KISHORE VAIGYANIK PROTSAHAN YOJANA (KVPY) | STREAM (SA) XI

FULL SYLLABUS TEST-1

KISHORE VAIGYANIK PROTSAHAN YOJANA

(KVPY) | STREAM (SA)_XI

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Time: 3 Hours Max. Marks : 100.

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

INSTRUCTIONS

1. Immediately fill the particulars on this page of the Test Booklet with Blue / Black Ball Point Pen. Use of pencil is strictly prohibited.

2. The question paper consists of two parts (both contain only multiple choice questions) for 100 marks. There will be four sections in Part-A (each containing 15 questions) and four sections in Part-B (each containing 5 questions).

3. There are Two parts in the question paper. The distribution of marks subjectwise in each part is as under for each correct response.

MARKING SCHEME :

PART-A :

MATHEMATICS

Question No. 1 to 15 consist of ONE (1) mark for each correct response & – 0.25 for incorrect response. PHYSICS Question No. 16 to 30 consist of ONE (1) mark for each correct response & – 0.25 for incorrect response. CHEMISTRY Question No. 31 to 45 consist of ONE (1) mark for each correct response & – 0.25 for incorrect response. BIOLOGY

Question No. 46 to 60 consist of ONE (1) mark for each correct response & – 0.25 for incorrect response. PART-B :

MATHEMATICS

Question No. 61 to 65 consist of TWO (2) marks for each correct response & – 0.5 for incorrect response.PHYSICS Question No. 66 to 70 consist of TWO (2) marks for each correct response & – 0.5 for incorrect response. CHEMISTRY Question No. 71 to 75 consist of TWO (2) marks for each correct response & – 0.5 for incorrect response. BIOLOGY

Question No. 76 to 80 consist of TWO (2) marks for each correct response & – 0.5 for incorrect response.

4. No deduction from the total score will be made if no response is indicated for an item in the Answer sheet. 5. No candidate is allowed to carry any textual material, printed or written, bits of papers, paper, mobile phone,

any electronic device, etc., except the Admit Card inside the examination hall/room. 6. Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the

bottom of each page. 7. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the

Room/Hall. However, the candidates are allowed to take away this Test Booklet with them. 8. Do not fold or make any stray marks on the Answer Sheet.

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(KVPY-STREAM-SA)_XI FULL SYLLABUS TEST PAPER-1

PART- A One Mark Questions

MATHEMATICS

1. The number of real roots of the polynomial equation x4 – x2 + 2x – 1 = 0 is (A) 0 (B) 2 (C) 3 (D) 4 2. Let P be an interior point of a convex quadrilateral ABCD and K, L, M, N be the midpoints of AB, BC,

CD, DA respectively. If area (PKAN) = 25, area(PLBK) = 36, and area(PMDN) = 41 then area(PLCM) is : (A) 20 (B) 29 (C) 52 (D) 54

3. The number of polynomials p(x) with integer coefficients such that the curve y = p(x) passes through (2, 2) and (4, 5) is

(A) 0 (B) 1 (C) more than 1 but finite (D) infinite

4. Let N1 = 255 + 1 and N2 = 165. Then

(A) N1 and N2 are coprime (B) the HCF (Highest Common Factor) of N1 & N2 is 55 (C) the HCF of N1 and N2 is 11 (D) the HCF of N1 and N2 is 33

5. Let s be the sum of the digits of the number 152 × 518 in base 10. Then (A) s < 6 (B) 6 s < 140 (C) 140 s < 148 (D) s 148 6. Let t be real number such that t2 = at + b for some positive integers a and b. Then for any choice of

positive integers a and b. t3 is never equal to (A) 4t + 3 (B) 8t + 5 (C) 10t + 3 (D) 6t + 5 7. Let p(x) = x2 + ax + b have two distinct real roots, where a, b are real numbers. Define g(x) = p(x3) for all

real number x. Then which of the following statements are true ? I. g has exactly two distinct real roots II. g can have more than two distinct real roots III. There exists a real number a such that g(x) for all real x

(A) Only I (B) Only I and III (C) Only II (D) Only II and III 8. Let x0, y0 be fixed real numbers such that 2 2

0 0x y > 1. If x, y are arbitrary real numbers such that

x2 + y2 1, then the minimum value of (x – x0)2 + (y – y0)2 is

(A) 22 20 0x y 1 (B) 2 2

0 0x y 1 (C) (|x0| + |y0| – 1)2 (D) (|x0| + |y0|)2 – 1

9. The number of non-negative integer solutions of the equations 6x + 4y + z = 200 and x + y + z = 100 is :

(A) 3 (B) 5 (C) 7 (D) Infinite 10. The number of pairs (a, b) of positive real numbers satisfying a4 + b4 < 1 and a2 + b2 > 1 is (A) 0 (B) 1 (C) 2 (D) more than 2



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11. Consider a triangle PQR in which the relation QR2 + PR2 = 5PQ2 holds. Let G be the point of

intersection of medians PM and QN. Then QGM is always (A) less than 45º (B) obtuse

(C) a right angle (D) acute and larger than 45º 12. Let x1, x2, …., x11 be 11 distinct positive integers. If we replace the largest of these integers by the

median of the other 10 integers, then (A) the median remains the same (B) the mean increases (C) the median decreases (D) the mean remains the same

13. All the vertices of a rectangle are of the form (a, b) with a, b integers satisfying the equation (a – 8)2 – (b – 7)2 = 5. Then the perimeter of the rectangle is (A) 20 (B) 22 (C) 24 (D) 26 14. Let N be the least positive integer such that whenever a non-zero digit c is written after the last digit of

N, the resulting number is divisible by c. The sum of the digits of N is : (A) 9 (B) 18 (C) 27 (D) 36

15. Let f be a function defined on the set of all positive integers such that f(xy) = f(x) + f(y) for all positive integers x,y. If f(12) = 24 and f(8) = 15, the value of f(48) is

(A) 31 (B) 32 (C) 33 (D) 34

PHYSICS

16. In an experiment on simple pendulum to determine the acceleration due to gravity, a student measures the length of the thread as 63.2 cm and diameter of the pendulum bob as 2.256 cm. The student should take the length of the pendulum to be

(A) 64.328 cm (B) 64.3 cm (C) 65.456 cm (D) 65.5 cm 17. A cylindrical vessel of base radius R and height H has a narrow neck of height h and radius r at one

end (see figure).The vessel is filled with water (density w) and its neck is filled with immiscible oil (density 0) Then the pressure at

H

M N

h

2r

2r

(A) M is g (h0 + Hw) (B) N is g (h0 + Hw) 2

2

R

r

(C) M is gHw (D) N is 22

20

2w

rR

hrHRg

18. A car around uniform circular track of radius R at a uniform speed v once in every T second. The

magnitude of the centripetal acceleration is ac. If the car now goes uniformly around a larger circular track of radius 2R and experiences a centripetal acceleration of magnitude 8ac, then its time period is.

(A)2T (B) 3T (C)T/2 (D) 3/2T



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19. A rigid ball rolls without slipping on a surface shown below.

Which on the following is the most likely representation of the distance travelled by the ball vs time

graph ?

(A)

time A.

distance

(B)

time B.

distance

(C)

time C.

distance

(D)

time D.

distance

20. A box filled water has a small hole on its side near the bottom. It is dropped from the top of a tower. As

it falls, a camera attached on the side of the box records the shape of the water stream coming out of the hole. The resulting video will show (A) the water coming down forming a parabolic stream. (B) the water going up forming a parabolic stream. (C) the water coming out in a straight line. (D) no water coming out.

21. If a ball is thrown at a velocity of 45 m/s in vertical upward direction, then what would be the velocity

profile as function of height ? Assume g = 10 m/s2.

(A)

(B)

45

V(m/s)

0 0 Height (m) 101

(C)

45

V(m/s)

0 0 Height (m) 101

(D)

45

V(m/s)

0 0 Height (m) 101



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22. A particle at a distance of 1m from the origin starts moving such that dr/d = r, where (r, ) are polar coordinates. Then the angle between resultant velocity and tangential velocity component is :

(A) 30 degrees (B) 45 degrees (C) 60 degrees (D) dependent on where the particle is 23. A student performs an experiment to determine the acceleration due to gravity g. The student throws a

steel ball up with initial velocity u and measures the height h traveled by it at different times t. The graph the student should plot on a graph paper to readily obtain the value of g is.

(A) h versus t (B) h versus t2 (C) h versus t (D) h/t versus t

24. Consider a bowl filled with water on which some black pepper powder have been sprinkled uniformly. Now a drop of liquid soap is added at the center of the surface of water. The picture of the surface immediately after this will look like

(A)

(B)

(C)

(D)

25. A solid cube of wood of side 2a and mass M is resting on a horizontal surface as shown in the figure.

The cube is free to rotate about a fixed axis AB. A bullet of mass m (<< M) and speed v is shot horizontally at the face opposite to ABCD at a height of 4a/3 from the surface to impart the cube and angular speed . It strikes the face and embeds in the cube. Then is close to (note : the moment of inertia of the cube about an axis perpendicular to the face and passing through the center of mass is

22Ma3

A

B

C

D

(A) Mvma

(B) Mv2ma

(C) mvMa

(D) mv2Ma

26. In Guericke's experiment to show the effect of atmospheric pressure, two copper hemispheres were

tightly fitted to each other to form a hollow sphere and the air from the sphere was pumped out to create vacuum inside. If the radius of each hemisphere is R and the atmospheric pressure is P, then the minimum force required (when the two hemispheres are pulled apart by the same force) to separate the hemispheres is

(A) 2R2P (B) 4R2P (C) R2P (D) R2P/2



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27. A ball is dropped vertically from height h and is bouncing elastically on the floor (see figure). Which of

the following plots best depicts the acceleration of the ball as a function of time.

t

h

(A)

t

acce

lera

tion

(B)

t

acce

lera

tion

(C)

t

acce

lera

tion

(D)

t

acce

lera

tion

28. A particle starts moving along a line from zero initial velocity and comes to rest after moving distance d. During its motion it had a constant acceleration f over 2/3 of the distance, and covered the rest of the distance with constant retardation. The time taken to cover the distance is (A) 2d/3f (B) 2 d/3f (C) 3d/ f (D) 3d/2f

29. A long cylindrical pipe of radius 20 cm is closed at its upper end and has an airtight piston of negligible

mass as shown. When a 50 Kg mass is attached to the other end of the piston, it moves down. If the air in the enclosure is cooled from temperature T to T – T, the piston moves back to its original position. Then T/T is close to (Assuming air to be an ideal gas, g = 10 m/s2, atmospheric pressure is 105 Pascal),

L

(A) 0.01 (B) 0.02 (C) 0.04 (D) 0.09 30. Force F applied on a body is written as ˆˆ ˆF (n.F)n G , where n̂ is a unit vector. The vector G is equal

to

(A) n̂ F (B) ˆ ˆn (n F) (C) ˆ(n F) F/ | F | (D) ˆ ˆ(n F) n



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CHEMISTRY

31. Fluoxymesterone, C20H29FO3, is an anabolic steroid. A solution is prepared by dissolving 10.0 mg of the steroid in 500 mL of water. How many moles of Fluoxymesterone are present in 1 mL of solutions.

