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Model question paper 10 standard ICSE.
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Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
Model Question Paper (P+C+M)
1. Calculate the mass of a thin disc of radius and of uniform surface density 0 .
(a) 2
0
2
R (b)
20
3
R (c) 2
0 R (d) 2
0
4
R
2. Calculate the potential energy of a thin rod of length L with constant linear mass density
held vertically on ground
(a) 2
2
gL (b)
2
4
gL (c) 2gL (d)
2
3
gL
3. A projectile is moving at 120ms at its highest point, where it breaks into two equal parts
due to an internal explosion. One part moves vertically up at 130ms with respect to the
ground. Then the other part will move at
(a) 110 13ms (b) 120ms (c) 130ms (d) 150ms
4. A body is projected vertically upwards. The times corresponding to height h while
ascending and while descending are 1t and 2t respectively. Then the velocity of
projection is (g is acceleration due to gravity)
(a) 1 2
2
g t t (b) 1 2
2
g t t (c) 1 2
1 2
g t t
t t (d) 1 2g t t
5. The angle between two vectors of magnitudes 12N and 3N on a body is 60 the resultant
w.r.t. 12 N force is about
(a) 30 (b) 60 (c) 10 (d) 75
6. A cube has a side of length 21.2 10 m . Calculate its volume.
(a) 6 31.7 10 m (b) 6 31.73 10 m (c) 6 31.70 10 m (d) 6 31.732 10 m
7. A wire has a mass 0.3 0.003 ,g radius 0.5 0.005 mm and length 6 0.06 cm . The
maximum percentage error in the measurement of its density is
(a) 1 (b) 2 (c) 3 (d) 4
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
8. The dimensions of a
b in the equation
2a tP
bx
where P is pressure, x is distance and t is
time are
(a) 2 3M LT
(b) 2MT
(c) 3 1ML T
(d) 3LT
9. In the relation; aZ
kp e
p is pressure, Z is distance, k is Boltzmann co9nstant and is
the temperature. The dimensional formula of will be
(a) 0 2 0M L T
(b) 1 2 1M L T
(c) 1 0 1M L T
(d) 0 2 1M L T
10. Square of the resultant of two equal vectors (forces) is equal to three times of its product.
The angle between them is
(a) 0 (b) 45 (c) 60 (d) 90
11. Given that P Q R and that R is perpendicular to P . If P R , then the angle between
P and Q is
(a) 4
rad
(b) 2
rad
(c) 34
rad
(d) rad
12. The horizontal range and maximum height attained by a projectile are R and H
respectively. If a constant horizontal acceleration 4
ga is imparted to the projectile due
to wind, then its horizontal range and maximum height will be
(a) ,2
HR H (b) ,2
2
HR H
(c) 2 ,R H H (d) ,R H H
13. The speed of a projectile when it is at its greatest height is 2
5times its speed at half the
maximum height. The angle of projection is;
(a) 30 (b) 60 (c) 45 (d) 1tan 3 /4
14. A projectile is thrown with a velocity of 110 2ms at an angle of 045 with horizontal. The
time interval between the moments when its speed is 1125ms is: 2(g 10m / s )
(a) 1.0 s (b) 1.5 s (c) 2.0 s (d) 0.5 s
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
15. A projectile is fired at an angle of 030 to the horizontal such that the vertical component
of its initial velocity is 180 ms . Its time of flight is T . Its velocity at t T / 4 has a magnitude
nearly equal to:
(a) 200 m / s (b) 300 m / s (c) 140 m / s (d) 100 m / s
16. After one second the velocity of a projectile makes an angle of 045 with the horizontal.
After another one second it is travelling horizontally. The magnitude of its initial
velocity and the angle of projection are:
2(g 10m / s )
(a) 014.62m / s,60 (b) 114.62m / s, tan (2) (c) 122.36m / s, tan (2) (d) 022.36m / s,60
17. A relief aeroplane is flying at a constant height of 1960 m with speed 600 km/hr above
the ground towards a point directly over a person struggling in flood water. At what
angle of sight i.e. angle made by the displacement vector of the kit with the vertical,
should the pilot release a survival kit if it is to reach the person in water? 2(g 9.8m / s ) .
