16 B 57 A 98 D 139 A 180 A 221 A17 B 58 D 99 C 140 A 181 B 222 C18 D 59 A 100 A 141 A 182 E 223 D19 D 60 C 101 A 142 D 183 A 224 A20 D 61 C 102 B 143 D 184 C 225 D21 B 62 B 103 D 144 C 185 B 226 D22 B 63 A 104 D 145 A 186 A 227 C23 C 64 C 105 B 146 D 187 B 228 B24 B 65 C 106 C 147 C 188 C 229 B25 C 66 C 107 B 148 D 189 B 230 B26 D 67 A 108 C 149 D 190 B 231 D27 B 68 B 109 A 150 D 191 C 232 A28 D 69 C 110 D 151 D 192 B 233 D29 A 70 B 111 D 152 C 193 D 234 D30 A 71 A 112 A 153 C 194 C 235 D31 B 72 B 113 D 154 B 195 B 236 B32 D 73 A 114 A 155 A 196 D 237 A33 D 74 B 115 D 156 B 197 D 238 C34 D 75 C 116 A 157 D 198 D 239 A35 D 76 B 117 A 158 C 199 A 240 A36 D 77 A 118 B 159 A 200 D 241 A37 A 78 C 119 C 160 C 201 A 242 C38 C 79 D 120 B 161 B 202 B 243 E39 C 80 B 121 B 162 C 203 A 244 A40 A 81 C 122 B 163 B 204 C 245 D41 D 82 B 123 D 164 C 205 A 246 A
239
Hint
1.
AVLLAV
yAFV
AyFL
2
22
AFVyAF
As cross sectional area of 2nd wire is 3 times therefore 9F force is required for same elongation.
2. Stress Lstressstrain
3. 2dL
ALL
AFY
2dL
(As F and Y are constant)
The ratio of 2dL
is maximum for case CD
4. Young modulus of wire does not vary with dimension of wire. It is the property of given material.5. Depression in beam.
2
3 3wL
4Y bd 1
Y
6. 2
1L r
If radius of the wire is doubled then increament in length will become 1/4 times, i.e. mm34
12
7.YAF
L
8.
A
F
Y Given stress 2
8
mN1018.3
F/A Y
l
9. Here 2k
KQ p
According to Hooke's law ppp xkF
Qp FF (Given) 1....kPkQ
xx
Q
p
240
Energy stored in a spring is
2kpkQ
21
pkQk
kQkp
QxPx
kk
UU
2
2
2
2
q
p
Q
p 2E
2UQUP
10. Breaking force area of crossection of 2r wire
If radius of wire is doubled then breaking force will become four times.
11.2
2LY
12. Young's modulus of material
strainalLongitudinstressLinear
If longitudinal strain is equal to unity, thenY = Linear stress produced
13. Energy stored per unit volume
strainstress21
14. Vstress.MaxstrainMax
StrainstressY
YA
mg
strainMax
15. m106.9
10528.95.1108
Y2dgL 11
8
222
16. stress 2r1stress
Areaforce
AB
22
A
B S4S2Br
rASS
17. Breaking force area of cross section of wire.i.e. load hold by the wire does not depend upon the length of the wire.
18. AYFL
2r1
(F, L & Y are constant)
19. 26 m10A IAY
20. 2r1
AYFL
(F, Land Y are same)
241
21.Steel
1
1
Steel
YYa
aAYFL
(F, Land Y are constant)
22. If length of the wire is doubled then strain = 1AreaForcestressY
23. dgPL
24. F = force developed AYA
25. lateral strainlongitudinal strain
26.strainYFA
strainAF
Yx
27. Strain & stress AF
28. Elongation in the wire AYTL
Elongation in wire tension in the wirein first case 1T wand in second case
www
w2T2
As Tension in the wire in both the cases are equal. Elongation in the wire will be equal.
29.FAYAl
30. 22 r1
yrF
(F, Land Y are constant)
31. Longitudinal strain stress= =L Yl
32. It is the specific property of a particular material at a given temperature which can be changed onlyby temperature variations.
33. When the length of wire is doubled then L and strain = 1
AFstrainY
34.LrF
LAYF
2
(y and L are constant)
242
35. 100 103
7rrY
36. Increase in tension of wire QYA
37.AYFstrain
strainAF
Y
38.L
ALY21W
2
39. Potential energy stored in the rubber cord calapult will be covered into kinetic energy of mass
LYAL
21mv
21 2
2
40. AYFL
A1
(F, L and Y are constant)
mm05.021.0
221
84
AA 1
22
1
1
2
41. ttanconsstrainstressY It depends only on nature of material.
42. AYFL
Final length = initial lengh + increament = 2L
43. AYFL
44. YroK
45. VYFL
YALFL
AYFL 22
2L If the volume of wire remains constant.
46. 22 rF
YrFL
AYFL
(Y = constant)
47. Breaking stress = Strain young's modulus48. In accordance with Hook's law
49. StrainYAF
243
50. Because strain is a dimensionless and unitless quantity.
51. areaforcestress
In the present case,force applied and area of cross-section of wires are same. Therefore stress has to be the same.
Strain = stress /YSince, the young's modulus of steel wire is greater than the copper wire, therefore strain in case ofsteel wire is less than that of in case of copper wire
52. Knowledge base
53. Breaking force 2r
If diameter becomes double then breaking force will become four times
i.e. N400041000
54. AYFL
YALFL
AYFL 22
If volume is fixed then 2L
55. Young's modulus strainstress
As the length of wire get doubled therefore strain = 1
28
mN1020strainY
56. Because dimension of invar does not vary with temperature.
57. yrFL
2
2rL
(Y & F are constant)
58. Q1LandQ1 111222
11221212 Qll
now 0,SoLL 11221212
59. Thermal stress = Qy
60. LYAF
61. 22
1Y
FL rr y
(F, Land l are constant)
244
62. yrFL
2
A1
(F, Land Y are constant)
2 1
1 2
AA
63. Y MgLA
64. AYFL
2rF
65.mgLLYA
66. Work done 2MgF
21
67. Volume
Ystress
21w
2
As F, A and Y are same - W volume (area is same)
w (V = A1)
21
2ww 1
2
1
2
1
68. F21w
69. 2k21k
21F
21wandFK
70. 2
2
rLU;
AY2FF
21U
(F and Y are constant)
71. F21U
72. Volume
Ystress
21U
2
73. When a wire is stretched through a length then work has to be done. This work is stored in the wirein the form of elastic potential energy.Potential energy of stretched wire is
strainstress21U Fx
21U6F
21U
245
74. 2strainYX21U
75. L2YAw
2
76.K2
TK2
FU22
77. Energy stored per unit volume
LAF
21
AL2FL
78.2
LYA
21U
79. At extension l1 the stored energy 21k
21 At extension l2 the stored energy 2
1k21
Work done in increasing its extension from l1 to 21
222 k
21
80. XFK
81. Isothermal elasticity ki = P
82. 1001%1
VVgiven
VBVP
YPVB
28
10
mN105.1
100105.1P
83. Ratio of adiabatic & isothermal elasticities
V
PP
CCV
PV
84. For triatomic gas 34V
85.
vv
Pk
86. PvvC
CC4.0
246
87.
vv
hsg
vvPk
88. If side of cube is L then LdL3
vdvLV 3
89.2
122
116
cmdynes10
mN10atm10
10001.0
100k
90. If the coefficient of volume expansion is and raise in temprature is then
vv
vv
volume elasticity
P
vv
P
91. If side of the cube is L then V = L2
LdL3
vdv
% change in voulume 3 (% change in length) %3%13
Bulk modulus 03.0vv
92. Adiabatic elasticity vp
For Argon EAr = 1.6 pFor hydrogon EH2 = 1.4 plAs elasticity of hydrogen & Argon are equal
Pl4.1P6.1
P78'P
93. Isothermal elasticity 25
mN10013.1013.1atm1KiP
94. Isothermal bulk modulus = pressure of gas
95. dv dL1 2v L
215.061041022
vdv 33
96. We know that L
dL21v
dv If 0
vdvthen
21
i.e. there is no change in volume
247
97. Y = 2N ( 1 + )
98. n
ny5.01n2Y
99. Twisting coulple
2rnC
4
If material and length of the wires A and B equal twisting coulple are applied then
4r1
4
1
2
2
1
rr
100. Value of possion's ratio die in range of 1 to 1/2.
101. 1n2y;21k3y
For Y=0 we get 021 also 01
lies between
102. 1n2n31n2y211
23
Now, substituting the value of in the following expression
213yk
103. 1n2Y104. There will be both shear stress and normal stress.
105. Angle of shear '12.0'30100
104Lr 1
106. For twisting, angle of shear L1
i.e. if L is more then will be small
107. 004.028.010Lr 2
108. 1n2Yand21K3Y Eliminating we got k3nnk9Y
109. Poisson's ratio varies between - 1 and 0.5
110. MTkn2T 2
If we draw a graph of between T2 & M then it will be straight line and for 0T;0M 2
i.e. graph should pass through the origin but from the graph it is not reflected it means the mass ofpan was neglected.
248
111. In the region OA, stress strain . i.e.Hooke's las hold good.112. As stress is shown on X axis and strain on Y axis so we can say that
slope1
tan1cotY
So elasticity of wire P is minimum and of wire R is maximum.113. Area of hysterisis loop gives the energy loss in the process of stretching and understretching of
rubber band and this loss will appear in the form of heating.
114.tantan
YY
B
A
B
A
115. i.e.for the same load thickest wire will show minimum elongation so graph D present the thickestwire.
116. From the graph N20F,M10 4 M1,M10A 26
46 1010120
AlFLY
21110
mN1021020
117. At point b yeilding of material starts.118. Graph between applied force and extension will be straight line because in elastic range.
Applied force extension. but the graph between extension and stored elastic energy will be parabolicin nature.
As 22 xUorkx21U
119.
dxdvF
In the region BC slope of the graph is positive. F negative i.e. force is attractive in nature.
In the region AB slope of the graph is negative. F positive i.e. force is repulsive in nature.
120. Force constant K = tan 300 3
1
121. In ductile materials, yeild point exist which in brittle material failure would occur without yeilding.
122. Young's modulus is defined strainstressY
123. Elasticity of wire decreases at high temperatare i.e. at higher temprature slope of graph will be less,So T1 > T2.
124. Attraction will be minimum when the distance b/w the molecule is maximum. Altraction will bemaximum at that point where the positive slope is maximum b'sc
duFdx
249
125. B'se stretching of coil simply chnages its shape without any change in the length of the wire used incoil Due to which shear of elasticity is prevolved.
126. A bridge during its use undergoes alternating strains for a large number of times each day dependingupon the movement of vehicle on it. When a bridge is used for long time. It losses its strength Due towhich the amount of strain in the bridge for a given stress will become large and ultimately the bridgemay collapse. This may not happen if the fridges are declared unsafe.
127. Ivory is more elastic than wet-clay. Hence, the ball of ivory will rise to a greater height. Infact the ballof wet day will not rise at all it will be same, what flattended permanently.
128. Young's modulus of a material.strainstressY
Here, stress force forceAreastaringRe
As restoring force is zero. 0Y
129. Work done 2strainY21strainstress
21
Since, elasticity of steel is more than copper,,
hence, more work has to be done in order to stretch the steel.
130.
yxAFt
133.
A
AW
Y
139. LL
AFY
250
Hint143. Soap helps to lower the surface tension of solution thus soap get stick to the dist partcles and grease
and these are removed by action of water.
145. Force required to separate the plates t
TA2F
147.
r2cosTcoaterdisplacedofweight
cosrT2w
148. mgTL2 L2
mgT
149. 21
22 RRT8w 22 rr2T8 Tr24 2
150. 21
22 rr8Tw
4d
4D8T
22
TdD2 22
151. Energy spent = T increase in surface ared
22 r4r242T JoulerT24 2
152. ATw
153. Surface energy = Surface tension surface ared
A2TE
New surface energy
2A2TF1
% decrase in surface energy 100E
EE i
154. The ration of the total surface energies before and after the change 1:n 21
1:231
155. TR8w 2
156. ATw A
wT
157. 3R34
251
158. As volume remains constant 33 r8000R r20R
1r48000TR4
dropsmall8000ofenergySurfacedropbigoneofenergySurface
2
2
159. Tension Areaenergysurface
AETor
160. Angle of contact is a 00
165. RT2p
167. BABA PPSor1pandrr
So air will flow from B to A i.e. size of A will increase
168. hdgRT4
4RhdgT
mN109.3 2
169. Mass of liquid in capillary tube
2RM
R1RM 2
RM If radius become double then mass will become twice.
172.rdg
T2h r1h 2211 hrhr
173. cm0.460coscos
h
174. When a capillary tube is broken at a height of 6 cm the height of water column will be 6 cm.
As ttanconscos
horrg
cos25h
43
80cos6cosor
cos6
0cos8 o
o
43cos 1
175. Rise in tempreture
R1
r1
JsdT3
R1
r1
JT3
(For water S = 1 and d = 1)
176. rdgcosT2h
r1h so the graph between h and r will be rectangular hyperbold
rT4p
r1p As radius of soap bubble increases with time t
1P
252
178. tIToTC
i.e. Surface tension decreases with increase in temperature179. Due to surface tension vertical force on drop
SmrTFv 2
Rr2T
RrrT
2
2
180. 9.R34T
Rr2 3
2
181. 23104.1411.0ATU
183. dgT2hR
RdgT2h constanthR
Hence when the tube is of insufficient length radius of curvature of the liquid meniscus increasses soas to maintain the product hR a finite constant.i.e. as h decreases R increases and the liquid meniscus becomes more and more flat but the liquiddoes not overflow.
184. The presence of impurities either on the liquid surface or dissolved in it considerably affect the forceof surface tension depending upon the degree of surface tension depending upon the degree ofcontamination. A highly soluable substance like sodium chioride when dissolved in water increasethe surface tension. But the sparing soluable or substance like phenol when dissolved in waterreduces the surface tension of water.
185. We know that the intermolecular distance between the gas moleculas is Large as compaired to thatof liquid Due to it the forces of cohesion in the gas moleculas are very small and these are quiteLarge for liquids. Therefor the concept of surface tension is applicable to Liquid but not to gasses.
186. The height of capillary rise is inversly propostional to radius (or diametre) of capillary tube
i.e. r1h so, for smaller r the value of his higher
187. When a drop of Liquid is poured on a glass, plate, the shape of the drop also is governed the forceof gravity for every small drops the potential energy due to gravity is insignification. Compared tothat due to surface tention. Hence, in this case the shape of the drop is determined by sufrace tensionalone and drope becomes spherical.
188. 2211 vpvp
303 r2
34pr
34ghp
253
189. Thrust on lamina = pressure at centroid area
1 A3 3
h g A gh
190. Bulk modulas
Bpvv
vpVB 00
Bp1vv 0
1
0 Bp1Density
Bp10
191. = 760 - 750 = mnat Hg = 4g192. Here, h = 20 m
Density of water 33
mkg10
Atmospheric pressure
pa1001.1Pa 5
ghpap
193. Weight of the ball = Buyoant force + viscous force194. For the floatation Vo dog = Vin dg
ddoVVin 0
ddoVVVinVVout 000
195. Weight of bodyWeight of water displaced Weight of oil displaced
34oilofgrvitySpecitic
w
0
196. Aschemedies principal explains buoyant force and bouant force depends on acceleration due togravity
204. If two drops of same radius r coalesce then radius of new drop is given by R
33333 r2Rr34r
34R
34
r2R 31
Is drop of radius r is falling in viscous medium then it acquire a critical velocity V and 2rV
2
31
2
1
2
rr2
rR
vv
132
2 v2v 52 32
sm45 3
1
254
206. According to Bernouli's theoram
cm30.0102
45.2h9vh
22
cm0.30
207. Time taken by outer to reach the bottom =
2 H Dt
g
and velocity of water coming out of hole
gD2v
Horizontal distance coveredtvx
209. Terminal speed ngr
92v
2
Where, Density of the substance Density of the liquid
If n and r are constant then
v210. Mass = volume Density
3r34M
As the density remains constant3rM
211. The terminal velocity of the spherical body of radius R density S falling through a liquid of density is given by
229t
RV g
n
where n is the coefficient of vescosity of the liquid.
1
21 1
T
29
RV g
n
2
22 2
t
2and
9R
V gn
According to the given problem
255
212. Diameter m108 3
sm4.0V1
v2 gh2v 21
A1v1 = A2v2
215. The volume of liquid displaced by floating ice
MVD
Volume of water formed by melting ice, wMVF
FD VVe.iw
ML
M
- Height of the blood column in the human body is more at feet than at the brain as P = hpg there face
the blood exeists more pressure at the feet than at the brain.- When to holes are made in the Hn air keeps on externing through the other hole Due to this the
pressure inside the tin does not become less than at mosphere pleassure which happen only whenhole is made.
218. F297F932F
5183
932F
5C o
219. Change in resistonce 99.071.270.3 to interval of temprature 900 C so change in resistance 55.071.226.3 corresponds to change in temprature.
C5055.099.0
90 o
220. Maxmimum density of water is at 40 C also
932F
5C
221. 932F
5C
222. The boiling point of mercury is 4000 C. Therefore the mercury thermeter can be used to measure therange upto 3600 C.
223. 932F
5C
225. LL.2
AALA 2
256
226. Coffcient of volume expansion
21
T.r
Hence, cofficent of linear expansion227. Water has maximum density at 40C so if the water is heated above 40C or Density cooled below
40C density decreases. In other words it expands so it overflows in both cases.
228. Increase in length LoL
Calorimetry229. Melting point of ice decreases with increase in pressuire.
230. capacityThermalmcthenk1if;c.m
231. Let final temprature of water bc heat taken = Heat given
7012201010101110
C50C8.48 oo
232. 1000023.0h0023.0 C23.0 o
233. At boiling point vapour pressure becomes equal to the external pressure.
234.mwmi
cwmiLiwmw
mix
Critical Thinking235. Thermostat is used in electric opporatas like refrigerator iron etc for automatic cut off Therefore for
metallic strips to bend on heating their coefficient of linear expansion should be different.236. Initially ice will absorb heat to rise its temprature to 00C then its melting takes place. If mi = Initial
mass of ice mi-1 = Mass of ice that melts and mw = Initial mass of water By low of mixture Heatgained by ice = Heat lost by water 20mwCLmi20cmi w
kg1mi201580mi205.02 7
So final mass water Initial mass of water + mass of ice that melts = 5 + 1 = 6 kg237. If mass of the bullet is m gm then total heat required for bullet to just melt down.
mLmc1
Now when bullet is stopped by the obstacle the loss in its mechanical energy 9v)10cm21 23
(As kg10mmg 3 )As 25% of this energy is absorbed by the
257
2 10mv8310mv
21
10075 3232 J
Now the bullet melt if 12
238. mctP
VIm4200
pm4200
pmct
Ckg14200water o
min3.6sec3814220
2010014200t
Graphical Optaions239. Initially on heating temprature rises from -100 C to 00 C Then ice melt and temprature does not rise
After the whole ice has melted temprature begins to rise until reaches 1000 C Then it becomesconstant as at the boiling point will not rise.
240. Density of water is maximum at 40 C and is less on either side of this temprature.
Assertion & Reason241. With rise in pressure melting point of ice decreases also ice contracts on melting.242. Celsius scale was the first temprature scale and fahrenheit is the smallest unit measuring temprature.243. Melting is associated with increasing of internal energy without change in temprature in view of the
reason being correct the amount of heat absorbed or given out during change of state is expressedwhere m is the mass of the substances and L is the latent heat of the substance.
244. If bath assertion and reason are true and the reason is the correct explanation of the assertion.245. The potential energy of water molecules is more. The heat given to melt the ice at 00C used up in
increasing the potentical energy of water molecules formed at 00 C246. Water has maximum density at 40 C on heating above 40 C or cooling below 40C density of water
decreases and its volume increases. Therefore water overflows in both the cases.
259
SUMMARY Thermal equilibrium, teroth law of thermodynamics, Concept of temperature. Heat, work and internal energy. First law of thermodynamics. Second law of thermodynamics, reversible and irrevessible processes.
Isothermal and adiabatic process. Carnot engine and its efficency Refrigerators.
0 0 0 0C 0 F 32 C F100 180 100 180
w P V (Isothesmal process)
RT n 1
2
vv
(Isothesmal process)
PV = Const (Isothesmal process)P v = Const (adiabasic process)P1- T = ConstTV-1 = Constant
r = cvcp
Monoatomic gases v p3 5C = R, = C = R,2 2
= 1.67
Diatomic gases v p5 7 γC = R, C = R, = 1.4 2 2
Polgatomic gases v p7R 9RC = , C = , 2 2
= 1.4
1 1 2 21W = (P V P V ) γ 1
(adiabafic Process)
1 2R (T T )= γ 1m
Q u W (First law of thermodynamics)
Q U Cv t (V = Const)
Q Cp t ( P = Const)
u 0 (isothesmal process cyelic process)
260
Coefricent of Performance of refrigesatos
1QWn
1 2
1
Q QQ
1 2
1
T TT
Coefcient of performance of refrigetors.
