kishore vaigyanik protsahan yojana (kvpy) | stream …

FULL SYLLABUS TEST-2

KISHORE VAIGYANIK PROTSAHAN YOJANA

(KVPY) | STREAM (SA)_XI

®

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


BIOLOGY


PART-B :

MATHEMATICS

Question No. 61 to 65 consist of TWO (2) marks for each correct response & – 0.5 for incorrect response.

PHYSICS


CHEMISTRY


BIOLOGY


4. No deduction from the total score will be made if no response is indicated for an item in the Answer sheet.

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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-2

PART- A One Mark Questions

MATHEMATICS

1. A 5-digit number abcde, when multiplied by 9, gives the 5-digit number edcba. The sum of the digits in

the number is (A) 18 (B) 27 (C) 36 (D) 45

2. Let x and y be two positive real numbers such that x + y = 1. Then the minimum value of y

1

x

1 is

(A) 2 (B) 2

5 (C) 3 (D) 4

3. Suppose p,q,r are positive rational numbers such that rqp is also rational. Then

(A) r,q,p are irrational

(B) qr,pr,pq are rational, but r,q,p are irrational

(C) r,q,p are rational

(D) qr,pr,pq are irrational

4. Let x, y be real numbers such that x > 2y > 0 and 2log(x – 2y) = logx + logy

Then the possible value(s) of y

x

(A) is 1 only (B) are 1 and 4 (C) is 4 only (D) is 8 only 5. Let p1(x) = x3 – 2020x2 + b1x + c1 and p2(x) = x3 – 2021x2 + b2x + c2 be polynomials having two common

roots and . Suppose there exist polynomials q1(x) and q2(x) such that p1(x)q1(x) + p2(x)q2 (x) = x2 – 3x + 2. Then the correct identity is

(A) p1(3) + p2(1) + 4028 = 0 (B) p1(3) + p2(1) + 4026 = 0 (C) p1(2) + p2(1) + 4028 = 0 (D) p1(1) + p2(2) + 4028 = 0 6. Let ABCD be a quadrilateral such that there exists a point E inside the quadrilateral satisfying

AE = BE = CE = DE. Suppose DAB, ABC,BCD is an arithmetic progression. Then the median of

the set {DAB, ABC, BCD} is

(A) 6

(B)

4

(C)

3

(D)

2

7. Let r be the remainder when 20212020 is divided by 20202. Then r lies between (A) 0 and 5 (B) 10 and 15 (C) 20 and 100 (D) 107 and 120 8. In the integers from 1 to 2021 are written as single integer like 123…91011…20202021, then the 2021st

digit (counted from the left) in the resulting number is (A) 0 (B) 1 (C) 6 (D) 9



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9. In a triangle ABC, a point D is chosen on BC such that BD : DC = 2:5. Let P be a point on the

circumcircle ABC such that PDB = BAC. Then PD : PC is

(A) 5:2 (B) 2 : 5 (C) 2 : 7 (D) 7:2

10. Let [x] be the greatest integer less than or equal to x, for a real number x. Then the following sum

16

16

15

15

14

14

13

13

12

122018

2020

2018

2020

2018

2020

2018

2020

2018

2020

is (A) 80 (B) 85 (C) 90 (D) 95 11. Let M = 230 – 215 + 1, and M2 be expressed in base 2. The number of 1's in this base 2 representation of

M2 is: (A) 29 (B) 30 (C) 59 (D) 60

12. In a triangle ABC, the altitude AD and the median AE divide A into three equal parts. If BC = 28, then

the nearest integer to AB + AC is (A) 38 (B) 37 (C) 36 (D) 33 13. The number of permutations of the letters a1, a2, a3, a4, a5 in which the first letter a1 does not occupy the

first position (from the left) and the second letter a2 does not occupy the second position (from the left) is

(A) 96 (B) 78 (C) 60 (D) 42 14. There are m books in black cover and n books in blue cover and all books are different. The number of

ways these (m + n) books can be arranged on a shelf so that all the books in black cover are put side by side is

(A) m!n! (B) m!(n + 1)! (C) (n + 1)! (D) (m + n)! 15. Let [x] be the greatest integer less than or equal to x, for a real number x. Then the equations [x2] = x + 1 has (A) two solution (B) one solution (C) no solution (D) more than two solution

PHYSICS

16. A frictionless wire AB is placed on a fixed smooth sphere of radius R. The length of the wire is adjustable along the surface of the sphere. A very small spherical bead slips on this wire as shown in

the figure. The time taken by the bead to slip from A to B after release varies with (if 0 < /2) as :