(a) 030 (b) 045 (c) 060 (d) not possible
18. There are two values of time for which a projectile is at the same height. The sum of
these two times is equal to:
(a) 3T / 2 (b) 4T / 3 (c) 3T / 4 (d) T
19. A ball is projected upwards from the foot of a tower. The ball crosses the top of the tower
twice after an interval of 6s and the ball reaches the ground after 12s. The height of the
tower is 2(g 10m / s ) :
(a) 120 m (b) 135 m (c) 175 m (d) 80 m
20. A v t graph is shown in the figure. We can draw the
following conclusions:
1. Between t 1 s to t 2 s speed of particle is
decreasing
2. Between t 2 s to t 3s speed of particle is increasing
3. Between t 5 s to t 6s acceleration of particle is negative
Between t 0 s tot 4s particle changes, its direction of motion twice
(a) only 1 and 2 are true (b) only 3 and 4 are true
O 1 2
3 4 5 6 t
v
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
(c) only 1 and 4 are true (d) only 2 and 3 are true
21. Trajectories of two projectiles are shown in figure. Let 1T and 2T be the time periods and
1u and 2u their speeds of projection. Then :
(a) 2 1T T (b) 1 2T T
(c) 1 2u u (d) 1 2u u
22. From an elevated point a body is dropped and another body is thrown horizontally at
the same instant. Choose the incorrect option:
(a) Time of descent of projected body is greater than the time of descent dropped
body
(b) Time of descent of projected body is equal to that of the time of descent dropped
body
(c) The ratio of vertical displacement in successive equal intervals of time is 1:3:5 for
both the bodies
(d) The velocity of striking of projected body is more than the dropped body
23. A particle is moving along a straight line with uniform acceleration and has velocities
7 /m s at P and 17 m/s at Q. R is the mid-point of PQ . Then
(a) the average velocity between R and Q is 12 /m s
(b) the ratio of time to go from P to R and that from R to Q is 3 : 5
(c) the velocity at R is 10 /m s
(d) the average velocity between P and R is 10m/s
24. A projectile projected with the same speed has the same range R for two angles of
projections. If 1T and 2T be the times fight in the two cases, then
(a) 1 2 T T R (b) 21 2 T T R (c) 1
2
sinT
T (d) 1
2
cotT
T
25. From the given system of equations, solve for T in terms of 1 2 3m ,m ,m andg.
1 1 1
2 2 2
3 3 3
1 2 3
T m g m ( a )......(i)
T m g m ( a )......(ii)
2T m g m ( a )......(iii)
a a 2a 0........(iv)
1
2
x
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
(a) 1 2 3
1 2 2 3 1 3
4m m mT g
4m m m m m m
(b) 1 2 3
1 2 2 3 1 3
m m mT g
4m m m m m m
(c) 1 2 3
1 2 2 3 1 3
4m m mT g
m m m m m m
(d) 1 2 3
1 2 2 3 1 3
m m mT g
m m m m m m
26. Choose the incorrect option in the following. The vectors A and B are drawn from a
common point and C A B , then angle between A and B is
(a) 90 if 2 2 2C A B (b) greater than 90 if 2 2 2C A B
(c) greater than 90 if 2 2 2C A B (d) less than 90 if 2 2 2C A B
27. If 2 2
3 2%, 4 0.12 and 5 4%, then the quanitity is given by,2
x yx y z
z
(a) 5 6.64% (b) 0.5 6.64% (c) 0.5 13.28% (d) 5 3.32%
28. Which of the following sets have different dimensions?
(a) Pressure, Young’s modulus, Stress
(b) Emf, Potential difference, Electric potential
(c) Heat, Work done, Energy
(d) Dipole moment, Electric flux, Electric field
29. Figure shows ABCDEF as a regular hexagon. The value of AB AC AD AE AF will be
(a) AO (b) 2AO (c) 4AO (d) 6AO
30. Many bullets with same speed u cover some area on ground.
The maximum area covered on ground is
(a) (b) 2 4
2
u
g
(c)
2u
g
(d)
4
2
u
g
Chemistry:
31. 41 gCH and 4g of compound X have equal number of moles. Thus, molar mass of X is:
(a) 116gmol (b) 132gmol (c) 14gmol (d) 164gmol
A B
C
DE
FO
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
32. In the following reaction, 3 2 3 4 3 4 2 3Ba(NO ) Na PO Ba (PO ) NaNO The number of moles of
3 4 2Ba (PO ) formed from 2 moles each of 3 2Ba(NO ) and 3 4Na PO is
(a) 1 mol (b) 1/3 mol (c) 2/3 mol (d) 4 mol
33. Oxidation numbers of Cl atoms in 2CaOCl (bleaching powder) are
Ca
Cl
ClO
*
* *
(a) Zero on each (b) 1 on *Cl and +1 on **Cl
(c) *1 onCl and 1 on **Cl (d) 1 on each
34. Which of the following changes requires a reducing agent?
(a) 2 24 2 7CrO Cr O (b) 3BrO BrO
(c) 23 3 4H AsO HAsO (d) 3 4Al(OH) Al(OH)
35. For the reaction between 22 2 4MnO (s) C O in basic solution, the unbalanced equation is:
2 24 2 4 2 3MnO C O MnO (s) CO in a balanced equation, the number of OH ions is:
(a) 0 (b) 4 on the right (c) 4 on the left (d) 2 on
the left
36. Which has maximum number of milliequivalents?
(a) 2 4100mLof 0.01NH SO (b) 100mLof 0.01NHCl
(c) 3 4100mLof 0.01N H PO (d) Equal
37. Sucrose solution is 1 molal. Mole fraction of sucrose in the aqueous solution is:
(a) 0.018 (b) 0.015 (c) 0.036 (d) 0.009
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
38. 1.06 g 2 3Na CO is dissolved in 100 mL solution. 10 mL of this solution can be neutralized
by:
(a) 10 mL of 0.1 N HCl (b) 10 mL of 0.1 M 3 4H PO
(c) 20 mL of 2 40.1MH SO (d) 20 mL of 0.1MHCl
39. 5.3 g of 2 3M CO is dissolved in 150 mL of 1 HCl. Unused acid required 100 mL of 0.5
N NaOH . Hence, equivalent weight of M is:
(a) 23 (b) 12 (c) 24 (d) 13
40. If the radius of first Bohr orbit is x , then de-Broglie wavelength of electron in 3rd orbit is
nearly:
(a) 2 x (b) 6 x (c) 9x (d) 3
x
41. 3 moles of mixture of CO and 2CO requires 80 g NaOH to convert all 2CO into 2 3Na CO .
How many additional grams of NaOH will be required for complete conversion into
2 3Na CO if CO is completely oxidized into 2CO
(a) 100 g (b) 120 g (c) 90 g (d) 160 g
42. If ,1 2 3
E E and E represent respectively the energies of an electron, an alpha particle and a
proton each having same de-broglie wavelength, then
(a) 1 3 2
E E E (b) 2 3 1
E E E (c) 1 2 3
E E E (d)1 2 3
E E E
43. A 3p orbital has:
(a) Two non-spherical nodes
(b) two spherical nodes
(c) One spherical and one non-spherical node
(d) one spherical and two non-spherical nodes
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
44. The radii of two of the first four Bohr’s orbits of the hydrogen atom are in the ratio 1:4.
The energy difference between them may be:
(a) Either 12.09eVor3.4eV (b) either 2.55eVor10.2eV
(c) either 13.6eV or3.4eV (d) either 3.4eV or0.85eV
45. An ion aMn has the magnetic moment equal to 4.9 BM the value of a is:
(a) 3 (b) 4 (c) 2 (d) 5
46. For which of the following, the radius will be same as for hydrogen atom having n 1?
(a) He ,n 2 (b) 2Li ,n 2 (c) 3Be ,n 2 (d) 2Li ,n 3
47. The circumference of thn orbit in H-atom can be expressed in terms of de Broglie
wavelength as:
(a) (0.529)n (b) n (c) 13.6 (d) n
48. The wave number of first line of Balmer series of hydrogen is 115200cm . The wave
number of first line of Balmer series of 2Li ion is:
(a) 115200cm (b) 160800cm (c) 176000cm (d) 1136800cm
49. An electron of H-atom from an excited state returns to ground state in one or more steps
shows emission of ten lines. How many of these belongs to visible spectrum?
(a) 6 (b) 5 (c) 4 (d) 3
50. The orbital represented by 4,2,0 is:
(a) 2z4d (b) x4P (c) z4P (d) xy4d
51. Which is incorrect about 34.25g NH ?
(a) it contains 0.25 mol of 3NH
(b) it contains 0.75 mol of H atoms
(c) it contains total of 1.0 mol of N and H atoms
(d) it contains 236 10 molecules of 3NH
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
52. Mole of fraction of ethanol 2 5(C H OH) in ethanol-water system is 0.25 . Thus, it has:
(a) 30% ethanol by weight of solution (b) 75% water by weight of
solution
(c) 46% ethanol by weight of solution (d) 60% water by weight of
solution
53. Which of the following represent redox reactions?
(a) 2 22 7 4 2Cr O 2OH 2CrO H O (b) 2 2
4 2 7 22CrO 2H Cr O H O
(c) 24 2 22MnO 3Mn 4OH 5MnO 2H O (d) 2 2
3 3 4Cu 4NH [Cu(NH ) ]
54. In the following cases, equivalent weight of the underlined may be equal to molecular
weight:
(a) 4 2 2 2 4 22CuSO 4KI Cu I 2K SO I (b) 2 2 3 2 4 62Na S O I 2NaI Na S O
(c) 4 2 2 2 3P 3NaOH 3H O 3NaH PO PH (d) 2 4 2 4 2H SO 2NaOH Na SO 2H O
55. If Aufbau rule and Hund’s rule are not considered, which of the following statements is
incorrect?