2Qw
2
1 2
QQ Q
2
1 2
TT T
I deal gas Cp Cv R
1. A difference of temperature of 250 C is equivalent to a difference of (A) 720 F (B) 450 F (C) 320 F (D) 250 F
2. What is the value of absolute temperature on the Celsius Scale ? (A) -273.150 C (B) 1000 C (C) -320 C (D) 00 C
3. The temperature of a substance increases by 270C What is the value of this increase of Kelvinscale? (A) 300K (B) 2-46K (C) 7 K (D) 27 K
4. At Which temperature the density of water is maximum? (A) 40 F (B) 420 F (C) 320 F (D) 39.20 F
5. The graph AB Shown in figure is a plot of a temperature of a body in degree Fahrenhit than slopeof line AB is
(A) 95
(B) 59
(C) 91
(D) 93
261
6. The temperature on celsius scale is 250 C What is the corresponding temperature on the FahrenheitScale? (A) 400 F (B) 450 F (C) 500 F (D) 770 F
7. The temperature of a body on Kelvin Scale is found to be x.K.when it is measured by Fahrenhitthesmometes. it is found to be x0F, then the value of x is . (A) 313 (B) 301.24 (C) 574-25 (D) 40
8. A Centigrade and a Fahrenhit thesmometes are dipped in boiling wates-The wates temperature islowered until the Farenhit thesmometes registered 1400 what is the fall in thrmometers (A) 800 (B) 600 (C) 400 (D) 300
9. A uniform metal rod is used as a bas pendulum. If the room temperature rises by 100C and theefficient of line as expansion of the metal of the rod is, 6 1
C2 10 0 what will have percentage increasein the period of the pendulum ? (A) -2 10-3 (B) 110-3 (C) -110-3 (D) 210-3
10. A gas expands from 1 litre to 3 litre at atmospheric pressure. The work done by the gas is about (A) 200 J (B) 2 J (C) 300 J (D) 2105 J
11. Each molecule of a gas has f degrees of freedom. The radio P
V
C γ = C
for the gas is
(A) 1 2f
(B) 11f
(C) 21f
(D) 1
13
f
12. Is the cyclic Process Shown on the V P diagram, the magnitude of the work done is
(A)2
12
2PP
(B) 2
12
2VV
(C) 2 2 1 1P V PV (D) 1212 VVPP4
13. If the ratio of specific heat of a gas at Consgant pressure to that at constant volume is , the Changein internal energy of the mass of gas, when the volume changes from V to 2V at Constant Pressurep, is
(A) γ 1PV (B) γ 1
R
(C) PV (D) γ PVγ 1
262
14. The change in internal energy, when a gas is cooled from 9270C ³ to 270C (A) 200% (B) 100% (C) 300% (D) 400%
15. For hydrogen gas Cp - Cv = a and for oxygen gas Cp - Cv = b, The relation between a and b isgiven by (A) a = 4b (B) a = b (C) a = 16b (D) a = 8b
16. In a thermodynamic process, pressure of a fixed mass of a gas is changed in such a manner that thegas release 20J of heat and 8J of work has done on the gas- If the inifial internal energy of the gaswas 30j, then the final internal energy will be (A) 58 J (B) 2 J (C) 42 J (D) 18 J
17. If for a gas cvcp
= 1.67, this gas is made up to molecules which are
(A) diatomic (B) Polytomic(C) monoatomic (D) mixnese of diatomic and polytomic molecules
18. An ideal monoatomic gas is taken around the cycle ABCDA as Shown in the P V diagram. Thework done during the cycle is given by
(A) PV (B) 21
PV (C) 2 PV (D) 4 PV
19. A given mass of a gas expands from state A to B by three different paths 1, 2 and 3 as shown in thefigure. If W1,W2 and W3 respectively be the work done by the gas along the three paths, then
(A) W1>W2>W3 (B) W1<W2<W3 (C) W1=W2=W3 (D) W1<W2, W1<W3
20. One mole of a monoatomic gas 5γ3
is mixed with one mole of A diatomic gas 7γ5
what
will be value of for mixture ?(A) 1-454 (B) 1-4 (C) 1-54 (D) 1-5
21. If du represents the increase in internal energy of a thesmodynamic system and dw the work doneby the system, which of the following statement is true ?(A) du = dw in isothermal process (C) du = - dw in an aidabadic process(B) du = dw in aidabadic process (D) da = - dw in an isothermal process
263
22. One mole of a monoatomic ideal gas is mixed with one mole of a diatomic ideal gas The molasspecific heat of the micture at constant volume is .......... (A) 4 R (B) 3 R (C) R (D) 2R
23. One mole of a monoatomic gas is heate at a constant pressure of 1 atmosphere from 0k to 100 k.If the gas constant R = 8.32 J/mol k the change in internal energy of the gas is approximate ? (A) 23 J (B) 1.25 103 J (C) 8.67 103 J (D) 46 J
24. A gas mixture consists of 2 mde of oxygen and 4 mole of argon at tempressure T.Neglecting allvibrational modes, the total internal energy of the system is (A) 11 RT (B) 9 RT (C) 15 RT (D) 4 RT
25. A monoatomic ideal gas, initially at temperature T1 is enclosed in a cylindes fitted with a frictionlesspiston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly
If L1 and L2 the lengths of the gas colum be fore and afters expansion respectively, then then 2
1
TT
is
given by
(A) 2
3
1
2
LL
(B) 2
3
2
1
LL
(C) 1
2
LL (D) 2
1
LL
26. Starting with the same intial Conditions, an ideal gas expands from Volume V1 to V2 in three differentways. The Work done by the gas is W1 if the process is purely isothermal, W2 if purely isobasic andW3 if purely adiabatic Then
(A) W2>W1>W3 (B) W2>W3>W1
(C) W1>W2>W3 (D)W1>W3>W2
27. In a given process on an ideal gas dw = 0 and dQ < 0. Then for the gas(A) the volume will increase (B) the pressure will semain constant(C) the temperature will decrease (D) the temperature will increase
28. Wafer of volume 2 filter in a containes is heated with a coil of 1kw at 270C. The lid of the containes
is open and energy dissipates at the late of J160S . In how much time tempreture will rise from
270 C to 770C. Specific heat of wafers is 4.2 KgKJ
(A) 7 min (B) 6 min 2s (C) 14 min (D) 8 min 20 S
264
29. 70 calorie of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressurefrom 300C to 350CThe amount of heat required to raise the temperature of the same gas through the same range atconstant volume is .................. calorie. (A) 50 (B) 30 (C) 70 (D) 90
30. When an ideal diatomic gas is heated at constant pressure, the Section of the heat energy suppliedwhich increases the infernal energy of the gas is..
(A) 73
(B) 53
(C) 52
(D) 75
31. Two cylinders A and B fitted with piston contain equal amounts of an ideal diatomic gas at 300 k.The piston of A is free to move, While that of B is held fixed. The same amount of heat is given to thegas in each cylindes. If the rise in temperature of the gas in A is 30K, then the rise in temperature ofthe gas in B is.(A) 30 K (B) 42 K (C) 18 K (D) 50 K
32. An insulated containes containing monoatomic gas of molas mass Mo is moving with a velocity, V.Ifthe container is suddenly stopped, find the change in temperature.
(A) 2Mov
5R(B)
2Mov4R
(C) 2Mov
3R(D)
2Mov2R
33. A Small spherical body of radius r is falling under gravity in a viscous medium. Due to friction themedium gets heated. How does the late of heating depend on radius of body when it attains terminalvelocity!(A) r2 (B) r3 (C) r4 (D) r5
34. The first law of thermodynamics is concerned with the conservation of(A) momentum (B) energy (C) mass (D) temperature
35. If heat given to a system is 6 k cal and work done is 6kj. The change in internal energy is .......... KJ.(A) 12.4 (B) 25 (C) 19.1 (D) 0
36. The internal energy change in a system that has absorbed 2 Kcal of heat and done 500J of work is(A) 7900 J (B) 4400 J (C) 6400 J (D ) 8900 J
37. Which of the following is not a thermodynamical function.(A) Enthalpy (B) Work done(C) Gibb's energy (D) Internal energy
38. Which of the following is not a thermodynamic co-ordinate.(A) R (B) P (C) T (D) V
39. The work of 62-25 KJ is performed in order to compress one kilo mole of gas adiabatically and in
this process the temperature of the gas increases by 50C The gas is _______ R = 8-3 J
molk(A) triatomic (C) monoatomic(B) diatomic (D) a mixture of monoatomic and diatomic
265
40. Cp and Cv denote the specific heat of oxygen per unit mass at constant Pressure and volumerespectively, then
(A) cp - cv = 16R
(B) Cp - Cv = R
(C) Cp - Cv = 32 R (D) Cp - Cv = 32R
41. When a System is taken from State i to State f along the path iaf, it is found that Q = 70 cal and w= 30 cal, along the path ibf. Q=52cal. W atoug the path ibf is
(A) 6 cal (B) 12 cal (C) 24 cal (D) 8 cal
42. One kg of adiatomic gas is at a pressure of 52
N5 10m
The density of the gas is 3
5kgm
what is the
energy of the gas due to its thermal motion ?(A) 2.5105 J (B) 3.5 105 J (C) 4.5 105 J (D) 1.5105 J
43. 200g of water is heated from 250 C0 450C Ignoring the slight expansion of the water the change in its
internal energy is (Specific heat of wafer C9cal1 0 )
(A) 33.4 KJ (B) 11.33 KJ (C) 5.57 KJ (D) 16.7 KJ44. During an adiabatic process, the pressure of a gas ifound to be propostional to the fifth power of its
absolute temperature. The radio cvcp
for the gas is
(A) 54
(B) 43
(C) 45
(D) 4
45. One mole of oxygen is heated at constant pressure stasting at 00 C. How much heat energy in cal
must be added to the gas to double its volume ? Take R = 2 cal
molk(A) 1938 (B) 1920 (C) 1911 (D) 1957
46. moles of a gas filled in a containes at temperature T is in equilibrium inidially - If the gas iscompressed slowly and is thesmally to half its initial volume the work done by the atmosphere on thepiston is
(A)µRT-
2 (B) 2RT
(C) RT ln(2 - 21
) (D) –RTln2
266
47. Heat capacity of a body depends on the .......... as well as on ..........(A) material of the body, its mass (C) mass of the body, itd temperature(B) material of the body, its temperature (D) Volume of the body, its mass
48. In thesmodynamics, the work done by the system is considered ......... and the work done on thesystem is Considered ...........(A) Positive, zero (B) nagative, Positive(C) zero, negative (D) Positive, negative
49. A thesmodynamic system goes from States(i) P,V to 2P, V (ii) P,V to P, 2V. Then what is work done in the two Cases.(A) Zero, PV (B) Zero, Zero (C) PV, Zero (D) PV,PV
50. For free expansion of the gas which of the following is true ?(A) Q = 0, W > 0 and Eint = -W (B) W = 0, Q >0 and Eint = Q(C) W > 0, Q < 0 and Eint = 0 (D) Q = W = 0 and Eint = 0
51. For an adiabatic process involving an ideal gas(A) P – 1 = T – 1 = constant (B) P1 – = T = constant(C) PT – 1 = constant (D) P – 1 T = constant
52. Figure shows four P V diagrams. which of these curves represent.isothermal and adiabatic processes ?
(A) A and C (B) A and B (C) C and D (D) B and D53. An ideal gas is taken through cyclic process as shown in the figure. The net work done by the gas is
(A) PV (B) 2 PV (C) 3 PV (D) zero54. moles of gas expands from volume VV1 to V2 at constant temperature T. The work done by the gas
is
(A) RT
1
2
VV
(B) RT ln
1
2
VVn (C) RT
1
VV
1
2(D) RT ln
1
VVn
1
2
267
55. A Cyclic Process ABCD is Shown in the P V diagam. which of the following curves representthe same Process ?
(A) (B) (C) (D)
56. A Cyclic process is Shown in the P T diagram.Which of the curve show the same process on a VT diagram ?
(A) (B)
(C) (D)
57. One mole of an ideal gas p
v
Cγ
C at absolute temperature T1 is adiabatically compressed from an
initial pressure P1 to a final pressure P2 The resulting temperature T2 of the gas is given by.
(A)
γγ 12
2 11
pT = T p
(B)
γ 1γ2
2 11
pT = T p
(C) γ
22 1
1
pT = T p
(D) γ 1
22 1
1
pT = T p
268
58. An ideal gas is taken through the cycle ABCA as shown in the figure. If the net heatsupplied to the gas in the cycle is 5J, the work done by the gas in the process CA is
(A) - 5 J (B) -10 J (C) -15 J (D) - 20 J59. In anisothermal reversible expansion, if the volume of 96J of oxygen at 270C is increased from 70
liter to 140 liter, then the work done by the gas will be
(A) 300 R loge(2) (B) 81 R loge
(2) (C) 210200 R og (D) 100 R log10
(2)
60. For an iso thermal expansion of a Perfect gas, the value of P
P
is equal to
(A)
12 ΔV- γ
V (B) ΔVγV
(C) 2 ΔVγ-V (D)
ΔVγV
61. For an aidabatic expansion of a perfect gas, the value of PP
is equal to
(A) Δr- γ v (B) v
v (C) 2 Δv-γ
v (D) Δvγ- v
62. If r denotes the ratio of adiabatic of two specific heats of a gas. Then what is the ratio of slope of anadiabatic and isothermal P V curves at their point of intersection ?
(A) 1γ
(B) -1 (C) (D) + 1
63. Work done permol in an isothermal ? change is
(A) 2
1log vRT e
v (B) 1
2
vv10logRT (C)
2
1
vv10logRT (D)
2
1
vvelogRT
64. The isothermal Bulk modulus of an ideal gas at pressure P is
(A) vP (B) P (C) 2p
(D) pv
65. The isothermal bulk modulus of a perfect gas at a normal Pressure is
(A) 1.013106 2mN
(B) 211
mN10013.1 (C) 2
5
mN10013.1 (D) 2
11
mN10013.1
269
66. An adiabatic Bulk modulus of an ideal gas at Pressure P is
(A) P (B) pγ (C) P (D) 2
P
67. What is an adiabatic Bulk mudulus of hydrogen gas at NTP? r = 1.4
(A) 1.4 2MN
(B) 1.4 105 2M
N(C) 110-8 2M
N(D) 1105
2MN
68. If a quantity of heat 1163.4 J is supplied to one mole of nitrogen gas, at room temprature at constant
pressure, then the rise intemperature is R = 8.31 J
m.l.k(A) 28 K (B) 65 K (C) 54 K (D) 40 K
69. One mole of O2 gas having a Volume equal to 22.4 liter at O c and 1 atmosiheric Pressure isCompressed isothermally so that its volume reduces to 11.2 lites. The work done in this Process is(A) 1672.4 J (B) -1728J (C) 1728J (D) -1572.4J
70. The Specific heat of a gas in an isothermal Process is
(A) zero (B) Negative (C) Infinite (D) Remairs
71. A Containes that suits the occurrence of an isothermal process should be made of
(A) Wood (B) Coppes (C) glass (D) Cloth
72. A thermodynamic Process in which temprature T of the system remains constant through out VariableP and V may Change is called(A) Isothermal Process (B) Isochoric Process(C) Isobasic Process (D) None of this
73. When 1g of wates O c and 5
2
N10m Pressure is Converted into ice of Volume 1.091 cm3 the external
work done be .......... J.(A) 0.0182 (B) -0.0091 (C) -0.0182 (D) 0.0091
74. The letent heat of Vaporisation of water is 2240 gJ
If the work done in the Process of expansion
of 1g is 168J. then increase in internal energy is .......... J(A) 2072 (B) 2408 (C) 2240 (D) 1904
75. The Volume of an ideal gas is 1 liter column and its Pressure is equal to 72 cm of Hg. The Volumeof gas is made 900 cm3 by compressing it isothermally. The stress of the gas will be ............. Hgcolumn.(A) 4 cm (B) 6 cm (C) 7 cm (D) 8 cm
270
76. In adiabatic expansion(A) u=0 (B) u = Positive (C) Ju = Nagative (D) w = 0
77. 1 mm3 Of a gas is compressed at 1 atmospheric pressure and temperature 27 C to 627 C Whatis the final pressure under adiabatic condition. r = 1.5
(A) 25
mN1080 (B) 2
5
mN1036 (C) 2
5
mN1056 (D) 2
5
mN1027
78. A monoatomic gas for it = 35
is suddenly Compressed to 18 of its original volume adiabatically
then the final Pressure of gas is ............. times its intial Pressure.
(A) 8 (B) 32 (C) 524
(D) 340
79. The Pressure and density of a diatomic gas = 75 Change adiabatically from (P,d) to (P1,d1) If
d'd =32 then
'pp
Should be
(A) 128 (B) 1281
(C) 32 (D) None of this
80. An ideal gas at 27 C is Compressed adiabatically, to 278
of its original Volume. If v = 53 , then the
rise in temperaure is(A) 225 k (B) 450 K (C) 375 K (D) 405 K
81. A diatomic gas intially at 18 C is Compressed adiabatically to one eight of its original volume. Thetemperature after Compression will be
(A) 10 C (B) 668 K (C) 887 C (D) 144 C
82. Work done by 0.1 mole of a gas at 27 C to double its volume at constant Pressure is
Cal. R = 2 KoCal
mol(A) 600 (B) 546 (C) 60 (D) 54
83. A gas expands 30.25m at Constant Pressure 103 2mN
the work done is
(A) 250 J (B) 2.5 erg (C) 250 W (D) 250 N
84. In an isochoric Process 1T 27 C and 2T 127 C then2
1
PP
will be equal to
(A) 599
(B) 32
(C) 34
(D) 43
85. If the temperature of 1 mole of ideal gas is changed from O C to 100 C at constant pressure, then
work done in the process is ............ J. R = 8-3 molkJ
(A) 8-310-3 (B) 8-3102 (C) 8-310-2 (D) 8-3103
271
86. A mono atomic gas is supplied the heat Q very slowly keeping the pressure constant The work doneby the gas.
(A) Q52
(B) Q32
(C) Q53
(D) Q51
87. The Volume of air increases by 5% in an adiabatic expansion. The percentange lecrease in itsPressure will br.(A) 5% (B) 6% (C) 7% (D) 8%
88. In PV diagram given below, the isochoric, isothermal and isobaric path respectively are
(A) BA,AD,DC (B) DC,CB,BA (C) AB,BC,CD (D) CA,DA,AB89. In the following indicatos diagram, the net amount of work done will be
(A) Negative (B) zero (C) Positive (D) Infinity90. Work done in the given PV diagram in the cyclic process is
(A) 4 PV (B) 3 PV (C) 2 PV (D) 2PV
91. In the Cyclic Process shown is the figure, the work done by the gas in one cycle
(A) 11VP40 (B) 11VP20 (C) 11VP10 (D) 11VP5
272
92. An ideal gas is taken V path ACBA as Shown in figure, The net work done in the whole cycle is
(A) 11VP3 (B) Zero (C) 11VP5 (D) 11VP393. A thermodynamic system is taken from state A to B along ACB and is brought back to A along BDA
as Shown in the PV diagram. The net work done during the complete cycle is given by the area.
(A) 121 PPACBP (B) ACBDA (C) 1 1ACBB A A (D) AAADBB 11
94. Two identical samples of a gas are allowed to expand (i) isothermally (ii) adiabatically work done is
(A) more inanisothermal process (B) more is an adiabatic process(C) equal in both process. (D) neithes of them
95. What is the relationship Pressure and temperature for an ideal gas under going adiabatic Change.(A) PT
= Const (B) PT-1 + = Const (C) P1 - T = Const (D) P - 1T = Const
96. For adiabatic Process which relation is true mentioned below ? p
v
Cγ =C
(A) pV = Const (B) TV = Const (C) TV =Const (D) TV–1 = Const97. For adiabatic Process which one is wrong statement?
(A) dQ = 0 (B) entropy is not constant(C) du = - dw (D) Q = constant
98. Air is filled in a motor tube at 27 C and at a Pressure of a atmosphere. The tube suddenly bursts.Then what is the temperature of air. given r of air = 1.5
(A) 150 K (B) 150 C (C) 75 K (D) 27.5 C
99. If v is the radio of Specigic heats and R is the universal gas constant, then the molar Specific heat atconstant Volume Cv is given by ............
(A) R
1vv (B) vR (C)
R1v (D)
1 Rr
v
273
100. A Car not engine operating between temperature T1 and T2 has efficiency 0.4, when T2 lowered by50K, its efficiency uncreases to 0.5. Then T1 and T2 are respectively.(A) 300 K and 100 K (B) 400 K and 200 K(C) 600 K and 400 K (D) 400 K and 300 K
101. A monoatomic gas is used in a car not engine as the working substance, If during the adiabaticexpansion part of the cycle the volume of the gas increases from V to 8V1 the efficiency of the engineis ..(A) 60% (B) 50% (C) 75% (D) 25%
102. A System under goes a Cyclic Process in which it absorbs Q1 heat and gives out Q2heat. Theefficiency of the Process is n and the work done is W. Which formula is wrong ?
(A) 21 QQW (B) 1
2
QQn (C)
1QWn (D)
1
2
QQ1n
103. A car not's engine whose sink is at a temperature of 300K has an efficiency of 40% By spaceshould the temperature of the source be increase the efficiency to 60%(A) 275 K (B) 325 K (C) 300 K (D) 250 K
104. An ideal gas heat engine is operating between 227 C and 127 C . It absorks 410 J Of heat at thehigher temperature. The amount of heat Converted into. work is .......... J.(A) 2000 (B) 4000 (C) 5600 (D) 8000
105. Efficiency of a car not engine is 50%, when temperature of outlet is 500K. in order to increaseefficiency up to 60% keeping temperature of intake the same what is temperature of out let.(A) 200 K (B) 400 K (C) 600 K (D) 800 K
106. A car not engine takes 63 10 cal of heat from a reservoir at 627 C , and gives to a sink at 27 C .The work done by the engine is
(A) 64.2 10 J (B) 616.8 10 J (C) 68.4 10 J (D) Zero
107. For which combination of working temperatures the efficiency of Car not's engine is highest.(A) 80 K, 60 K (B) 100 K, 80 K (C) 60 K, 40 K (D) 40 K, 20 K
108. An ideal heat engine working between temperature T1 and T2 has an efficiency n. The new efficiencyif both the source and sink temperature are doubled, will be
(A) n (B) 2n (C) 3n (D) 2n
109. An ideal refrigerator has a freetes at a temperature of 13 C , The coefficent of perfomance of theengine is 5. The temperature of the air to which heat is rejected will be.
(A) 325 C (B) 39 C (C) 325 K (D) 320 C
110. An engine is supposed to operate between two reservoirs at temperature 727 C and 227 C . Themaximum possible efficiency of such an engine is
(A) 43
(B) 41
(C) 21
(D) 1
274
111. A car not engine Convertsm one sixth of the heat input into work. When the temperature of the sinkis reduces by 62 C the efficiency of the engine is doubled. The temperature of the source and sinkare(A) 800 C, 370 C (B) 950 C, 280 C (C) 900 C, 370 C (D) 990 C, 370 C
112. Car not engine working between 300 K and 600 K has work output of 800J per cycle. What isamount of heat energy supplied to the engine from source per cycle
(A) J1600
cycle(B) J
cycle2000 (C) J1000cycle (D) J1800
cycle
113. What is the value of sink temperature when efficiency of engine is 100%(A) 300 K (B) 273 K (C) 0 K (D) 400 K
114. A car not engine having a efficiency of n = 101
as heat engine is used as a refrigerators. if the work
done on the system is 10J. What is the amount of energy absorbed from the reservoir at lowestemperature !(A) 1 J (B) 90 J (C) 99 J (D) 100 J
115. The temperature of sink of car not engine is 27 C . Efficiency of engine is 25% Then find thetemperature of source.(A) 2270 C (B) 3270 C (C) 270 C (D) 1270 C
116. The efficiency of car not's engine operating between reservoirs, maintained at temperature 27 C
and 123 C is ............(A) 0.5 (B) 0.4 (C) 0.6 (D) 0.25
117. If a heat engine absorbs 50KJ heat from a heat source and has efficiency of 40%, then the heatreleased by it in heat sink is ...........(A) 40 KJ (B) 30 KJ (C) 20 J (D) 20 KJ
118. The efficiency of heat engine is 30% If it gives 30KJ heat to the heat sink, than it should haveabsorbed ........... KJ heat from heat source.(A) 42.8 (B) 39 (C) 29 (D) 9
119. If a heat engine absorbs 2KJ heat from a heat source and release 1.5KJ heat into cold reservoir,then its efficiency is ...........(A) 0.5% (B) 75% (C) 25% (D) 50%
120. If the doors of a refrigerators is kept open, then which of the following is true?(A) Room is cooled (B) Room is eithers cooled or heated(C) Room is neither cooled nor heated (D) Room is heated
275
Assertion-ReasonInstructions :-
Read the assertion and reason carefully to mask the correct option out of the options givenbelow.(A) If both assertion and reason are true and the reason is the correct explanation of the assertion.(B) If both assertion and reason are true but reason is not be correct explanation of assertion.(C) If assertion is true but reason is false.(D) If the assertion and reason both are false.
121. Assertion : The melting point of ice decreases with increase of PressureReason : Ice contracts on melting.(A) C (B) B (C) A (D) D
122. Assertion : Fahrenhit is the smallest unit measuring temperature.Reason : Fahrenhit was the first temperature scale used for measuring temperature.(A) A (B) C (C) B (D) D
123. Assertion : A beakes is completely, filled with water at 4 C . It will overlow, both when heated orcooled.Reason : These is expansion of water below 40 C(A) A (B) B (C) C (D) D
124. Assertion : The total translation kinetic energy of all the molecules of a given mass of an ideal gas is1-5 times the product of its Pressure and its volume.Reason : The molecules of a gas collide with each other and velocities of the molecules change dueto the collision(A) D (B) C (C) A (D) B
125. Assertion : The car not is useful in understanding the perfomance of heat engineReason : The car not cycle provides a way of determining the maximum possible efficiency achivablewith reservoirs of given temperatures.(A) A (B) B (C) C (D) D
276
Match column126. Heat given to process is positive, match the following column I with the corresponding option of
column I1
Colum-i Colum-ii(A) JK (p) W >0(B) KL (q) Q <0(C) LM (r) W <0(D) MJ (s) Q >0
(A) A-p, B-q, C-r, D-s (C) A-r, B-s, C-p, D-q(B) A-q, B-p, C-s, D-r (D) A-s, B-r, C-q, D-p
127. In Column I different Process is given match corresponding option of column I1
Column - I Column - II (A) adiabatic process (p) p = 0 (B) Isobaric process (a) u = 0 (C) Isochroic process (r) Q = 0 (D) Isothermal process (s) W = 0
(A) A-p, B-s, C-r, D-q (C) A-r, B-p, C-s, D-q(B) A-s, B-q, C-p, D-r (D) A-q, B-r, C-q, D-p
Comprehehsion TypeIn a containes of negligible heat capacity, 200g ice at 0 C and 100g steam at 100 C are added to200g of water that has temperature 55 C .Assume no heat is lost to the surroundings and the pressurein the container is constant 1 atm.