O

A

C

B

(A)

t

(B)

t

(C)

t

(D)

t



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17. A ball is thrown upward at an angle of 30° with the horizontal and lands on the top edge of a building

that is 20 m away. The top edge is 5m above the throwing point. The initial speed of the ball in metre/second is (take g = 10 m/s2) :

(A) 10 m/s (B) 20 m/s (C) 25 m/s (D) 30 m/s 18. A swimmer wants to cross a river and reach point B directly from A. The speed of the swimmer in still

river and that of river flow are same. The dotted line AN is normal to flow direction of river. For swimmer

to reach point B, the angle his velocity relative to river will make with line AN is given by . Then the

value of is :

(A) 60° (B) 30° (C) 45° (D) 0°

19. Two blocks of masses m1 = 8 kg and m2 = 7 kg are connected by a light string passing over a light frictionless pulley. The mass m1 is at rest on the inclined plane and mass m2 hangs vertically. The angle of inclination is 30º. Therefore, the force of friction acting on m1 is

(A) 30 N up the plane (B) 30 N down the plane (C) 40 N up the plane (D) 40 N down the plane

20. Kinetic energy of a particle moving in a straight line varies with time as K = 4t2. The force acting on the particle.

(A) is constant (B) is increasing (C) is decreasing (D) first increases and then decreases 21. Two blocks A and B each of same mass are attached by a thin inextensible string through an ideal

pulley. Initially block B is held in position as shown in figure. Now the block B is released. Bolck A will slide to right and hit the pulley in time tA. Block B will swing and hit the surface in time tB. Assume the surface as frictionless.

(A) tA = tB (B) tA < tB (C) tA > tB (D) data are not sufficient to get relationship between tA and tB.

22. Two identical disks are positioned on a vertical axis as shown in the fig. The bottom disk is rotating at

an angular velocity o and has rotational kinetic energy k

o. The top disk is initially at rest. It allowed to

fall and sticks to the bottom disk. The change in the rotational kinetic energy of the system after the collision is:

(A) ko/2 (B) (1/2) k

o (C) (1/4) k

o (D) k

o/4



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23. The figure shows four progressive waves A, B, C & D. It can be concluded from the figure that with respect to wave A:

(A) the wave C is ahead by a phase angle of /2 & the wave B lags behind by a phase angle /2

(B) the wave C lags behind by a phase angle of /2 & the wave B is ahead by a phase angle of /2

(C) the wave C is ahead by a phase angle of& the wave B lags behind by the phase angle of

(D) the wave C lags behind by a phase angle of & the wave B is ahead by a phase angle of 24. Which of the following will have maximum average total kinetic energy at temperature 300 K. (A) 1 kg of H

2 (B) 1 kg of He

(C) 1

2 kg of H

2 +

1

2 kg of He (D)

1

4 kg of H

2 +

3

4 kg of He

25. An ideal gas follows a process described by PV2 = C from (P1, V1, T1) to (P2, V2, T2) (C is a constant). Then

(A) if P1 > P2 then T2 > T1 (B) if V2 > V1 then T2 < T1 (C) if V2 > V1 then T2 > T1 (D) If P1 > P2 then V1 > V2

26. In the figure shown a hole of radius 2 cm is made in a semicircular disc of radius 6 cm at a distance 8 cm from the centre C of the disc. The distance of the centre of mass of this system from point C is:

(A) 4 cm (B) 8 cm (C) 6 cm (D) 12 cm

27. The intensity of sound during the festival season increased by 100 times. This could imply a decibel

level rise from (A) 20 to 120 dB (B) 70 to 72 dB (C) 100 to 10000 dB (D) 80 to 100 dB 28. A man stands at rest in front of a large wall. A sound source of frequency 400 Hz is placed between him and the wall. The source is now moved towards the wall at a speed of 1 m/s. The number of beats heard per second will be(speed of sound in air is 345 m/s) (A) 0.8 (B) 0.58 (C) 1.16 (D) 2.32

29. During the head on collision of two masses 1 kg and 2 kg the maximum energy of deformation is 100

3J.

If before collision the masses are moving in the same direction, then their velocity of approach before the collision is :

(A) 10 m/sec. (B) 5 m/sec. (C) 20 m/sec. (D) 10 2 m/sec.