(a) 2Fe will have configuration as Ar
(b) 2Cu is colourless ion
(c) Mn has magnetic moment of 3 BM
(d) K belongs to d block
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
56. Of the following quantum state designations, which is/are not allowed state for an
electron in an atom:
n l m
(a) 3 2 -2
(b) 3 1 0
(c) 3 0 -1
(d) 3 2 0
57. For the reaction 3 4 2 2 4 2H PO 2Ca(OH) Ca HPO 2H O . Select the incorrect statement.
1mole 1mole
(a) the equivalent weight of 3 4H PO is 98
(b) the resulting solution is neutralized by 1 mole of KOH
(c) 1 mole of 3 4H PO is completely neutralized by 1.5 mole of 2Ca(OH)
(d) None of these
58. The set of quantum numbers not possible for an electron is
(a) 1,1,1, 1 2 (b) 1,0,0, 1 2 (c) 1,0,0, 1 2 (d) 2,0,0, 1 2
59. The number of d-electrons in 2 26Fe Z is not equal to that of the
(a) p electrons in . 10Ne At no
(b) s electrons in . 12Mg At no
(c) d -electrons in . 26Fe At no
(d) p electrons in . . 17Cl At no of Cl
60. Without looking at the periodic table, select the elements of III A group of the periodic
table.
(Atomic numbers are given)
(a) 3, 11, 19, 37 (b) 5, 13, 21, 39 (c) 7, 15, 31, 49 (d) 5,13, 31, 49
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
Maths:
61. If 24 36 48log 12 , log 24 , log 36a b c then 1 abc
(a) 2ab (b) 2bc (c) 2ca (d) 0
62. If 2.3 0.23 1000x y then
1 1
x y
(a) 3 (b) 2 (c) 1
3 (d)
1
2
63. If coslog sin 0ecx x then
(a) 0x (b) 0x (c) 1 1x (d) Not possible
64. If sin sin2 , cos cos2x y then 2 2 2 2 3x y x y
(a) 2x (b) 2y (c) x (d) y
65. If 21 sin sin .....to 4 2 3 thenx x x
(a) 6
(b)
4
(c)
3
(d)
3
or
2
3
66. For 133 ,2 cos2
AA
(a) 1 sin 1 sinA A (c) 1 sin 1 sinA A
(c) 1 sin 1 sinA A (d) 1 sin 1 sinA A
67. If
3
4then
2
12 cot
sin
(a) 1 cot (b) 1 cot (c) 1 cot (d) 1 cot
68. cot16 cot 44 cot 44 cot76 cot76 cot16
(a) 0 (b) 1 (c) 2 (d) 3
69. If 1 tan 2 cot 62
3tan 152 cot 88
k
then the value of k is
(a) 1 (b) -1 (c) 1
2 (d)
1
2
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
70. Which of the following is true
(a) If 3 3 3 30 then a b c a b c abc
(b) If 4 4 4 0 then 2a b c a b c ab bc ca
(c) If 0a b then a b ab
(d) None of these
71. In a right angled triangle, if the hypotenuse is 4 times the length of the perpendicular
drawn to it from its opposite vertex, then one of the angles of the triangle is
(a) 30 (b) 45 (c) 60 (d) 75
72. If , ,A B C are angles of a triangle such that tan tan tan 6A B C , tan tan 2A B then which
of the following is true.
(a) tan 1A
(b) tan 2B
(c) tan 3C
(d) None of these
73. If
22 2
3 5log log
4 4 2x x
x
then the value of x can be
(a) 1
32 (b) 1
32
(c) 2
32 (d) 1
42
74. If tan , tan are roots of the equation 2 0 0x px q p then
(a) 2 2sin sin cos cosp q q
(b) tan1
p
q
(c) cos 1 q
(d) sin p
75. If 2log.log5 xa a then x
(a) 125 (b) 2a (c) 25 (d) 5
76. If 222 cba then cc baba )()( log
1
log
1
(a) 2
1 (b) 1 (c) 2 (d) 3
77. If zyx 7553 then zyxy
(a) 2xz (b) xz (c) 3xz (d) 4xz
78. If 3 2 3 0a x b x c then 3323 cxbxa
(a) 0 (b) 3abcx (c) abc (d) 1
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
79. If 13
5)sin(,
5
4)cos( and , lies between 0 and
4
then 2tan
(a) 33
56 (b) 56
33 (c) 65
16 (d) 61
60
80. If qec cotcos , then the value of eccos
(a) q
q1
(b) q
q1
(c)
1
2
1 (d) none of these
81. If 41
41
33
x , 41
41
33
y then 2223 yx
(a) 1 (b) 64 (c) 49 (d) 81
82. If 001.0log 1.0x , 81log9y , then yx 2
(a) 223 (b) 23 (c) 12 (d) 22
83. 40tan20tan340tan20tan is equal to
(a) 2
3 (b) 3 (c)
4
3 (d) 1
84. 22
tantantantan1
(a) 22 tantan (b) 22 secsec (c) 22 cottan (d)
22 cossec
85. If is any angle then 24 secsec is equal to
(a) 24 tantan (b) 24 tantan (c) 42 tantan (d) 2tan2
86. The value of 2266 sincos3cossin is
(a) 0 (b) 1 (c) 2 (d) 3
87.
ac
ac
cacb
bc
cbba
ba
ab
x
x
x
x
x
x
22
22
22
(a) 32
1
ax
(b) 1 (c) x (d) none of these
88.