(A) 1.16 × 10–10 (B) 1.19 × 10–17 (C) 5.95 × 10–8 (D) 2.38 × 10–11

32. What volume of 2 10–4 M Ba(OH)2 must be added to 300 mL of a 1 10–4 M HCl solution to get a solution in which the molarity of hydronium (H3O

+) ions is 5 10–11 M ? (A) 375 mL (B) 300 mL (C) 225 mL (D) 450 mL 33. An electron is accelerated from rest and it has wavelength of 1.41 Å by how much amount potential

should be dropped so that wavelength associated with electron becomes 1.73 Å. (A) 25 V (B) 50 V (C) 75 V (D) 12.5 V

34. In which of the followng arrangements, the sequence is not strictly according to the property written against it ?

(A) BeCO3 < MgCO3 < CaCO3< SrCO3 – Thermal stability.

(B) BeSO4 > MgSO4 > CaSO4 > SrSO4 – Solubility in water.

(C) Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2 – Basic strength.

(D) MgF2 < CaF2 < SrF2 < BaF2 – Lattice energy. 35. The hybridisation and shape of BrF3 molecule are: (A) sp3d and T shaped (B) sp2d2 and tetragonal (C) sp3d and bent (D) none of these 36. Rate of a chemical reaction does not depend upon : (A) temperature of reaction (B) concentration of reactants (C) moles of reactants (D) catalyst

37. Standard enthalpy of vapourisation vap Hº for water at 100ºC is 40.66 kJ mol–1. The internal energy of vaporisation of water at 100ºC (in kJmol–1) is :

(Assume water vapour to behave like an ideal gas). (A) + 37.56 (B) – 43.76 (C) + 43.76 (D) + 40.66

38. Which species among the following have smallest ionic radius :Al3+, K+ , Mg2+ , Rb+ (A) Al3+ (B) K+ (C) Mg2+ (D) Rb+ 39. A gas cylinder contains 1 mole oxygen gas at 2.46 atm pressure and 27ºC. The mass of oxygen that

would escape, if first the cylinder was heated to 127ºC and then the valve was held open until the gas pressure was reduce to 1 atm, the temperature being maintained at 127ºC, is :

(A) 22.4 g (B) 11.2 g (C) 20.8 g (D) 9.6 g 40. Which one of the following alkali metals gives hydrated salts? (A) Li (B) Na (C) K (D) Cs 41. Which of the following statement is correct ? (A) The complete transfer of electron takes place in the inductive effect. (B) Inductive effect increases with increase in distance. (C) The resonance structures are hypothetical structure and they do not represent any real molecule. (D) The energy of resonance hybrid is always more than that of any resonating structure.



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42. Which of the following structures constitute resonance structures.

(A) & CH3 – O – N = O

(B) &

(C) (CH3)2 CO &

(D) CH3 – CH = CH – CH3 & CH3 – CH2 – CH = CH2

43. RX + Mg Ether RMgX 3CH OH n-Butane

What can be R in the above reaction sequence ?

(A) n-Butyl (B) sec-Butyl (C) n-Propyl (D) Isopropyl

44.

Order of rate of electrophilic addition reaction with HBr will be : (A) IV> I > III > II (B) I > II > III > IV (C) I > III > II > IV (D) IV > I > II > III 45. Write the IUPAC name of following compound

(A) 2-Methyl-3-bromo-4-sulphohexanoic acid (B) 3-Bromo-2-methyl-4-sulphohexanoic acid (C) 4-Bromo-5-carboxyhexane-3-sulphonic acid (D) 5-Carboxy-4-bromo-3-sulphonic acid

BIOLOGY

46. Prokaryotes possess

(A) Nucleus (B) Nucleoid (C) Nucleolus (D) None of the above

47. Endosperm of angiosperm is a ........ structure

(A) Haploid (B) Diploid (C) Tetraploid (D) Triploid

48. Primary roots and its branches constitute

(A) Adventitious root system (B) Tap root system

(C) Fibrous roots (D) Seminal roots

49. Living cells that provide mechanical support to the plant are -

(A) Parenchyma (B) Sclerenchyma (C) Collenchyma (D) chlorenchyma

50. A cell organelle containing hydrolytic enzymes is

(A) mesosome (B) lysosome (C) microsome (D) ribosome



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51. Guttation is the result of :

(A) Diffusion (B) Transpiration (C) Osmosis (D) Root pressure

52. The C4 plants are photosynthetically more efficient than C3 plants because

(A) The CO2 efflux is not prevented

(B)There is no photorespiration

(C) The CO2 compensation point is more

(D) CO2 generated during photorespiration is trapped and recycled through PEP carboxylase

53. Anaerobic products of fermentation are

(A) Alcohol and lipoprotein (B) Ether and nucleic acid

(C) Protein and nucleic acid (D) Alcohol & lactic acid

54. The body of______is unsegmented with a distinct head, muscular foot and visceral mass

(A) Asterias (B) Ophiura (C) Balonoglossus (D) Devil fish

55. The enzyme amylase present in animals can break glycosidic linkages between glucose monomers

only if the monomers are in the -form. Which of the following can be broken by amylase?

(A) Glycogen, starch and amylopectin (B) Glycogen and cellulose

(C) Cellulose and chitin (D) Starch, amylopectin and cellulose

56. Liver is the largest gland and is associated with various functions, choose the option which is not

correct

(A) Metabolism of carbohydrate (B) Digestion of fat

(C) Formation of bile (D) Secretion of hormone called gastrin

57. Which statement about the mammalian circulatory system is correct

(A) The average diameter of lumen of arteries is greater than that of veins

(B) The order of decreasing velocity of blood flow is Arteries > Capillaries> Veins

(C) The total surface area of the capillaries is considerably greater than that of all of the arteries,

arterioles, venules and veins

(D) In order to return blood to the heart, the blood pressure in veins is higher than it is in the capilaries



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58. Blood carries the CO2 in 3 forms the correct percentages of CO2 in these forms are

As Carbamino-haemoglobin in RBC As bicarbonates Dissolved form in plasma

(A) 20-25% 70% 7%

(B) 70% 20-25% 7%

(C) 20-25% 7% 70%

(D) 7% 20-25% 70%

59. Hypothalamus is the important part of the brain concerned with homeostasis of the body. The functions

of hypothalamus are

(A) Secretion of posterior pituitary hormones (B) Regulation of body temperature

(C) Regulation of water balance (D) All of these

60. Benedict's solution is not reduced by

(A) Glucose (B) Fructose (C) Maltose (D) Sucrose

PART- B

Two Mark Questions

MATHEMATICS 61. The number of solid cones with integer radius and integer height each having its volume numerically

equal to its total surface area is (A) 0 (B) 1 (C) 2 (D) infinite

62. Consider a semicircle of radius 1 unit constructed on the diameter AB, and let O be its centre. Let C be a point on AO such that AC : CO = 2 : 1. Draw CD perpendicular to AO with D on the semicircle. Draw OE perpendicular to AD with E on AD. Let OE and CD intersect at H. Then DH equals

(A) 1

5 (B)

1

3 (C)

1

2 (D)

5 – 12

63. What is the sum of all natural numbers n such that the product of the digits of n (in base 10) is equal to

36n10n2 ? (A) 12 (B) 13 (C) 124 (D) 2612

64. Let ABC be a triangle in which AB = BC. Let X be a point on AB such that AX : XB = AB : AX. If AC =

AX, then the measure of ABC equals

(A) 18° (B) 36° (C) 54° (D) 72°



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65. If a 3-digit number is randomly chosen, what is the probability that either the number itself or some permutation of the number (which is a 3-digit number) is divisible by 4 and 5 ?

(A) 1

45 (B)

29180

(C) 1160

(D) 14

PHYSICS

66. Persons A and B are standing on the opposite sides of a 3.5 m wide water stream which they wish to cross. Each one of them has a rigid wooden plank whose mass can be neglected. However, each plank is only slightly longer than 3m. So they decide to arrange them together as shown in the figure schematically. With B (mass 17 kg) standing, the maximum mass of A, who can walk over the plank is close to,

3.5 m

3 m A B

(A) 17 kg. (B) 65 kg (C) 80 kg (D) 105 kg 67. A uniform metal plate shaped like a triangle ABC has a mass of 540 gm. The length of the sides AB,

BC, and CA are 3 cm, 5 cm and 4 cm, respectively. The plate is pivoted freely about the point A. What mass must be added to a vertex, so that the plate can hang with the long edge horizontal ?

(A) 140 gm at C (B) 540 gm at C (C) 140 gm at B (D) 540 gm at B 68. A 20 gm bullet whose specific heat is 5000 J(kg–°C) and moving at 2000 m/s plunges into a 1.0 kg

block of wax whose specific heat is 3000 J (kg–°C) . Both bullet and wax are at 25°C and assume that (i) the bullet comes to rest in the wax and (ii) all its kinetic energy goes into heating the wax. Thermal temperature of the wax in °C is close to.

(A) 28.1 (B) 31.5 (C) 37.9 (D) 42.1



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69. On a pulley of mass M hangs a rope with two masses m1 and m2 (m1 > m2) tied at the ends as shown in

the figure. The pulley rotates without any friction, whereas the friction between the rope and the pulley is large enough to prevent any slipping. Which of the following plots best represents the difference between the tensions in the rope on the two sides of the pulley as a function of the mass of the pulley ?

m1 m2

M

T1 T2

(A)

M 0

T1

– T

2

(B)

M 0

T1

– T

2 (C)

M 0

T1

– T

2

(D)

M 0

T1

– T

2

70. A smaller with side b (depicted by dashed lines) is excised from a bigger uniform cube with side as shown below such that both cubes have a common vertex P. Let X = a/b. If the centre of mass of the remaining solid is at the vertex O of smaller cube then X satisfies.

b

P

O

(A) x3 – x2 – X – 1 = 0 (B) x2 – x –1 = 0 (C) x3 – X2 – X – 1 = 0 (D) X3 – X2 – X + 1 = 0

CHEMSITRY

71. 10 g of a hydrocarbon (not necessarily alkane) on analysis was found to contain 1 g hydrogen. If all H-atoms from 1 mole of hydrocarbon are removed and converted into H2 gas, then the gas produced has mass of 8 g. Then, molecular formula of hydrocarbon is :

(A) C2H4 (B) C3H4 (C) C4H8 (D) C6H8 72. Given below are a set of half-cell reactions (acidic medium) along with their E° (V with respect to normal hydrogen electrode) values.

2 + 2e– 2¯ E° = 0.54 Cl2 + 2e– 2Cl¯ E° = 1.36 Mn3+ + e– Mn2+ E° = 1.50 Fe3+ + e– Fe2+ E° = 0.77 O2 + 4H+ + 4e– 2H2O E° = 1.23 Among the following, identify the correct statement : (A) Chloride ion is oxidised by O2 (B) Fe2+ is oxidised by iodine (C) Iodide ion is oxidised by chlorine (D) Mn2+ is oxidised by chlorine



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73. The bond dissociation energy of gaseous H2, Cl2 and HCl are 104, 58 and 103 kcal mol–1 respectively. The enthalpy of formation for HCl gas will be

(A) – 44.0 kcal (B) – 22.0 kcal (C) 22.0 kcal (D) 44.0 kcal 74. The correct stability order of following is :

(I) (II) (III) (IV)

(A) I > II > III > IV (B) III > IV > II > I (C) II > IV > III > I (D) IV > III > II > I 75. Member of which of the following pair of isomers are not position isomers ?