128. What is the final temperature the System ?
(A) 72 C (B) 48 C (C) 100 C (D) 94 C
129. At the final temperature, mass of the toal water present in the system is(A) 493.6g (B) 483.3g (C) 472.6 g (D) 500 g
130. Amount of the Sm left in the system, is equal to(A) 16-7 g (B) 8-4 g (C) 12 g (D) 0 g
277
KEY NOTE1 B 26 A 51 B 76 C 101 C2 A 27 C 52 C 77 D 102 B3 D 28 D 53 A 78 B 103 D4 D 29 A 54 B 79 A 104 A5 A 30 D 55 C 80 C 105 B6 D 31 B 56 D 81 B 106 C7 C 32 C 57 B 82 C 107 D8 C 33 D 58 A 83 A 108 A9 B 34 B 59 C 84 D 109 B
10 A 35 C 60 D 85 B 110 C11 C 36 A 61 D 86 A 111 D12 D 37 B 62 C 87 C 112 A13 A 38 A 63 A 88 D 113 C14 C 39 C 64 B 89 C 114 B15 B 40 D 65 C 90 A 115 D16 D 41 B 66 A 91 B 116 A17 C 42 A 67 B 92 D 117 B18 A 43 D 68 D 93 B 118 A19 B 44 C 69 D 94 A 119 C20 D 45 C 70 C 95 C 120 D21 C 46 B 71 B 96 D 121 C22 D 47 A 72 A 97 B 122 B23 D 48 D 73 D 98 A 123 A24 A 49 A 74 A 99 C 124 D25 B 50 D 75 D 100 D 125 A
126 B127 C128 C129 B130 A
278
HINT1. 180
F10025
180F
100C
00
2. T = 273.15 + 0 C0 = 273.15 + 0 C
3. equal4. 40 C
932F
5c
5. 932F
5c
y = m x + c with comparision m = 95
6.932f
5c
7. 5273K
932F
8.ΔTc ΔTF = 100 180
ΔTc100 C40TC 0
9. 6 51 12 0 2 10 10 102 2
TT
% in increase = %10110010100TT 35
10. W = pΔV
11. pC 1 R, C2 2vf f R
p
v
1 1C 2 22 2 = 1C
2 2
f fRf
f f f fR
279
12. W = area inside the closed curve
1212 VV21)pp(
21
[Treat Circle as an cllipse]
13. 2 1P(V -V ) PVΔu = = γ γ - 1 -1
14.1 1 1 2 1 2
2 2 2 2u T u T u u T T
u T u T
15. For all gases Cp - Cv = R
16. wuQ
17.p
v
C 2 = +C f
18. W = area enclose by PV graph
PVV)(2VP)(2PBCXAB
19. W = area under the PV curveSmallest for curve 1, largest for curve 3
20. monoatomic gas ,R23cv diatomic gas ,R
25cv
one mole of each gas is mixed Cv (mix) = R2R25R
23
21
Cp (mix) = R + Cv
= R + 2R = 3R
r = p
v
C mix 3R = = 1.5C mix 2R
21. dQ = du + dw
22. Cv mix = 21
2211
nnCvnCvn
23. vγΔU = C ΔT
24. u = f2
n Rt
n = number of mole of the gasf = number of degree of freedom
Total 02 ABU = U U
280
25. TV = [an adiabatic process]
1 1T V 2 2= T T
1 2
2 1
T V = T V
26. W = area under the curve and volume axis on the P V diagram.27. dQ = du + dw28. energy required in heating water = ms
from coil, energyavilable Power of coil Power lost
S
For 840 J time required = 1S 4.2104 J = ?
54.2 10 500 8.33min 8min 205840
t J
29. Cp = Q P
r T
Rcpcv
= 7-2 = 5 Kmolcal
30.1 5
7uQ
u = n Cv T
Q = n Cp T
1 57
v
p
CuQ C r
31. For A (Q) P = r Cp (T) A (Isobaric process)For B (Q) V = Cv (T) B (Isochroic process)(Q )P = (Q)V
32. Decrease in K.E = increase in I.E 2v
1 mv = γ C ΔT2
33. Rate of heat produced = (Viscous force F) (Velocity V)
2dQ nrdt
v 22
o(ρ ρ ) g2 g
rn
281
34. Q = U +W36. Q = U + W39. Q = U +W
u = u - 62.25 (adiabatic process)u = 62.25 x 103 J
Cv = u
n t
for monoatomic gas Cv = KmolJ45.1238
23R
23
40. If CP1 and CV1 is a molar specific heat CP1 - CV1 = R ------------ (i)
1 1Cp CvCp ,Cv32 32
[molar mass of O2 = 32]
From equation (i) 32Cp - 32Cv = R
Cp - Cv = 32R
41. (i) Q = w + u
(ii) Q W u u( equal)
42. 52
52
u RT
PV RT
u PV
[diatomic gas]
3mass 1V mdensity 5
43. u mc T
44. given 5P T
an adiabatic process 1P T
45. 51
282
46. W = Patm V
= Patm V2 (i)
Intially Patm V = rRT RTV
pat
49. (i) Volume Constant 0VPW
(ii) Pressure Constant 2 V
V
W P dv PV
50. For free expansion Q = W = O
T = const Eint o 51. PV v = Const (an adiabatic process)
PV = m Rt µ RTV
P
52. For both solpe is negative & Slope of adiabatic curve is more
53. W area of P V diagarm
= PV)V2(P21
54. AB Constant P, increasing V, increasing TTBC Constant T, increasing V, decreasing PCD Constant V, decreasing P, decreasing TTDA Constant T, decreasing V, increasing P Also BC is at highes temperature than AD
57. PV = const [adiabatic process]
ideal gas PV = RT (m = 1) PRTV
RTPP
= const -1T
P
r
r = const
58. AB BC CA
CA
Q = W +W +W5 = 10+0+W
59. 2 210
1 1
V M VW = µRT oge 2-3 RT og V Mo V
283
60. PV = constant (isothermal Process)
ΔP -ΔVPΔV - VΔP = 0 = P V
61. PV = Constant (an adiabatic Process)Pr v-1 V + V P = 0
62.
PPP
P
adiabatic
isothermal
=
VVV
V
- γ -
= g
63. W = mRT ln 1
2
VV
= R T loge 1
2
VV
64. isothermal bulk modulus B = P
64. isothermal bulk modulus B = P = 1.013105 2mN
66. an adiabatic bulk modulus B = vP
67. an adiabatic bulk modulus B = 1.4 1 105 = 1.4 1052m
N
68. dQ = Cp dT69. W = - RT log e (isothermally compressed)
70. Ciso = Tm
Q = [ T = 0, isothermal process]
71. For isothermal process base is conductory
73. W = PV (isothemal process)74. u = Q - w74. P1V2 = P2V2 (isothermal process)76. Q = u+W
0 = u +W
77. -1T
P
= Constant [adiabatic change]
23
21
1
2
TT
1P2P
1 32 2
5P2 900=
30010
284
78. PV = Constant (adiabatic compressed)
82 1
1 2
P V = P V
79. PV = Constant (adiabatic process
80.-1
2 1
1 2
T V = T V
81. TV - 1 = Constant
T2 = T1
-11
2
VV
82. K600TT
300V2
VTT
VV
222
1
2
1
W = PV
83. W = PV = 1030.25 = 250 J84. PT (constant volume)
2
1
2
1
TT
PP
85. W = PV
86. u)Q(W P
VP )Q()Q(
PQVQ1)Q( P
CPCV1)Q( P
87.dp dv × 100 = - × 100p V
(adiabatic expansion)
89. cyclic process 1 nagative work2 net positive
90. W area of closed PV diagram= (3V-V) (3P-P) = 4PV
91. W = area under curve
285
92. W = area under curve94. (Area)iso >(Area)adi Wiso > wad:95. PV = Con96. PV = Con
97. Q = O, Q = const, du = - dw (aidabatic process)
98.
1
2 2
1 1
T P = T P
99. CP = CP = CV CV
but CP - CV = R
100. n = 1- 1
2
TT 100.
101. -1 -11 1 2 2T V = T V
(qdiabafic process)
-12
1 21
VT =T V
103.1
2
TT1n
104. 2
1
T 1-400 1n = 1 - = = T 500 5
1W = n Q
105.1
2
1
2
TT,
TT1n Should be minimum
106.1
2
TT1n)i(
nTT1
T2T21n)ii(
1
2
1
21
109. 2
1 2 T
T T
286
110. 21
10005001
TT1n
1
2
111.2
1 1
T W 1 1(i) n = 1 - = n = - (1)T Q 6 6
1
21
T62T1n)ii(
11
2
T62
TT1
)2(T62n
1
Now, n1 = 2n
112.2 1
1 1 1 2
T TWn = 1 - = Q = wT Q T -T
113. 2
11 Tn
T
114.2
11 Tn
T
12
2
TW = Q -1T
115.2
1
Tn = 1 - T
116.2
1
Tn = 1 - T
117.2
1
Qn = 1 - Q
118.2
1
Qn = 1 - Q
287
119.2
1
Qn = 1 - Q
121. with rise in pressure melting point of ice decreases. Also ice contracts on melting.122. celcius scale was the first temperature scale and Fahrenhit is the smallest unit measuring.123. Water has maximum density at 4 C on heating above 4 C or cooling below 4 C density of
water decreases and its volume increases, therefore, water overflows in the both cases.
124. 21 3 m ( ) = RT2 2
υ
125. car not cycle has maximum efficiency126. (a) isochoric process uQ0w
PT P decrease, T also decrease u negative 0Q
(b) isobasic process, volume increase 0W (c) isochoric process uQ0W
P T, P increase T increase 0Q
(d) Volumedecrease 0W 127. adiabatic process 0Q
Isobasic process P = const 0P Isochroic process V = const 0W Isothermal process T = const 0u
128 to 130.head rewuired by ice and water to go upto 1000 C = m1L + m1sw T + mw sw T= 20080+2001100+200145= 16,000+20,000+9,000= 45,000 cal= give by ms mass of steam= ms L
ms = g3.83540
000,45 convert into waters of 1000 C
Total water = 200 + 200 + 83.3= 483.3 g
steam left = 100 - 83.3 = 16.79
289
SUMMARY* Gas Laws:1. Boyle's law: For a given mass of an ideal gas at constant temperature, the volume of a gas is
inversely proportional to its pressure, i.e.
P1αV or PV = constant P1V1 = P2V2
mPPV)i( = constant
P = constant or
2
2
1
1 PP
Where density, mV
, and m = constant
(ii) As number of molecules per unit volume
NnV
NVn
also N = const.
(iii)
nNPPV = constant n
P = constant or
2
2
1
1
nP
nP
(iv) Graphical representation: (If m and T are constant)
2. Charle's law: At constant pressure, the volume of the given mass of a gas is directly proportionalto its absolute temperature.
i.e. VV TT
= constant = 2
2
1
1
TV
TV
(i) Tm
TV
= constant V
ρm
or T = constant = 2211 TT
290
(ii) Graphical representation: (If m and P are constant)
3. Gay-Lussac`s Law or pressure law:According to it for a given mass of an ideal gas at constant volume, pressure of a gas isdirectly proportional to its absolute temperature.i.e.
P T or TP
= constant 1 2
1 2
P PT T
(i) Graphical representation: (If m and V are constant)
* Avogadro's law: Equal Volume of all the gases under similar conditions of temperature andpressure contain equal number of molecules.i.e. N1 = N2
Avogadro Number: The number of particles (atoms or molecules) in one mole of substance(gas) is called Avogadro number (NA) which has a magnitudeNA = 6.023 1023 mol-1
291
* Equation of State OR Ideal Gas EquationThe equation which relates the pressure (P), volame (V) and temperature (T) of the givenstate of an of an ideal gas is known as ideal gas equation or equation of state.
Ideal gas equation is PV = µRTwhere µ = number of mole
R = universal gas constant = 8.314 J.mol-1 K-1
* Different forms of Ideal gas-state-equation
(i) µRT PV B BA A
N R RT Nk T kN N
where kB = Boltzmann`s constant= 1.38 10-23 JK-1
(ii) B BNP k T nk TV
where VNn = number density of gas
= number of molecules per unit volume
(iii)o o
M MPV RT µM M
where Mo = molar mass of the gas
o o
M RT ρRT P V M M
= ( m ρV
density of the gas)
* The work done during the change in volume of the gas:
It can be obtained from the graph of P – V f
i
V
VW P dν
* Assumption of Ideal gases (or kinetic theory of gases)Kinetic theory of gases relates the macroscopic properties of gases (such as pressure, temperatureetc.) to the microscopic properties of the gas molecules (such as momentum, speed, kineticenengy of molecule etc.)
Assumptions:1. Gas is made up of tiny particles, These particles are called molecules of the gas.2. The molecules of a gas are identical, spherical, rigid and perfectly elastic point masses.3. The molecules of a gas perform incessant random motion.4. The molecules of a gas follow Newton's laws of motion.5. The number of molecules in a gas is very large.6. The volume of molecules is negligible in comparision with the volume of gas.
292
7. Inter molecular forces act only when two molecules come close to each other or collide.8. The time spent in a collision between two moleules is negligible in comparision to time
between two successive collisions. The collisions between the molecules and between amolecule and the wall of a container are elastic.
* Pressure of an Ideal gas
2 21 13 3
rmsP ρ ν ρ ν
where = density of the gasv2
rms = <2> = mean squre velocity of molecule* Relation between pressure and kinetic energy
2rms
1P ρ (i)3
ν
K.E. per unit volume is
E = 1 M2 V 2
rms = 12 2
rms ( )ii
From equation (i) and (ii)
23
P E
* Root mean square speed (vrms) :It is defined as the square root of mean of squares of the speed of the speed of different
molecules.
i.e. rms = 2 2 2 21 2 3 N........
Nν ν ν ν
From the expression of pressure
P = 13 2
rms rms = 3Pρ =
3PVM
= 3RTM
= 3 Bk T
m
where MρV
= density of the gas
AN
Mm mass of each molecule
- rms Tν
- with increase in molecular weight rms speed of gas molecule decreases as rms1M
v
293
- rms speed of gas molecules does not depend on the pressure of gas (If temperature remainsconstant)
- At T= 0 K , vrms= 0, i.e. the rms speed of molecules of a gas is zero at O K. This temperatureis called absolute zero.
* Kinetic interpretation of temperatureKinetic energy of of 1 mole ideal gas
21 1 3 3E 2 2 2rms
RTM ν M RTM
- For 1 molecule B3E k T2
, kB = Boltzmann`s constant
- For N molecule B3E N k T2
- Kinetic energy per nolecule of gas does not depend upon the mass of the molecule but onlydepends on the temperature.
* Mean free path:-The distance travelled by a gas molecule between two successive collisions is known as free path.The average of such free paths travelled by a molecule is called mean free path.mean free path,
= Total distance travelled by a gas molecule between two successive collisionsTotal number of collisions
- 2dn21
- BB2 2
B
1 k T PP nk T nk T2 n d 2 Pdπ π
- Collision frequency = number of collisions per second.
rmsν
* Degrees of Freedom :The term degrees of freedom of a molecule or gas are the number of independent motions thata molecule or gas can have.The independent motion of a system can be translational, rotational or vibrational or any combinationof these.Degress of freedom,f = 3A - B; where A = Number of independent particles,
B = Number of independent restrictions- monoatomic gas 3 degrees of freedom
( All translational)
294
- Diatomic gas 5 degrees of freedom( 3 translational + 2 rotational)
- triatomic gas 6 degrees of freedom (at room temperature)(Non-linear) ( 3 translational + 3 rotational)
8 degrees of freedom (at very high temperature) (3 translational + 3 rotational + 2 vibrational)
* Law of equipartition of energy (Boltzmann law)According to this law, for any system in equilibrium, the total energy is equally distributed
among its various degrees of freedom and each degrees of freedom is associated with energy
B1 k T2 where kB = Boltzmann`s constant.
- At a given temperataue T, all ideal gas molecules no matter what their mass have the same average
translational kinetic energy = B3 k T2
- At same temperature gases with different degrees of freedom (i.e. H2 and He) will have different
average energy Bk T2f
( f = degress of freedom different for different gases.)
- The total energy associated with each modlec Bk T2f
* Specific heat of a gas :- Specific heat at constant volume (Cv)- The amount of heat required to change the temperature of l mole of gas by 1 K, keeping its
volume constant, is called specific heat of the gas at constant volume.- Specific heat at constant volume (Cp)
The amount of heat required to change the temperature of l mole of gas by 1 K, keeping itspressure constant, is called specific heat of the gas at constant pressure.Molar specific heat:
- The quantity of heat required to change the temperature of 1 mole of gas by 1 K (or 1o C) iscalled molar specific heat of the gas.Ratio of CP and CV is .
PV P
V
1 RC 2 12 1 C R , C 1 RC 2 2R
2
ffff f
295
MCQChoose the correct alternative from given options.1. Volume, pressure and temperature of an ideal gas are V, P and T respectively. If mass of molecule
is m, then its density is [ kB = Boltzmann`s constant]
(A) B
Pk T (B)
B
Pmk T (C)
B
.Pk TV (D) Bmk T
2. The temperature of an ideal gas at atmospheric pressure is 300 K and volume 1 m3. If temperatureand volume become double, then pressure will be(A) 4 105 Nm–2 (B) 2 105 Nm–2 (C) 1 105 Nm–2 (D) 0.5 105 Nm–2
3. At 100 K and 0.1 atmospheric pressure, the volume of helium gas is 10 litres. If volume andpressure are doubled, its temperature will change to(A) 127 K (B) 400 K (C) 25 K (D) 200 K
4. What is the mass of 2 litres of nitrogen at 22.4 atmospheric pressure and 273 K.(R = 8.314 Jmol k-1)(A) 14 22.4 g (B) 56 g (C) 28 g (D) None of these.
5. An electron tube was sealed off during manufacture at a pressure of 1.2 10-7 mm of mercury at270C. Its volume is 100 cm3. The number of molecules that remain in the tube is _______(density of mercury is 13.6 gcm–3)(A) 3.9 1011 (B) 3 1016 (C) 2 1014 (D) 7 1011
6. A vessel contains 1 mole of O2 gas (relative molar mass 32) at a temperature T. The pressure ofthe gas is P. An identical vessel containing 1 mole of He gas (relative molar mass 4) at atemperature 2T has pressure of ..........
(A) 8 P (B) p8 (C) 2 P (D) P
7. The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying avolume V, will be
(A) RT325PV
(B) RT
165PV
(C) RT
25PV
(D) PV = 5 RT
8. A gas at 1 atmosphere and having volume 100 ml is mixed with another gas of equal moles at0.5 atm and having volume 50 ml in flask of one litre, what is the final pressure?(A) 0.125 atm (B) 0.75 atm (C) 1 atm (D) 0.5 atm
9. The quantity B
PVk T represents
(A) mass of gas (B) number of moles of gas(C) number of molecules in gas (D) K. E. of gas
10. Equation of gas in terms of pressure (P), absolute temperature (T) and density () is
(A) 2
22
1
11
ρTP
ρTP
(B) 22
2
11
1
ρTP
ρTP
(C) 2
12
1
21
TρP
TρP
(D) 2
22
1
11
TρP
TρP
296
11. 02 gas is filled in a vessel. If pressure is doubled, temperature becomes four times, how many timesits density will become.
(A) 4 (B) 41
(C) 2 (D) 21
12. At a given volume and temperature the pressure of a gas(A) Varies inversely as the square of its mass (B) Varies inversely as its mass(C) is independent of its mass (D) Varies linearly as its mass
13. If pressure of a gas contained in a closed vessel is increased by 0.4% when heated by 10 C theinitial temperature must be.(A) 2500 C (B) 250 C (C) 250 K (D) 2500 K
14. To decrease the volume of a gas by 5% at constant temperature the pressure should be(A) Incseased by 5.26% (B) Decreased by 5.26%(C) Decreased by 11% (D) Increased by 11%
15. A gas at the temperature 250 K is contained in a closed vessel. If the gas is heated through1 K, then the percentage increase in its pressurse will be(A) 0.4% (B) 0.1 % (C) 0.8% (D) 0.2%
16. The product of the pressure and volume of an ideal gas is(A) A constant(B) Directly proportional to its temperature.(C) Inversely proportional to its temperature.(D) Approx. equal to the universal gas constant.
17. At OoC the density of a fixed mass of a gas divided by pressure is x. At 100o C, the ratio willbe
(A) x (B) x373273
(C) x273373
(D) x273100
18. Air is pumped into an automobile tube upto a pressure of 200 kPa in the morning when the airtemperature is 22o C. During the day, temperature rises to 42o C and the tube expands by 2%The pressure of the air in the tube at this temperature will be approximately.(A) 209 kPa (B) 206 kPa (C) 200 kPa (D) 212 kPa
19. The volume of a gas at 20 C is 200 ml. If the temperature is reduced to –20o C at constantpressure, its volume will be.(A) 172.6 ml (B) 17.26 ml (C) 19.27 ml (D) 192.7 ml
20. 2g of O2 gas is taken at 27o C and pressure 76 cm Hg. Find out volume of gas (ln litre)(A) 3.08 (B) 44.2 (C) 1.53 (D) 2.44
21. 1 mole of gas occupies a volume of 100 ml at 50 mm pressure. What is the volume occupied bytwo moles of gas at 100 mm pressure and at same temperature(A) 50 ml (B) 200 ml (C) 100 ml (D) 500 ml
297
22. A partition divides a container, having insulated walls, into twocompartments, I and II. The same gas is filled the compartments.The ratio of number of molecules in compartments I and II is.(A) 6:1 (B) 1:6 (C) 4:1 (D) 1:4
23. A cylinder contains 10 kg of gas at pressure of 107 N/m2. The quantity of gas taken out of the
cylinder, if final pressure is 26 Nm105.2 , will be ______ (temperature of gas is constant)
(A) 5.2 kg (B) 3.7 kg (C) 7.5 kg (D) 1 kg24. The volume of a gas at pressure 21 104 Nm–2 and temperature 27oC is 83 Litres.
If R = 8.3 J mol–1K–1. Then the quantity of gas in g-mole will be(A) 42 (B) 7 (C) 14 (D) 15
25. The pressure and temperature of an ideal gas in a closed vessel are 720 kPa and 40o C respectively.
If th1
4 of the gas is released from the vessel and the temperature of the remaning gas is raised
to 353o C, final pressure of the gas is(A) 1440 kPa (B) 540 kPa (C) 1080 kPa (D) 720 kPa
26. Suppose ideal gas equation follows 3VP = constant, Initial temperature and volume of the gas areT and V respectively. If gas expand to 27 V, then temperature will become
(A) 9 T (B) 27 T (C) T9 (D) T
27. The temperature of a gas at pressure P and volume V is 270 C Keeping its volume constant if itstemperature is raised to 9270 C, then its pressure will be(A) 3 P (B) 2 P (C) 4 P (D) 6 P
28. Air is filled in a bottle at atmospheric pressure and it is corked at 35o C, If the cork can comeout at 3 atmospheric pressure then upto what temperature should the bottle be heated in order toremove the cork.(A) 325.5o C (B) 651o C (C) 851o C (D)None of these
29. At what temperature volume of an ideal gas becomes triple(A) 819o C (B) 182o C (C) 646o C (D) 546o C
30. To double the volume of a given mass at an ideal gas at 270 C keeping the pressure constant onemust raise the temperature in degree centigrade(A) 54o (B) 600o (C) 327o (D) 270o
31. At constant temperature on incerasing the pressure of a gas 5% its volume will decrease by(A) 5% (B) 5.26% (C) 4.76% (D) 4.26%
32. At on 0o C pressure measured by barometer is 760 mm. what will be pressure at 100o C(A) 780 mm (B) 760 mm (C) 730 mm (D) None of these.