30. For a particle in S.H.M., if the amplitude of displacement is ‘a’ and the amplitude of velocity is ‘v’ the

amplitude of acceleration is

(A) va (B) 2v

a (C)

2v

2a (D)

v

a



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CHEMISTRY

31. Na+, Mg2+, Al3+ and Si4+ ions are isoelectronic. The value of ionic radii of these ions would be in

the order :

(A) Na+ > Mg2+ > Al3+ > Si4+ (B) Na+ < Mg2+ < Al3+ < Si4+

(C) Na+ > Mg2+ > Al3+ < Si4+ (D) Na+ < Mg2+ > Al3+ > Si4+

32. The heat of combustion of C, S and CS2are –393.3 kJ, –293.7 kJ, and –1108.76 kJ. What will be the

heat of formation of CS2 ?

(A) –128.06 kJ (B) +970 kJ (C) +1108.7 kJ (D) +12 kJ

33. The correct order of increasing pH values of the aqueous solutions of baking soda, rock salt,

washing soda and slaked lime is:

(A) Baking Soda < Rock Salt < Washing Soda < Slaked lime

(B) Rock Salt < Baking Soda < Washing Soda < Slaked lime

(C) Slaked lime < Washing Soda < Rock Salt < Baking Soda

(D) Washing Soda < Baking Soda < Rock Salt < Slaked lime

34. The weight of oxalic acid required to neutralize 100 mL of normal NaOH is : (A) 6.3 g (B) 126 g (C) 530 g (D) 63 g 35. Which has the highest e/m ratio ? (A) He2+ (B) H+ (C) He+ (D) D+

36. Which of the following compound is planar and non-polar? (A) XeO4 (B) SF4 (C) XeF4 (D) CF4

37. The oxidant which cannot act as reducing agent is: (A) SO2 (B) NO2 (C) CO2 (D) C1O2

38. The value of Planck's constant is 6.63 × 10–34 Js. The velocity of light is 3.0 × 108 m s–1. Which value is

closest to the wavelength in nanometers of a quantum of light with frequency of 8 × 1015 s–1 ? (A) 3 × 107 (B) 2 × 10–25 (C) 5 × 10–18 (D) 4 × 101 39. The closed containers of the same capacity and at the same temperature are filled with 44 g of H2

in one and 44 g of CO2 in the other. If the pressure of carbon dioxide in the second container is 1atm, then pressure of hydrogen in the first container would be:

(A) 1 atm (B) 10 atm (C) 22 atm (D) 44 atm 40. The bond order in N2

+ is: (A) 1.5 (B) 3.0 (C) 2.5 (D) 2.0 41. Which of the underlined atoms in the molecules shown below have sp-hybridization?

(u) CH2CHCH3

(w) CH3CH2+

(y) CH3CN

(v) CH2CCHCl

(x) H — C C — H (z) (CH3)2CNNH2

(A) x and z (B) x, y and z (C) u, w and x (D) v, x and y 42. What is the IUPAC name of the compound ?

(A) 2 - Chloro - 2 - butane (B) 3 - Chloro - 1 - butene (C) 3 - Methyl - 3 - chloropropene - 1 (D) 3 - Chloro - 3 - methyl - 1 - propene



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43. Cyclohexene on reaction with cold alkaline KMnO4 forms, (A) trans-hexanediol (B) hexadiketone (C) cis-hexanediol (D) pentadiketone

44. Which of the following describes the best relationship between the methyl groups in the chair

conformation of the substance shown below?

(A) Trans (B) Anti (C) Gauche (D) Eclipsed

45. The IUPAC name of 2 2 2

||O

OHC CH CH C CH COOH is

(A) 1-formyl-3-oxopentanoic acid (B) 5-formyl-3-oxopentanoic acid (C) 3-oxo-5-formylpentanoic acid (D) 3-oxo-1-formylpentanoic acid

BIOLOGY

46. Function of non-kinetochore microtubules is ;

i. To help the chromosomes to get arranged at the equator.

ii. To elongate the cell during anaphase.

iii. To help the separation of chromatids during anaphase.

Which of the statements/s is/are correct?

(A) i only (B) ii only (C) iii only (D) i and iii

47. Chemiosynthetic autotropic bacteria oxidise various inorganic substances such as nitrates, nitrites and

ammonia for

(A) DNA replication (B) RNA synthesis (C) ATP production (D) Protein production

48. White coat color in guinea pigs is recessive (b) to black (B). Ovary from black homozygous guinea pig

is transplanted into a white ovarectomized female. If this white female is mated with a white male the

offspring will be :

(A) black with the genotype BB. (B) white with the genotype bb.

(C) black or white with genotype Bb OR bb. (D) black with the genotype Bb.

49. In Maize, the stilt roots develop from

(A) Lower nodes (B) Upper nodes (C) Upper internodes (D) None of the above.