(A) & (B) &

(C) & (D) &

BIOLOGY

76. Select the mismatched pair (A) Crossing over - Recombinase (B) Alignment of chromosomes at the equatorial plate - Metaphase (C) Separation of sister chromatids - Anaphase-I (D) G2 - Period of cytoplasmic growth and

DNA content is 4C

77. Apical dominance in higher plants is due to (A) Presence of Auxin (B) Enzyme activity and metabolism (C) Carbohydrates (D) Photoperiodism

78. Which of the following statement is absolutely correct? (A) Human kidneys can produce urine nearly fourteen times concentrated than the initial filtrate formed. (B) ADH can affect the kidney function by its dilator effects on blood vessels (C) ANF can cause vasodilation (dilation of blood vessels) and thereby decrease the blood pressure (D) NaCl is transported by the descending limb of Henle’s loop which is exchanged with the ascending

limb of vasa recta

79. Fight or flight reaction causes activation of the (A) Parathyroid glands, leading to increased metabolic rate (B) Kidney, leading to suppression of renin angiotensin-aldosterone pathway (C) Adrenal medulla, leading to increased secretion of epinephrine and norepinephrine (D) Pancreas leading to a reduction in the blood sugar levels 80. ATP is (A) Adenosine D-ribose -3- phosphate (B) Adenosine L-ribose -3- phosphate (C) Adenine D-ribose -3- phosphate (D) Adenine L-ribose -3- phosphate



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FULL SYLLABUS TEST-1 KISHORE VAIGYANIK PROTSAHAN YOJANA

(KVPY) | STREAM (SA)_XI_

ANSWER KEY

PART- A 1. (B) 2. (C) 3. (A) 4. (D) 5. (B) 6. (B) 7. (B)

8. (A) 9. (C) 10. (D) 11. (C) 12. (C) 13. (A) 14. (A)

15. (D)

16. (B) 17. (A) 18. (C) 19. (B) 20. (D) 21. (A) 22. (B)

23. (D) 24. (C) 25. (D) 26. (C) 27. (B) 28. (C) 29. (C)

30. (D)

31. (C) 32. (D) 33. (A) 34. (D) 35. (A) 36. (C) 37. (A)

38. (A) 39. (A) 40. (A) 41. (C) 42. (B) 43. (A) 44. (D)

45. (B)

46. (B) 47. (D) 48. (B) 49. (C) 50. (B) 51. (D) 52. (B)

53. (D) 54. (D) 55. (A) 56. (D) 57. (C) 58. (A) 59. (D)

60. (D)

PART- B 61. (B) 62. (C) 63. (B) 64. (B) 65. (B)

66. (D) 67. (C) 68. (C) 69. (C) 70. (A)

71. (D) 72. (C) 73. (B) 74. (D) 75. (A)

76. (C) 77. (A) 78. (C) 79. (C) 80. (C)



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HINTS & SOLUTIONS

PART- A MATHEMATICS

1. x4 – x2 + 2x – 1 = 0 X2 – (x –1)2 = 0 (x2 + x – 1) (x2 – x + 1) = 0 x2 + x – 1 = 0 has two real roots. 2.

M

N

B

D

C

L

K

A

P

Note : Area of APN = Area of PDN Area of APK = Area of PBK Area PCL = Area of PBL Area of PCM = Area of PDM Hence . Area (PKAN) + Area (PLCM) = Area (PMDN) + Area (PLBK) Hence Area (PLCM) = 36 + 41 – 25 = 52 3. y = P(x) = a0 + a1 a2 x2 + ……+anxn

a0,a1,a2,a3,……….an I 2 = P(2) ….(1) 5 = P(4) ….(2) By (1) & (2)

3 = a1 (4 – 2) + a2 (42 – 22) + a3(43 –23) + …….. + an(4n – 2n) Clearly RHS is even and LHS is odd no polynomial exists.

4. 165 = 3 × 5 × 11 x + y divides xn + yn

32 + 1 divides 3211 + 111

Hence N1 is multiple of 33, simultaneously unit digit in N1 is 9 so it not the multiple of 5

Hence HCF of N1 & N2 is 33 5. 152 × 518

= 9 × 520

2010log 9 5 = 3

10 102log 20log 5

= 2 4771 20 1– 0.3010

= 14 {characteristic} Hence the number have 15 digits At worst all digit can have value 9 Hence sum should less then 135 And the last digit should be 5 Hence Sum should greater than Or equal to 6

6. t2 = a + b

t3 = t2 . t = at2 + bt

= a(at + b) + bt t3 = (a2 + b) .t + ab

t3 = 4t + 3 when a = 1 and b = 3 t3 = 10t + 3 when a = 3 and b = 1 t3 = 6t + 5 when a = 1 and b = 5

but t3 can never be equal to 8t + 5 for any point of a and b.

7. If , are roots of p(x) = 0 Roots of g(x) = 0 are x3 = , Hence only 2 real roots

further g(x) b– 4

a2

xR.

Hence I & III are correct

8. 1yx 20

20 x0 – y0 fixed

x, y arbitrary

1yx 22 , Let x = con

y = sin for men z = (x – x0)2 + (y – y0)2

z = x2 + x02 + y2 + y02 – 2 (x x0 + y y0) put x - cos , y = sin z = 2

020 yx – 2 (x0 cos + y0 sin)

d

dz 0 – 2 (–x

0 sin + y

0 cos)

0ddz

–x0 sin = – y0 cos

tan = 0

0

xy

G y0

x0

20

20 yx

sin = 20

20

0

yx

y

cos = 20

20

0

yx

x

x = 20

20

0

yx

x

, y =

20

20

0

yx

y

z =

2

020

20

0

2

020

20

0 yyx

yx

yx

x

2

20

20

0

2

20

20

0 1yx

12y1

yx

12x



®




2

20

20

20

20

20

20

yx

yx1yx

2

20

20

220

20 1yxyx1

9. Subtracting the given equations we get 5x + 3y = 100

x = 20 – 5y3

y is multiple of 5, let y = 5k x = 20 – 3k k = 0, 1, 2, ……6 Hence numbers of solutions are 7. 10. Let a2 = m & b2 = N then m > 0 and N > 0

Now given condition is M + N > 1 and M2 + N2 < 1

(M,N) lies inside circle x2 + y2 < 1 and above line x + y > 1 (M, N) lies in shaded region and number of points in shaded region are infinite, so number of pair (a,b) are also infinite.

11.

G

M R Q

P

N

L

Let QR= p, PR = q, PQ =r Given p2 +q2 = 5r2

Now 22

22

2PM

3QN

GMQG

= 9

PM9

QN4

22

=

222222 pq2r241

qp2r241

.491

= 22

QM4

p

Hence Angle QGM is 90º. 12. Let x1 < x2 < x3 ……x11

median of x1, x2 …..x10 is 2

xx 65

Now the new set of number are x1, x2,

….x5, 2

xx 65 , x6, ….x10

Hence median is 2

xx 65 < x6

median decreases 13. (a – 8)2 – (b – 7)2 = 5 (a – b –1) (a + b – 15) = 5 I1 I2 Four cases I1 I2 5 1 1 5 –5 –1 –1 –5 Case-1 a – b – 1 = 5 & a + b – 15 = 1 a = 11, b = 5 Case-2

a – b – 1 = –5 & a + b – 15 = –1 a = 5, b = 9 Case-3

a – b – 1 = 1 & a + b – 15 = 5 a = 11, b = 9 Case-4

a – b – 1 = –1 & a + b – 15 = –5 a = 5, b = 5

Perimeter = 4 + 4 + 6 + 6 = 20 14. Let the number be N If we write c after the last digit now number

is 10N + c Now c|10N + c c = 1, 2, ….., 9 c|10N c = 1, 2, ….., 9 c|N for c = 4, 7, 9 Hence N is LCM of (4, 7, 9) = 252

(0,1)

(1,0)

(M,N)



®




so sum of digit = 9 15. f(xy) = f(x) + f(y)

f(x) = knx

Now, 24 = k(n12)

15 = k(n8) 5 = k(n2)

Let y = kn48

= k[n12 + n4]

= k 24 2 5k k

y = 34

PHYSICS

16. r = 1.128cm leff = l + r = 63.2 + 1.128 = 64.3

17.

H

M N

h

2r

2r

A

Pressure at Point A PA = 0hg Pressure at M PM = PA + W Hg 0h + W H)g

18. (i) 2

c

Va

R and

2 RT

V

(ii) ac' = 8ac

2

1V2R

= 2V

8R

V1 = 4V

T' = 2 (2R)

4V

= RV

= T2

19. During slope speed will be increase, so

distance covered will increase parabolically

with the time. Then speed will remain constant so further graph will be a straight line with greater slope than the first straight line portion of graph. So if change in speed at the sharp corners is not considered, the best answer should be (B) rather than C.

V

t

time

distance

However if kink is considered and Collision is inelastic then graph will be (C). But in either case answer cannot be D

20. Since bucket and water both are in state of free fall so water will not come out of the hole. (D)

21. v2 = v2 – 2gh parabola (A)

22. dr

rd

dr / dt

rd / dt

dr / dt

r

dr

rdt

Vt=r Vnet=r

2

=45°

Vr=r

23. R = 21Ut – gt2

h 1u – gtt 2



®




2ug

t

ht

1– g2

u

slope =

24. Surface tension of soap solution is less than water.

25.

C D

B

A

Q a

MI about AB

I = 2 2

22Ma 8MaM(a 2)

3 3

Li = Lf

mv 4a3

= I

(mv) 4a3

= 28ma

3

(m << M) can be

neglected

(mv) 4a3

= 2M8a

3

(mv)(4a)x33x8M

=

mv2Ma

=

26.

P P

P P

F = 2px R

= 2R p

27. a = –g except during collision. 28.

t2

Vmax

maxVf

= t1 t

v

tan = maxVt

t1 = maxVf

max 21

xV x t d2

...(1)

max 11 2d

xV x t2 3

...(2)

2max

4dfV

3

Vmax = 4dt3

...(3)

By (1) & (3)

2

1 4dft d

2 3

t2 = 3df

29.

P0, T0, V0

P

50kg

P0, T0, V0 P, T0, V P, T, V0

P0 V0 = PV

0

0

VVT T

0

0 0 0

VT P mg1–T V P AP

4

50

T 50x10x7x(10)1–T 22x400x(10)

0

T 7T 22x8

= 1– 78x22

= 0.04



®




30. F = ( ˆˆ ˆ(n.F)n a )

(nxF)xn = (nxn)F – (n.F) n̂

Ans- (D)

CHEMISTRY

31. Number of moles in 500 mL = 310 10

336

Number of moles in 1 mL = 310 10

336 500

=

5.95 × 10–8 .

32. [OH–] =wK

[H ] =

14

11

10

5 10

4 4V 2 10 2 300 1 10300 V

= 2 10–4

[OH–] V = 450 ml

33. 1 = 1/ 2150

V

V1 = 21

150

V1 = 2

150

(1.41) = 75 V

and V2 = 2

150

(1.73) = 50 V

Hence potential should be dropped by 25 V.

34. Lattice energy decreases as the size of the metal increases.

MgF2 = – 2906 kJ mol– ;

CaF2 = –2610 kJ mol–1

SrF2 = – 2459 kJ mol– ;

BaF2 = – 2367 kJ mol–1

37. H = E + n(g) RT 40.66 × 1000 = E + (1) × 8.314 × 373. E = 37.56 kJ mol–1

38. Rb+ and K+ has more number of shells than Mg2+ and AI3+ . AI3+ and Mg2+ are isoelectronic but AI3+ has higher nuclear charge so AI3+ < Mg2+. Hence Al3+ is the smallest one.