3
1
3
432
23
31525
15
9863
(a) 228 (b) 325 (c) 229 (d) 320
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
89. The moon’s distance from the earth is 0000,35 km and its distance subtends an angle of 31
’ at the eye of the observer. The diameter of the moon is:
(a) 27
113157 km (b)
27
11315 km (c)
27
113050 km (d) none of these
90.
3
10
5log
1
1.0log
15 7
(a) 0 (b) 1 (c) 2 (d) 3
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
Answers:
Physics:
1. c 2. a 3. a 4. a 5. c 6. a 7. d 8. b 9. a 10. c
11. c 12. d 13. b 14. a 15. c 16. c 17. c 18. d 19. b 20. b
21. d 22. a 23. d 24. a 25. a 26. c 27. b 28. d 29. d 30. c
Chemistry:
31. d 32. c 33. b 34. b 35. c 36. d 37. a 38. d 39. a 40. b
41. d 42. a 43. c 44. b 45. a 46. c 47. d 48. d 49. d 50. a
51. d 52. c 53. c 54. b 55. b 56. c 57. a 58. a 59. d 60. d
Maths:
61. B 62. c 63. d 64. b 65. d 66. b 67. b 68. d 69. b 70. b
71. d 72. c 73. b 74. a 75. c 76. c 77. a 78. b 79. a 80. c
81. b 82. c 83. b 84. b 85. b 86. b 87. a 88. a 89. a 90. c
Solutions: (P+C+M)
Physics:
1. 02dm rdr
Integrate to get M
20 M R
2. Conceptual
3. Here 2 2 220 30 v i.e., 11300 10 13v ms
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
4. Time of ascent = time of descent i.e., 1 2t t
Using, v u at , we get
Velocity of projection 1 10 . .u gt i e u gt
1t can be replaced with 1 21 2
2 2
g t tt tu
5. Using 2
1 2
sintan
cos
F
F F
we get
3sin 60 3 3 / 2 3 3tan
12 3 1/ 2 2712 3cos60
5.19620.19245 10.9
2
6. 3
3 21.2 10V l m
6 31.728 10 m
Length l has two significant figures; the volume V will also have two significant figures.
Therefore, the correct answer is 6 31.7 10V m .
7. Density 2
m
r L
100 2 100m r L
m r L
8. 2a t
Pbx
2Pbx a t
2Pbx a T
2 2
1 4
1 2or
T Tb M T
P x ML T L
2
2
1 4
TaMT
b M T
9. 0 0 0ZM L T
k
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
k
Z
Further P
k
P ZP
Dimensions of k are that to energy.
Hence, 2 2
0 2 0
1 2
ML TM L T
LML T
10. Using 2 2 2 2 cosR A B AB we get
2 2 22 cosR A A A A B is given
i.e., 2 22 2 cosR A A (given)
but 23R A
2 22 2 2
2
3 2 13 2 2 cos cos . . 60
22
A AA A A or i e
A
11. Here P Q R
and Q R P
Since R P , therefore 2 2 2Q R P
Also, R P , therefore 2 22Q P or 2Q P
Now, 2 2 22 cosP Q PQ R i.e., 2 2 2 22 2 2 cosP P P P
Which gives, 1
cos2
i.e., 3
4
12. 22
, and2
y yx
u uT H R u T
g g Where T time of flight of the projectile and xu and yu are
respectively the horizontal and vertical components o initial velocity.
When a horizontal acceleration is also given to the projectile yu and hence T and H will remain
unchanged while the range will become
21
2xR u T aT
2
2
41
2 4
yugR R H
g
13. Let be the angle of projection and u its initial speed. The maximum height will be:
2 2 2 2sin sin
2 2
u uH gH
y
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
Now, cosHv u
or 2 2 2cosHv u …(1)
2 2 2/2 2
2H
Hv u g u gH
2 2
2 sin
2
uu
… (2)
Now it is given that
or 2 2/2 /2
2 2or
5 5H H H Hv v v v
Substituting the values from Equations (1) and (2), we get
2 2
2 2 22 sincos
5 2
uu u
or 2
2 sin5cos 2 1
2
or 2 2 2 35 1 sin 2 sin or sin
4
or 3
sin or 602
14. 0 1xu ucos45 10ms
0 1yu u sin45 10ms
2 2 2y xV v v
125= 2 1y yv 100 v 5ms
y yv u gt
5 10 10t
1
2
1t sec
t 1S2
t 3 / 2sec
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
15. 0xx
y
cot30 3 u 80 3m / su
u
x2uT T 16s
g
At x
Tt 4s,v 80 3m / s
4
yv 80 10 4 40m / s 2 2v (80 3) (40) 140m / s
16. Total time of flight is T 4s and if u its initial speed and the angle of projection. Then
2usinT 4
g
Or usin 2g
After 1 s velocity vector makes an angle of 045 with horizontal i.e,
Or ucos (usin ) (gt)
Or ucos 2g g(t 1)
Or ucos g
Squaring and adding Eqs. (1) and (2), we get
2 2 2 2 2u 5g 5(10) m / s u 22.36m / s
Dividing eq. (1) by (2), we get, tan 2 or 1tan (2)
17. As the plane is flying at a speed of 600 (5 /18), i.e.,
(500 / 3)m / s horizontally (at a height of 1960 m above
the ground), the taken by the kit to reach the ground
2h 2 1960t 20s
g 9.8
And in this time the kit will move horizontal by
x ut (500 / 3) 20 (10,000 / 3)m
So the angle of sight x 10000 10
tan 1.7 3h 3 1960 5.88
1 0tan ( 3) 60
x
h
A
P
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
C
B
u A
h
18.