33. Hydrogen gas is filled in a ballon at 200 C. If temperature is made 400 C, pressure remaining thesame what fraction of haydrogen will come out(A) 0.75 (B) 0.07 (C) 0.25 (D) 0.5
P, V, T
I
2P, 2V, T
II
298
34. When the pressure on 1200 ml of a gas is increased from 70 cm to 120 cm of mercury at constanttemperature, the new volume of the gas will be(A) 400 ml (B) 600 ml (C) 700 ml (D) 500 ml
35. A gas at 270 C temperature and 30 atmospheric pressure is allowed to expand to the atmosphericpressure if the volume becomes two times its initial volume, then the final temperature becomes(A) 2730 C (B) -1730 C (C) 1730 C (D) 1000 C
36. A gas at 270 C has a volume V and pressure P. On heating its pressure is doubled and volumebecomes three times. The resulting temperature of the gas will be(A) 15270 C (B) 6000 C (C) 1620 C (D) 18000 C
37. A perfect gas at 270 C is heated at constant pressure to 3270 C. If original volume of gas at 27o Cis V then volume at 327o C is
(A) 2 V (B) V (C) V2 (D) 3 V
38. A vessel contains 1 mole of 2O gas (molar mass 32) at a temperature T. The pressure of the gasis P. An identical vessel containing one mole of He gas (molar mass 4) at temperature 2T has apressure of
(A) 2 P (B) P (C) 8 P (D) P8
39. The pressure and temperature of two different gases P and T having the volumes V for each. Theyare mixed keeping the same volume and temperature, the pressure of the mixture will be,
(A) P (B) P2 (C) 4 P (D) 2 P
40. Air is filled at 600 C in a vessel of open mouth. The vessel is heated to a temperature T so thatth1
4 part of air escapes. Assuming the volume of the vessel remaining constant the value of T is.
(A) 3330 C (B) 1710 C (C) 4440 C (D) 800 C41. A gas is filled in a cylinder, its temperature is incresecd by 20% on kelvin scale and volume is
reduced by 10%. How much percentage of the gas will leak out(A) 15% (B) 25% (C) 40% (D) 30%
42. The pressure is exerted by the gas on the walls of the container because(A) It sticks with the walls(B) It is accelerated towards the walls(C) It loses kinetic energy(D) On collision with the walls there is a change in momentum
43. The relation between the gas pressure P and average kinetic energy per unit volume E is
(A) E32P (B) E
23P (C) EP (D) EP =
244. The root mean square speed of hydrogen molecules of an ideal hydrogen kept in a gas chamber
at 00C is 3180 ms–1. The pressure on the hydrogen gas is (Density of hydrogen gas is2 3 5 28.99 10 kg m , 1 atm 1.01 10 Nm )
(A) 1.0 atm (B) 3.0 atm (C) 2.0 atm (D) 1.5 atm
299
45. Gas at a pressure Po is contained in a vessel. If the masses of all the molecules are halved andtheir speeds are doubled, the resulting pressure will be equal to
(A) 2 P0 (B) 4 P0 (C) 2P0 (D) P0
46. A cylinder of capacity 20 litres is filled with 2H gas. The total average kinetic energy of translatory
motion of its molecules is 51.5 10 J. The pressure of hydrogen in the cylinder is(A) 4 106 Nm–2 (B) 3 106 Nm–2
(C) 5 106 Nm–2 (D) 2 106 Nm–2
47. The average kinetic energy per molecule of a gas at -230 C and 75 cm pressure is erg105 14 for
2H . The mean kinetic energy per molecule of the 2O at 227 C and 150 cm pressure will be
(A) erg1080 14 (B) erg1010 14
(C) erg1020 14 (D) erg1040 14
48. The ratio of mean kinetic energy of hydrogen and oxygen at a given temperature is(A) 1:8 (B) 1:4 (C) 1:16 (D) 1:1
49. The ratio of mean kinetic energy of hydrogen and nitrogen at temperature 300 K and 450 Krespectively is(A) 2:3 (B) 3:2 (C) 4:9 (D) 2:2
50. Pressure of an ideal gas is increased by keeping temperature constant what is the effect on kineticenergy of molecules.(A) Decrease (B) Increase(C) No change (D) Can`t be determined
51. A sealed container with negligible co-efficient of volumetric expansion contains helium (a monoatomicgas) when it is heated from 200 K to 600 K, the averagy K.E. of helium atom is
(A) Halved (B) Doubled (C) Unchanged (D) Increased by factor 252. The mean kinetic energy of a gas at 300 K is 100J. mean energy of the gas at 450 K is equal to
(A) 100 J (B) 150 J (C) 3000 J (D) 450 J
53. At what temperature is the kinetic energy of a gas molecule double that of its value at 27 C
(A) 540 C (B) 1080 C (C) 3270 C (D) 3000 C54. The average kinetic energy of a gas molecule at 270 C is 6.21 10–21 J. Its average kinetic energy
at 2270 C will be
(A) J1022.5 21 (B) J1035.11 21 (C) J102.52 21 (D) J1042.12 2155. The average translational energy and rms speed of molecules in sample of oxygen gas at 300 K
are 216.21 10 J and 484 m s respectively. The corresponding values at 600 K are nearly(assuming ideal gas behaviour)(A) 216.21 10 J , 968 m s (B) 2112.42 10 J , 684 m s(C) 2112.42 10 J , 968 m s (D) 218.78 10 J , 684 m s
300
56. The average translational kinetic energy of O2 (molar mass 32) molecules at a particular temperatureis 0.068 eV. The translational kinetic energy of N2 (molar mass 28) molecules in eV at the sametemperature is(A) 0.003 eV (B) 0.068 eV (C) 0.056 eV (D) 0.678 eV
57. At O K which of the follwing properties of a gas will be zero(A) Kinetic energy (B) Density (C) Potential energy (D) Vibrational energy
58. The kinetic energy of one mole gas at 300 K temperatue is E. At 400 K temperature kinetic enrgy
is E'. The value of E'E is
(A) 2 (B) 916
(C) 1.33 (D) 34
59. The average kinetic energy of hydrogen molecules at 300 K is E. At the same temperature theaverage kinetic energy of oxygen molecules will be
(A) E16 (B) E (C) 4 E (D) E
460. The temperature at which the average translational kinetic energy of a molecule is equal to the
energy gained by an electron accelerating from rest through a potential differencc of 1 volt is
(A) K106.4 3 (B) K107.7 3 (C) K106.11 3 (D) K102.23 3
61. At a given temperature the rms velocity of molecules of the gas is(A) Proportional to molecular weight(B) Inversely proportional to molecular weight(C) Inversely proportional to square root of molecular weight(D) Proportional to square of molecular weight
62. According to the kinetic theroy of gases the r.m.s velocity of gas molecules is directly proportionalto(A) 2T (B) T (C) T (D) T
1
63. The speeds of 5 molecules of a gas (in arbitrary units) are as follws: 2, 3, 4, 5, 6, The root meansqure speed for these molecules is(A) 4.24 (B) 2.91 (C) 4.0 (D) 3.52
64. To what temperature should the hydrogen at room temperature (27o C) be heated at constantpressuse so that the rms velocity of its molecule becomes double of its previous value
(A) 927 C (B) 600 C (C) 108 C (D) 1200 C
65. Root mean square velocity of a molecule is at pressure P. If pressure is increased two times,then the rms velocity becomes(A) 3 (B) 2 (C) 0.5 (D)
66. The rms speed of gas molecules is given by
(A) oM2.5RT
(B) o
RT2.5M (C)
o
RT1.73M (D) oM1.73
RT
301
67. A sample of gas is at Oo C. To what temperature it must be raised in order to double the rmsspeed of molecule.(A) 270o C (B) 819o C (C) 100o C (D) 1090o C
68. If the ratio of vapour density for hydrogen and oxygen is 1
16 , then under constant pressure the
ratio of their rms velocities will be(A) 4:1 (B) 1:16 (C) 16:1 (D) 1:4
69. The molecules of a given mass of a gas have a rms velocity of 200m s at 27 C and5 21.0 10 Nm pressure when the temperature is 127 C and pressure is 5 20.5 10 Nm , the
rms velocity in m s will be
(A) 2100 (B) 3
2100(C) 3
400(D) None of these
70. If the molecular weight of two gases are 1M and 2M , then at a given temperature the ratio of
root mean square velocity 1 and 2 will be
(A) 2
1
MM
(B) 1
2
MM
(C) 21
21
MMMM
(D) 21
21
MMMM
71. To what temperature should the hydrogen at 327 C cooled at constant pressure, so that the rootmean square velocity of its molcules become half of its previous value
(A) 100 C (B) 123 C (C) 0 C (D) 123 C
72. At what temperature is the root mean square velocity of gaseous hydrogen molecules equal to thatof oxygen molecules at 47 C ?(A) -73 K (B) 80 K (C) 20 K (D) 3 K
73. The root mean square velocity of the molecules in a sample of helium is th57 that of the molecules
in a sample of hydrogen. If the temperature of hydorgen sample is 00C, then the temperature ofthe helium sample is about(A) 273 C (B) 0 C (C) O K (D) 100 C
74. At room temperature ( 27 C ), the rms speed of the molecules of certain diatomic gas is found tobe 1930 m/s. The gas is
(A) O2 (B) 2C (C) H2 (D) F2
75. If three molecules have velocities 0.5, 1 and 2 the ratio of rms speed and average speed is (Thevelocities are in km/s)(A) 0.134 (B) 1.34 (C) 1.134 (D) 13.4
302
76. At what temperature pressure remaining constant will the rms speed of a gas molecules increaseby 10% of the rms speed at NTP?(A) 57.3 K (B) 57.30 C (C) 557.3 K (D) -57.30 C
77. When temperature of an ideal gas is increased from 27o C to 227o C, its rms speed changed from400 ms–1 to Vs. The Vs is(A) 516 ms–1 (B) 746 ms–1 (C) 310 ms–1 (D) 450 ms–1
78. At what temperature the molecules of nitrogen will have the same rms. velocity as the moleculesof Oxygen at 127oC.(A) 2730 C (B) 3500 C (C) 770 C (D) 4570 C
79. The temperature of an ideal gas is increased from 270 C to 1270 C, then percentage increase inrms is(A) 33% (B) 11% (C) 15.5% (D) 37%
80. Let A and B the two gases and given : B
A
B
A
MM4
TT
. Where T is the temperature and M is
molecular mass. If A and B are the r.m.s speed, then the ratio A
B
will be equal to __________
(A) 2 (B) 4 (C) 0.5 (D) 181. The rms. speed of the molecules of a gas in a vessel is 400 ms-1 . If half of the gas leaks out,
at constant temperature, the r.m.s speed of the remaining molecules will be
(A) 800 ms-1 (B) 200 ms-1 (C) 2400 ms-1 (D) 400 ms-1
82. At which temperature the velocity of 02 molecules will be equal to the rms velocity of N2 moleculesat 00 C(A) 930 C (B) 400 C(C) 390 C (D) can not be calculated.
83. The rms speed of the molecules of a gas at a pressure 105 Pa and temperature 00 C is0.5 km/s. If the pressure is kept constant but temperature is raised to 8190 C, the rms speedbecomes(A) 1.5 kms-1 (B) 2 kms-1 (C) 1 kms-1 (D) 5 kms-1
84. The root mean square velocity of a gas molecule of mass m at a given temperature is proportionalto(A) m0 (B) m-1/2 (C) m1/2 (D) m
85. The ratio of the vapour densities of two gases at a given temperature is 9:8, The ratio of the rmsvelocities of their molecule is
(A) 22:3 (B) 3:22 (C) 9:8 (D) 8:9
86. At what temperature, pressure remaining unchanged, will the rms velocity of a gas be half its valueat OoC ?(A) 204.75 K (B) 204.750 C (C) -204.75 K (D) -204.750 C
87. The rms velocity of gas molecules is 300 ms-1. The rms velocity of molecules of gas with twicethe molecular weight and half the absolute temperature is(A) 300 ms-1 (B) 150 ms-1 (C) 600 ms-1 (D) 75 ms-1
303
88. Calculate the temperature at which rms velocity of S02 molecules is the same as that of O2molecules at 270 C. Molecular weights of Oxygen and SO2 are 32 g and 64 g respectively(A) 3270 C (B) 327 K (C) 1270 C (D) 2270 C
89. For a gas, the rms speed at 800 K is(A) Four times the value at 200 K(B) Twice the value at 200 K(C) Half the value at 200 K(D) same as at 200 K
90. A mixture of 2 moles of helium gas (atomic mass = 4 amu), and 1 mole of argon gas (atomic mass
= 40 amu) is kept at 300 K in a container. The ratio of the rms speeds rms
rms
(helium)(argon)
vv
is
(A) 0.45 (B) 2.24 (C) 3.16 (D) 0.3291. The temperature of an ideal gas is increased from 270 C to 9270 C. The root mean square speed
of its molecules becomes(A) Four times (B) One-fourth (C) Half (D) Twice
92. At a given temperature the root mean square velocities of Oxygen and hydrogen molecules are inthe ratio(A) 1:4 (B) 1:16 (C) 16:1 (D) 4:1
93. If mass of He atom is 4 times that of hydrogen atom, then rms speed of the is(A) Two times of H rms speed.(B) Four times of H rms speed.(C) Same as of H rms speed.(D) half of H rms speed.
94. At temperature T, the rms speed of helium molecules is the same as rms speed of hydrogenmdecules at normal temperature and pressure. The value of T is(A) 5460 C (B) 00 C (C) 2730 C (D) 136.50 C
95. The root mean square speed of hydrogen molecules at 300 K is 1930 m/s. Then the root meansquare speed of Oxygen molecules at 900 K will be
(A) 836 m/s (B) 643 m/s (C) 1930 3 m/s (D) 1930 m/s
396. If rms speed of a gas is rms = 1840 m/s and its density = 8.9910-2 kg/m3 , the pressure of
the gas will be(A) 1.01 103 Nm–2 (B) 1.01 105 Nm–2 (C) 1.01 107 Nm–2 (D) 1.01 Nm–2
97. When the temperature of a gas is raised from 270 C to 900 C , the percentage increase in therms velocity of the molecules will be(A) 15% (B) 17.5% (C) 10% (D) 20%
98. The rms speed of a gas at a certain temperature is 2 times than that of the Oxygen moleculeat that temperature, the gas is_____(A) SO2 (B) CH4 (C) H2 (D) He
304
99. The temperature at which the rms speed of hydrogen molecules is equal to escape velocity on earthsurface will be(A) 5030 K (B) 10063 K (C) 1060 K (D) 8270 K
100. What is the meanfree path and collision frequency of a nitrogen molecule in a cylinder containingnitrogen at 2 atm and temperature 17oC ? Take the radius of nitrogen molecule to be 1A
.
M olecular mass of nitrogen = 28 , kB = 1.38 10–23 JK–1, 1 atm = 1.013 105 Nm–2
(A) 97 1058.2,m102.2 (B) 87 1058.4,m101.1
(C) 97 1058.4,m101.1 (D) 97 1058.3,m102.2
101. The radius of a molecule of Argon gas is 1.780A . Find the mean free path of molecules of Argon
at 00 C temperature and 1 atm pressure.23 1
Bk 1.38 10 JK
(A) m1065.6 8 (B) m1065.3 7 (C) m1065.6 7 (D) m1065.3 8
102. A monoatomic gas molecule has (A) Three degrees of freedom (B) Five degrees of freedom(C) Six degrees of freedom (D) Four degrees of freedom
103. A diatomic molecule has how many degrees of freedom (For rigid rotator) (A) 4 (B) 3 (C) 6 (D) 5
104. The degrees of freedom for triatomic gas is ______ (At room temperature) (A) 8 (B) 6 (C) 4 (D) 2
105. If the degrees of freedom of a gas are f, then the ratio of two specific heats p
v
CC
is given by
(A) 21f
(B) 11f
(C) 2 1f (D) 31
f
106. A diatomic gas molecule has translational, rotational and vibrational degrees of freedom. The
p
v
CC
is
(A) 1.29 (B) 1.33 (C) 1.4 (D) 1.67107. The value of Cv for one mole of neon gas is
(A) R23
(B) R27
(C) R21
(D) R25
108. The relation between two specific heats of a gas is
(A) V PRC C J (B) P VC C J
(C) P VRC C J (D) V PC C J
305
109. The molar specific heat at constant pressure for a monoatomic gas is
(A) R23
(B) R25
(C) 4 R (D) R27
110. For a gas 67.0CR
V
. This gas is made up of molecules which are
(A) Diatomic (B) monoatomic(C) polyatomic (D) mixture of diatomic and polyatomic molecules
111. The specific heat of an ideal gas is(A) Proportional to T2 (B) Proportional to T3
(C) Proportional to T (D) Independent of T112. The specific heats at constant pressure is greater than that of the same gas at constant volume
because (A) At constant volume work is done in expanding the gas.(B) At constant pressure work is done in expanding the gas.(C) The molecular vibration increases more at constant pressure.(D) The molecular attraction increases more at constant pressure.
113. One mole of ideal monoatomic gas 53
is mixed with one mole of diatomic gas 73
.
What is for the mixture? denotes the ratio of specific heat at constant pressure to that atconstant volume.
(A) 3523 (B)
2315 (C) 3
2 (D) 43
114. For a gas if ratio of specific heats at constant pressure and volume is , then value of degreesof freedom is
(A) 2
1 (B) 25 12
(C) 3 12 1 (D) 9 1
2
115. The molar specific heat at constant pressure of an ideal gas is R27
. The ratio of specific heat
at constant pressure to that ratio at constant volume is
(A) 57 (B) 9
7 (C) 87 (D) 7
5
116. For a gas 75
, the gas may probably be
(A) Neon (B) Argon (C) Helium (D) Hydrogen117. From the following P - T graph, what inference can be drawn
(A) V 2< V1 (B) V2= V1
(C) V 2>V1 (D) none of these
306
118. The figure shows the volume V versus temperature T graphs fora certain mass of a perfect gas at two constant pressure ofP1 and P2. What inference can you draw from the graphs.(A) P1< P2
(B) P1>P2
(C) P1= P2
(D) No inference can be drawn due to insufficient information.119. Which one the following graphs represents the behaviour of an ideal gas
(A) (B) (C) (D)
120. Under constant temperature, graph between p and V1 is
(A) Hyperbola (B) Circle (C) Parabola (D) Straight line
Direction:-(a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.(b) If both Assertion and Reason are true, but Reason is not correct explanation of the Assertion.(c) If Assertion is true; the Reason is false.(d) If Assertion is false, but the reason is true.
121. Assertion : 300 cc of a gas at 270 C is cooled at -30 C at constant pressure. The final volumeof the gas would be 270 cc
Reason : This is as per charle's law 1
2
1
2
TT
VV
(A) a (B) b (C) c (D) d122. Assertion : The time of collision of molecules is of the order of 10-8 s, which is very very small
compared to the time between two successive collisions.Reason : This is an experimental fact.(A) a (B) b (C) c (D) d
123. Assertion : Mean free path of gas varies inversly as density of the gas.Reason : Mean free path varies inversely as pressure of the gas.(A) a (B) b (C) c (D) d
307
KEY NOTE1 B 26 A 51 B 76 B 101 A2 C 27 C 52 B 77 A 102 A3 B 28 B 53 C 78 C 103 D4 C 29 D 54 D 79 C 104 B5 A 30 C 55 B 80 A 105 D6 C 31 C 56 B 81 D 106 B7 A 32 D 57 A 82 C 107 A8 A 33 B 58 C 83 C 108 C9 C 34 C 59 B 84 B 109 B10 B 35 B 60 B 85 B 110 B11 D 36 A 61 C 86 D 111 D12 D 37 A 62 B 87 B 11213 C 38 A 63 A 88 A 113 C14 A 39 D 64 A 89 B 114 A15 A 40 B 65 D 90 C 115 D16 B 41 B 66 C 91 D 116 D17 B 42 D 67 B 92 A 117 C18 A 43 A 68 A 93 D 118 B19 A 44 B 69 C 94 C 119 C20 C 45 C 70 B 95 A 120 D21 C 46 B 71 D 96 B 121 A22 D 47 D 72 C 97 C 122 A23 C 48 A 73 B 98 B 123 B24 B 49 A 74 C 99 B25 C 50 C 75 C 100 C
1 B 26 A 51 B 76 B 101 A2 C 27 C 52 B 77 A 102 A3 B 28 B 53 C 78 C 103 D4 C 29 D 54 D 79 C 104 B5 A 30 C 55 B 80 A 1056 C 31 C 56 B 81 D 106 B7 A 32 D 57 A 82 C 107 A8 A 33 B 58 C 83 C 108 C9 C 34 C 59 B 84 B 109 B
10 B 35 B 60 B 85 B 110 B11 D 36 A 61 C 86 D 111 D12 D 37 A 62 B 87 B 11213 C 38 A 63 A 88 A 113 C14 A 39 D 64 A 89 B 114 A15 A 40 B 65 D 90 C 115 D16 B 41 B 66 C 91 D 116 D17 B 42 D 67 B 92 A 117 C18 A 43 A 68 A 93 D 118 B19 A 44 B 69 C 94 C 119 C20 C 45 70 B 95 A 120 D21 C 46 71 D 96 B 121 A22 D 47 72 C 97 C 122 A23 C 48 73 B 98 B 123 B24 B 49 A 74 C 99 B25 C 50 C 75 C 100 C
C
B
ACBD
308
Hint / Solution1. (B) PV = µRT
O
MPV RTM
oPM MRT V
o A
B
A
PM P m N Pm PmRT RT k TR T
N
2. (C)VTPRTPV
If T and V both doubled, then pressure remains same
3. (B) PV μRT PV α T
If P and V doubled, then T becomes four times,
4. (C) PVPV µRT µRT
Mass of litre nitrogen = oµ M
5. (A) Gas equation for N molecules, PV = NkBT
B
PVNk T
6. (C) VµRTPµRTPV
For same Rµ, and V, P T
7. (A) molecular weight of oxygen = 32 (g)
number of moles in 5 g of oxygen = 325
Equation of state is PV = µRT
RT325 PV
8. (A) Total number of moles is conserved,
1 1 2 2PV PV PVRT RT RT
1 1 0 0 0 .5 5 0 P 1 0 0 0+ =R T R T R T
(1 L = 100 m )
atm125.0P
309
9. (C) The ideal gas equation is
µRTPV
PV RT µ 1
B A BA
RPV k N T kN
AB
PV Nk T
Avogadro's number..
10. (B). µRTPV
o
MPV RTM
o
M RTV M
o
RT MM V
o
P RT M
= const.
11. (D) µRTPV
MP µRT
o
M M PP RTM T
12. (D) µRTPV
o
MPV RT P MM
( V, T T Constant )
13. (C). µRTPV
P T ( Closed vessel i.e. volume is constant)
1 1
2 2
P TP T
14. (A). µRTPV = Constant ( temp. is const.)
2211 VPVP
1 1 2 1 2 195 95%
100PV P V V V
310
2 11.0526P P
11 P0526.0P
11 P%26.5P
Pressure 5.26% increases.15. (A) Closed Vessel . i.e. volume remains constant.
From, µRTPV
2 2
1 1
P TP α TP T
1
12
1
12
TTT
PPP
16. (B) PV T ( µ,R Constant)
17. (B)o
MPV µRT RTM
oMM
PV RT
densityP
= oMRT
At 0 C
densityP
=
M x...........................(i)R 273
At 100 C
densityP
=
M .........................(ii)R 373
At 100 C
densityP
273 x373
18. (A)PV =µRT = constant 1 1 2 2
1 2
P V P VT T
19. (A) PV=µRT
since P is const. V T
1 1
2 2
V TV T
20. (C) PV= µRT
o o
M MRT= RT V =M M P
311
21. (C) PV= µRT
1 1 1
2 2 2
P V µP V µ
(T const.)