K+ = 1.38 Å, Mg2+ = 0.72 Å, Al3+ = 0.535 Å and Rb+ = 1.64 Å

39. 1 1

1

n TP =

2 2

2

n TP

1 3002.46

=

2n 4001

n2 = 0.3

Mass of oxygen left = 0.3 × 32 = 9.6 g

Mass of oxygen escaped = 1 × 32 – 9.6 = 22.4 g.

42. Position of nuclei does not changes in case of resonating structures. 43. R is CH3–CH2–CH2–CH2– or CH3 – –

CH2 – CH3

44. Rate of electrophilic addition reaction stability of carbocation.

45.

3-Bromo-2-methyl-4-sulphohexanoic acid

PART- B MATHEMATICS

61. Leh Height of cone = h

Radius of base = r

And slant height = ; = = 22 hr Given volume = surface area.

22 rrhr31

r33rh r3–rh31

3–h3r

hr 22

9h6–h9r

hr 22

22

3hr2–

9hr

h222

2

9–r

546

9–r

r6h

22

2

Since h and r must be integers, and r2 – g must be a factor of 54

r2 – 9 must be divisible by 3 .k3r

h = 1–k

66

9–k9

546

22



®




Since O < k2 – 1 < 6 k = 2 is only such value; for which h is integer. So, r = 2 × 3 = 6i

h = 6 + 36 = 8

10 is the only possible values for r and h.

62.

H E

AB

D r = 1

O C

ACCO

= 21

OC = 13

AC = 23

OD = 1

OCD : OD2 = OC2 + CD2

1 = 19

+ CD2

CD = 89

Let ADC =

tan = ACCD

= 2

8 =

1

2

cos = DEDH

= DA2DH

DH =DA

2cos { cos =

CDAD

}

= 2 2

3 2 1 1

12

= 1

2

DH = 1

2

63. Product of digits of natural number will be a non negative integer so, n2 – 10n – 36 0 n (–, 5 – 61 ] (5 + 61 , ) but n IN so n 13; where n N case-1 for all 2 digit natural numbers max value of product of digits = 9 × 9 = 81

so n2 – 10n – 36 81 n [5 – 142 , 5 + 142 ] but n is taken as a 2 digit natural no.; so 13 n < 17; product of digits = 3, 4, 5 or 6 for 13, 14, 15 and 16 respectively checking n = 12 product of digits = 1 × 3 = 3 and 132 – 10 × 13 – 36 = 3 so 13 satisfies the given condition Hence it is a solution chcking for n = 14 product = 1 × 4 = 4 142 – 10 × 14 – 36 = 196 – 140 – 36 = 20 > 6 and n2 – 10n – 36 is increasing function for n > 5; rest of the 2 digit integers won't satisfy the given condition case-2 for all 3-digit integers max product = 9 × 9 × 9 = 729 The smallest 3 digit no. is 100 f(n) = n2 – 10n – 36; f(100) = 1002 – 10 × 100 – 36 = 8964 > 729 and f(n) is increasing Hence no 3 digit Integers and similary any higher integer will not satisfy n = 13 is the only answer.

64.

B

x

D

C b A

a–b

b

Let AC = b & AB = BC = a

Given ba

bab

b2 = a2 – ab

2

51ab

Let D be foot of perpendicular from B

sin (ABD) = a2

b=

415

ABD = 18º ABC = 36°

65. Let abc be required numbers which is

divisible by 20 Clearly one digit must be 0 and one digit must be even, other can be any

Case-I when two digits are 0,0

x 0 0 x = {1,2,.....,9} 9 ways



®




Case-II when one O and remaining two

are distinct x y Hence number of ways = 4 × 8 × 2 × 2 = 128 ways

Case-III when two are even 0 and same and other 0.

4 × 1 × 2 = 8 ways Total favorable ways = 9 + 128 + 8 = 145 Total ways = 900

Probability = 145900

= 29

180

PHYSICS

66.

0.5 m 2.5 m

Ma A

B

3.5 m

17 Kg

3 m 3 m

A170x2.5 M x10x0.5 MA = 85 67.

C 4 A

5 3

4/3 37º

C1

1

B

37º 53º

C1

5/3

3cos53º6 = 9/5 B C

4

A

3

here = 53º – 37º

AC1 = 2

2 41

3

AC1 = 53

For equilibrium, the combine centre of mass should be vertically below the point of suspension 'A' for this : m1x1 = m2x2

(540) 5

sin3

= (mB) 95

B

5 9(540) sin(53º 37º ) (m )

3 5

solving mB = 140 gm So answer should be (C)

68. 12

mV2 = m1s1T + m2s2T

21 20 20(2000) 500 (1 3000) T

2 1000 1000

T = 12.9ºC Tf = 25ºC + 12.9ºC = 37.9ºC 69.

m1g – T1 = (m1)(a) T2 – m2g = (m2) (a)

(T1 –T2) = (I)2

a

R

(m1 – m2)g = (m1 + m2 + 2i

R)a

a = 1 2

1 2 2

(m – m )gI

(m m )R

(T1 – T2) = 2

2M(R ) a

2 R

a = constant (T1 – T2) = 1 2

1 2

(m – m )g(M)2 M

m m2

(T1 – T2) =

1 2

1 2

(m – m )g2(m – m )

1M

M = 0 (T1 – T2) = 0; and M , T1 – T2 = (m1 – m2)g

70. Assume vertex P as origin. All the three

coordinates are symmetrical, so we can deal only with x-coordinate

xcm = 1 1 2 2

1 2

m x m xm m



®




b =

3 3

3 3

a b(a ) (b )

2 2a b

2b =

4 4

3 3

a ba b

2b

=

22

22

22

22

baba

baba

babab–a

bab–aba

a3 – a2b –ab2 – b3 = 0, dividing by b3

01–ba–

ba–

ba

23

x3 – x2 – x – 1 = 0

CHEMSITRY

71. C : H = 9

12 :

11

= 3 : 4

Empirical formula = C3H4 So, molecular formula = (C3H4)n = C3nH4n

Moles of H-atoms formed after removal = 1 × 4n

Moles of H2 formed = 4n2

= 2n

Mass of H2 formed = 2n × 2 = 4n = 8 (given)

n = 2 Mol. formula = C3 × 2H4 × 2 = C6H8 72. According to SRP values chlorine can oxidise iodide ion.

73. Given H2 (g) HCl (g); H = 104 kcal ...(1) Cl2 (g) 2Cl(g) ; H = 58 kcal ...(2) HCl (g) H(g) + Cl(g) ; H = 103 kcal ...(3) Heat of formation for HCl

12

H2 (g) + 12

Cl2 (g) HCl (g) ; H = ?

Divide equation (1) and (2) by 2, and then add

12

H2 (g) + 12

Cl2 (g) H(g) + Cl(g) ;

H = 81 kcal ...(4) Subtracting equation (3) from equation (4)

12

HCl (g) H(g) + Cl(g) ;

H = 103 kcal ...(3) – – – – ––––––––––––––––––––––––––––––––––––– 12

H2 (g) + 12

Cl2 (g) HCl(g) ; H = – 22.0

kcal

Enthalpy of formation of HCl gas = – 22.0 kcal

74. Stability resonance Hyperconjugation length of conjugation is equal in all I, II, III &

IV. But Hyperconjugation is IV > III = II > I and

Hyperconjugative structure of III is more stable than II since it is more delocalised.

75. & are metamers



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(KVPY-STREAM-SA)_XI FULL SYLLABUS TEST PAPER-1

PART- A

One Mark Questions

xf.kr

1. cgqin lehdj.k x4 – x2 + 2x – 1 = 0 ds okLrfod ewyksa dh la[;k gS% (A) 0 (B) 2 (C) 3 (D) 4

2. fcUnq P ,d voeq[k prqHkqZt (convex quadrilateral) ABCD ds vanj gS ,oa fcUnq K, L, M ,oa N Hkqtk,¡ Øe'k% AB, BC, CD ,oa DA ds e/; fcUnq gSA ;fn {ks=kQy (PKAN) = 25, {ks=kQy(PLBK) = 36 rFkk {ks=kQy (PMDN) = 41 gS rc {ks=kQy(PLCM) D;k gksxk ? (A) 20 (B) 29 (C) 52 (D) 54

3. cgqin p(x) ds lHkh xq.kkad (coefficients) iw.kk±d gSA ;fn oØ js[kk y = p(x) fcUnqvksa (2, 2) ,oa (4, 5) ls xqtjrh gS] rc ,sls cgqinksa dh la[;k gksxh

(A) 0 (B) 1 (C) 1 ls vf/kd] ijarq lhfer (D) vuar

4. ;fn N1 = 255 + 1 rFkk N2 = 165 gS] rc (A) N1 rFkk N2 vlgHkkT; la[;k,¡ gSA (B) N1 rFkk N2 ds egÙke lekioZrd dk eku 55 gSA (C) N1 rFkk N2 egÙke lekioZrd dk eku 11 gSA (D) N1 rFkk N2 egÙke lekioZrd dk eku 33 gSA

5. Lka[;k 152 × 518 dks ;fn vk/kkj (base) 10 esa fy[kk tk,] rc blds vadksa dk ;ksx S gSA rc (A) s < 6 (B) 6 s < 140 (C) 140 s < 148 (D) s 148



®




6. dqN /kukRed iw.kk±d la[;kvksa a vkSj b ds fy, ;fn t ,d okLrfod la[;k bl izdkj gS fd t2 = at + b rc fdlh

/kukRed iw.kk±d a vkSj b ds fy,] t3 fuEufyf[kr esa fdlds cjkcj ugha gS ?

(A) 4t + 3 (B) 8t + 5 (C) 10t + 3 (D) 6t + 5

7. eku ysa fd p(x) = x2 + ax + b, tgka a ,oa b okLrfod la[;k,¡ gS] ds nks fHkUu okLrfod ewy gSA lHkh okLrfod la[;kvksa x ds fy, g(x) = p(x3) dks ifjHkkf"kr dhft,A rc fuEu esa ls dkSulk dFku lR; gS ?