50. A man with Haemophilia has a daughter of normal phenotype. She marries a man who is not

haemophilic. If the couple has four sons, what is the possibility that all four will be born haemophilic?

(A) 1/4 (B) 1/32 (C) 1/16 (D) 1/8

51. Which is feature of bacterial cell

(A) Presence of nucleolus (B) Absence of cytoplasmic streaming movement

(C) Presence of single stranded DNA (D) Presence of cellulosic cell wall



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52. It is significant to note that in the 24 hours average duration of cell cycle of human cell, cell division

proper lasts for only about :-

(A) Four hours (B) 90 minutes (C) An hour (D) 10 hours

53. Path of water movement from soil to xylem is

(A) metaxylem protoxylem cortex soil root hair.

(B) cortex root hair endodermis pericycle protoxylem metaxylem.

(C) soil root hair cortex endodermis pericycle protoxylem metaxylem.

(D) pericycle soil root hair cortex endodermis protoxylem metaxylem.

54. Which statement is true-

(A) All chordata are viviparous (B) All lizards are poisonous

(C) All amphibians lay their eggs in the water (D) All aves are capable to fly

55. Amphibians and Reptiles, the left atrium receives oxygenated blood from the gills/lungs/skin and the

right atrium receives the deoxygenated blood from other body parts. However, they get mixed up in the

single ventricle which pumps out mixed blood called:-

(A) Incomplete single circulation (B) Complete single circulation

(C) Incomplete double circulation (D) Complete double circulation

56. Lungs are covered with covering of

(A) Mucous membrane (B) Pleural membrane

(C) Pericardium (D) Peritoneum

57. Which part of nephron is not situated in the cortical region of the kidney?

(A) Malpighian corpuscle (B) PCT

(C) DCT (D) Loop of Henle

58. Which of the following can’t be associated with chemical synapse?

(A) Pre-synaptic membrane (B) Post-synaptic membrane

(C) Neurotransmitter (D) Two way transmission of impulse

59. The receptor for protein hormones are present on

(A) Cell surface (B) Nucleus

(C) Endoplasmic reticulum (D) Cytoplasm

60. Two elements needed for building healthy teeth and bones are:

(A) iron and calcium (B) phosphorous and iron

(C) calcium and phosphorous (D) CO2 and H2O



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PART- B

Two Mark Questions

MATHEMATICS 61. Consider a regular 10-gon with its vertices on the unit circle. With one vertex fixed. draw straight lines to

the other 9 vertices. Cell them L1, L2, ….., L9 and denote their lengths by 1, 2, ,,,,,9 respectively then

the product 1, 2, ,,,,,9 is

(A) 10 (B) 10 3 (C) 3

50 (D) 20

62. Let ABC be triangle such that AB = 4. BC = 5 and CA = 6. Choose points D, E, F on AB, BC, CA

respectively. such that AD = 2, BE = 3, CF = 4. Then ABCarea

DEFarea

is

(A) 4

1 (B)

15

3 (C)

15

4 (D)

30

7

63. For [0,]. Let f() = sin(cos and g() = cos(sin) . Let a = fmax

0, b =

fmin

0 ,

c = gmax

0and d =

gmin

0 . The correct inequalities satisfied by a, b, c, d are

(A) b < d < c < a (B) d < b < a < c (C) b < d < a < c (D) b < a < d < c 64. A bottle in the shape of a right-circular cone with height h contains some water. When its base is placed

on a flat surface, the height of the vertex from the water level is a units. When it is kept upside down,

the height of the base from the water level is 4

a units. Then the ratio

a

h is

(A) 4

851 (B)

8

851 (C)

4

651 (D)

8

651

65. Consider the following two statements: I. If n is a composite number, then n divides (n –1)!. II. There are infinitely many natural numbers n such that n3 + 2n2 + n divides n!. Then (A) I and II are true (B) I and II are false (C) I is true and II is false (D) I is false and II is true

PHYSICS

66. At the instant a motor bike starts from rest in a given direction, a car overtake the motor bike, both moving in the same direction The speed time graphs for motor bike and the car are represented by OAB and CD respectively. Then

C

A B

D

60

40

20

O 27 18 12

speed

(m/s)

s.

(A) at t = 18 s the motorbike and car are 180 m apart.

(B) at t = 18 s the motor bike and car are 720 m apart

(C) the relative distance between motor bike and car always remains same.