22. (D) BB
PVPV Nk T Nk T
Now, B B
2P 2V PVN' 4 4 Nk T k T
N 1N' 4
23. (C) PV=µRT
o
MPV = RT P MM
(V, R, T - Constant)
71 1
62 2 2
P M 10 10P M 2.5 10 M
M2 = 2.5 kg
Hence mass of the gas taken out of the cylinder = 10 – 2.5 = 7.5 kg
24. (B)PVPV =µRT = RT
25. (C)o
MPV = µRT PV = RT P MTM
2 2 2
1 1 1
P M TP M T
26. (A) VP3 = constant = k 31
3
1 1P PV V
13
kPV
13
kPV = µRT V µRTV
23k V µRT
23 µRTV
k
Hence
2 23 3
1 1
2 2 2
V T V TV T 27 V T
22
1 T T 9T9 T
312
27. (C) Using Gay-lussac's law 1 1
2 2
P TP T
28. (B) At constant volume
1 2 22 1
1 2 1
P P PT TT T P
29. (D) At constant Pressure V T
2 2 22 1
1 1 1
V T VT TV T V
30. (C) 11
2 2
TVV TV T
31. (C) 2 12 1 1
1 2
1 V P 100 100P V V 0.9524 VV V P 105 105
2 1V 1 0.0476 V
1 1V 0.0476V
1 1V 4.76% V
32. (D) 2 2 22 1
1 1 1
P T TP T P PP T T
33. (B) 2 2 2 1 2 1
1 1 1 1
V T V V T TV TV T V T
273 40 273 20VV 273 20
313 293 0.07293
34. (C) At constant Pressure PV = constant
1 1 2 2P V P V 1 2
2 1
P VP V
35. (B)1 1 2 2 2 2
2 11 2 1 1
P V P V P VT TT T P V
36. (A) 1 1 2 2
1 2
P V P VT T
2 22 1
l 1
P VT TP V
37. (A) 1 1
2 2
V TV TV T
313
38. (A) PV =µRT P=µT ( V and R = constant )
2 2 2
1 1 1
P µ TP µ T
39. (D) 1PVPV=µRT µRT
and 2PVµRT
1 2µ µ RT 2 PV RTP ' 2PV RT V
40. (B) For open mouth vessel, pressure is constant.volume is also given constant.
Hence from PV µRT
o
M 1PV RT TM M
1 2
2 1
T MT M
14
th part escapes, so remaining mass in the vessel is
2 13M M4
1
1
3 M273 60 4T M
0T 444 K 171 C
41. (B) Let initial conditions = V, Tfinal conditions = V', T'By Charle's law, V T ( P remains constant )
V V ' V V' V ' 1.2 VT T' T 1.2 T
But as per question, volume is reduced by 10%means V' = 0.9 Vso percentage of volume leaked out
%1.2 0.9 V100 25%
1.2 V
42. (D) Pressure F 1 PPA A t
( P = change in momentum)
43. (A) 2P3
( Energy per unit volume )
2 E 2 E3 V 3
44. (B)2
rmsrms
3P P3
314
45. (A) 1 1 2rms rms
2 2 1
P M3P 3PV P M M P M
46. (C)3E PV2
2EP3V
47. (B) The average kinetic energy
1 1B
2 2
3 E TE k T2 E T
48. (D) Kinetic energy is a function of temperature.
49. (A) 1 1
2 2
E TE TE T
50. (C) Kinetic energy of ideal gas depends only on its temperature. Hence, it remains constantwhether pressure is increased or decreased.
51. (B) Kinetic energy is directly proportional to temperature. Hence if temperature is doubled,kinetic energy will also be doubled.
52. (B) Average kinetic energy Temperature 1 1
2 2
E TE T
53. (C)1 1
2 2
E TE TE T
54. (D) 1 1
2 2
E TE TE T
55. (B) Average translational K.E. of a molecule is = B3 k T2
At 300 K, average K.E. = 216.21 10 J
At 600 K average K.E. = 212 6.21 10
= 2112.42 10 J
We know that rms = B3k Tm
At 300 K, rms = 484 ms–1
At 600 K, rms = 12 484 684 ms
56. (B) Average translational K.E. of a molecules B3 k T2
(Where, kB = Boltzmann's constant )This is same, for all gases at same temperature.
57. (A) At 0 K Kinetic energy is zero.
315
58. (C)3E RT E T2
E' T 'E T
59. (B) E T
60. (B) B3 k T 1 eV2
19
323
B
2 eV 2 1.6 10T 7.7 10 K3 k 3 1.38 10
61. (C) rms = o
3RTM rms
o
1M
62. (B) rms T
63. (C) rms = 2 2 2 2 2
1 2 3 4 5
5
= 4.24
64. (A) rms T
rms 22
rms 11
TT
65. (D) rms velocity does not depend on pressure.
66. (C) rms = o o o
3RT RT RT3 1.73M M M
67. (B) rms T , To double the rms speed temperature should be made four times i.e.
2 1T 4T
68. (A) 1 2
2 1
3P 16 4 :11rmsv
69. (C) rms velocity doesn't depend on pressure, it depends upon temperature only.
2rms rms rms
3RT= T TMo
1 1
2 2
TT
70. (B) rms 1o 1
3RT 1=M M
and 22
1M
1 2
2 1
MM
71. (D) 2rms rms rms
3RT= T TMo
72. (C) rms o rmso
3RT T M ( , R constant)M
316
73. (B) rms rmso o
3RT TM M
2
2 2
He HHe
H H He
T M57 T M
He
25 4T 27349 2
0273 K 0 C
74. (C) rmso
3RTM
o 22rms
3 RT 3 8.3 300M1920
32 10 kg 2g
Gas is hydrogen.
75. (C) rms speed, rms = 2 2 2
1 2 3
3
average speed, 1 2 3
3
76. (B) As
rms t
rms 0O
TT
and rms speed increases by 10%
273 t1.1273
or 2273 t 1.1 1.21273
0t 273 1.21 1 57.3 C
77. (A) rms T 2 2
1 1
TT
78. (C) rms oo
3RT T MM
( rms , R constant)
2 2
2 2
ON N
O O O
MTT M
79. (C) rmso
3RTM
% increase in
2 1
o orms
1
o
3RT 3 RTM M 20 17.32100% 100% 15.5%
17.323RTM
317
80. (A)A B A B
A B A B
T T T T4 2M M M M
A B
A B
3 RT 3 RT2M M
AA B
B
2 2
81. (D) Since temperature is constant.so vrms remains same.
82. (C) rms oo
3 RT= T MM
( rms , R Const. )
2 2
2 2
oO O
N o N
MTT M
83. (C) rms T
rms 11
rms 22
TT
84. (B) Brms
3k Tm
-1
2rms m
85. (B) At a given temperature rms1
86. (D) rms T and rms rms2 1
12
rms 2
rms 01
T 1T 2
273 t 1273 0 2
273 t 1 273t 273273 4 4
= 68.25-273 = -204.750 C
87. (B) rms3 RTMo
T2
rms rms1 2o o
3R3 RT ;M 2M
o
3RT4M
o
1 3RT2 4M
1
2rmsv
= 150 ms–1
318
88. (A) rmso
3 RTM
1rms -3
3 8.314 2903RT= = 508.24 msMo 28×10
here rms rms1 2
2 2 2
2 2
O SO SO
o oO SO
T T T300M M 32 64
2Tso 600 K 0327 C
89. (B) rms T 1 1
2 2
TT
vv
90. (C)
rms He
rms Ar
o He
o Ar
3RTM 40 10 3.16
43RTM
91. (D) rms T 2 2
1 1
TT
92. (A) rmso
1M
So
2 2
2 2
rms oO H
rms oH O
M=
M
93. (D) He Hrms
H He
1 m m m
94. (C) rms = o0
3RT T MM
oHe He
H o H
MT =T M
95. (A) rms =
22 2
22 2
rms O oO H
o rms H oH O
T × M3 RT =M T × M
96. (B) rms3P=
or 2
rmsP =3
319
97. (C) 2 2rms
o 1 1
3RT T 273+90= = = =1.1M T 273+27
% increase = 2
1
1 ×100% = 0.1×100% =10%
98. (B)
o1 2rms
o 2 o 1
M1 =M M
o 2
o 2
M1 = M = 16322
Hence the gas is 4CH
99. (B) Escape velocity from the earth's surface is 11.2 kms–1
So, rms = Vescape 2
escape o
o
M3RT TM 3R
100. (C) B2
k T=2πPd
-23
2-5 -10
1.38×10 290=
2 3.14 2.026×10 2×10 -7= 1.11 10 m
Collision Frequency = no. of collision per second rms
9-7
508.24= = 4.58×101.11×10
1rms -3
o
3 8.314 2903RT= = 508.24 msM 28×10
101. (A) B2 2
1 k T=2πnd 2 πPd
86.65 10 m
102. (A) A monoatomic gas molecule has only three translational degrees of freedom.103. (D) A diatomic molecule has three translational and two rotational degrees of freedom. Hence
total degrees of freedom, f = 3+2 = 5104. (B) For a triatomic gas f = 6 ( 3 translation + 3 rotational )
105. (C)21p
v
CC f
106. (B) Degrees of freedom = 3 ( translatory ) + 2 ( rotatory ) + 1 ( vibratory ) = 6
21p
v
CC f
= 216
= 1 41 1.333 3
107. (A) Neon gas is mono atomic and for mono atomic gases v
3C R2
108. (C) When Cp and Cv are given with caloric and R with Joule then Cp – Cv = RJ
320
109. (B) p v p vC C R C R C 3R = R+ R =32 2f R f
5 R2
110. (B) vR 3C = = 1.5 R = R
0.67 2This is the value for mono atomic gases
111. According to the equilibrium theorem, the molar heat capacities should be independent of temperatureHow ever, variations in Cv and Cp are observed as the temperature changes. At very hightemperatures, vibrations are also inportant and that affects the values of Cv and Cp for diatomicand poly atomic gases. Here in the question according to given information (D) may be correctanswer.
112. (B)
113. (C)
5 73 51 1 2 2
5 73 51 2
mix1 2
5 71 2 3 5
1 1µ µ1 1 31 1 1.5µ µ 1 1 2
1 1 1 1
114. (A) 2 2 1 21
2 1 1f f
f f
115. (D) molar specific heat at constant pressure, p7C R2
Since p v v pC C R C C R 7 R R2
5 R2
116. (D)75
for a diatomic gas.
117. (C) 2 1 2 1As tan > tan 2 1
T TP P
Also from PV µRT , T VP 2 1V V
118. (B) 1 2 1 2As tan < tan 1 2
V VT T
from PV = µRT ; V 1T P Hence 1 2
1 2
1 1 P PP P
119. (C) For an ideal gas PV = constanti.e PV doesn't vary with V
120. (D) At constant temperature, PV = constant 1PV
321
Unit - 10Ocsillations And
Waves
322
SUMMARY
1. Waves : The motion of the disturbance in the medium (or in free space) is called wave pulse orgenerally a wave.
2. Amplitude of a wave : Amplitude of oscillation of particles of the medium is called the amplitude ofa wave.
3. Wavelength and frequency : The linear distance between any two points or particles having phasedifference of 2 rad is called the wavelength (λ) of the wave.
Frequency of wave is just the frequency of oscillation of particles of the medium. Relation betweenwavelength and frequency :
v = f, = k
where, v is the speed of wave in the medium.
4. Mechanical waves : The waves which require elastic medium for their transmission are calledmechanical waves, e.g. sound waves.
5. Transverse and longitudinal waves : Waves in which the oscillations are in a direction perpendicularto the direction of wave propagation are called the transverse wave.Waves in which the oscillations of the particles of medium are a!cng the direction of wave propagationare called longitudinal waves.
6. Wave Equation : The equation which describe the displacement for any particle of medium at a requiredtime is called wave equation. Various forms of wave equations are as follows :
(i) y A sin ( t - kx) (ii) t xy A sin –T
(iii) xy A sin 2 t –v
(iv) 2y A sin vt x
The above equations are for the wave travelling in the direction of increasing value of x. If the waveis travelling in the direction of decreasing value of x then put '+' instead of '—' in above equations.
7. The elasticity and inertia of the medium are necessary for the propagation of the mechanical waves.
8. The speed of the transverse waves in a medium like string kept under tension, Tv
where, T = Tension in the string and (I = mass per unit length of the string -- y
9. Speed of sound waves in elastic medium, Ev
where, E = Elastic constant of a medium, = Density of the medium.
Speed of longitudinal waves in a fluid, B Pv
323
where, B — Bulk modulus of a medium p
V
Cy
C = 1.41 (for air)
Speed of longitudinal waves in a linear medium like a rod, v
where, = Young modulus, = Density of a medium
At constant pressure and constant humidity, speed of sound waves in gas is directly proportional tothe square root of its absolute temperature.
RTv v TM
The speed of sound in a gas does not depend on the pressure variation.10. Principle of Superposition : When a particle of medium comes under the influence of two or more
waves simultaneously, its net displacement is the vector sum of displacement that could occur underthe influence of the individual waves.
11. Stationary Waves : When two waves having same amplitude and frequency and travelling in mutuallyopposite directions are superposed the resultant wave formed loses the property of propagation. Sucha wave is .called a stationary wave.
Equation of stationary wave : y = – 2 A sinkx cos tAmplitude of stationary wave : 2 A sin kx
Position of nodes in stationary wave xn = n2
where, n = 1,2, 3.....At all these points the amplitude is zero.Position of antinodes in stationary wave's,
xn = (2n – 1) 4
where , n — 1, 2, 3,....
The amplitude of all these points is 2A.12. Frequencies corresponding to different normal modes of vibration in a stretched string of length L
fixed at both the ends are given by,
nnv n Tf 2L 2L
where n — 1, 2, 3......
13. In a closed pipe the values of possible wavelengths required for stationary wave pattern are given by.
4Ln (2n - 1)
and possible frequencies, n 1vf (2n – 1) (2n – 1) f
4L
where, n = 1, 2, 3,..... and L = length of pipe.In a closed pipe only odd harmonics f1, 3f1, 5f1 .... are possible.
324
14. In an open pipe the values of possible wavelength required for stationary waves are given by,
2Ln n
and possible frequencies, n 1nvf nf2L
where, n — 1, 2, 3,......and
L - length of pipe.In open pipe of the harmonics like f1, 2f1, 3f1 ..... are possible.
15. Beat: The phenomenon of the loudness of sound becoming maximum periodically due to superpositionof two sound waves of equal amplitude and slightly different frequencies is called the 'beats'.Number of beats produced in unit time = f1 – f2.
16. Doppler Effect : Whenever there is a relative motion between a source of sound and a listener withrespect to the medium in which the waves are propagating the frequency of sound experienced by thelistener is different from that which is emitted by the source. This phenomenon is called Dopplereffect.
Frequency listened by the listener, L
LS
S
v vf fv v
Where, v = velocity of sound, vL = velocity of a listener,vS = velocity of a source, fS = frequency of sound emitted by the source.
17. If a body repeats its motion along a certain path, about a fixed point, at a definite interval of time, it issaid to have periodic motion.
18. If a body moves to and fro, back and forth, or up and down about a fixed point in a fixed interval oftime, such a motion is called an oscillatory motion.
19. When a body moves to and fro repeatedly about an equilibrium position under a restoring force, whichis always directed towards equilibrium position and whose magnitude at any instant is directlyproportional to the displacement of the body from the equilibrium position of that instant then such amotion is known as simple harmonic motion.
20. The maximum displacement of the oscillator on cither side of mean position is called amplitude of theoscillator.
21. The time taken by the oscillator to complete one oscillation is known as periodic time or time period orperiod (T) of the oscillator.
22. The number of oscillation completed by the simple harmonic oscillator in one second is known as itsfrcquency(f).
23. 2 times the frequency of oscillator is the angular frequency CO of that oscillator..
24.1 2 1 2T or f or = f T T
25. For simple harmonic motion, the displacement y(t} of a particle from its equilibrium position is representedby sine, cosine or its linear combination like
y( t ) = A sin ( t + )y(t) = B cos ( t + )y( t ) A' sin t + B' cos twhereA' Acos and B' = Bsin
325
26. The velocity of SHO is given by 2 2v A – y
27. The acceleration of SHO is given by a = -2y
28. A particle of mass m oscillating under the influence of Hook's Law exhibits simple harmonic motionwith
k ;m
mT 2k
29. Differential equation for SHM is 2
2d y + y = 0dth
30. For scries combination of n spring of spring constants k1, k2, k3,..., kn, the equivalent spring constant is
1 2 n
1 1 1 1 = = ...k k k k
the periodic time mT 2k
31. For parallel combination of n springs of spring constants k ky k ...., kn, the equivalent springconstant is
k = k1 + k2 + k3 + .... + kn and period mT 2k
32. The kinetic energy of the SHO is K = 21 m2
(A2 - y2)
33. The potential energy of the SHO is U = 21 ky2
34. The total mechanical energy of SHO is E = K + U = 21 m2
A2 = 21 kA2
35. For SHO, at y — 0, the potential energy is minimum (U = 0) and the kinetic energy is maximum
21( K kA E )2
36. For SHO, at y = A, the potential energy is maximum 21(U kA E )2
and the kinetic energygy
is minimum (K = 0)
37. Simple harmonic motion is the projection of uniform circular motion on a diameter of the referencecircle.
326
38. For simple pendulum, for small angular displacement
1T = 2 andg
2 g 2 f = T l
39. For simple pendulum, T is independent of the mass of the bob as well as the amplitude of the oscillaions.40. The differential equaiton for damped harmonic oscillation is
with the displacement
2
2d y dym b = + ky = 0
dtdt
and angular frequency 2
2k b' – m 4m
41. 2 –etlm1E( t ) kA e2
gives the mechanical energy of damped oscillation at time t.
42. A system oscillates under the influence of external periodic force are forced oscillations.43. The differential equation for forced oscillations is
02
2d y b dy k F + y = sin t
m dt m mdt
44. The amplitude for forced oscillation is
01
2 2 2 2 2 20
FA [m ( – ) + b ]
327
MCQFor the answer of the following questions choose the correct alternative from among the given ones.
SECTION – I:
1. If the equation for a particle performing S.H.M. is given by y = Sin2t + 3 Cos2t, its periodic time
will be ………..s.
(A) 21 ( B ) p (C) 2p (D) 4p.
2. The distance travelled by a particle performing S.H.M. during time interval equal to its periodic timeis ……..
(A) A (B) 2A (C) 4A (D) Zero.
3. A person standing in a stationary lift measures the periodic time of a simple pendulum inside the liftto be equal to T. Now, if the lift moves along the vertically upward direction with an acceleration of
3g
, then the periodic time of the lift will now be ………
(A) T3 (B) T23
(C) 3T
(D) 3T
4. If the equation for displacement of two particles executing S.H.M. is given by y1 = 2Sin(10t+è)and y2 = 3Cos10t respectively, then the phase difference between the velocity of two particleswill be ………..
(A) – è (B) è (C) 2 (D) 2
.
5. When a body having mass m is suspended from the free end of twosprings suspended from a rigid support, as shown in figure, its periodictime of oscillation is T. If only one of the two springs are used, thenthe periodic time would be ………
(A) 2T
(B) 2T
(C) T2 (D) 2T
6. If the maximum velocity of two springs ( both has same mass ) executing S.H.M. and having forceconstants k1 and k2 respectively are same, then the ratio of their amplitudes will be ……….
(A) 2
1
kk
(B) 1
2
kk
(C) 2
1
kk
(D) 1
2
kk
328
7. As shown in figure, two masses of 3.0 kg and 1.0 kg are attached at the two ends of a spring havingforce constant 300 N m- 1 . The natural frequency of oscillation for the system will be …………hz.( Ignore friction )
(A) ¼ (B) 1/3(C) 4 (D) 3
8. The bob of a simple pendulum having length ‘ l’ is displaced from its equilibrium position by an angleof è and released. If the velocity of the bob, while passing through its equilibrium position is v, thenv = ………..
(A) )1(2 Cosgl (B) )1(2 Singl
(C) )1(2 Singl (D) )1(2 Cosgl
9. If 41
of a spring having length l is cutoff, then what will be the spring constant of remaining part?
(A) k (B) 4k (C) 34k
(D) 43k
10. The amplitude for a S.H.M. given by the equation x = 3Sin3pt + 4Cos3pt is ………m.(A) 5 (B) 7 (C) 4 (D) 3.
11. When an elastic spring is given a displacement of 10mm, it gains an potential energy equal to U. Ifthis spring is given an additional displacement of 10 mm, then its potential energy will be ……..(A) U (B) 2U (C) 4U (D) U/4.
12. The increase in periodic time of a simple pendulum executing S.H.M. is ………….when its lengthis increased by 21%.(A) 42 % (B) 10% (C) 11% (D) 21%.
13. A particle executing S.H.M. has an amplitude A and periodic time T. The minimum time required by
the particle to get displaced by 2A
from its equilibrium position is …….. s.
(A) T (B) T/4` (C) T/8 (D) T/16.14. If a body having mass M is suspended from the free ends of two springs A and B, their periodic time
are found to be T1 and T2 respectively. If both these springs are now connected in series andif the same mass is suspended from the free end, then the periodic time is found to be T.Therefore …………..
(A) T = T1 + T2 (B) 21
111TTT
(C) T2 = T12 + T2
2 (D) 22
21
2
111TTT
.
3kg 1kg
329
15. The displacement of a S.H.O. is given by the equation x = A Cos ( ùt + 8
). At what time will it attain
maximum velocity?
(A)
83
(B)
38
(C)
163
(D)
16 .
16. At what position will the potential energy of a S.H.O. become equal to one third its kinetic energy?
(A) 2A
(B) 2A
(C) 3A
(D) A3 .
17. Three identical springs are shown in figure. When a 4 kgmass is suspended from spring A, its length increases by1cm. Now if a 6 kg mass is suspended from the free endof spring C, then increase in its length is ………cm.(A) 1.5 (B) 3.0(C) 4.5 (D) 6.0.
18. For particles A and B executing S.H.M., the equation for displacement is given by y1 =0.1Sin(100t+p/3) and y2 = 0.1Cospt respectively. The phase difference between velocity ofparticle A with respect to that of B is …………
(A) 3
(B) 6
(C) 6
(D) 3
19. The periodic time of a simple pendulum is T1. Now if the point of suspension of this pendulumstarts moving along the vertical direction according to the equation y = kt2, the periodic time
of the pendulum becomes T2. Therefore, 22
21
TT
= …( k = 1 m/s2 & g= 10 m/s2 )
(A) 6/5 (B) 5/6 (C) 4/5 (D) 1
20. A hollow sphere is filled with water. There is a hole at the bottom of this sphere. This sphere issuspended with a string from a rigid support and given an oscillation. During oscillation, thehole is opened up and the periodic time of this oscillating system is measured. The periodictime of the system………….
(A) will remain constant
(B) Will increase upto a certain time
(C) Increases initially and then decreases to attain its initial periodic time
(D) Initially decreases and then will attain the initial periodic time value.
330
21. The periodic time of a S.H.O. oscillating about a fixed point is 2 s. After what time will the kineticenergy of the oscillator become 25% of its total energy?
(A) 1/12 s (B) 1/6 s (C) ¼ s (D) 1/3 s.
22. A body having mass 5g is executing S.H.M. with an amplitude of 0.3 m. If the periodic time of the
system is 10
s, then the maximum force acting on body is ……….
(A) 0.6 N (B) 0.3 N (C) 6 N (D) 3 N
23. As shown in figure, a body having mass m is attached with two springs having spring constants k1and k2. The frequency of oscillation is f. Now, if the springs constants of both the springs areincreased 4 times, then the frequency of oscillation will be equal to ………….
(A) 2f (B) f/2
(C) f/4 (D) 4f
24. The figure shows a graph of displacement versus time for a particle executing S.H.M. The acceleration
of the S.H.O. at the end of time t = 34
second is ………..cm.s – 2
(A) 2
323 (B)
32
2
(C) 32
2(D) 2
323
25. As shown in figure, the object having mass M is executing S.H.M. with an amplitude A. Theamplitude of point P shown in figure will be …….
(A) 2
1
kAk
(B) 1
2
kAk
(C) 21
1
kkAk (D)
21
2
kkAk
26. A particle is executing S.H.M. between x = - A and x = +A. If the time taken by the particle to travelfrom x = 0 to A/2 is T1 and that taken to travel from x = A/2 to x = A is T2, then ……….(A) T1 < T2 ( B ) T1 > T2 (C) T1 = 2T2 (D) T1 = T2
27. For a particle executing S.H.M., when the potential energy of the oscillator becomes 1/8 the maximumpotential energy, the displacement of the oscillator in terms of amplitude A will be ……….