OA OB BC OBt t t t T
19. BCt 6 / 2 3s
AC a AC
ut 12 / 2 6s t t
g
160u ms
6 3 3AB AC BCt t t S
Further 2AB AB
1h ut gt
2
21(60)(3) (10)(3)
2 135m
20. Acceleration is slope of v t graph is negative between 5 to 6t t . Between t 0 to t 4s particle
changes its direction of motion at t 1s and at t 3s
21. Maximum height and time of flight depends on the vertical component of initial velocity
1 2 y1 y2H H u u
Hence, 1 2T T
Range 2u sin 2 2(usin )(ucos )
Rg g
x y2u u
g
2 1 xz x1R R u u or 2 1u u
22. (b) for dropped body d
2Ht ;
g for projected body d
2Ht
g they are suffering same vertical
displacements
(c) Ratio of displacements 1:3:5 from nS u a(n 1/ 2)
(d) 1V 2gh dropped body 22V u 2gh for projected body
So (a) is incorrect.
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
23. Let the distance between P and Q is 2s .
Then 2 2 2 2v u a s
or 2 2
17 7 4as
2 24 240 /as m s
or 2 22 120 /as m s
velocity of particle at R is
2,2 2 2 7 120 169v u as
, 13 /v m s
Now, let 1 2andPR RQt t t t
Then ,1v u at
or 113 7 at
or 1 6 /at m s …(1)
Similarly, ,2v v at
or 217 13 at
or 2 4 /at m s …(2)
From equations (1) and (2) 1
2
6 3
4 2
t
t
Average velocity between andR Q is
, 22 2
,2
2 2
1122
RQ
v t ats
v v att t
1
13 4 15 /2
m s
Similarly, average velocity between p and R is
P R Q
s s
7 /u m s
,v 17 /v m s
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
21 1
11 1
1122
PR
ut ats
v u att t
1
7 6 10 /2
m s
24. We know that 2 2sin 2 2 sin cosu u
Rg g
…(1)
Also 90
R R
Now, 12 sinu
Tg
and
2
2 sin 90 2 cosu uT
g g
2 2
1 2 2
4 sin cos 2 sin cos 2 2u u RT T
g g gg
or 1 2T T R
25. Substituting 1 2a ,a and 3a from eqn. (i), (ii) and (iii) respectively in eqn. (iv) we have
1 2 3
T T 2Tg g 2 g 0
m m m
This gives 1 2 3
1 2 2 3 1 3
4m m mT g
4m m m m m m
26. From the addition of two vectors, we know that 2 2 2 2 cosC A B AB
From this expression it is clear that
(a) 2 2 2 when 90C A B
(b) 2 2 2 when 90C A B
(d) and 2 2 2 when 90C A B ; Hence C is incorrect
27. 2 2 9 4% 16 6% 9 0.36 16 0.96 25 1.32 25 5.28% 5 2.64% x y
2 10 4% z . Therefore, 2 2
5 2.64%0.5 6.64%
2 10 4%
x y
z Answer is b.
28. Dipole Moment = (Charge) × (Distance)
Electric Flux = (Electric Field) × (Area)
29. It is equal to 5 4 3 2AB BC CD DE EF . But AB DE and BC EF
So 3 3 6AB BC CD AD AO .
30. Area
22 4
2 u uR
g g
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
Chemistry:
31. Equating No. of moles 1 4
x 6416 x
32. 3 2Ba(NO ) is limiting reagent 3moles 3 2Ba(NO ) 1mole 3 4 2Ba (PO )
2 12moles ___________?
3
33.