I. g ds dsoy nks fHkUu&fHkUu okLrfod ewy gSA II. g ds nks ls vf/kd fHkUu&fHkUu okLrfod ewy gks ldrs gSA III. ,d okLrfod la[;k bl izdkj gS fd lHkh okLrfod x ds fy, g(x) gSA

(A) dsoy I (B) dsoy I ,oa III (C) dsoy II (D) dsoy II ,oa III

8. eku ysa fd x0, y0 vpj (fixed) okLrfod la[;k,¡ gSa] 2 20 0x y > 1 ekU; gSA ;fn x, y dksbZ okLrfod la[;k,¡ gSa]

ftuds fy, x2 + y2 1 ekU; gS] rc (x – x0)2 + (y – y0)2 dk U;wure eku gksxk%

(A) 22 20 0x y 1 (B) 2 2

0 0x y 1 (C) (|x0| + |y0| – 1)2 (D) (|x0| + |y0|)2 – 1

9. lehdj.kksa 6x + 4y + z = 200 ,oa x + y + z = 100 ds v_.kkRed (non-negative) iw.kk±d gyksa dh la[;k D;k gksxh (A) 3 (B) 5 (C) 7 (D) vaur

10. /kukRed okLrfod la[;kvksa ds ,sls ;qXeksa (a, b) tks a4 + b4 < 1 ,oa a2 + b2 > 1 dks larq"B djrs gSa] dh la[;k gksxh% (A) 0 (B) 1 (C) 2 (D) 2 ls vf/kd

11. ,d f=kHkqt PQR dh Hkqtkvksa ds fy, laca/k QR2 + PR2 = 5PQ2 ekU; gSA ;fn ekf/;dk;sa (medians) PM ,oa QN fcUnq G ij foPNsfnr (intersect) djrh gSa] rc QGM ges'kk% (A) 45º ls de gksxk (B) vf/kdks.k (obtuse) gksxk

(C) ledks.k gksxk (D) U;wudks.k ,oa 45º ls vf?kd

12. eku ysa fd x1, x2, …., x11 11 fHkUu /kukRed iw.kk±d gSA ;fn buds lcls cM+s iw.kk±d dks 'ks"k 10 iw.kk±dksa dh ekf/;dk ls ixfrLFkkfir fd;k tkrk gS] rc (A) ekf/;dk vifjofrZr jgsxhA (B) vkSlr c<+ tk,xkA (C) ekf/;dk ?kV tk,xhA (D) vkSlr vifjofrZr jgsxkA

13. ,d vk;r dh lHkh 'kh"kZ (a, b) izdkj ds gS] tgk¡ a, b iw.kk±d gSA ;fn a, b lehdj.k (a – 8)2 – (b – 7)2 = 5 dks larq"B djrk gS] rc bl vk;r dh ifjf/k gksxh%

(A) 20 (B) 22 (C) 24 (D) 26

14. eku ysa fd N ,d ,slk U;wure /kukRed iw.kk±d bl izdkj gS fd tc Hkh N ds vafre vad ds ckn v'kwU; vad c fy[k fn;k tk, rks ifj.kkeh la[;k c ls HkkT; gks tkrh gSA rc N ds lHkh vadksa ds ;ksx dk eku D;k gksxk ? (A) 9 (B) 18 (C) 27 (D) 36

15. ,d Qyu f lHkh /kukRed iw.kk±d la[;kvksa ds leqPp; ds fy, bl izdkj ifjHkkf"kr gS : f(xy) = f(x) + f(y), tgk¡

x vkSj y /kukRed gS] ;fn f(12) = 24 rFkk f(8) = 15 gS] rks f(48) dk eku gksxk&

(A) 31 (B) 32 (C) 33 (D) 34

HkkSfrd foKku

16. ,d ljy yksyd ds iz;ksx }kjk xq:Roh; Roj.k ekius ds fy, ,d Nk=k yksyd ds /kkxs dks 63.2 cm rFkk yksyd ls yxs xksyd ds O;kl dks 2.256 cm ekirk gSA Nk=k dks yksyd dh yEckbZ D;k ysuh pkfg;s ?

(A) 64.328 cm (B) 64.3 cm (C) 65.456 cm (D) 65.5 cm



®




17. fp=kkuqlkj ldjh xnZu okys ,d csyukdkj crZu ftlds vk/kkj dh f=kT;k R rFkk ÅpkbZ H gS xnZu dh Å¡pkbZ h rFkk xnZu dh f=kT;k r gSA crZu dks ikuh ls rFkk xnZu dks vfeJ.kh; (immiscible) rsy ls Hkjk tkrk gSA ikuh rFkk rsy ds ?kuRo Øe'k% w rFkk 0 gSA fp=k esa fn, x, fcUnqvksa ij nkc dk D;k eku gksxkA ?

H

M N

h

2r

2r

(A) M ij g (h0 + Hw) (B) N ij g (h0 + Hw) 2

2

R

r

(C) M ij gHw (D) N ij 22

20

2w

rR

hrHRg

18. ,d dkj R f=kT;k ,dleku o`rh; iFk ij ,dleku pky v ls xfr djrs gq, T lSd.M esa ,d iwjk pDdj yxkrh gSA

bl nkSjku vfHkdsUnzh; Roj.k dk eku ac gSA ;fn ge dkj ,d cMs o`rh; iFk] ftldh f=kT;k 2R gS] ij ,dleku pky ls pyrh gq, vfHkdsUnzh; Roj.k 8ac eglwl djrh gS] rks vkorZdky gksxkA

(A)2T (B) 3T (C)T/2 (D) 3/2T

19. tSlk fd n'kkZ;k x;k gS] ,d <`< xsan ,d lrg ij fcuk fQlys yq<+d jgh gSA

fuEu esa dkSulk vkjs[k xsan }kjk pyh x;h nwjh vkSj le; ds lEcU/k dks fu:fir djrk gS ?

(A)

time A.

distance

(B)

time B.

distance

(C)

time C.

distance

(D)

time D.

distance

20. ,d ikuh ls Hkjs cDls dh fupyh lrg ds Nksj ij ,d fNnz gSA bl cDls dks ,d Åaps ehukj dh Nr ls fxjk;k

tkrk gSA ckWDl ds fxjrs le;, bldh lrg ij yxk dSejk cDls ls ckgj vkrs gq;s ikuh ds iFk dks vfHkfyf[kr (record) djrk gSA dSejs }kjk vfHkfyf[kr pyfp=k esa fn[ksxk fd (A) ty ,d ijoy;kdkj iFk ds vuqlkj uhps fxjrk gSA (B) ty ,d ijoy;kdkj iFk ds vuqlkj Åij tkrk gSA



®




(C) ty ,d lh/kh js[kk esa ckgj vk;sxkA (D) cDls ls ckgj ty ugha vk;sxkA

21. ,d xsan dks 45 m/s ds osx ls Å/okZ/kj fn'kk esa Åij dh vksj Qsadk tkrk gSA osx dk Å¡pkbZ ds lkis{k esa lgh vkjs[k D;k gksxk ? eku yhft, g = 10 m/s2.

(A)

(B)

45

V(m/s)

0 0 Height (m) 101

(C)

45

V(m/s)

0 0 Height (m) 101

(D)

45

V(m/s)

0 0 Height (m) 101

22. ,e d.k tks fd ewy fcUnq ls 1m dh nwjh ij gS bl izdkj pykuk izkjEHk djrk gS fd dr/d = r, tgkW (r, ) /kqzoh;

funsZ'kkad gSa] rc ifj.kkeh osx rFkk osx ds Li'kZjs[kh; Hkkx ds chp dk dks.k (A) 30º gSA (B) 45º gSA (C) 60º gSA (D) d.k dh fLFkfr ij fuHkZj djrk gSA

23. ,d fo|kFkhZ xq:Roh;Roj.k g dks Kkr djus ds fy, ,d iz;ksx djrk gSA fo|kFkhZ ,d LVhy dh xsan dks izkjfEHkd xfr u ls Åij Qsadrk gS vkSj vyx&vyx le;ksa t ij xsan }kjk r; dh x;h Å¡pk;h h dks ekirk gSA g ds eku dks 'kh?kzrk ls Kkr djus ds fy, fo|kFkhZ dks ,d xzkQ isij ij dkSulk xzkQ vkjsf[kr djuk pkfg, \

(A) h ds lkis{k t (B) h ds lkis{k t2 (C) h ds lkis{k t (D) h/t ds lkis{k t

24. ikuh ls Hkjh ,d dVksjh esa dkyh ehpZ ds pw.kZ dks ,d leku :i ls fNM+dk x;k gSA vc dVksjh esa fLFkr ikuh ds lrg ds chpksa chp lkcqu ?kqys nzo dh ,d cw¡n fxjkbZ tkrh gSA blds rqjUr ckn nzo dh lrg dSlh fn[kkbZ nsxh\

(A)

(B)

(C)

(D)

25. fp=k esa 2a Hkqrk okyk ,oa M nzO;eku ds ,d ycM+h dk Bksl ?ku ,d {kSfrt ry ij foJkekoLFkk esa gSA ;g ?ku vius

AB ds lkis{k eqDr :i ls ?kw.kZu dj ldrk gSA m (<< M) nzO;keku dh ,d xksyh {kSfrt pky v ls ABCD ds Bhd foijhr okyh lrg ls 4a/3 ÅWpkbZ ij Vdjkdj ?ku dks dks.kh; osx iznku dj nsrk gSA Vdjkus ds ckn xksyh ds



®




vUnj ?kl tkrh gSA fuEu esa ls dkSu ds djhc gksaxs \ (/;ku nhft, fd lrg ds yEcor~ ,oa nzO;eku dsanz (center

of mass) ls xqtjrs gq, v{k ds lkis{k ?ku dk tM+Ro vk?kw.kZ 22Ma

3 gSA

A

B

C

D

(A) Mvma

(B) Mv2ma

(C) mvMa

(D) mv2Ma

26. xq,fjd (Guericke's) ds iz;ksx esa ok;qeaMyh; nkc ds vlj dks fn[kkus ds fy, rkacs ds nks v/kZo`Ùkh; xksyksa dks ,d

nwljs ds lkFk tksM+dj ,d [kks[kyk xksyk cuk;k tkrk gS] vkSj bl xksys esa fuokZr iSnk djus ds fy, gok fudky nh tkrh gSA ;fn izR;sd v/kZxksys dh f=kT;k R gS vkSj ok;qeaMyh; nkc P gS] rc v/kZxksyksa dks vyx djus ds fy, U;wure fdruk cy yxkuk gksxk \

(A) 2R2P (B) 4R2P (C) R2P (D) R2P/2 27. tSlk fd layXu vkjs[k esa n'kkZ;k x;k gS] ,d xsan ftls h Å¡pkbZ ls fxjk;k x;k gS] Q'kZ ij izR;kLFk :i ls mNyrh

gSA fuEu esa ls dkSu lk vkjs[k xsan ds Roj.k (acceleration) dks le; ds Qyu ds :i esa n'kkZrk gS

t

h

(A)

t

acce

lera

tion

(B)

t

acce

lera

tion

(C)

t

acce

lera

tion

(D)

t

acce

lera

tion



®




28. ,d d.k ,d ljy js[kk ij 'kwU; izkjfEHkd osx (initial velocity) ls pyrk gqvk d nwjh r; dj ds :d tkrk gSA bl xfr ds nkSjku] 2/3 nwjh rd mldk Roj.k fu;r :i ls f jgrk gS vkSj og ckdh dh nwjh ,d fu;r eanu ls r; djrk gSA iwjh nwjh dks r; djus esa dqy le; fdruk yxk

(A) 2d/ 3f (B) 2 d/ 3f (C) 3d/ f (D) 3d/ 2f

29. tSlk fd layXu vkjs[k esa n'kkZ;k x;k gS] ,d yEcs csyukdkj uyh] ftdlh f=kT;k 20 cm gS] ds Åijh Nksj dks can j[kk x;k gSA blds vUnj ,d Hkkjghu ok;q:) eqlyh gSA ;fn bl eqlyh ds nwjh Nksj ij 50 Kg Hkkj yVdk fn;k tk,] rc ;g uhps tkus yxrk gSA ;fn ?ksjs ds vUnj dh gok dks rkieku T ls T – T rx BaM+k fd;k tkc] rks eqlyh okil fiNyh ewy txg ij igq¡p tkrh gSA rc T/T dk eku buesa ls fdlds fudV gS (ok;q dks vkn'kZ xSl eku fy;k tk, g = 10 m/s2, ok;qeaMyh; nkc 105 ikldy gSA)?