(A) 2A
(B) 22A
(C) 2A
(D) 23A
.
331
28. The average values of potential energy and kinetic energy over a cycle for a S.H.O. will be……………….. respectively.
(A) 0 , 22
21 Am (B) 22
21 Am , 0
(C) 22
21 Am , 22
21 Am (D) 22
41 Am , 22
41 Am .
29. The ratio of force constants of two springs is 1:5. The equal mass suspended at the free ends of bothsprings are performing S.H.M. If the maximum acceleration for both springs are equal, the ratio ofamplitudes for both springs is ………
(A) 51
(B) 51
(C) 15
(D) 15
30. When a mass M is suspended from the free end of a spring, its periodic time is found to be T. Now,if the spring is divided into two equal parts and the same mass M is suspended and oscillated, theperiodic time of oscillation is found to be T’. Then ………..
(A) T < T’ (B) T = T’ (C) T > T’ (D) Nothing can be said.
31. The periodic time of two oscillators are T and 45T
respectively. Both oscillators starts their oscillation
simultaneously from the mid point of their path of motion. When the oscillator having periodic time Tcompletes one oscillation, the phase difference between the two oscillators will be ………
(A) 900 (B) 1120 (C) 720 (D) 450
32. A rectangular block having mass m and cross sectional area A is floating in a liquid having density r.If this block in its equilibrium position is given a small vertical displacement, its starts oscillating withperiodic time T. Then in this case…..
(A) mT 1 (B) T (C) A
T 1 (D) 1T
33. As shown in figure, a spring attached to the ground vertically has a horizontalmassless plate with a 2 kg mass in it. When the spring ( massless ) is pressedslightly and released, the 2 kg mass, starts executing S.H.M. The force constantof the spring is 200 N m- 1. For what minimum value of amplitude, will the massloose contact with the plate? ( Take g = 10 ms – 2 )
(A) 10.0 cm (B) 8.0 cm
(C) 4.0 cm (D) For any value less than 12.0 cm.
332
34. Which of the equation given below represents a S.H.M.?
(A) acceleration = - k ( x + a ) (B) acceleration = k ( x + a )
(C) acceleration = kx (D) acceleration = - k0x + k1x2
{ Here k, k0 and k1 are force constants and units of x and a is meter }
35. The displacement for a particle performing S.H.M. is given by x = A Cos( ùt + Ô). If the initialposition of the particle is 1 cm and its initial velocity is p cm s- 1 , then what will be its initial phase? Theangular frequency of the particle is p s-1.
(A) 42
(B) 47
(C) 45
(D) 43
36. Two simple pendulums having lengths 144 cm and 121 cm starts executing oscillations. At sometime, both bobs of the pendulum are at the equilibrium positions and in same phase. After how manyoscillations of the shorter pendulum will both the bob’s pass through the equilibrium position andwill have same phase?
(A) 11 (B) 12 (C) 21 (D) 20
37. The maximum velocity and maximum acceleration of a particle executing S.H.M. are 1 m/s and 3.14m/s2 respectively. The frequency of oscillation for this particle is ……..
(A) 0.5 s- 1 (B) 3.14 s- 1 (C) 0.25 s- 1 (D) 2 s- 1
38. A particle having mass 1 kg is executing S.H.M. with an amplitude of 0.01 m and a frequency of 60hz. The maximum force acting on this particle is ……….. N
(A) 144p2 (B) 288p2 (C) 188p2 (D) None of these.39. A simple pendulum having length l is given a small angular displacement at time t = 0 and released.
After time t, the linear displacement of the bob of the pendulum is given by ………………
(A) x = aSin2p tgl
(B) x = aCos2p tlg
(C) x = aSin tlg
(D) x = aCos tlg
40. Two masses m1 and m2 are attached to the two ends of a massless spring having force constant k.When the system is in equilibrium, if the mass m1 is detached, then the angular frequency of mass m2
will be ………….
(A) 1m
k(B)
2mk
(C) 12
mmk
(D) 21 mm
k
41. When the displacement of a S.H.O. is equal to A/2, what fraction of total energy will be equal tokinetic energy? { A is amplitude }
(A) 2/7 (B) ¾ (C) 2/9 (D) 5/7
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42. The speed of a particle executing motion changes with time according to the equation y = aSinùt +bCosùt, then ……..(A) Motion is periodic but not a S.H.M.(B) It is a S.H.M. with amplitude equal to a+b(C) It is a S.H.M. with amplitude equal to a2 + b2
(D) Motion is a S.H.M. with amplitude equal to 22 ba .
43. A body is placed on a horizontal plank executing S.H.M. along vertical direction. Its amplitude ofoscillation is 3.92 x 10 – 3m. What should be the minimum periodic time so that the body does notloose contact with the plank?(A) 0.1256 s (B) 0.1356 s (C) 0.1456 s (D) 0.1556 s
44. If the kinetic energy of a particle executing S.H.M. is given by K = K0 Cos2ùt, then the displacementof the particle is given by ……….
(A) tSinmK
2/1
20
(B) tSinm
K
2/1
202
(C) tSinmK
2/1
0
22
(D) tSin
mK
2/102
45. The equation for displacement of two identical particles performing S.H.M. is given byx1 =4Sin(20t+p/6) and x2 =10Sinùt. For what value of ù will both particles have same energy?(A) 4 units (B) 8 units (C) 16 units (D) 20 units
46. A spring having length l and spring constant k is divided into two parts having lengths l1 and l2. If l1
= nl2, the force constant of the spring having length l2 is ……….
(A) k(1+n) (B) k
nn1
(C) k (D) )1( nk
47. When a mass m is suspended from the free end of a massless spring having force constant k, itsoscillates with frequency f. Now if the spring is divided into two equal parts and a mass 2m issuspended from the end of anyone of them, it will oscillate with a frequency equal to ………….
(A) f (B) 2f (C) 2f
(D) f2
48. A mass m on an inclined smooth surface is attached to two springs as shown in figure. The otherends of both springs are attached to rigid surface. If the force constant of both spring is k, then theperiodic time of oscillation for the system is ………
(A) 2/1
22
kM (B)
2/122
kM
(C) 2/1
22
kMgSin (D)
2/122
kMg
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49. A body of mass 1 kg suspended from the free end of a spring having force constant 400 Nm-1 isexecuting S.H.M. When the total energy of the system is 2 joule, the maximum acceleration is………ms – 2 .(A) 8 ms – 2 (B) 10 ms – 2 (C) 40 ms – 2 (D) 40 cms – 2
50. When a block of mass m is suspended from the free end of a massless spring having force constantk, its length increases by y. Now when the block is slightly pulled downwards and released, it startsexecuting S.H.M with amplitude A and angular frequency ù. The total energy of the system comprisingof the block and spring is ……….
(A) 22
21 Am (B) 222
21
21 kyAm (C) 2
21 ky (D) 222
21
21 kyAm .
51. A spring is attached to the center of a frictionless horizontal turn table and at the other end a body ofmass 2 kg is attached. The length of the spring is 35 cm. Now when the turn table is rotated with anangular speed of 10 rad s – 1 , the length of the spring becomes 40 cm then the force constant of thespring is ……….. N/m.(A) 1.2 x 103 (B) 1.6 x 103 (C) 2.2 x 103 (D) 2.6 x 10 3
52. As shown in figure (a) and (b), a body of mass m is attachedat the ends of the spring system. All springs have the samespring constant k. Now when both systems oscillates alongvertical direction, the ratio of their periodic time is ……..(A) ¼ (B) ½(C) 2 (D)4
53. A simple pendulum is executing S.H.M. around point O between the end points B and C with aperiodic time of 6 s. If the distance between B and C is 20 cm then in what time will the bob movefrom C to D? Point D is at the mid-point of C and O.(A) 1 s (B) 2 s (C) 3 s (D) 4 s
54. A small spherical steel ball is placed at a distance slightly away from the center of a concave mirrorhaving radius of curvature 250 cm. If the ball is released, it will now move on the curved surface.What will be the periodic time of this motion? Ignore frictional force and take g = 10 m/s2.
(A) s4
(B) p s (c) s2
(D) 2p s
55. Two identical springs are attached at the opposite ends of a rod having length l and mass m. Therod could rotate about its mid-point O as shown in figure. Now, if the point A of the rod ispressed slightly and released, the rod starts executing oscillatory motion. The periodic time of thismotion is ………
(A) k
m2
2 (B) km22
(C) km
32
(D) km
23
l2
l2
k
k
335
56. A simple pendulum having length l is suspended at the roof of a train moving with constant acceleration‘a’ along horizontal direction. The periodic time of this pendulum is ………
(A) glT 2 (B) ag
lT
2 (C) aglT
2 (D) 222ag
lT
.
57. A trolley is sliding down a frictionless slope having inclination è. If a simple pendulum is suspended
on top of this trolley, its periodic time is given by effglT 2 , where gefff = ………..
(A) g (B) g sinθ (C) g cosθ (D) g tanθ58. One end of a massless spring having force constant k and length 50 cm is attached at the upper end
of a plane inclined at an angle è = 300.When a body of mass m = 1.5 kg is attached at the lower endof the spring, the length of the spring increases by 2.5 cm. Now, if the mass is displaced by a smallamount and released, the amplitude of the resultant oscillation is ………..
(A) 7
(B) 72
(C) 5
(D) 52
59. Two blocks A and B are attached to the two ends of aspring having length L and force constant k on ahorizontal surface. Initially the system is in equilibrium.Now a third block having same mass m, moving withvelocity v collides with block A. In this situation………
(A) During maximum contraction of the spring, the kinetic energy of the system A-B will be zero.
(B) During maximum contraction of the spring, the kinetic energy of the system A-B will be mv2 / 4
(C) Maximum contraction of the spring is kmv
(D) Maximum contraction of the spring is kmv 2
60. The displacement of a particle executing S.H.M. is given by y = 4Cos2(t/2)Sin1000t. This displacementis due to superposition of ………… S.H.M.’s.
(A) 2 (B) 3 (C) 4 (D) 5
61. The displacement of a particle is given by x = ACost. Which of the following graph representsvariation in potential energy as a function of timet and displacement x.
(A) I, III (B) II, IV
(C) II, III (D) I, IV
PE
Ot
v
336
62. A system is executing S.H.M. The potential energy of the system for displacement x is E1 and for adisplacement of y, the potential energy of the system is E2. The potential energy for a displacementof (x+y) is ………
(A) E1 + E2 (B) 22
21 EE (C) E1 + E2 + 2 21EE (D) 21EE
63. A system is executing S.H.M. with a periodic time of 4/5 s under the influence of force F1. When aforce F2 is applied, the periodic time is (2/5) s. Now if F1 and F2 are applied simultaneously alongthe same direction, the periodic time will be ………
(A) 554
(B) 545
(C) 548
(D) 558
64. The periodic time of a simple pendulum is 3.3 s. Now if the point of support of the pendulum startsmoving along the vertically upward direction with a velocity v = kt ( where k = 2.1 m/s2 ), then thenew periodic time is …………… s. { Take g = 10 m/s2 }
(A) 3 (B) 2.5 (C) 3.33 (D) 2.33
65. A block is placed on a horizontal table. The table executes S.H.M. along the horizontal plane with aperiod T. The coefficient of static friction between the table and block is µ. The maximum amplitudeof oscillation should be ………so that the block does not slide off the table.
(A) 5gT
(B) 2
2
4gT
(C) 2gT
(D) µgT
66. As shown in figure, a block A having mass M is attached to one end of a massless spring. The blockis on a frictionless horizontal surface and the free end of the spring is attached to a wall. Anotherblock B having mass ‘m’ is placed on top of block A. Now on displacing this system horizontallyand released, it executes S.H.M. What should be the maximum amplitude of oscillation so that Bdoes not slide off A? Coefficient of static friction between the surfaces of the block’s is µ.
(A) Amax = kmg
(B) A max = kgMm )(
( C) Amax = kgmM )(
(D) AAmax = kgmM )(2
337
U(x)U(x)
U(x)U(x)
67. A particle is executing S.H.M. about the origin at x = 0. Which of the following graph showsvariation in potential energy with displacement?
(A) (B)
(C) (D)
68. A horizontal plank is executing SHM along the vertical direction with angular frequency ù. A coin isplaced on top of this plank. If the amplitude of oscillation is increased gradually, for what maximumamplitude will the coin be on the verge of loosing contact with the plank?
(A) When is plank is at its maximum height (B) When the plank is at the midpoint.
(C) When the amplitude is 2g
(D) When the amplitude is 2
2
g
SECTION : II
Assertion – Reason type questions :
Note:
For the following questions, statement as well as the reason(s) are given. Each questions has four options.Select the correct option.
(a) Statement – 1 is true, statement- 2 is true; statement-2 is the correct explanation of statement – 1 .
(b) Statement – 1 is true, statement- 2 is true but statement-2 is not the correct explanation of statement– 1 .
(c) Statement – 1 is true, statement- 2 is false
(d) Statement – 1 is false, statement- 2 is true
(A) a (B) b (C) c (D) d
69. Statement – 1 : If a spring having spring constant k is divided into equal parts, then the springconstant of each part will be 2k.
338
Statement – 2 : When the length of the elastic spring is increased ( stretched ) by x, then the amount
of work required to be done is 2
21 kx
(A) a (B) b (C) c (D) d
70. Statement – 1 : The periodic time of a S.H.O. depends on its amplitude and force constant.Statement – 2 : The elasticity and inertia decides the frequency of S.H.O.
(A) a (B) b (C) c (D) d
71. Statement – 1 : For small amplitude, the motion of a simple pendulum is a S.H.M. with
periodic time glT 2 . For large amplitudes, periodic time is greater than g
l2 .
Statement – 2 : For large amplitude, the speed of the bob is more when it passes through themid-point ( equilibrium point ).
(A) a (B) b (C) c (D) d
72. Statement – 1 : Periodic time of a simple pendulum is independent of the mass of the bob.Statement – 2 : The restoring force does not depend on the mass of the bob.
(A) a (B) b (C) c (D) d
73. Statement – 1: The periodic time of a simple pendulum increases on the surface of moon.Statement – 2 : Moon is very small as compared to Earth.
(A) a (B) b (C) c (D) d
74. Statement – 1: If the length of a simple pendulum is increased by 3%, then the periodic timechanges by 1.5%.
Statement – 2 : Periodic time of a simple pendulum is proportional to its length.
(A) a (B) b (C) c (D) d
75. Statement – 1: For a particle executing S.H.M. with an amplitude of 0.01 m and frequency30 hz, the maximum acceleration is 36p2 m/s2.
Statement – 2 : The maximum acceleration for the above particle is + ù2A, where A is amplitude.
(A) a (B) b (C) c (D) d
76. Statement – 1 : The periodic time of a stiff spring is less than that of a soft spring.
Statement – 2 : The periodic time of a spring depends on its force constant value and for a stiffspring, it is more.
(A) a (B) b (C) c (D) d
339
77. Statement – 1 : The amplitude of an oscillator decreases with time.
Statement – 2 : The frequency of an oscillator decreases with time.
(A) a (B) b (C) c (D) d
78. Statement – 1 : For a particle executing SHM, the amplitude and phase is decided by its initialposition and initial velocity.
Statement – 2 : In a SHM, the amplitude and phase is dependent on the restoring force.(A) a (B) b (C) c (D) d
SECTION – IIICOMPREHENSION BASED QUESTIONS
NOTE: Questions 79 to 81 are based on the following passage.Passage – 1:As shown in figure, two light springs having force constants k1 = 1.8 N m – 1 and k2 = 3.2 N m – 1 and ablock having mass m = 200 g are placed on a frictionless horizontal surface. One end of both springs areattached to rigid supports. The distance between the free ends of the spring is 60 cm and the block ismoving in this gap with a speed v = 120 cm s – 1 .
v
79. When the block is moving towards spring k2, what will be the time taken for the spring to getmaximum compressed from point D?(A) p s (B) (p/2) s (C) (p/3) s (D) (p/4) s
80. When the block is moving towards k1, what will be the time taken for it to get maximum compressedfrom point C?(A) p s (B) (2/3) s (C) (p/3) s (D) (p/4) s
81. What will be the periodic time of the block, between the two springs?(A) 1+ (5p/6) s (B) 1+ (7p/6) s (C) 1+ (5p/12) s (D) 1+ (7p/12) s
NOTE: Questions 82 to 84 are based on the following passage.Passage – 2 :A block having mass M is placed on a horizontal frictionless surface. This mass is attached to one end ofa spring having force constant k. The other end of the spring is attached to a rigid wall. This systemconsisting of spring and mass M is executing SHM with amplitude A and frequency f. When the block ispassing through the mid-point of its path of motion, a body of mass m is placed on top of it, as a result ofwhich its amplitude and frequency changes to A’ and f’.
340
82. The ratio of frequencies ff '
= ………..
(A)
MmM
(B)
Mmm
(C)
'mAMA
(D)
mA
AmM ')(
83. If the velocity before putting the mass and after putting it is v and v1 respectively, then vv1
= ……….
(A)
MmM
(B)
MmM
(C) AA
mMmM 1
(D) AA
mMmM 1
.
84. The ratio of amplitudes AA1
= …….
(A)
mmM
(B)
mMm
(C)
mMM
(D)
MmM
NOTE: Questions 85 to 90 are based on the following passage.
Passage – 3:
The equation for displacement of a particle at time t is given by the equation y = 3Cos2t + 4Sin2t. .
85. The motion of the particle is ………
(A) Damped motion (B) Periodic motion (C) Rotational motion (D) S.H.M.
86. The periodic time of oscillation is ………
(A) 2 s (B) p s (C) (/2) s (D) 2p s
87. The amplitude of oscillation is ……cm
(A) 1 (B) 3 (C) 5 (D) 7
88. The maximum acceleration of the particle is ……..cm / s2.
(A) 4 (B) 12 (C) 20 (D) 28
89. If the mass of the particle is 5 gm, then the total energy of the particle is ……erg.
(A) 250 (B) 125 (C) 500 (D) 375
90. The frequency of the particle is ………s- 1 .
(A) (1/p) (B) p (C) (1/2p) (D) (p/2)
341
Waves
SECTION – I :
91. Equation for a harmonic progressive wave is given by y = A sin ( 15pt + 10px + p/3) where x is inmeter and t is in seconds. This wave is ……….
(A) Travelling along the positive x direction with a speed of 1.5 ms – 1 .
(B) Travelling along the negative x direction with a speed of 1.5 ms – 1 .
(C) Has a wavelength of 1.5 m along the – x direction.
(D) Has a wavelength of 1.5 m along the positive x – direction.
92. If the velocity of sound wave in humid air is vm and that in dry air is vd, then……
(A) vm > vd ( B ) vm < vd ( C ) vm = vd ( D ) vm >> vd
93. The ratio of frequencies of two waves travelling through the same medium is 2:5. The ratio of theirwavelengths will be ………
(A) 2:5 (B) 5:2 (C) 3:5 (D) 5:3
94. If the maximum frequency of a sound wave at room temperature is 20,000 hz then its minimumwavelength will be approximately ……… ( v = 340 ms – 1 )
(A) 0.2 A 0 ( B ) 5 A 0 ( C ) 5 cm to 2 m (D) 20 mm
95. If the equation of a wave in a string having linear mass density 0.04 kg m – 1 is given by y = 0.02
Sin
50.004.02 xt , then the tension in the string is …………..N. ( All values are in mks )
(A) 6.25 (B) 4.0 (C) 12.5 (D) 0.5
96. If the equation for a transverse wave is y = A Sin2p
x
Tt
, then for what wavelength will the
maximum velocity of the particle be double the wave velocity?
(A) 4A
(B) 2A
(C) pA (D) 2pA
97. Consider two points lying at a distance of 10 m and 15 m from an oscillating source. If the periodictime of oscillation is 0.05 s and the velocity of wave produced is 300 m/s, then what will be thephase difference the two points?
(A) p (B) p/6 (C) p/3 (D) 2p/3
342
98. A string is divided into three parts having lengths l1, l2 and l3 each. If the fundamental frequencyof these parts are f1, f2 and f3 respectively, then the fundamental frequency of the originalstring f = ……….
(A) 321 ffff (B) f = f1 + f2 + f3
(C) 321
1111ffff
(D) 321
1111ffff
99. Waves produced by two tuning forks are given by y1 = 4Sin500pt and y2 = 2Sin506pt. The numberof beats produced per minute is …….
(A) 360 (B) 180 (c) 60 (D) 3
100. Equation for a progressive harmonic wave is given by y = 8Sin2p( 0.1x – 2t), where x and y are incm and t is in seconds. What will be the phase difference between two particles of this waveseparated by a distance of 2 cm?
(A) 180 (B) 360 (C) 720 (D) 540
101. As shown in figure, two pulses in a string having center to center distance of 8 cm are travelling alongmutually opposite direction. If the speed of both the pulse is 2 cm/s, then after 2 s, the energy ofthese pulses will be ………
(A) zero
(B) totally kinetic energy
(C) totally potential energy
(D) Partially potential energy and partially kinetic energy.
102. Two waves are represented by y1 = ASinùt and y2 = aCosùt. The phase of the first wave, w.r.t. tothe second wave is ……….
(A) more by radian (B) less by p radian (C) more by p/2 (D) less by p/2
103. If the resultant of two waves having amplitude b is b, then the phase difference between the twowaves is …….
(A) 1200 (B) 600 (C) 900 (D) 1800
104. If two antinodes and three nodes are formed in a distance of 1.21 A0, then the wavelength of thestationary wave is ……….
( A ) 2.42 A0 (B) 6.05 A0 (C) 3.63 A0 (D) 1.21 A0
105. The function Sin2(ùt ) represents……
(A) A SHM with periodic time p/ù. (B) A SHM with a periodic time 2p/ù.
(C) A periodic motion with periodic time p/ù. (D) A periodic motion with period 2p/ù.
343
106. If two almost identical waves having frequencies n1 and n2, produced one after the other superposesthen the time interval to obtain a beat of maximum intensity is ……..
(A) 21
1nn (B)
21
11nn
(C) 21
11nn
(D) 21
1nn
107. When two sound waves having amplitude A , angular frequency ù and a phase difference of p/2superposes, the maximum amplitude and angular frequency of the resultant wave is ………
(A) A2 , ù (B) 2A
, 2
(C) 2A
, ù (D) ,2A 2
108. The amplitude of a wave in a string is 2 cm. This wave is propagating along the x-direction with aspeed of 128 m/s. Five such waves are accommodated in 4 m length of the string. The equation forthis wave is …….(A) y = 0.02Sin(15.7x – 2010t ) m (B) y = 0.02Sin(15.7x + 2010t ) m(C) y = 0.02Sin(7.85x – 1005t ) m (D) y = 0.02Sin(7.85x + 1005t ) m
109. A string of length 70 cm is stretched between two rigid supports. The resonant frequency for thisstring is found to be 420 hz and 315 hz. If there are no resonant frequencies between these twovalues, then what would be the minimum resonant frequency of this string?(A) 10.5 hz (B) 1.05 hz (C) 105 hz (D) 1050 hz
110. Sound waves propagates with a speed of 350 m/s through air and with a speed of 3500 m/s throughbrass. If a sound wave having frequency 700 hz passes from air to brass, then itswavelength ………….(A) decreases by a fraction of 10 (B) increases 20 times(C) increases 10 times (D) decreases by a fraction of 20
111. A transverse wave is represented by y = ASin (ùt-kx). For what value of its wavelength will thewave velocity be equal to the maximum velocity of the particle taking part in the wave propagation?(A) 2pA (B) A (C) pA (D) pA/2
112. Two monoatomic ideal gases 1 and 2 has molecular weights m1 and m2. Both are kept in twodifferent containers at the same temperature. The ratio of velocity of sound wave in gas1 and 2 is ……….