(D) None of these



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67. A bullet of mass m moving with velocity v0 just grazes (bullet does not slip on the disc during the contact

with the disc) the top of a uniform circular disc of mass M and radius R resting on a rough horizontal surface as shown in figure. Assuming that the disc rolls without slipping, find the velocity of the bullet just after it grazes the disc :

(A) M3m8

mu4

(B)

M3m8

mu8

(C)

M3m4

mu8

(D)

M3m4

mu4

68. Two pendulums differ in lengths by 22 cm. They oscillate at the same place so that one of them makes

30 oscillations and the other makes 36 oscillations during the same time. The lengths (in cm) of the pendulum are :

(A) 72 and 50 (B) 60 and 38 (C) 50 and 28 (D) 80 and 58 69. Four identical beakers contain same amount of water as shown below. Beaker 'a' contains only water.

A lead ball is held submerged in the beaker 'b' by string from above. A same sized plastic ball, say a table tennis (TT) ball, is held submerged in beaker 'c' by a string attached to a stand from outside. Beaker 'd' contains same sized TT ball which is held submerged from a string attached to the bottom of the beaker. These beakers (without stand) are placed on weighing pans and register reading Wa, Wb, Wc and Wd for a, b, c and d, respectively. (Effects of the mass and volume of the stand and string are to be neglected.)

a

b

lead

c

TT

d

TT

(A) Wa = Wb = Wc = Wd (B) Wb = Wc > Wd > Wa

(C) Wb = Wc > Wa > Wd (D) Wb > Wc > Wd > Wa

70. 540 g of ice at 0ºC is mixed with 540 g of water at 80ºC. The final temperature of the mixure is (A) 0ºC (B) 40ºC (C) 80ºC (D) 100°C

CHEMSITRY

71. Which of the following metal sulphides has maximum solubility in water?

(A) CdS (Ksp = 36 × 10–30) (B) FeS (Ksp= 11 × 10–20)

(C) HgS (Ksp = 32 × 10–54) (D) ZnS (Ksp = 11 × 10–22)

72. Diborane is a potential rocket fuel which undergoes combustion according to the equation

B2H6(g) + 3O2(g) B2O3(s) + 3H2O(g)

Calculate the enthalpy change for the combustion of diborane. Given

(i) 3B(s) +2

3O2(g) B2O3(s) ; H = – 1273 kJ per mol

(ii) H2(g) +2

1O2(g) H2O (l) ; H = – 286 kJ per mol

(iii) H2O(l) H2O (g) ; H = 44 kJ per mol

(iv) 2B(s) + 3H2(g) B2H2 (g) ; H = 36 kJ per mol

(A) + 2035 kJ per mol (B) – 2035 kJ per mol (C) + 2167 kJ per mol (D) – 2167 kJ per mol



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73. Select the incorrect graph for velocity of e– in an orbit vs. Z, n

1and n :

(A) (B)

(C) (D)

74. Arrange the following substances in order of decreasing heat of combustion (maximum minimum).

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

75. Arrange in the order of increasing acidity

(A) III < I < II (B) I < III < II (C) III < II < I (D) II < I < III

BIOLOGY

76. Which statements is/are true for dorsiventral leaves?

(a) The size of the vascular bundles is dependent on the size of veins

(b) Vascular bundles are surrounded by a layer of thick walled cells, called bundle sheath

(c) Mesophyll is not differentiated

(A) a and b (B) b and c (C) a, b and c (D) a and c

77. Read the following statement and select the correct option that bears true statements

(i) After S-phase DNA number become twice but chromosomes remain same i.e. 2n

(ii) A cell in G0 phase in metabolically active and can enter in division phase depending on the

requirement of organism

(iii) Cell division is a progressive process that shows the distinct boundaries between various stages

(iv) At the end of prophase GB, ER, Nucleolus and nuclear envelope get disappear

(A) i, ii & iv (B) i, ii & iii (C) ii, iii & iv (D) all of these



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78. Excessive loss of water causes wilting of leaves which can be prevented by

(A) keeping the plant in bright light. (B) applying vaseline on the leaf surface.

(C) adding high amount of fertilizers to the soil. (D) spraying the plant with alcohol.