L

(A) 0.01 (B) 0.02 (C) 0.04 (D) 0.09 30. ,d fiaM ij vkjksfir cy dks ˆˆ ˆF (n.F)n G ls fu:fir fd;k x;k gS] tgk¡ n̂ bdkbZ lfn'k G dk eku fuEufyf[kr

esa ls D;k gksxk\

(A) n̂ F (B) ˆ ˆn (n F) (C) ˆ(n F) F / | F | (D) ˆ ˆ(n F) n

jlk;u foKku 31. ¶yksvkWfDle LVSjkWu C20H29FO3, ,d nnZ fuokjd LVSjkbM gS 500 mL ty esa 10.0 mg LVSjkbM dks ?kksydj ,d

foy;u dks cuk;k tkrk gSA bl foy;u ds 1.0 mL esa ¶yksvkWfDle LVSjkWu ds fdrus eksy gksxsa (A) 1.16 × 10–10 (B) 1.19 × 10–17 (C) 5.95 × 10–8 (D) 2.38 × 10–11

32. 2 10–4 M Ba(OH)2 foy;u dk fdruk vk;ru 1 10–4 M HCl ds 300 mL foy;u esa feyk;k tk;s ftlls ifj.kkeh foy;u esa gkbMªksfu;e vk;u (H3O

+) dh lkUnzrk 5 10–11 M gks tk;s ? (A) 375 mL (B) 300 mL (C) 225 mL (D) 450 mL

33. fojkekoLFkk ls ,d bysDVªkWu] ftldh rajx ðS?;Z 1.41 Å gS] dks Rofjr djrs gSA fdruh ek=kk esa foHko dks de fd;k tk;s fd bysDVªkSu ls lEcfU/kr rajx ðS/;Z 1.73 Å gks tk;s &

(A) 25 V (B) 50 V (C) 75 V (D) 12.5 V

34. fuEu nh xbZ O;oLFkkvksa esa ls fdlesa] fn;s x;s xq.kksa ds vuqlkj Øe dk vuqlj.k n`<+rk ls ugha gksrk gS ?

(A) BeCO3 < MgCO3 < CaCO3< SrCO3 – rkih; LFkkf;Ro

(B) BeSO4 > MgSO4 > CaSO4 > SrSO4 – ty esa foys;rk

(C) Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2 – {kkjh; lkEF;Zrk

(D) MgF2 < CaF2 < SrF2 < BaF2 – tkyd ÅtkZ

35. BrF3 v.kq dk ladj.k rFkk vkd`fr gS : (A) sp3d rFkk T vkdkj (B) sp2d2 rFkk prq"Qydh;

(C) sp3d rFkk csUV (D) buesa ls dksbZ ugha

36. ,d jklk;fud vfHkfØ;k dh nj fuEu ij fuHkZj ugha djrh gS %



®




(A) vfHkfØ;k ds rkieku ij (B) vfHkdkjdksa dh lkUnzrk ij (C) vfHkdkjdks ds eksyks ij (D) mRizsjd ij

37. 100ºC ij ty dh ok"ihdj.k dh ekud ,UFkSYih vap Hº 40 kJ mol–1 gSA blh rki ij (100ºC ij) ty ds ok"ihdj.k dh vkUrfjd ÅtkZ (kJmol–1 esa) gksxh %

(;g ekudj fd ty ok"i ,d vkn'kZ xSl dh Hkk¡fr O;ogkj djrh gS) (A) + 37.56 (B) – 43.76 (C) + 43.76 (D) + 40.66

38. Al3+, K+ , Mg2+ , Rb+ esa U;wure vk;fud f=kT;k fdl Lih'kht dh gS % (A) Al3+ (B) K+ (C) Mg2+ (D) Rb+

39. 27ºC rki rFkk 2.46 atm nkc ij ,d csyukdkj ik=k 1 eksy vkWDlhtu xSl j[krk gSA tc ik=k dks 127ºC rki rd xeZ fd;k tk,] rFkk okYo dks rc rd [kksyk tk;s] tc rd fd xSl dk nkc de gk sdj 1 atm ugha gks tk,] rc bl ifjfLFkfr esa vkWDlhtu dk fdruk Hkkj fu"dkflr gks tk,xk] ;fn rki 127ºC ij fLFkj j[kk tkrk gS %

(A) 22.4 g (B) 11.2 g (C) 20.8 g (D) 9.6 g

40. fuEu esa ls dkSulh {kkjh; /kkrq tyh; yo.k nsrk gS \ (A) Li (B) Na (C) K (D) Cs 41. fuEu esa dkSulk dFku lgh gS \ (A) izsjf.kd izHkko esa bysDVªkWuksa dk iw.kZr;k LFkkukUrj.k gksrk gSA (B) izsjf.kd izHkko nwjh c<+us ds lkFk c<+rk gSA (C) vuquknh lajpuk;sa dkYifud lajpuk;sa gS rFkk ;s okLrfod lajpuk dks iznf'kZr ugha djrh gSA (D) fdlh Hkh vuquknh lajpuk dh rqyuk esa vuquknh ladj dh ÅtkZ vf/kd gksrh gSA

42. fuEu esa ls vuquknh lajpuk,sa gSA

(A) & CH3 – O – N = O

(B) &

(C) (CH3)2 CO &

(D) CH3 – CH = CH – CH3 & CH3 – CH2 – CH = CH2

43. RX + Mg bZFkj RMgX 3CH OH

n- C;wVsu mijksDr vfHkfØ;k Øe esa R gks ldrk gS ? (A) n-C;wfVy (B) f}rh;d - C;wfVy (C) n-izksfiy (D) vkblksizksfiy

44.

HBr ds lkFk bysDVªkWuLusgh ;kSxkRed vfHkfØ;k dh nj dk Øe gksxk % (A) IV> I > III > II (B) I > II > III > IV (C) I > III > II > IV (D) IV > I > II > III

45. fuEu ;kSfxd dk IUPAC uke gksxk



®




(A) 2-esfFky-3-czkseks-4-lYQksgsDlsuksbd vEy (B) 3-czkseks-2-esfFky-4-lYQksgsDlsuksbd vEy (C) 4-czkseks-5-dkcksZDlhgsDlsu-3-lYQksfud vEy (D) 5-dkckZsDlh-4-czkseks-3-lYQksfud vEy

tho foKku

46.

PART- B

Two Mark Questions

xf.kr

61. iwa.kkZd f=kT;k ,ao iw.kkZd ÅapkbZ okys izR;sd 'kdqa dk vk;ru dk l[;kRed eku blds dqy i`"Bh; {ks=kQy ds cjkcj gSA ,sls 'kadqvksa dh la[;k gSA

(A) 0 (B) 1 (C) 2 (D) vuar

62. ,d bdkbZ f=kT;k okyk v/kZo`Ùk js[kk AB dks O;kl eku dj cuk;k x;k gS] bl o`Ùk dk dsUnz O gS] js[kk AO esa C ,d

,slk fcUnq gS fd AC : CO = 2 : 1 js[kk AO ij yEc CD ,sls Mkysa fd D v/kZo`Ùk ij gks] js[kk AD ij ,d yEc OE

,sls Mkysa fd fcUnq E, AD ij fLFkr gks] OE vkSj CD fcUnq H ij dkVrs gS] rc DH dk eku gksxk&

(A) 1

5 (B)

1

3 (C)

1

2 (D)

5 – 12

63. ;fn vk/kkj 10(base 10) esa izkd`frd l[;kvksa n ds vdksa dk xq.kuQy 36n10n2 –gS rc ,slh lHkh izkdfrd l[;kvksa dk ;ksxQy gksxk (A) 12 (B) 13 (C) 124 (D) 2612

64. eku ysa fd ABC ,d f=kHkqt gS ftlds fy, AB = BC gSA ekuysa fd X AB ij ,d ,slk fcUnq gS ftlds fy, AX :

XB = AB : AX ;fn AC = AX, rc dks.k ABC dk eku gS&

(A) 18° (B) 36° (C) 54° (D) 72°

65. ;fn fdlh rhu vadksa dh la[;k dks ;kn`PN :i ls (randomly) p;fur fd;k tk;s] rc bl ckr dh izkf;drk

(probability) D;k gksxh fd la[;k Lo;a ;k bldk dksbZ Øelap; (permutation) tks rhu vadksa dh la[;k gS] 4 ;k 5

ls foHkkT; gS ?

(A) 1

45 (B)

29180

(C) 1160

(D) 14

HkkSfrd foKku

66. nks O;fDr A rFkk B ,d pkSM+h ty/kkjk ds foijhr rV ij [kM+s gS vkSj ,d 3.5 ehVj pkSM+h ty/kkjk dks ikj djuk pkgrs gSA nksuksa ds ikl ydM+h dh n`<+ r[rh gS ftuds nzO;eku ux.; gSA ;s nksuksa r[rs 3 ehVj (meter) ls FkksM+h lh T;knk yEckbZ ds gSA mUgksus nksuksa r[rksa dks fp=kkuqlkj O;ofLFkr fd;k gSA ;fn B ftldk nzO;eku 17 kg gS] r[rs ij [kM+k gks rks r[rs ij pyrs gq, ij djus ds fy, A dk egÙke nzO;eku fuEufyf[kr esa ls fdlds djhc gksxk\



®




3.5 m

3 m A B

(A) 17 kg. (B) 65 kg (C) 80 kg (D) 105 kg

67. ,dleku /kkrq dh iV~Vh] tks fd ,d f=kHkqt ABC ds vkdkj esa gS] dk nzO;eku 540 gm gSA Hkqtkvksa AB,BC, ,ao CA dh yEckbZ Øe'k% 3cm ,5cm ,oa 4cm gSA ;g iV~Vh eqDr :i ls fcUnq A ij /kqjkxzLr (pivotted) gSA ,d 'kh"kZ ij D;k nzO;eku tksMk tk, ftlls fd iV~Vh bl rjg yVds fd mldh yEch Hkqtk iw.kZr% {kSfrt gks tk,

(A) C ij 140 gm (B) C ij 540 gm (C) B ij 140 gm (D) B ij 540 gm



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68. ,d 20 gm dh xksyh] ftldh fof'k"V Å"ek 5000 J(kg–°C) dh pky ls ,d 1.0 kg ekse ds fiaM] ftldh Å"ek 3000 J (kg–°C) gS] esa izos'k dj tkrh gSA xksyh vkSj ekse ds fiaM dk ekieku 25°C gSA eku ysa% ;fn (i) xksyh ekse ds vUnj vk dj :i tkrh gS rFkk (ii) xksyh dh laiw.kZ ÅtkZ ekse dks Åf"er djrh gS] rks ekse ds fiaM dk vafre rkieku °C esa blds utnhd gksxk %

(A) 28.1 (B) 31.5 (C) 37.9 (D) 42.1

69. fp=k esa n'kkZ, vuqlkj M nzO;eku dh ,d f?kjuh ls m1 rFkk m2 nzO;eku ds nks Hkkj (m1 > m2) ,d jLlh ls yVds

gq, gSA f?kjuh fcuk ?k"kZ.k ds ?kwerh gS] tcfd jLlh ,oa f?kjuh ds chp dk ?k"kZ.k bruk T;knk gS fd jLlh f?kjuh ij ugha fQlyrhA fuEu esa ls dkSu lk vkjs[k f?kjuh ds nksuksa rjQ dh jfLl;ksa ds chp ds rukoksa ds vUrj dks f?kjuh ds nzO;eku ds lkis{k lgh&lgh fu:fir djrk gSA

m1 m2

M

T1 T2

(A)