(A) 1
2
mm
(B) 2
1
mm
(C) 2
1
mm
(D) 1
2
mm
113. A wire having length L is kept under tension between x = 0 and x = L. In one experiment, the
equation of the wave and energy is given by y1 = ASin
Lx
Sinùt and E1 respectively. In another
experiment, it is y2 = ASin
Lx2
Sin2ùt and E2. Then ……
(A) E2 = E1 (B) E2 = 2E1 (C) E2 = 4E1 (D) E2 = 16E1
344
114. Twenty four tuning forks are arranged in such a way that each fork produces 6 beats/s with thepreceding fork. If the frequency of the last tuning fork is double than the first fork, then the frequencyof the second tuning fork is ………(A) 132 (B) 138 (C) 276 (D) 144
115. If two SHM’s are given by the equation y1 = 0.1 Sin(100pt + p/3) and y2 = 0.1 Cospt, then thephase difference between the velocity of particle 1 and 2 is ………(A) p/6 (B) - p/3 (C) p/3 (D) - p/6
116. The wave number for a wave having wavelength 0.005 m is …….. m – 1 .(A) 5 (B) 50 (C) 100 (D) 200
117. An listener is moving towards a stationary source of sound with a speed ¼ times the speed of sound.What will be the percentage increase in the frequency of sound heard by the listener?(A) 20% (B) 25% (C) 2.5% (D) 5%
118. When the resonance tube experiment, to measure speed of sound is performed in winter, the firstharmonic is obtained for 16 cm length of air column. If the same experiment is performed in summer,the second harmonic is obtained for x length of air column. Then ….(A) 32 > x > 16 (B) 16 > x (C) x > 48 (D) 48 > x > 32
119. What should be the speed of a source of sound moving towards a stationary listener, so that thefrequency of sound heard by the listener is double the frequency of sound produced by the source?{ Speed of sound wave is v }(A) v ( B ) 2v (C) v/2 (D) v /4
120. A metal wire having linear mass density 10 g/m is passed over two supports separated by a distanceof 1 m. The wire is kept in tension by suspending a 10 kg mass. The mid point of the wire passesthrough a magnetic field provided by magnets and an a.c. supply having frequency n is passedthrough the wire. If the wire starts vibrating with its resonant frequency, what is the frequency of a.c.supply?(A) 50 hz (B) 100 hz (C) 200 hz (D) 25 hz
121. If the listener and the source of sound moves along the same direction with the same speed, then……..
(A) s
L
ff
< 1 (B) s
L
ff
=0 (C) s
L
ff
= 1 (D) s
L
ff
>1
122. A wire of length 10 m and mass 3 kg is suspended from a rigid support. The wire has uniform crosssectional area. Now a block of mass 1 kg is suspended at the free end of the wire and a wave havingwavelength 0.05 m is produced at the lower end of the wire. What will be the wavelength of thiswave when it reached the upper end of the wire?(A) 0.12 m (B) 0.18 m (C) 0.14 m (D) 0.10 m
123. If the mass of 1 mole of air is 29 x 10 – 3 kg, then the speed of sound in it at STP is …….. ( ã=7/5).{ T = 273 K, P = 1.01 x 105 Pa }
(A) 270 m/s (B) 290 m/s (C) 330 m/s (D) 350 m/s
345
124. A wave travelling along a string is described by y = 0.005Sin(40x – 2t) in SI units. The wavelengthand frequency of the wave are………
(A) (p/5) m ; 0.12 hz (B) (p/10) m ; 0.24 hz (C) (p/40) m ; 0.48 hz (D) (p/20) m ; 0.32 hz
125. Two sitar strings A and B playing the note “Dha” are slightly out of time and produce beats offrequency 5 hz. The tension of the string B is slightly increased and the beat frequency is found todecrease to 3 hz. What is the original frequency of B if the frequency of A is 427 hz?
(A) 432 (B) 422 (C) 437 (D) 417
126. A rocket is moving at a speed of 130 m/s towards a stationary target. While moving, it emits a waveof frequency 800 hz. Calculate the frequency of the sound as detected by the target. ( Speed ofwave = 330 m/s)
(A) 1320 hz (B) 2540 hz (C) 1270 hz (D) 660 hz
127. Length of a steel wire is 11 m and its mass is 2.2 kg. What should be the tension in the wire so thatthe speed of a transverse wave in it is equal to the speed of sound in dry air at 200 C temperature?
(A) 2.31 x 104 N (B) 2.25 x 104 N (C) 2.06 x 104 N (D) 2.56 x 104 N
128. A wire stretched between two rigid supports vibrates with a frequency of 45 hz. If the mass of thewire is 3.5 x 10 – 2 kg and its linear mass density is 4.0 x 10 - 2 kg/m, what will be the tension in thewire ?
(A) 212 N (B) 236 N (C) 248 N (D) 254 N
129. Tube A has both ends open while tube B has one end closed, otherwise they are identical. The ratioof fundamental frequency of tube A and B is ……..
(A) 1:2 (B) 1:4 (C) 2:1 (D) 4:1
130. A tuning fork arrangement produces 4 beats/second with one fork of frequency 288 hz. A little waxis applied on the unknown fork and it then produces 2 beats/s. The frequency of the unknown forkis ……….hz.
(A) 286 (B) 292 (C) 294 (D) 288
131. A wave y = aSin(ùt – kx ) on a string meets with another wave producing a node at x = 0. Thenthe equation of the unknown wave is ………
(A) y = a Sin(ùt + kx ) (B) y = - aSin( ùt+kx) (C) y = aSin(ùt – kx ) (D) y = - aSin(ùt – kx)
132. When temperature increases, the frequency of a tuning fork…….
(A) Increases (B) Decreases
(C) remains same (D) Increases or decreases depending on the material.
133. A tuning fork of known frequency 256 hz makes 5 beats per second with the vibrating string of apiano. The beats frequency decreases to 2 beats/s when the tension in the piano string is slightlyincreased. The frequency of the piano string before increase in the tension was ……..hz.
(A) 256 + 2 (B) 256 – 2 (C) 256 – 5 (D) 256 + 5.
346
134. An observer moves towards a stationary source of sound with a velocity one – fifth the velocity ofsound. What is the percentage increase in the apparent frequency?(A) 5% (B) 20% (C) zero (D) 0.5%
135. The speed of sound in Oxygen ( O2) at a certain temperature is 460 m/s. The speed of sound inhelium at the same temperature will be …….…….ms- 1 . (Assume both gases to be ideal)(A) 330 (B) 460 (C) 5002 (D) None of these
136. In a longitudinal wave, pressure variation and displacement variation are………(A) In phase (B) 900 out of phase (C) 450
out of phase (D) 1800 out of phase137. A tuning fork of frequency 480 hz produces 10 beats/s when sounded with a vibrating sonometer
string. What must have been the frequency of the string if a slight increase in tension produces fewerbeats per second than before?(A) 480 (B) 490 hz (C) 460 hz (D) 470 hz
138. Which of the following functions represents a wave?
(A) ( x – vt)2 (B) ln( x + vt ) (C) 2)( vtxe (D) vtx 1
139. Two sound waves are represented by y = aSin(ùt-kx) and y = aCos(ùt-kx). The phase differencebetween the waves in water is ……..
(A) 2
(B) 4
(C) (D) 43
140. A string of linear density 0.2 kg/m is stretched with a force of 500 N. A transverse wave of length4.0 m and amplitude 1/l meter is travelling along the string. The speed of the wave is …….. m/s.(A) 50 (B) 62.5 (C) 2500 (D) 12.5
141. Two wires made up of same material are of equal lengths but their radii are in the ratio 1:2. Onstretching each of these two strings by the same tension, the ratio between their fundamental frequencyis ……..(A) 1:2 (B) 2:1 (C) 1:4 (D) 4:1
142. The tension in a wire is decreased by 19%, then the percentage decrease in frequency will be………(A) 19% (B) 10% (C) 0.19% (D) None of these
143. An open organ pipe has fundamental frequency 100 hz. What frequency will be produced if its oneend is closed?(A) 100, 200, 300, …. (B) 50, 150, 250….(C) 50, 100, 200, 300… (D) 50, 100, 150, 200,…..
144. A closed organ pipe has fundamental frequency 100 hz. What frequencies will be produced if itsother end is also opened? (A) 200, 400, 600, 800,….. (B) 200, 300, 400, 500, …..(C) 100, 300, 500, 700,…… (D) 100, 200, 300, 400, …….
347
145. A column of air of length 50 cm resonates with a stretched string of length 40 cm. The length of thesame air column which will resonate with 60 cm of the same string at the same tension is ……..
(A) 100 cm (B) 75 cm (C) 50 cm (D) 25 cm
146. Two forks A and B when sounded together produce 4 beats/s. The fork A is in unison with 30 cmlength of a sonometer wire and B is in unison with 25 cm length of the same wire at the same tension.The frequencies of the fork are …….
(A) 24 hz, 28 hz (B) 20 hz, 24 hz (C) 16 hz , 20 hz (D) 26 hz, 30 hz
147. A tuning fork of frequency 200 hz is in unison with a sonometer wire. The number of beats heard persecond when the tension is increased by 1 % is …….
(A) 1 (B) 2 (C) 4 (D) 0.5
148. A bus is moving with a velocity of 5 m/s towards a huge wall. The driver sounds a horn of frequency165 hz. If the speed of sound in air is 335 m/s, the number of beats heard per second by thepassengers in the bus will be …….
(A) 3 (B) 4 (C) 5 (D) 6
149. A vehicle with a horn of frequency n is moving with a velocity of 30 m/s in a direction perpendicularto the straight line joining the observer and the vehicle. The observer perceives the sound to have afrequency ( n + n1 ). If the sound velocity in air is 300 m/s, then……
(A) n1 = 10n (B) n1 = 0 (C) n1 = 0.1 n (D) n1 = - 0.1n
150. In a sine wave, position of different particles at time t = 0 is shown in figure. The equation for thiswave travelling along the positive x – direction can be …….
(A) y = Asin ( ω t – kx )
(B) y = Acos (kx – ω t)
(C) y = Acos( ω t – kx )
(D) y = Asin(kx – ω t)
151. Which of the following changes at an antinode in a stationary wave?
(A) Density only (B) Pressure only
(C) Both pressure and density (D) Neither pressure nor density
152. A sonometer wire supports a 4 kg load and vibrates in fundamental mode with a tuning fork offrequency 416 hz. The length of the wire between the bridges is now doubled. In order to maintainfundamental mode, the load should be changed to …….
(A) 1 kg (B) 2 kg (C) 8 kg (D) 16 kg
153. In brass, the velocity of a longitudinal wave is 100 times the velocity of a transverse wave. If Y = 1x 1011 N/m2, then stress in the wire is …………
(A) 1 x 1013 N/m2 (B) 1 x 109 N/m2 (C) 1 x 1011 N/m2 (D) 1 x 107 N/m2.
y
0 x
348
154. The frequency of tuning fork A is 2% more than the frequency of a standard fork. Frequency oftuning fork B is 3% less than the frequency of the standard fork. If 6 beats per second are heardwhen the two forks A and B are excited, then frequency of A is …… hz.(A) 120 (B) 122.4 (C) 116.4 (D) 130
155. Fundamental frequency of a sonometer wire is n. If the length and diameter of the wire are doubledkeeping the tension same, the new fundamental frequency is ………
(A) 22n
(B) 22n
(C) n2 (D) 4n
156. A car blowing its horn at 480 hz moves towards a high wall at a speed of 20 m/s. If the speed ofsound is 340 m/s, the frequency of the reflected sound heard by the driver sitting in the car will beclosest to ……..hz.
(A) 540 (B) 524 (C) 568 (D) 480
157. A cylindrical tube open at both ends has a fundamental frequency f in air. The tube is dipped verticallyin water so that half of it is in water. The fundamental frequency of the air column is now………(A) f/2 (B) f (C) 3f/4 (D) 2f
158. Three sound waves of equal amplitudes have frequencies (v-1 ), v, ( v+1). They superpose to givebeats. The number of beats produced per second will be …….
(A) 3 (B) 2 (C) 1 (D) 4
159. A wave travelling along the x-axis is described by the equation y(x,t) = 0.005Cos(áx - ât). If thewavelength and the time period of the wave are 0.08 m and 2.0 s respectively, then á and â inappropriate units are ………(A) á = 12.50p , â= p/2.0 (B) á = 25p , â= p
(C) á = 0.08/p , â= 2.0/p (D) á = 0.04/p , â= 1.0/p
160. A wave travelling along a string is described by the equation y = ASin(ùt-kx). The maximum particlevelocity is ………..
(A) Aù (B) ù/k (C) dù/dk (D) x161. A string is stretched between fixed points seperated by 75 cm. It is observed to have a resonant
frequencies of 420 hz and 315 hz. There are other resonant frequencies between these two. Thenthe lowest frequency for this string is ……..hz.
(A) 1.05 (B) 1050 (C) 10.5 (D) 105
162. Two tuning forks P and Q when set vibrating gives 4 beats/second. If the prong of fork P is filed, thebeats are reduced to 2/s. What is the frequency of P, if that of Q is 250 hz.?
(A) 246 hz (B) 250 hz (C) 254 hz (D) 252 hz
163. The length of a string tied across two rigid supports is 40 cm. The maximum wavelength of a stationarywave that can be produced in it is ………. cm.
(A) 20 (B) 40 (C) 80 (D) 120
349
164. A stationary wave of frequency 200 hz are formed in air. If the velocity of the wave is 360 m/s, theshortest distance between two antinodes is …….m
(A) 1.8 (B) 3.6 (C) 0.9 (D) 0.45
165. A tuning fork produces 8 beats per second with both 80 cm and 70 cm of stretched wire of asonometer. Frequency of the fork is …….hz.
(A) 120 (B) 128 (C) 112 (D) 240
166. An open pipe is in resonance in 2nd harmonic with frequency f1. Now one end of the tube is closedand frequency is increased to f2, such that the resonance again occurs in the nth harmonic. Choosethe correct option.
(A) n = 3, f2 = (3/4)f1 (B) n = 3, f2 = (5/4)f1 (C) n = 5, f2 = (5/4)f1 (D) n = 5, f2 = (3/4)f1
SECTION : II
Assertion – Reason type questions :
Note:
For the following questions, statement as well as the reason(s) are given. Each questions has four options.Select the correct option.
(a) Statement – 1 is true, statement- 2 is true; statement-2 is the correct explanation of statement – 1 .
(b) Statement – 1 is true, statement- 2 is true but statement-2 is not the correct explanation of statement– 1 .
(c) Statement – 1 is true, statement- 2 is false
(d) Statement – 1 is false, statement- 2 is true
167. Statement – 1: Two waves moving in a uniform string having uniform tension cannot have differentvelocities.
Statement – 2 : Elastic and inertial properties of string are same for all waves in same string.Moreover speed of wave in a string depends on its elastic and inertial properties only.
(A) a (B) b (C) c (D) d
168. Statement – 1: When a sound source moves towards observer, then frequency of sound increases.
Statement – 2 : Wavelength of sound in a medium moving towards the observer decreases.
(A) a (B) b (C) c (D) d
169. Statement – 1: Newton’s equation for speed of sound was found wrong because he assumed theprocess to be isothermal.
Statement – 2 : When sound propagates, the compressions and rarefactions happen so rapidlythat there is not enough time for heat to be distributed.
(A) a (B) b (C) c (D) d
350
170. Statement – 1 : When pressure in a gas changes, velocity of sound in gas may change.
Statement – 2 : Velocity of sound is directly proportional to square root of pressure.
(A) a (B) b (C) c (D) d
171. Statement – 1 : If wave enters from one medium to another medium then sum of amplitudes ofreflected wave and transmitted wave is equal to the amplitude of incident wave.
Statement – 2 : If wave enters from one medium to another medium some part of energy istransmitted and rest of the energy is reflected back.
(A) a (B) b (C) c (D) d
SECTION – IIICOMPREHENSION BASED QUESTIONS
NOTE: Questions 172 to 174 are based on the following passage.
Passage – 1
A string 25 cm long and having a mass of 2.5 g is under tension. A pipe closed at one end is 40 cm long.When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, 8beats per second is heard. It is observed that decreasing the tension in the string decreases the beatfrequency. The speed of sound in air is 320 ms – 1 .
172. The frequency of the fundamental mode of the closed pipe is ….. hz
(A) 100 (B) 200 (C) 300 (D) 400
173. The frequency of the string vibrating in its 1st overtone is …… hz
(A) 92 (B) 108 (C) 192 (D) 208.
174. The tension in the string is very nearly equal to ……
(A) 25 N (B) 27 N (C) 28 N (D) 30 N
NOTE: Questions 175 to 178 are based on the following passage.
Passage – 2
Standing waves are produced by the superposition of two waves y1 = 0.05Sin(3pt – 2x) and y2 =0.05Sin(3pt + 2x) where x and y are in meters and t is in seconds.
175. The speed ( in ms – 1 ) of each wave is ……
(A) 1.5 (B) 3.0 (C) 3p/2 (D) 3p
176. The distance ( in meters ) between two consecutive nodes is …….
(A) p/2 (B) p (C) 0.5 (D) 1.0
351
177. The amplitude of a particle at x = 0.5 m is ……
(A) 1.08 x 10 – 1 m (B) 5.4 x 10 –2 m (C) (p/2) x 10 – 1 m (D) p x 10 – 1 m
178. The velocity ( in ms – 1 ) of a particle at x = 0.25 m at t = 0.5 s is ……
(A) 0.1p (B) 0.3p (C) zero (D) 0.3
NOTE: Questions 179 to 181 are based on the following passage.
Passage – 3
When two sound waves travel in the same direction in a medium, the displacement of a particle located atx at time t is given by y1 = 0.05Cos(0.50px - 100pt ) & y2 = 0.05Cos ( 0.46px - 92pt ), where y1, y2 andx are in meter and t is in seconds.
179. What is the speed of sound in the medium?
(A) 332 m/s (B) 100 m/s (C) 92 m/s (D) 200 m/s
180. How many times per second does an observer hear the sound of maximum intensity?
(A) 4 (B) 8 (C) 12 (D) 16
181. At x = 0, how many times between t = 0 and t = 1 s does the resultant displacement become zero?
(A) 46 (B) 50 (C) 92 (D) 100
NOTE: Questions 182 to 183 are based on the following passage.
Passage – 4
The equation y = 10Sin 4x
Cos10t represents a stationary wave where x and y are in centimeter and t is
in seconds.
182. The amplitude of each component wave is …….
(A) 5 cm (B) 10 cm (C) 20 cm (D) between 5 cm and 10 cm.
183. The separation between two consecutive nodes is ………(A) 2 cm (B) 4 cm (C) 5 cm (D) 8 cm
352
1 (B) 41 (B) 81 (D) 121 (C) 161 (D)2 (C) 42 (D) 82 (A) 122 (D) 162 (A)3 (B) 43 (A) 83 (A) 123 (C) 163 (C)4 (C) 44 (B) 84 (C) 124 (D) 164 (C)5 (A) 45 (B) 85 (D) 125 (B) 165 (A)6 (D) 46 (A) 86 (B) 126 (A) 166 (C)7 (D) 47 (A) 87 (C) 127 (A) 167 (D)8 (A) 48 (A) 88 (C) 128 (C) 168 (A)9 (C) 49 (C) 89 (A) 129 (C) 169 (A)10 (A) 50 (B) 90 (A) 130 (B) 170 (B)11 (C) 51 (B) 91 (B) 131 (B) 171 (D)12 (B) 52 (B) 92 (A) 132 (B) 172 (B)13 (C) 53 (A) 93 (B) 133 (C) 173 (D)14 (C) 54 (B) 94 (D) 134 (B) 174 (B)15 (A) 55 (C) 95 (A) 135 (D) 175 (C)16 (A) 56 (D) 96 (C) 136 (D) 176 (A)17 (B) 57 (C) 97 (D) 137 (D) 177 (B)18 (C) 58 (A) 98 (C) 138 (C) 178 (C)19 (A) 59 (B) 99 (B) 139 (A) 179 (D)20 (C) 60 (B) 100 (C) 140 (A) 180 (A)21 (D) 61 (A) 101 (B) 141 (B) 181 (D)22 (A) 62 (C) 102 (D) 142 (B) 182 (A)23 (A) 63 (A) 103 (A) 143 (B) 183 (B)24 (D) 64 (A) 104 (D) 144 (A)25 (D) 65 (B) 105 (A) 145 (B)26 (A) 66 (B) 106 (A) 146 (B)27 (B) 67 (D) 107 (A) 147 (A)28 (D) 68 (C) 108 (C) 148 (C)29 (C) 69 (B) 109 (C) 149 (B)30 (C) 70 (D) 110 (C) 150 (D)31 (C) 71 (B) 111 (A) 151 (D)32 (C) 72 (C) 112 (A) 152 (D)33 (A) 73 (B) 113 (C) 153 (D)34 (A) 74 (C) 114 (B) 154 (A)35 (D) 75 (B) 115 (B) 155 (D)36 (B) 76 (A) 116 (D) 156 (A)37 (A) 77 (C) 117 (B) 157 (B)38 (A) 78 (C) 118 (C) 158 (C)39 (D) 79 (D) 119 (C) 159 (B)40 (B) 80 (C) 120 (A) 160 (A)
KEY NOTE
353
HINT1. y sin 2t 3 cos 2t
1 3y 2 sin 2t cos 2t2 2
2 cos sin 2t sin cos 2t 2sin(2t )
Compaling than with y A sin(wt ), we get
2w 2 2 t sT
2.
3. T 2g
When lift moves up with accleratim g3 the effective graritatianl acclenations
in 1 g 4gg g 3 3
new peliodic time '
'
lT ' 2g
4. 11
dy 2 10cos(10t )dt
22 3 10sin t 30cos(10t )
Phase diffdence = 10t 10t 2 2
5. For series combination, 1 2
S 1 21 2
K k kK ( k k K)K k 2
now '
'sk1 TT = 2 T = 2 TT kk
6. For maximum velocity, 1 1 2 2A A
1 2 2 2 21 2
2 1 1 1 1
A k / m k ( m m )A k / m k
354
7. Reduced mass of system 1 2
1 2
m m 0.75 kgm m
freq of oscillation 1 k 20f 3Hz2 m
8. KE at o = PE at A
21 m mgh mg 1 cos2
2g (1 cos )
9. 1 1 2 2K K k
1 23K K k
4 4
force constant of spring having lenght 34 l in
24k k3
10. Amplitnde of SHM given by x = a sin t+b cos t in
2 2A a + b = 1
2 2 2(3 4 ) 5m
11.2
2 2 22
1 1
u yu y u 4uu y
12. 1T l because 2 and g are constants
2 2 1
1 1 1
1.21 1.1T l lT l l
2 1
1
T T%increase 100 10%T
13. y A sin( t )
A 1Asin t 0 sin t 42 2
2 .tT 4
Tt 8
355
14.1 2 2
1 1 22 21 1 2
m 4 .M 4 MT 2 k and kk T T
for Series connection; 1 2
1 2
k kMT 2 where kk k k
12 2
1 22
MT 2 (T T )4 M
2 2 2 2 21 2 1 2T T T T T T
15. cos( ) – sin ( )8 2dxx A t v A tdt
If sin( t ) 1,8 then velocity will be maximam
3 3t t t8 2 8 8
16. 3u = k
2 2 21 1 A3 ky k(A y ) y 2 2 2
17. For spring bA restaning face F=kx
displacement Fx k
figure (b) if the resultant spring contant
1m k , then 1 1 2
1 2
k kkk k
(Series)
k2
in figure (b) if on applying a force F`, if
displacement in x`, then F 'x ' k '
x ' F ' k 6 2 3x k ' F 4
x ' 3x 3 1cm 3cm
18. 11
dydt
22 dyand v
dt
356
19. Here 2y kt
22
2
dy d y2kt 2k 2 msdt dt
the point of support in moving upwards with an accelaration of 2m2 s
effective nacc l . 2g ' g a 12 m / s
Now 1 2T 2 and T 2g g '
20. '
T 2g
as water leaks, the center of gravity moves down and hence “ ” increases.
T increases initiallyWhen all the water has leaked, the center of gravity moves up and hence “ ” decreases andhence T decreases Finally the centre of gravity steady at the center of sphde and so T will remainconstant.