79. Choose the incorrect statement:

(A) Some division of labour (activities) occur among the cells in the members of phylum porifera

(B) Division of labour (activities) is completely absent among the cells in poriferans

(C) Open circulatory system is found in Tunicates, hemichordates, arthropods and non-cephalpod

molluscs

(D) In closed circulatory system, the exchange of materials between blood and body cells occur at the

level of blood capillaries

80. Which of the following statement is not correct?

(A) Goblet cells are present in the mucosa of intestine and secrete mucus

(B) Oxyntic cells are present in the mucosa of stomach and secrete HCl

(C) Acini are present in the pancreas and secrete carboxypeptidase

(D) Brunner's glands are present in the submucosa of stomach and secrete pepsinogen



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FULL SYLLABUS TEST-2 (KFST-2)



ANSWER KEY

PART- A

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

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

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

22. (B) 23. (B) 24. (A) 25. (B) 26. (B) 27. (D) 28. (D)

29. (A) 30. (B)

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

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

45. (B)

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

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

60. (C)

PART- B

61. (A) 62. (C) 63. (C) 64. (B) 65. (D) 66. (A) 67. (B)

68. (A) 69. (B) 70. (A)

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

76. (A) 77. (A) 78. (B) 79. (B) 80. (D)



®




HINTS & SOLUTIONS

PART- A MATHEMATICS

1. Given 9 × abcde = edcba

(10–a) abcde = edcba

a b c d e o - a b c d e _________ e d c b a a = 1, e = 9, b = 0, c = 9, d = 8

abcde = 10989

sum of digits = 27 2. x + y = 1, x, y > 0

Let x = sin2& y = cos2

Hence, y

1

x

1 = cosec + sec2

= 2+ tan2 + cot2 4 3. p,q,r are positive rational

and rationalrqp

So p + q + r +

2 rp2qr2pq rational

rationalrpqrpq2

4. 2 log (x – 2y) = log x + log y

log (x – 2y)2 = log xy

(x – 2y)2 = xy x2 + 4y2 – 5xy = 0

2

2

y

x + 4 – 5

y

x = 0

2

y

x

–

y

x5 + 4 = 0

1

y

x

4

y

x = 0

y

x = 1, 4

x > 2y y

x = 4

5. P1(x) = x3 – 2020x2 + bi x + c1

P2(x) = x3 – 2021 x2 + b2x + c2 P1(x) q1(x) + P2(x) q2(x) = x2 – 3x + 2

as P1(x) q1 (x) + P2 (x) = 0 has , as common roots

P1(x) q(x) + P2(x) q2(x) has two factors

(x – ), (x – )

x2 – 3x + 2 = 0 has factors x – , –

= 1 and = 2

P1(x) = (x – 1) (x – 2) (x – 2017) and P2(x) = (x – 1) (x – 2) (x – 2018)

P1(3) = (2) (1) (–2014) = – 4028 and P2(1) = 0

P1(3) = (2) (1) (–2014) = – 4028 and P2(1) = 0

P1(3) + P2(1) + 4028 = 0 6.

•

D C

B A

y

w

x

z E

x + y + z + w = 360º

+ + + = 180º

+ + + = 2( + ) using A.P.

+ = + = 180º – ( + )

+ = + = 90º

So, median of angles

= 3

BCDABCDAB

= 3

= 3

)º90(2º90 = 90º =

2

7. (2021)2020 = (1 + 2020)2020 = 1 + 2020C1. 2020 + 2020C0. (2020)2 +...... = 1 + (2020)2 + (2020)2 I So reminder = 1 8.

710,709........101100

610

99........121110

90

9......321

18302909

189

189 + 1830 = 2019 upto 709 total number of digits = 2019 next number is 710 so 2021 digit is 1

9. Let PCB =

A

B C

P

D

DPC = –

BPC =



®




DPB = – ( – ) = In DPB & PCB

DPB = PCB

and DPB = PBC

So DPB = PBC

then BC

PB

PB

DB

PC

DP

Now BC

PB

PB

DB

PC

PD2

So 7

2

PC

PD

10. ?

16

16

15

1

14

14

13

13

12

122018

2020

2018

2020

2018

2020

2018

2020

2018

2020

22020 +1 = 22 (22018) + 1 = 22 (22018+1)–3 = 4(22018+1)–3

1212

12 2

2018

2020

……………..(1)

Similarly 1313

13 2

2018

2020

1414

14 2

2018

2020

1515

15 2

2018

2020

& 1616

16 2

2018

2020

Sum = (22 + 32 +42 + 52 + 62)–5

=

66

1376

= 91–6 = 85

11. M = 230 – 215 + 1

M2 = 260 – 246 + 231 + 230 – 216 + 1 = 246 (214 – 1) + 231 + 216 (214 – 1) +1

246 112

)12(22

12

)12( 141631

14

246 ( 1 + 2 + 22 + .....+ 213) + 231 + 1 216 (1 + 2 + 22 +.....+213) +1

14

)2+........ + 2 + (2 594746

+ 231 +

0291716

214

)2+........ + 2 + (2

There will be 30 times digit 1 12.