34. That which undergoes reduction, requires reducing agent for that process
3BrO BrO under goes reduction
35. 7 3 4 4
2 24 4 2(s) 3MnO C O MnO C O
2 2
4 34 2 2 24OH 2MnO 3C O 2MnO 6CO 2H O
36. meq N V(ml)
37. Sucrose 21molin1000gH O
moles of 1000
water18
moles of sucrose =1
total moles 1000
118
mole fraction of sucrose 1 18
0.0181000 1018
118
38. 100 ml soln has-1.06 gm
10ml_______?0.106gm
2 3meq of of Na CO meq of acid
Ca
+2 Cl-1
O-Cl-2 +1
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
0.106
1000 2 meq53
acid (NVml)
39. 2 3 2 2M CO 2HCl 2MCl H O CO
M is a monovalent cation.
Thus, equivalent weight of 2 3
molar massM CO M 30
2
Let unused HCl be VmL
V mL of 1N HCl 100mL of 0.5 N NaOH
V 50mL
1 N HCl used 150 50 100mL 0.1 equiv.
Equiv. of 2 3
5.3M CO 0.1
M 30
M=23
40. n
nhmvr
2
20n a nh
mv :Z 2
0a x Bohr’s radius of H-atom n 3,z 1
9x 3hmv
1 2
hmv
6 x
h6 x
mv
41. 2 2 3 2CO 2NaOH Na CO H O
1mole 2 40gm The mixture has 2moles of co So it requires 4 moles of
NaOH
42. As mass decreases, kinetic energy increases to have same wave length.
43. Radial node n 1 angular node .
44. The two orbits are either I and II or II and IV
n2
n1
r 4
r 1 and 2
nr n
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
Thus, 2 1
13.6E E 13.6 10.2ev
4
and 4 2
13.6 13.6E E 2.55eV
16 4
45. Magnetic moment n(n 2) : where n is no. of unpaired electron n(n 2) or n 4
Thus, electronic configuration of aMn having 4 unpaired electron is
3 2 2 6 2 6 425Mn : ls ,2s ,2p ,3s 3p 3d
46. 3 212
r Hr Be 2
4
2 3 n2 1 n
r H( r H r H 2 andr Be )
n
47. h
mu and
nhmur
2
2 r n i.e., circumference of an orbit is integer multiple of
48. H 2H 12 2
1 1 1R
n n
2H 2
12 2Li
1 1 1R 9
n n
2
H
Li9
49. 10 lines means electron must be in 5th orbit
(5 1) 4 3 2 1 3 lines in balmer senes
50. General representation is n,l,m
51. Moles of 3
wt 4.25NH 0.25
mol.wt 17
moles of H-atoms = 0.25 3 0.75
total moles of atoms= 0.75 0.25 1
No. of molecules 230.25 6.023 10 231.5 10
52. Let 2 5C H OH in the mixture 1w g
and 2H O 2w g
moles of 2 5C H OH ,
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
11
22 2
wn
46
wMole of H O,n
18
Total moles 1 21 2
w wn n
46 18
Mole fraction of 12 5
1 2
n 1C H OH 0.25
n n 4
1
2
n 1mol 46g
n 3mol 54g
Total mass of solution =46+54=100g
Thus, 100 g contains 46 g 2 5C H OH and 254gH O Thus, (c) and (d) are correct.
53. There should be increase and decrease in oxidation state
54. Eq . wt=mol .wt
Change in oxidation. State per molecule Change in oxidation state must be one
2 2.5
2 2 3 2 2 4 6Na S O I NaI Na S O
55. Conceptual
56. In ‘C’ m-value cannot be -1
57. 3 4 2 4 2H PO Ca(OH) CaHPO 2H O
1 mole 1 mole
(a) Eq. wt of 3 4
98H PO 49
2
(b) One mole of 2Ca(OH) can neutralize only two gm equivalent of base is verified
(c) No. of gm equivalents are equal.
58. ‘ l ’ value never greater then ‘n'
59. In 2 6Fe d electrons are present
60. Conceptual
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
Maths:
61. Ans: (b) 2
48 48 48 481 log 48 log 12 log 24 2log 24 2abc bc
62. Ans: (c) 310
10
32.3 10 , .log 2.3 3
log 2.3
xx x
310
10
30.23 10 , log 0.23 3
log 0.23
yy y
63. Ans: (d) Since cos 1 sin 1ecx x which is impossible.
64. Ans: (b) Use verification by taking 0 .
65. Ans: (d) Sum of infinite terms of G.P1
a
r
1 1 4 2 3 4 2 3 34 2 3 1 sin 1
1 sin 4 24 2 3 4 2 3x
x
3 2sin (or) .
2 3 3x x
66. Ans: (b) 3
, 1 sin cos sin ,4 2 4 2 2
A A AA
cos sin 0,cos sin 0
2 2 2 2
A A A A .
2cos cos sin cos sin cos sin cos sin 1 sin 1 sin2 2 2 2 2 2 2 2 2
A A A A A A A A AA A
67. Ans: (b) Put 5
6
and verify.