21. Kinetic energy = 25 % E
1K E4
22.2
2max max 2
4F ma mA mA 0.6 N.T
23. For parallel combination p 1 2k k k
freqrency l 2k k1 k 1f (1)2 m 2 m
Now when 1 2k and k in increased 4 times,
1 1 24(K K )1f 2f .(from(1))2 m
24. From graph A=1 cm T=8s
2 3y Asin t Asin t y cmT 2
2 1 12 2
2 2 2
-4 3 3 cma w y .T 2 3 s
357
25. Now 1k k
contant
1 1 2 2 2 1 1k k k
Now, A = l11 2 2
1 =2 2 2
k +k k l Ak k k
26. 2 2w A x ,
the velocity for moving form x=o to A2x will ge more them for A
2x tox A
1 2.T T
27. max1U u8
22 2 21 1 1 Aky kA y
2 8 2 8
28. In the expression for both Kinetic and potential energy, We have the square of the halmonicfunctions (sine or cisine).
The average of which over a cycle is 12
2 2E 1u K m A2 4
29. Angnlar freqvency km
Since ‘m’ is constant, 1
K
Now, 2 maxmax
aa AA
max maxa ak AA K
1Ak
30. For a spring, m 1T 2 T ( mk k
is comtant)
358
31. Phase of st1 oscillater 1 11
2t tT
For nd2 oscillater, 2 22
2t t+
T
phase diff 1 2θ -θ32. Restoring force F = – Ayg = – (Ag)y = – ky
1
m 1k A g T 2 Tk A
33. To loose comtact, the condition in ;2m A mg
2gA 2mg ( k mw )
k
34. In SHM, accelelation and displacement are opposite in direction Also a y..
35. Here t = 0, x = 1 cm and 1 1 cm s , w s
Now, x A cos( t ) ---- (1)
Velocity dx A sin( t )dt
----- (2)
Solved the equation (1) and (2)
36.1 2
144 121T 2 and T 2g g
1 2T T
When the shorter pendulum completes n oscillations, the longer one completes (n-1)oscilla tions (when in same phase).
2 1nT (n 1)T
37.2r 3.14
r
13.142 f 3.14 f 0.5s2
38. Maximum force 2 2 2m A m4 f A
39. Periodic time l 2 gT 2 and g T l
Linear displacement x = a cos t
359
40.1 2
km m
or removing m, angular frequency
2
k'm
41. Kinetic energy 2 2 21K m (A y )2
Now total energy 2 21E m A2
42. y = a sin t + b cos t,
Taking a = Acosθ and b = Asinθ,
y A cos sin t A sin cos t
A sin( t )
Now 2 2 2 2 2.a b A A a b
43. The body will not loose contact with the surface,
if 2
22
m4mg m r rT
{where r is amplitude} rT 2g
44. Maximum kinetic energy 2 20
1K m A2
1
202
2KAm
Equation for displacement is ;
12
o2
2Ky Asin t sin tm
45. 2 2 2 21E m A E A2
2 2 21 1 2 2E (A ) ( A ) ( A )
46. 2k be the spring constant of the spring having lenght 2.
Now, 1 2.
2 2.n
47.1 k 1 2kf and f ' { k ' 2k}
2 m 2 2m
f ' f
360
48. Here both springs are in parallel. The restoring force on the system in only due to spring and not dueto gravitational force We can ignore the slope.Equivalent spring cantant =k+k=2k
Periodic time 1MT 2
2k
49. Enelgy stoved =Work done
21E kA2
Now maximum accelaration2
maxa A
50. Potential energy gainad by the spring on suspending mass “m” 21is ky .2
When system executes SHM, the energy gained by the system 2 21 m A2
total final energy of the system 2 2 21 1m A ky .2 2
51. Radius of the rotational motion r =0.4 mWhen the turn table rotates, the restoring forcedeveloped in the spring = centrifugal force
2 2restoreF m r 2(10) 0.4 80N
Now increase in lenght of spring = 40-35 = 5 cm
Force constant 3F 80k 1.6 10 N / m.x 0.05
52. In case-I, springs are connected in parallel.
equivalent force constant p 1 2k k k 2k.
Peliodic time pm mT 2 2kp 2k
In case-II, spring are connected in series.
Equivalent force constant 1 2
s1 2
k k kkk k 2
periodic time ss
m 2mT 2 2k k
p
s
T 1T 2
.
361
53. Here T=6 s
Amplitude 1OB OC BC 10cm2
OD =5 cmNow driplacement x A sin(wt ) (1)
where A=10 gm, 2
3T rad
Now if at t=0, oscillator is at C.i.e. at t=0, x=A
A A sin( 0 ) (OR) A=A sin sin 1
2
Putting this in eqn. (1)
x A sin( t ) A cos t2 = 10 cos t
for x = 5 cm,
5=10cos t 1cos t 2
t 3
t 1 S 54. Force responsible for oscillation in
F mg sin mg { is small}
xmgR
Comparing this withF= - kx ;
mgkR
55. Let the rod be pressed down by “x” at pointA and released.
both spring gets displaced by “x” Restoring torgue produced
= kx kx2 2
=kx
Now tan 2xx
2
BO D C
362
If in small; tan 2x
x2
2ktorque k2 2
Now moment of inertia of rod with reference to O is if I, then2 2
2
Id kdt 2
2 2
2
d kdt 2I
Comparing with 2
22
d ;dt
2 2k 2 mwhere and I2I T 12
2mT3k
56. Here 2 acceleration vectors g. and a are acting along mutually prependicular direction .n 2 2
effeffective acceleratioin l g g a
eff
T 2g
57. 22 2off xg a (g ay) here, ax
2gsinθ cosθ, ay = gsin θ2 2 2
x y ya g a 2ga
= 2 2 2 2 2 2 2 2a sin cos g g sin 2g sin 2 2g (1 sin ) 2 2g cos
effg g cos 58. When the length of spring increases by x=2.5 cm
force F mg sin
force constant F mgsinkx x
m xT 2 2 s.k gsin 7
363
59. Since the spring is massless, when Ccollides with A, both A and B will gain equal momentum. Also, since A and B have equal mass,both will have same velocity. Let this velocity be u. Acc. to the law of conservation of momentum,m mu mu 2mu
2vu
Now if the compression produced in the spring is x, then acc. to law of conservation of energy,
2 2 2 21 1 1 1m mu mu kx2 2 2 2
2 2 22 22 2
4kx v kxv um m
42 2
(1)2 2
kx v mx vm k
Now block A and B will have equal kinetic energy.
2 2 2 21 1 1kx mu mu mu2 2 2
During max imum contraction,
kinetic energy of the system A-B is
mu2 = 2.
4m v
60. Displacement
2 ty = 4 cos sin100t2
61. Here x = A cos t
Now potential energy 2 21 m x {taking P.E. as a function of x}2
when x=0, potential energy=0graph (b) III
Also, potential energy 2 21 m (A cos t)2
{taking P. E. as a function of time}
At t=0 potential energy 2 21 m A2
graph -I
364
62. 2 2 21 1
1 1 (1)2 2
E m x E x m
2 2 22 2
1 1E m y E y m (2)2 2
2 2 21 1E m (x y) E (x y) m (3)2 2
From (1), (2) {(3),
1 2E E E
or 1 2 1 2E E E 2 E E
63. 2 11
k kx F (1)m mx mx
Similarly, 2 22
F (2)mx
If F1 and F2 acts simaltaneously, then angular frequency
1 22
F Fw (3)mx
From (1), (2) and (3); 2 221 2 Now, use equ.
2 = T
64. Initial periodic time 1T 2 (1)g
When pendulum moves along vertical direction, effictive acceleration effg g a
where ‘a’ in accleration of pendulum.
Now, 2d d (kt)a k 2.1 m.sdt dt
New periodic time 2eff
T 2 (2)g
1
2
1 eff
T gT g
365
65. Block will not slide if mg ma
g a
To prevent the block from sliding the maximum acceleration
of table must be maxa g
Now maximum accleration 2maxa A
2maxA g
2
max 2 2
g gTA4
66. Angular frequency of system
12k (1)
(M m
Now to prevent B from sliding off A, the maximum force acting on B should not be more thanthe frictional force mg .
2max max maxf ma m A (2)
From (1) & (2)
max maxkf m A
m M
To prevent block from sliding, maxmax
mkAf mg mgm M
67. Restoring force F=- kx
Now, duFdx
dukxdx
du k.x.dx
x 2
0
kxU(x) k.x.dx C2
Where C in contant of integration.Now in a SHM, potential energy at the equilibrium position is zero.
u(x 0) 0 C 0
21u(x) kx2
in an equation for a parabola.
366
68. At the upper most end,
when 2mg R m A, coin will loose contact.Taking R=0
2m A mg
2gA
69. Force required to increase the lenght by x in F=kx-----(1)After spring is divided into 2 equal parts,
xF k 'x ' where x ' 2
xk ' (2)2
from (1) & (2); k ' 2k70. Frequency of SHM depends on elasticity & inertia.
71. Restoring force F mg sin OR
e eF mg where g g sin
If is small, sinθ θ
Effective value of g is eg θ
For large oscillation , g sin g ( sin )
1T 2 g
72. Restoring force F=- mg sin which depends on “m”
73. “g” in less on moon
form the equation 1
T 2 ,g
T will increaseAs compared to earth, moon in small
367
74. Periodic time T
1T2
{ On differentiation}
T 1.5%T
75. 2 2 2 2 2maxa A 4 f A 4 (30) 0.01
22
m36 s
Since the oscillator moves between + A & -A,
maximum acceleration 236 77. Energy dissipates & so amplitnde decreases.
Statement -2 in false.78. Statement -1 in true. statement -2 in false. In a SHM, amplitnde & phase does not depend on
restornig force.
79. Time taken by the spring 2k to get maximum compressed from point D= half period of oscillationof the block.(if block in attached at the frce end of spring)
i.e. 2
22
T 1 m 1 0.2t 2 2 s2 2 k 2 0.3 4
80. Similarly 11
Tt s2 3
81. Time period of Block T=Time taken by the block to move from C to D and D to C
82.1 k 1 kf and f '
2 M 2 m M
83. According to the law of conservation of momentum,MV (M m) '
84. According to the law of conservation of energy,Kinetic Energy at mid point = potential Energy at the end points
2 21 12 2
Mv kA
And 2 21 11 1( )
2 2M m v kA
368
85. y = 3 cos 2t + 4 sin 2t
A sin 3 And A cos 4
y A sin cos 2t A cos sin 2t
y Asin ( t )
Which shows that the motion is simple harmonic motion
86.2 2T s
2
87. Amplitude 2 2A 3 4 5 cm
88. Maximum Accelaration of2 2 2a particle A 5(2) 20 cms
89. Mechanical Energy 2 21 m A 250 erg2
90. frequency of the particle 11 1f ST
91. On comparing y = A sin (15 10 )3
t x
with y = A sin ( )t kx
92. At constant pressure density of water vapour is less than dry air.
with increase in humidity according to the equation pv
the
velocity of sound increases.
93.1 1 2
2 1
f f
f
94. From the equation v = fλ, minmax
λ =f
v 17 mm which is nearer to 20 mm
95. On comparing with the wave equation
y = A sin 2 t x we getT
, T = 0.04 s, =0.5 m 1252
v ms
2 6.25 NT v
369
96. Maximum velocity of particle = A
wave velocity = fλ Maximum velocity of particle = 2 x wave velocity
A = 2f A
97. Putting values in vT
if phase diff.. = in the interval x is then
2 2 2x (15 10)15 3
98. Freq. of a wave in a string 1f
1 2 3 1 2 3
1 1 1 1f f f f
99. On comparing 1y 4 sin 500 t with 1 1y A sin t
we get 1 1 12 f = 500 f 250Hz
Similarly 2y 2sin 506 t
2 2 22 f 506 f 253 Hz
Freq. of beats 2 1f – f 3
No.of beats heard per minute 3 60 180
100. y 8sin 2 (0.1x 2t)
y 8sin 2 (2t 0.1x) comparing with t xy AsinT
We get 1
=0.1 10cm
Now path diffrence between 2 particles 2 .x kx
2 1 8 0 21 0
= 72o
101. Distance covered by the pulse = speed x time = 4 cm in 2 seconds both will cover 4 cm & thecentre of both will superpose & potential energy will be zero.
Total energy will be in the from of kinetic energy..
370
102. 1y asin t & 2πy a cos t a sin ( t 2 )
1st wave is ilagging behind in phase by π 2103. Here A is the amplitude of resultant wave formed by 2 waves having amplitude A1 and A2
respectively.2 22
1 2 1 2A A A 2A A cos Also in the phase 1 2A , A
Now puttng 1 2A A b & A b , We get
2 2b 2b 1 cos
01cos 1202
104. As seen from fig., distance between 3 nodes in λ
105. 2 1 cos 2 t 1 1y sin t cos 2 t ____________ 12 2 2
1υ = 2 sin 2 t sin 2 t2
2a 2 cos 2 t
2 1= 2× 2 -Y2
{From eng. (1)}
= –42 1 y2
a α -y {SHM}
Now, 2 π π2 TT
106. No.of beats produced per second = 1 2n n
Time interval between 2 consecutive beats 1 2
1n n
107. Since the phase difference between the 2 waves in π2 they are oscillating along mutually
perpendicalal direction.
Resultant ampltude 2 2= A A 2 A
Angular freq. will remain same.
371
108. 1A 0.02 m, υ = 128 msk
4 2π5λ =4 = m, k 2.5 π 7.855 λ
128 k 128 7.85 1005
y A sin kx t
y = 0.02 sin (7.85x - 1005 t)109. Let the number of loops obtained for 315Hz and 420Hz n and (n+1) respectively.
n 1f nf 315
n 1 1f n 1 f 420
n 1 n 1f f f 105 Hz
110. When, sound waves travel fromone medium to another, its frequency does not change.
υfλ
= consant
a b
a b
υ υλ λ
bb a a
a
υλ λ 10λυ
111. Wave velocity = max. velo.of particle
Ak A =
1k 2
λ = 2πA
112. Speed of sound in an ideal gas RTυm
1 2
2 1
υ mυ m
1m
113. 2 21E m A2
2 2 21 m 4π f A2
2 22 1 1
2 2
E f f 1E α fE f 2f 4
2 1E 4E
372
114. Let hte freq. of 1st fork be f1
frequency of 2nd fork = 1 1f 6 f 6 2 1
freq. of th 24th fork = 1f 6 24 1 1f 138
Now, freq. of 24th fork = 2 x freq. of 1st fork (given) 1 1f 138 2 f 1f 138 Hz
115. Differentabing 1y 0.1 sin π100 t +3
w.r.t time,
1πυ (0.1) (100) cos 100 π t +3
Similarly differentiating 2y 0.1 cos t w.r.t. time,
2πυ 0.1 sin π t2
phase difference between the 2 velocities is
π π πδ π t + π t rad3 2 6
116. Wave number 11 1 200 mλ 0.005
117. Frequency heard by the listener
LL s
υ + υf fυ
sυ = 0
L 2
s
f υ + υf υ
υυ + 54
υ 4
% increase L S
S
f f 5 4100 100 25%f 4
118. From RTυ =M
, υ α T
In summer, velocity increases & hence decreases and so L increases.
The length of 2nd halmonics 1x = 3L 3 16 48 cm
In summer, velocity being more, 1x 3L x 48
373
119. In LL s
s
υ+ υf fυ υ
putting L L sυ 0, f 2f , υ υ, s sυ υ
s ss
υ2f fυ υ
s2υ υ sυυ2
120. For resonance, the frequency of a.c. supply should be same as fundamnetal freq. of wire.
1 Tf2 L µ
= 50 Hz
121.L L
S S
f υ + υf υ υ
or
L L
S S
f υ υf υ - υ
but, L Sυ υ L
S
f 1f
122. Since the rope is heavy, the tension at the lower end & top end of the rope will be different.Mass of rope 2m 3kg
Mass of block 1m 1 kg
tension at the lower end 1 1T m g 1 g N &
at the upper end in 2 1 2T m m g 4 g N
Now speed of wave in rope υ= λ =µ µT Tf
λ = T ( f ,μ are constants)
Wave length at lower and 1 1λ = T & at
the upper end 2 2λ = T
2 2
1 1
λ Tλ T
2
21
TλT
= 1λ = 2λ 0.1m
123. Speed of sound Pρ
mass of 1 mole airρVolume of 1 mole air
3
3 3
29 10 kg 1.322.4 10 m
speed 5
17 1.01 10 330 ms5 1.3
374
124 From the phase angle (40-2t), we get k = 40 OR 20402
and = 2 OR Hzff 122
125 Increase in tension of string increases its frequency. If the original frequency of B(fB) weregreater than that of A(fA), further the increase in fB should have resulted in increase in the beatfrequency. But the beat frequency is found to decrease. This shows that fA- fB = 5 Hz andfA=427 Hz, we getfB = 422 Hz
126L
L Sυ υ 330 0f f 800 1320 Hzυ υs 330 130
127 222 (340)112.2υ
LMμυT
µTυ
4T 2.31 10 N
128 µ4lfTµT
2l1f 22
2 2 22 M 4f MT 4f µ 248 N
µ µ
129 In tube A, lA 2
In tube B, lB 4
2lυ
λυυA
A 12
υυ
4lυ
λυυ
B
A
BB
130 The was decreases the frequency of unknown fork. The possible unknown frequencies are, (288+4)Hz and (288-4) Hz. Wax reduces 284 Hz and so beats should increases. It is not given in thequestion. This frequency is ruled out. Wax reduces 292 Hz and so beats should decrease. It isgiven that the beats decrease from 2 to 4. Hence the unknown fork has frequency 292 Hz.consider option (a)
131 Stationary wave : Y = a sin (wt-kx) + a sin(wt+kx)When x = 0, Y 0. The option is not acceptableconsider option (b)stationary wave : Y = a sin (wt-kx) - a sin(wt+kx)At x = 0, Y = 0. This option holds good.Option (c) gives Y = 2a sin(wt - kx)At x = 0, Y 0Option (d) gives Y = 0.Hence option (b) holds good.
375
132 When temperature increases, l increases. Hence frequency decreases.133 The possible frequency of piano are (256 + 5)Hz and (256 - 5)Hz.
For a piano string µT
2l1υ When tension T increases v increases.
(i) If 261 Hz increases, beats / second increase. This is not given.(ii) If 251 Hz increases due to tension, beats / second decrease. This is given.
134 By Doppler’s effect,
S
L
S
L
υυυυ
ff
L L
S
υυf υ υ 65f υ υ 5
511
561
ff
ff - f increase Fractional
S
L
S
SL
%205
100 increase percentage
135 MRTP
2o
He
1
2
γ32
4γ
υυ
Hence option (d) is correct.136 In a longitudinal wave, pressure is maximum where displacement is minimum. Therefore pres-
sure and displacement variations are 180 out of phase
137 Frequency of tuning fork f1 = 480 Hz. Number of beats s-1, n = 10Frequency of string f2 = (480 + 10)Hz. A slight increase in tension increase f2
f2 = 480 - 10 = 470 Hz.138 (c) is the correct choice because its value is finite at all times.
139 As cos)90sin( The phase difference between the two waves is 2
140150ms
0.2500
µTυ
141 Here, 212
121 TT,
21
rr,ρρ
1
1
11 ρ2
1f
T
rl , 2
2
22 ρ2
1f
T
rl 12
rr
ff
2
1
2
1
376
142 109
10081
1
2
1
2 TT
ff
%101001
21
f
ff
143 When one end is closed f100
250H1 z
2 1 3 1f =3f =150Hz,f =5f =250Hz and so on...
144 When other end of pipe is opened, its fundamental frequency becomes 200Hz. The overtonehave frequencies 400, 600, 800 Hz..
145 As2
''
6040
'50
' 75cm2 2
1
22
146 65
3025
ff
1
2
2
1
2 1 2f -f =4. on solving, weget f =24 Hz
Hz20f1
147 20011
10011
100101
ff 2
1
1
2
200fff 1
12 1200fffsabbeofnumbers 1
121
148L L
S S
f υ+υ= ,f υ+υ Here Hz165f,ms5,ms5 S
1S
1L
Lf =170 Hz 5165170sabbeofNumber 1
149 As the source is moving perpendicular to straight line joining the observer and source, (as ifmoving along a circle), apparent frequency is not affected n1 = 0
150 As is clear from figure, at t = 0, x = 0, displacement-y = 0 Therefore option (A) OR (D) may becorrect. In case of (D) y = A sin (kx - wt)
dy = -ω A cos(kx- t)dt
dy = kA cos(kx-ωt)dx
dydy dydt – v - v dy dt dx
dx
i.e. particle velocity = - (wave speed) xslope and slope at x = 0 and t = 0 is positive, in figureTherefore, particle velocity is in negative y - direction
151 At a displacement antinode, a pressure node is present. Since pressure does not change at itsnode, nor does density.
377
152 For a sonometer fundamental
T221f
To maintatin the fundamental mode, in doubling the length, tension must be quadrupled.
153 velocity of transverse waves T 2
T Tm r
velocity of longitudinal waves LY
L2
T
Y YT / r stress
154 Let the frequency of standard fork = x
x10097f,x
100102f BA , x
10097fB
Now xxff BA 10097
100102
Hzx 120
155 If the length of the wire between the two bridges is , then the frequency of vibration is
dπrT
2l1
mT
2l1n 2
If the length and diameter of the wire are doubled keeping the tension same, then new funda-mental frequency will be n/4
156S
L
S
L
υυυυ
ff
using this equation the frequency of reflected sound heard by the girl,
L
L
S
υ υfυ – υ Sf
157 f open pen
υ2 o
openclosed closed
losed open
υ υf As4 4 /2 2c
openopen
υ f2
i.e. frequency remains unchanged.158 If we assume that all the three waves are in same phase at t = 0, we shall hear only 1 beat s-1
378
159 y (x, t) = 0.005 cos )( tx compare it with standard equation
y (x, t) = A cos (kx - wt) = A cos
t
T22
2 2 and =
T
160 Given that the displacement of particle is y = A sin (t - kx) ..........(i)
The particle velocity dyp ................(ii)dt
v
Now, on diffrentiating eqn.1 with respect to t dy A cos( t kx)dt
From eqn.(2 mental mode of the colsed pipe is
Hz 2000.404
3204Lυf1
173 Since the beat frequency is 8, the frequncy of the string vibrating in its first Overtone is 192 Hzor 208 Hz.
Where for 1st Overtone frequency mT1'f1
........(1)
It is given that the beat frequency decreases if the tension in the string is decreased.
11 f'f Hence 'f1 = 208Hz and not 192Hz
174 substituting the values of m. and 'f1 in equation 1 we get T = 27.04 N
1752π 2 λ mλ
12πf 3λ3π υ msλ 2
176 Distance between two consecutive nodes = m22
177 The resultant displacement is given by,y = 0.1 cos 2x sin3 t Or y = A sin 3 tWhere A is the Amplitude of standing waves given by 0.1 cos 2x
At x = 0.5m, cos 2x = cos (1rad) = πcos cos57.3 0.054 m
3.14
Amplitude A at (x = 0.5m) = 0.1 0.54 = 0.54m
178 Particle velocity dy dυ (0.1 Cos 2x sin 3πt) = 0.1 x 3 cos 2x sin 3 tdt dt
at x = 0.25m and t = 0.5 s, v = 0
379
179 The two displacements can be written as
1 1 1y A cos (k x t)
and 2 2 2y A cos (k x t) compare this equation with given equation and get solution.
180 Beat frequncy = 21 ff = 1 2–2 2
181 The resultant displacement is given by
21 yyy
1 1 2 2A cos ( k x t) A cos (k x t) For x = 0 we have
1 2y=A cos ω t+A cos ω ty 0.10 cos (96 t) cos (4 t)
Between t = 0 and t = 1 s, Cos 96 π t becomes zero 96times and cos 4 π t becomes zero 4times Hence the resultant displacement Y at x = 0 becomes zero 100 timesbetween t = 0 and t = 15.
182 1y = A sin(kx+ t)
ry A sin (kx t) y yi yr
y 2A sin kx cos t Here 2A=10 5A
380
N O T E S