A

B C D E

c b

14 7 7

BC = 28

AB + AC = c + b

tan2 =

21, tan =

7

2

49–1

27

=

21 37 so = 30º

Now c + b =

)31(143

22114

º60sin

21

º30sin

7

13. O O O O O

a2 4 = 24

O O O O O a1

3C1 3C1. 3 = 54

Total = 24 +54 = 78 14. Total books = m + n All black books put together

Total books

= n + 1 (consider m books at 1) Total arrangement = m.1n

15. [x2] = x + 1

x = [x2] – 1

clearly x [as [x2] and both integer)

x = x2 – 1

x2 – x – 1 = 0

x2 – x – 1 = 0 PHYSICS

16. As s = 1

2 at2

2R cos = 1

2 g cos t2

‘t’ is independent of ‘’

17. y = x tan – 2

2 2

gx

2u cos

5 = 20 tan30° – 2 2

10 400

2u cos 30

u 20 m/s 18. Since the speed of river flow and the

person is same, the resultant direction AB is



®




such that 90

2

= 60 = 30°

19.

30° m2

80×sin30° m1

T = 70N 70N

= 40 N

f = 30 N

20. 2 21mv 4.t

2

1 dvm 2 v

2 dt = 4 × 2t

m. dv

dt =

8t

v =

2

8t8m

8t

m

21.

Tcos

Tsin

A

T

T sin < T

tA < tB

22. I = 2I'

' = 2

k = ki – kf =

2

2 ik1 1(2I) I

2 2 2 2

23. In figure, 'C' reaches the position where 'A'

already reaches after t = /2 and 'A' reaches the position where 'B'

already reaches after t = /2 Hence (B).

24. TKE = U = 2

nRT

In case of H2 degree of freedom is greatest

and number of moles n is highest So this is the case of maximum kinetic

energy. 26. Taking C as origin and x & y–axes as

shown in figure. Due to symmetry about y–axis

xcm = 0

ycm = 1 1 2 2

1 2

m y m y

m m

=

22

22

(6 ) 4(6 ) –[ (2) (8)]2 3

(6 )(2)

2

(m Area)

= 2

2

8(18 4)

(18 4)

= 8 cm.

27. Sound intensity = 10 log10

0I

I,

I = 100 I0 so

log10 0

0100

i

i

= 20 So level rises by 20 dB

28. f = 2 2

c c cf2u 2uf.f f

c u c u cc u

f = 2.1.400 800

345 345 = 2.32

29. U = 1 2

1 2

1 m m

2 (m m ) (V1 – V2)2 =

100

3

(V1 – V2)2 × 2m .m

2 (m 2m) =

100

3

Putting m = 1 kg : (V1 – V2) = 10 m/sec. AlternateSolution: When deformation is maximum

both the particles are moving with same velocity. So applying momentum conservation.

m1v1 + m2v2 = m1v 1’ + m2v1’

v1’ = 1 1 2 2

1 2

m v m v

m m

Applying energy conservation:

1

2 m1v1

2 + 1

2m2v2

2 = 1

2 (m1 + m2) (v1´ )2

+ Udeformation

U deformation= 1

2 1 2

1 2

m m

m m

× (v1 – v2)2 = 100

3

v1 – v2 = 10m/sec. 30. Maximum velocity v = a

Maximum acceleration f = 2 a f = 2v

a

vf/kdre osx v = a

vf/kdre Roj.k f = 2 a f = 2v

a

CHEMISTRY



®




31. Na+ Mg2+ Al3+ Si4+

Protons 11 12 13 14

Electrons 10 10 10 10

Size of isoelectronic cations decreases with

increase in magnitude of nuclear charge

Order of decreasing size is

Na+ > Mg2+ > Al3+ > Si4+

32. C + 2S CS2

H = H – HR

= –1108.76 – [–393.3 – 2 ´ (293.7)]

= –1108.76 + 393.3 + 587.4

= –128.06 kJ

33. Higher the acidic character of a compound,

lesser will be its pH value.

Rock Salt (NaCl) < Baking Soda (NaHCO3)

< Washing Soda (Na2CO3) < Slaked lime

(Ca(OH)2)

34. Number of Eq. of NaOH = Number of

Eq. of oxalic acid

Then 63

acidoxalicof.wt

1000

1100

W= 6.3 g

35. e/m for He2+ = 2/4 e/m for H+ = 1/1

e/m for He+ = 1/4 e/m for D+ = 1/2

Value of e/m is highest for H+.