M 0

T1

– T

2

(B)

M 0

T1

– T

2

(C)

M 0

T1

– T

2

(D)

M 0

T1

– T

2

70. Hkqtk a okys ,d cM+s ?ku ls Hkqtk b dk ,d NksVk ?ku bl izdkj dkVk tkrk gS fd nksuksa ?kuksa dk ,d mHk;fu"B 'kh"kZ P gS] eku yhft;s fd X = a/b ;fn cps gq, Bksl dk nzO;eku dsUnz O ij gks rks X fuEUk esa ls fdl lehdj.k dks larq"B djrk gS \

b

P

O

(A) x3 – x2 – X – 1 = 0 (B) x2 – x –1 = 0 (C) x3 – X2 – X – 1 = 0 (D) X3 – X2 – X + 1 = 0

jlk;u foKku

71. ,d 10 g gkbMªksdkcZu (,Ydsu gksuk vko';d ugha) dk fo'ys"k.k djus ij ik;k tkrk gS fd og 1 g gkbMªkstu j[krk gSA ;fn 1 eksy gkbMªksdkcZu ls lHkh H-ijek.kq gVk;s tkrs gSa o bUgsa H2 xSl esa ifjofrZr fd;k tkrk gS] rks mRikfnr xSl dk nzO;eku 8 g ik;k tkrk gSA rc gkbMªksdkcZu dk v.kqlw=k fuEu gS %

(A) C2H4 (B) C3H4 (C) C4H8 (D) C6H8

72. ;gka ij dqN v)Z lsy vfHkfØ;k ¼vEyh; ek/;e½ esa nh x;h gS ,oa muds E° ds eku (lkekU; gkbMªkstu bysDVªksM ds lkis{k V esa) fn;s x;s gSA



®




2 + 2e– 2¯ E° = 0.54 Cl2 + 2e– 2Cl¯ E° = 1.36 Mn3+ + e– Mn2+ E° = 1.50 Fe3+ + e– Fe2+ E° = 0.77 O2 + 4H+ + 4e– 2H2O E° = 1.23

fuEufyf[kr esa dkSulk oDrO; lgh gS \ (A) DyksjkbM vk;u dk vkWDlhdj.k O2 ls gksrk gSA (B) Fe2+ dk vkWDlhdj.k vk;ksMhu ls gksrk gSA (C) vk;ksMkbM vk;u dk vkWDlhdj.k Dyksjhu ls gksrk gSA (D) Mn2+ dk vkWDlhdj.k Dyksjhu ls gksrk gS

73. xSlh; H2, Cl2 o HCl dh ca/k fo;kstu ÅtkZ Øe'k% 104, 58 o 103 kcal mol–1 gSa HCl xSl ds fuekZ.k dh ,UFkSYih gksxh (A) – 44.0 kcal (B) – 22.0 kcal (C) 22.0 kcal (D) 44.0 kcal

74. fuEu ds LFkkf;Ro dk lgh Øe gS %

(I) (II) (III) (IV)

(A) I > II > III > IV (B) III > IV > II > I (C) II > IV > III > I (D) IV > III > II > I

75. fuEu esa ls dkSulk leko;oh ;qXe fLFkfr leko;ork ugha n'kkZrk gS \

(A) rFkk (B) rFkk

(C) rFkk (D) rFkk

tho foKku

76.



®




FULL SYLLABUS TEST-1

KISHORE VAIGYANIK PROTSAHAN YOJANA

(KVPY) 2023 | STREAM (SA)_XI

ANSWER KEY

PART- A

1. (B) 2. (C) 3. (A) 4. (D) 5. (B) 6. (B) 7. (B) 8. (A) 9. (C) 10. (D) 11. (C) 12. (C) 13. (A) 14. (A) 15. (D) 16. (B) 17. (A) 18. (C) 19. (B) 20. (D) 21. (A) 22. (B)

23. (D) 24. (C) 25. (D) 26. (C) 27. (B) 28. (C) 29. (C) 30. (D)

31. (C) 32. (D) 33. (A) 34. (D) 35. (A) 36. (C) 37. (A) 38. (A) 39. (A) 40. (A) 41. (C) 42. (B) 43. (A) 44. (D) 45. (B)

PART- B

61. (B) 62. (C) 63. (B) 64. (B) 65. (B)

66. (D) 67. (C) 68. (C) 69. (C) 70. (A)

71. (D) 72. (C) 73. (B) 74. (D) 75. (A)



®




HINTS & SOLUTIONS

PART- A

xf.kr

1. x4 – x2 + 2x – 1 = 0 X2 – (x –1)2 = 0 (x2 + x – 1) (x2 – x + 1) = 0

x2 + x – 1 = 0 nks okLrfod ewy gSA 2. uksV: APN dk {ks=kQy = PDN dk {ks=kQy APK dk {ks=kQy = PBK dk {ks=kQy PCL dk {ks=kQy = PBL dk {ks=kQy PCM dk {ks=kQy = PDM dk {ks=kQy (PKAN) dk {ks=kQy + (PLCM) dk {ks=kQy = (PMDN) dk {ks=kQy + (PLBK) dk {ks=kQy (PLCM) dk {ks=kQy = 36 + 41 – 25 = 52 3. y = P(x) = a0 + a1 a2 x2 + ……+anxn

a0,a1,a2,a3,……….an I 2 = P(2) ….(1) 5 = P(4) ….(2) By (1) & (2)

3 = a1 (4 – 2) + a2 (42 – 22) + a3(43 –23) + …….. + an(4n – 2n) Li"Vr;k RHS le gS rFkk LHS fo"ke gSA cgqin fo/keku ugha gSA

4. 165 = 3 × 5 × 11 x + y divides xn + yn

32 + 1 divides 3211 + 111

Hence N1 is multiple of 33, simultaneously unit digit in N1 is 9 so it not the multiple of 5

Hence HCF of N1 & N2 is 33 5. 152 × 518

= 9 × 520

2010log 9 5 = 3

10 102log 20log 5

= 2 4771 20 1– 0.3010

= 14 {characteristic} Hence the number have 15 digits At worst all digit can have value 9 Hence sum should less then 135 And the last digit should be 5 Hence Sum should greater than Or equal to 6

6. t2 = a + b t3 = t2 . t = at2 + bt = a(at + b) + bt t3 = (a2 + b) .t + ab t3 = 4t + 3 when a = 1 and b = 3 t3 = 10t + 3 when a = 3 and b = 1 t3 = 6t + 5 when a = 1 and b = 5 but t3 can never be equal to 8t + 5 for any point of a and b.

7. ;fn , ds ewy p(x) = 0 g(x) = 0 ds ewy x3 = ,

vr% dsoy 2 okLrfod ewy gSA

iqu% g(x) b– 4

a2

xR.

8. 1yx 20

20 x0 – y0 fixed

x, y vpj

1yx 22 , ekuk x = con

y = sin z = (x – x0)2 + (y – y0)2

z = x2 + x02 + y2 + y02 – 2 (x x0 + y y0) x - cos , y = sin z = 2

020 yx – 2 (x0 cos + y0 sin)

d

dz 0 – 2 (–x0 sin + y

0 cos)

0ddz

–x0 sin = – y0 cos

tan = 0

0

xy

G y0

x0

20

20 yx

sin = 20

20

0

yx

y

cos = 20

20

0

yx

x

x = 20

20

0

yx

x

, y =

20

20

0

yx

y

z =

2

020

20

0

2

020

20

0 yyx

yx

yx

x

2

20

20

0

2

20

20

0 1yx

12y1

yx

12x

2

20

20

20

20

20

20

yx

yx1yx

2

20

20

220

20 1yxyx1

9. Subtracting the given equations we get 5x

+ 3y = 100



®




x = 20 – 5y3

y is multiple of 5, let y = 5k x = 20 – 3k k = 0, 1, 2, ……6 vr% gyksa dh la[;k 7 gSA

10. ekuk a2 = m vkSj b2 = N rc m > 0 rFkk N > 0 vc fn;k x;k izfrcU/k M + N > 1 vkSj M2 + N2 < 1

(M,N) lies inside circle x2 + y2 < 1 and above line x + y > 1 (M, N) lies in shaded region and number of points in shaded region are infinite, so number of pair (a,b) are also infinite.

11.

G

M R Q

P

N

L

ekuk QR= p, PR = q, PQ =r fn;k x;k p2 +q2 = 5r2

vc 22

22

2PM

3QN

GMQG

= 9

PM9

QN4

22

=

222222 pq2r241

qp2r241

.491

= 22

QM4

p

vr% QGM dk dks.k 90º. 12. ekuk x1 < x2 < x3 ……x11

ekf/;dk x1, x2 …..x10 = 2

xx 65

Now the new set of number are x1, x2,

….x5, 2

xx 65 , x6, ….x10

vr% ekf/;dk 2

xx 65 < x6

ekf/;dk gkkleku 13. (a – 8)2 – (b – 7)2 = 5 (a – b –1) (a + b – 15) = 5 I1 I2 pkj fLFkfr I1 I2 5 1 1 5 –5 –1 –1 –5 fLFkfr-1 a – b – 1 = 5 & a + b – 15 = 1 a = 11, b = 5 fLFkfr-2

a – b – 1 = –5 & a + b – 15 = –1 a = 5, b = 9 fLFkfr-3

a – b – 1 = 1 & a + b – 15 = 5 a = 11, b = 9 fLFkfr-4

a – b – 1 = –1 & a + b – 15 = –5 a = 5, b = 5

ifjf/k = 4 + 4 + 6 + 6 = 20 14. ekuk la[;k N gSA If we write c after the last digit now number

is 10N + c vc c|10N + c c = 1, 2, ….., 9 c|10N c = 1, 2, ….., 9 c|N for c = 4, 7, 9 Hence N is LCM of (4, 7, 9) = 252 blfy, vadksa dk ;ksx = 9 15. f(xy) = f(x) + f(y) f(x) = knx vc, 24 = k(n12)

15 = k(n8) 5 = k(n2) ekuk y = kn48

= k[n12 + n4]

(0,1)

(1,0)

(M,N)



®




= k 24 2 5k k

y = 34

HkkSfrd foKku

16. r = 1.128cm leff = l + r = 63.2 + 1.128 = 64.3

17.

H

M N

h

2r

2r

A

Pressure at Point A PA = 0hg Pressure at M PM = PA + W Hg 0h + W H)g

18. (i) 2

c

Va

R and

2 RT

V

(ii) ac' = 8ac

2

1V2R

= 2V

8R

V1 = 4V

T' = 2 (2R)

4V

= RV

= T2

19. During slope speed will be increase, so

distance covered will increase parabolically with the time. Then speed will remain constant so further graph will be a straight line with greater slope than the first straight line portion of graph. So if change in speed at the sharp corners is not considered, the best answer should be (B) rather than C.

V

t

time

distance

However if kink is considered and Collision is inelastic then graph will be (C). But in either case answer cannot be D

20. Since bucket and water both are in state of free fall so water will not come out of the hole. (D)

21. v2 = v2 – 2gh parabola (A)

22. dr

rd

dr / dt

rd / dt

dr / dt

r

dr

rdt

Vt=r Vnet=r

2

=45°

Vr=r

23. R = 21Ut – gt2

h 1u – gtt 2

2ug

t

ht

1– g2

u

slope =



®




24. Surface tension of soap solution is less

than water. 25.