68. Ans: (d) cot cot 60 cot 60 cot 60 cot cot 60 3 ,R n n z .
69. Ans: (b)
1 cot 90 2 cot 62 1 cot88 cot 62 cot88 cot 62 13
cot 62 cot88 cot 62 cot88tan 90 62 cot88k
cot 88 62 cot150 3
1k
70. Ans: (b) 3 3 30 3x y z x y z xyz
(a) Put 3 3 3 3 3 3 3, , 0 3x a y b z c a b c a b c abc .
(b) 4 4 4 4 4 40a b c a b c .
Squaring on both sides 4 42. 2.a b ab c a b c ab
Squaring on both sides 2 2 2 4a b c ab bc ca ab
2 2 2a b c ab bc ca
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
71. Ans: (d) sin ,sin 904 4
r r
y x
2 22 2
2 2sin cos 1 1
16 16
r r
y x
2 2
2 2 2 2
2 2
1 11 16
16
rx y x y
y x
22 2 4 4 1 0
y yx y xy
x x
tan 2 3 15 (or) 75
y
x
72. Ans: (c) In 1, tan tan tan tan tan tan 6 2tan tan 3 tan 3ABC A B C A B C C C C
Now tan tan 3, tan tan 2A B A B tan , tan 1,2A B
73. Ans: (b)Put
23 5 24 4
23 5 1
log . Now 2 log 24 4 2
tt
xt
x t x tt
.
23 23 4 5 1
3 4 5 2 04 2
t tt t t
t
1
2 32
1log 1 (or) 2 (or) - 2 (or) 2 (or) 2
3t x x
74. Ans: (a) tan tan
tan tan , tan tan tan1 tan tan 1 1
p pp q
q q
2 2sin sin cos cosp q
22 22
2 22 2
1 1 1. 1 . 1
1 1
p q q qp p p q q qq
p q p q
31. Ans: (c) 25 5log log 2 log 2 5 25aa x x x
32. c 2 2 2 2 2 2a b c a b c
1 1
log loglog log
c ca b a b
a b a bc c
2 2 2log log 2c ca b c
33. Ans: (a) 3 5 75x y z k 11 1
3 ,5 ,75yx zk k k
2 1125 5 3 y xzk k k
1 2 12
xyx y
z y x z 2 2xy xz zy xy zy xz
34. Ans: (b) 3 2 3 0a x b x c
33 2 3 3 2 33 . .a x b x c a x b x c 3 2 3 3 3a x b x c abcx
90
4
r
r
y
x
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
35. a 4 5
cos ,sin5 13
3 5
tan , tan4 12
tan tantan 2 tan
1 tan tan
3 5
4 12tan 23 5
14 12
36 20 56
48 15 33
36. Ans: (c) qec cotcos , q
ec1
cotcos
qqec
qqecec
1
2
1cos
1cotcoscotcos
37. Ans (b) 233 21
21
2
x , 233 21
21
2
y ,
LHS2
21
21
2
21
21
21
21
33122332333
3
15123323312 2
12
11
641643
11512
38. Ans: (c) 31.0log31.0log 1.03
1.0 , 29log29log 92
9 y
Then 122232 yx
ba,xyyxyxba thatsuch 2
39. Ans: (b) 360tan 0 0tan 40 20 3 0 0tan 40 tan 20
31 tan 40 tan 20
0 0 0 0tan 20 tan 40 3 tan 20 tan 40 3 .
40. Ans: (b) LHS 2 2
1 tan tan tan tan
tantan2tantantantan2tantan1 2222
22 tan1tan1 22 secsec
85. Ans: (b) 4 2 2 2 2 2sec sec sec sec 1 sec tan 2 2 2 41 tan tan tan tan .
86. Ans: (b) We have, 6 6 2 2sin cos 3sin cos (1)
222266 cossincossin3cossin 1cossin322
87. Ans: (a) 2 2 2 2 2 2
ab a b a b b c bc b c ca c a c ax x x
Subject Topic Test # Date
P + C + M Term Examination IIT – XI - TE
JUL 2013
3 3 3 33 3 3 3 3 3 3 3 3
3
2
2
1 a b c ab c a b b c c a a
ax x x x x
x
88. Ans: (a)
1
3 2 2 2
4 4 112 2 3 3 33
3 3 2 2 7
5 5 3 5 3
4 1 3 4 11 53 2 2
3 3 32 2
2 4 6 2 42
3 3 3
3 2 7 3 2 7
5 5
228
5
2283
0
0
.
89. Ans: (a) Let d be the diameter of moon and t its distance from the earth. If the moon
subtend an angle at the eye of the observer, then 27
113157
18060
31350000
ld cm.
90. Ans: (c) 5 5log 7 log 7 33
310
10
1 15 5 7 1
log 101log
10
283