C D

B

A

Q a

MI about AB

I = 2 2

22Ma 8MaM(a 2)

3 3

Li = Lf

mv4a3

= I

(mv) 4a3

= 28ma

3

(m << M) can be

neglected

(mv)4a3

= 2M8a

3

(mv)(4a)x33x8M

=

mv2Ma

=

26.

P P

P P

F = 2px R

= 2R p

27. a = –g except during collision. 28.

t2

Vmax

maxVf

= t1 t

v

tan = maxVt

t1 = maxVf

max 21

xV x t d2

...(1)

max 11 2d

xV x t2 3

...(2)

2max

4dfV

3

Vmax = 4dt3

...(3)

By (1) & (3)

2

1 4dft d

2 3

t2 = 3df

29.

P0, T0, V0

P

50kg

P0, T0, V0 P, T0, V P, T, V0

P0 V0 = PV

0

0

VVT T

0

0 0 0

VT P mg1–T V P AP

4

50

T 50x10x7x(10)1–T 22x400x(10)

0

T 7T 22x8

= 1– 78x22

= 0.04

30. F = ( ˆˆ ˆ(n.F)n a )

(nxF)xn = (nxn)F – (n.F) n̂

Ans- (D)

jlk;u foKku



®




31. 500 mL esa eksyksa dh la[;k = 310 10

336

1 mL esa eksyksa dh la[;k = 310 10

336 500

= 5.95

× 10–8 .

32. [OH–] = wK

[H ] =

14

11

10

5 10

4 4V 2 10 2 300 1 10300 V

= 2 10–4

[OH–] V = 450 ml

33. 1 = 1/ 2150

V

V1 = 21

150

V1 = 2

150

(1.41) = 75 V

rFkk V2 = 2

150

(1.73) = 50 V

(vr% foHko 25 V ls de fd;k tkuk pkfg,A)

34. /kkrq ds vkdkj esa o`f) ds lkFk tkyd ÅtkZ esa deh gksrh gSA

MgF2 = – 2906 kJ mol– ; CaF2 = –2610 kJ

mol–1

SrF2 = – 2459 kJ mol– ; BaF2 = – 2367 kJ

mol–1

37. H = E + n(g) RT 40.66 × 1000 = E + (1) × 8.314 × 373. E = 37.56 kJ mol–1

38. Rb+ rFkk K+ esa Mg2+ rFkk AI3+ dh vis{kk dks'kksa dh la[;k vf/kd gksrh gSA AI3+ rFkk Mg2+ lebysDVªkWfud Lih'kht gSa fdUrq AI3+ esa mPp ukfHkdh; vkos'k mifLFkr gS] blfy, AI3+ < Mg2+ gksxkA vr% Al3+vkdkj esa lcls NksVk gksxkA

K+ = 1.38 Å, Mg2+ = 0.72 Å, Al3+ = 0.535 Å rFkk Rb+ = 1.64 Å

39. 1 1

1

n TP

= 2 2

2

n TP

1 3002.46

=

2n 4001

n2 = 0.3

cph gqbZ vkWDlhtu dk Hkkj = 0.3 × 32 = 9.6 g vkWDlhtu dk fu"dkflr Hkkj = 1 × 32 – 9.6 =

22.4 g.

42. vuquknh lajpuk esa ijek.kq dh fLFkfr ifjofrZr ugha gksrh gSA

43. R = CH3–CH2–CH2–CH2– ;k CH3 – –

CH2 – CH3

44. bysDVªkWuLusgh ;ksxkRed vfHkfØ;k dh nj dkcZ/kuk;u dk LFkkf;RoA

45.

(3-czkseks-2-esfFky-4-lYQksgsDlsuksbd vEy)

PART- B

xf.kr 61. ekuk 'kadq dh ÅapkbZ = h

f=kT;k = r

ÅapkbZ = ; = = 22 hr vk;ru = i`"Bh; {ks=kQy

22 rrhr31

r33rh r3–rh31

3–h3r

hr 22

9h6–h9r

hr 22

22

3hr2–

9hr

h222

2

9–r

546

9–r

r6h

22

2

Since h and r must be integers, and r2 – g must be a factor of 54

r2 – 9 must be divisible by 3 .k3r

h = 1–k

66

9–k9

546

22

Since O < k2 – 1 < 6 k = 2 is only such value; for which h is integer. blfy, r = 2 × 3 = 6 i

h = 6 + 36 = 8



®




10 is the only possible values for r and h.

62.

H E

AB

D r = 1

O C

ACCO

= 21

OC = 13

AC = 23

OD = 1 OCD : OD2 = OC2 + CD2

1 = 19

+ CD2

CD = 89

Let ADC =

tan = ACCD

= 2

8 =

1

2

cos = DEDH

= DA2DH

DH =DA

2cos { cos =

CDAD

}

= 2 2

3 2 1 1

12

= 1

2

DH = 1

2

63. Product of digits of natural number will be a

non negative integer blfy, , n2 – 10n – 36 0

n (–, 5 – 61 ] (5 + 61 , ) ijUrq n IN blfy, n 13; tgk¡ n N case-1 for all 2 digit natural numbers max value of product of digits = 9 × 9 = 81 so n2 – 10n – 36 81 n [5 – 142 , 5 + 142 ] but n is taken as a 2 digit natural no.; so 13 n < 17; product of digits = 3, 4, 5 or 6 for 13, 14, 15 and 16 respectively checking n = 12

vadksa dk xq.ku = 1 × 3 = 3 vkSj 132 – 10 × 13 – 36 = 3 so 13 satisfies the given condition vr% gy gSA chcking for n = 14 xq.kuQy = 1 × 4 = 4 142 – 10 × 14 – 36 = 196 – 140 – 36 = 20 > 6 and n2 – 10n – 36 is increasing function for n > 5; rest of the 2 digit integers won't satisfy the given condition case-2 for all 3-digit integers max product = 9 × 9 × 9 = 729 The smallest 3 digit no. is 100 f(n) = n2 – 10n – 36; f(100) = 1002 – 10 × 100 – 36 = 8964 > 729 and f(n) is increasing Hence no 3 digit Integers and similary any higher integer will not satisfy n = 13 dsoy mÛÙkj gSA

64.

B

x

D

C b A

a–b

b

ekuk AC = b rFkk AB = BC = a

fn;k gSba

bab

b2 = a2 – ab

2

51ab

Let D be foot of perpendicular from B

sin (ABD) = a2b

=4

15

ABD = 18º ABC = 36° 65. Let abc be required numbers which is

divisible by 20 Clearly one digit must be 0 and one digit must be even, other can be any

Case-I when two digits are 0,0

x 0 0 x = {1,2,.....,9}

9 rjhds Case-II when one O and remaining two are distinct x y

Hence number of ways = 4 × 8 × 2 × 2 = 128 ways

Case-III when two are even 0 and same and other 0.

4 × 1 × 2 = 8 ways Total favorable ways = 9 + 128 + 8 = 145 dqy rjhds = 900



®




izkf;drk = 145900

= 29

180

HkkSfrd foKku

66.

0.5 m 2.5 m

Ma A

B

3.5 m

17 Kg

3 m 3 m

A170x2.5 M x10x0.5 MA = 85 67.

C 4 A

5 3

4/3 37º

C1

1

B

37º 53º

C1

5/3

3cos53º6 = 9/5 B C

4

A

3

here = 53º – 37º

AC1 = 2

2 41

3

AC1 = 53

For equilibrium, the combine centre of mass should be vertically below the point of suspension 'A' for this :

m1x1 = m2x2

(540) 5

sin3

= (mB) 95

B

5 9(540) sin(53º 37º ) (m )

3 5

solving mB = 140 gm So answer should be (C)

68. 12

mV2 = m1s1T + m2s2T

21 20 20(2000) 500 (1 3000) T

2 1000 1000

T = 12.9ºC Tf = 25ºC + 12.9ºC = 37.9ºC 69.

m1g – T1 = (m1)(a) T2 – m2g = (m2) (a)

(T1 –T2) = (I)2

a

R

(m1 – m2)g = (m1 + m2 + 2i

R)a

a = 1 2

1 2 2

(m – m )gI

(m m )R

(T1 – T2) = 2

2M(R ) a

2 R

a = constant (T1 – T2) = 1 2

1 2

(m – m )g(M)2 M

m m2

(T1 – T2) =

1 2

1 2

(m – m )g2(m – m )

1M

M = 0 (T1 – T2) = 0; and M , T1 – T2 = (m1 – m2)g

70. Assume vertex P as origin. All the three

coordinates are symmetrical, so we can deal only with x-coordinate

xcm = 1 1 2 2

1 2

m x m xm m

b =

3 3

3 3

a b(a ) (b )

2 2a b

2b =

4 4

3 3

a ba b

2b

=

22

22

22

22

baba

baba

babab–a

bab–aba

a3 – a2b –ab2 – b3 = 0, dividing by b3



®




01–ba–

ba–

ba

23

x3 – x2 – x – 1 = 0

jlk;u foKku

71. C : H = 9

12 :

11

= 3 : 4

ewykuqikrh lw=k = C3H4 blfy,] v.kqlw=k = (C3H4)n = C3nH4n

gVkus ds i'pkr~ cus H-ijek.kqvksa ds eksy = 1 × 4n

cuuss okys H2 ds eksy = 4n2

= 2n

cuuss okys H2 dk nzO;eku = 2n × 2 = 4n = 8 (fn;k x;k gS)

n = 2 v.kqlw=k = C3 × 2H4 × 2 = C6H8

72. SRP eku ds vuqlkj Dyksjhu vk;ksMhu vk;u dks vkWDlhd`r dj ldrk gS

73. ¼fn;k x;k gS½ H2 (g) HCl (g); H = 104 kcal ...(1) Cl2 (g) 2Cl(g) ; H = 58 kcal ...(2) HCl (g) H(g) + Cl(g) ; H = 103 kcal ...(3) (HCl ds fy, lEHkou dh Å"ek½

12

H2 (g) + 12

Cl2 (g) HCl (g) ; H = ?

¼(1) o (2) dk 2 ls Hkkx nsus ij o fQj tksM+us ij½

12

H2 (g) + 12

Cl2 (g) H(g) + Cl(g) ;

H = 81 kcal ...(4) ¼lehdj.k (4) ls lehdj.k (3) dks ?kVkus ij½

12

HCl (g) H(g) + Cl(g) ;

H = 103 kcal ...(3) – – – – –––––––––––––––––––––––––––––––––––––– 12

H2 (g) + 12

Cl2 (g) HCl(g) ; H = – 22.0

kcal ¼HCl xSl ds lEHkou dh ,UFkSYih = – 22.0 kcal½

74. LFkkf;Ro vuqukn vfrla;qXeu lHkh I, II, III o IV esa la;qXeu dh yEckbZ leku

gSA ysfdu vfrla;qXeu IV > III = II > I vkSj III dh

vfrla;qfXer lajpuk II dh rqyuk esa vf/kd LFkk;h gSA pwafd ;g vf/kd foLFkkuhd`r gSA

75. rFkk e/;ko;oh gSA



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KISHORE VAIGYANIK PROTSAHAN YOJANA (KVPY) | STREAM (SA) XI