36.

37. CO2 has the maximum oxidation state of

+4. So it cannot go to a higher oxidation

state.

38.

39.

40. N2

+ = 7 + 7 – 1 = 13 electrons

Electronic Configuration is : 1s2 *1s2

2s2 *2s2 2px2 2py

2 2pz1

Bond order = 2

1 [No. of bonding electrons

– No. of antibonding electrons]

=2

1(9 – 4) =

2

1× 5 = 2.5

41.

42. It has chain of 4C atoms and double bond

is given preference over Cl atom.

43. Cyclohexene + KMnO4 (cold & alkaline)

cis-Hexanediol.

44.

45. 2 2 2

||O

OHC CH CH C CH COOH 12345

5-formyl-3-oxo pentanoic acid

PART- B MATHEMATICS

61.

10

28l

9l10A

1l1A 2A

3A

4A

3l

2l

II

O

cos)1)(1(2

–11

5

21

22



®




5

cos2–221

5

cos–12

10

sin4 2 1 = 2sin

10

= 9

Similarly 2 = 2 sin 810

2

10

2cos2

10

3sin2 73

,

10

5sin2,

10cos2

10

4sin2 564

1.2.3 ………..9

=2222

10cos2.

10

2cos2.

10

2sin2

10sin2

.

10

5sin2

= 2.10

4sin2.

10

2sin2

22

= 32. 2

22

4

5210

4

52–1032

5

2sin.

5sin

= 10 62.

A

C B E

F D

3 2

2

2 6

4

5

2

4

Let area of ABC =

= 2

1× 6 × 4sinA = 12sinA

Also = 2

1 × 4 × 5sinB = 10sinB

Also = 2

1× 5 × 6sinC = 15sinC

Let Area of ADF

= 1 = 2

1 × 2 × 2sinA = 2sinA

Let Area of BDE

= 2 = 2

1 × 2 × 3sinB = 3sinB

Let Area of ECF

= 3 = 2

1 × 4 × 2sinC = 4sinC

321 –––

ABCofArea

DEFofArea

321 –––1

= 1 – Asin12

Asin2 –

Bsin10

Bsin3 –

Csin15

Csin4

= 1 – 6

1 –

10

3 –

15

4 =

15

4 Ans.

63. cos [-1,1]

sin (cos) [–sin1, sin1]

a = sin1, b = –sin1

sin [0,1] [0, ]

cos(sin) [cos1, cos0]

cos(sin) [cos1,1]

b < d < a < c

64.

B C

A

mr

a r1

h

a

r

h

h

a

h

arr1

V of water = a.h

ra

3

1hr

3

12

222

4

a

m

h

4

ahr

rr

r

4

ah

h1

1

V of water =

4

ah.

h

4

ahr

3

12

22

2

2

2

2

32

h

4

ah

.r3

1

h

ahr

3

1

3

33

4

ahah

2

3

3

4

1

a

h1

a

h

64

1

16

1.t3

4

1.t3t1t 233



®




–64 = 48t2 + 12t – 1

48t2 – 12t – 63 = 0

96

1224012

96

1209614412t

= 96

851212

8

851

a

ht

65. n3 + 2n2 + n = n(n +1)2

So, 2)1n(n

!n

=

2)1n(

)!1–n(

if (n +1) is prime number, then (n –1)! will not divisible by n + 1

Ex. = 7 If (n +1) is a composite number, then

(n –1)! may be divisible by (n +1). Ex. 24, 25, 27, .................................... So, statement I is false and statement II is

true.

PHYSICS

66. at t = 18 second separation between bike and car is difference of area bounded by two curves

18 ×4 0 – ½(18 × 60) = 180 m At 27 second area bounded by each is

equal to 27x40=1080 m relative distance reduced to zero when

areas are equal 1/2 × 18 × 60 + 9 × 60 – 27 × 40 = 0

67. Conserving the angular momentum of (bullet + disc) system about O.

mv02R = mv(2R) +

0

v = 2R ; 0 =

2MR2

3

v = M3m8

mu8

68. T1 = 12

g

T2 = 22

g

1

t30

T 2

1

T 5

T 6

2

t

T = 36 6T

2 = 5T

1

T1

2 = 88 36 6 2

100 11 10

=

6 2

5

6 2

5 = 12

g

36 2

25

= 4 × 10 × 1

10 Ans. (A)

69.

a Wa = Ww = Weight of Water

b

lead

Wb = Ww + Fb ;

Fb = Buoyant Force

c

TT

Wc = Wb + Fb ;

Fb = Buoyant Force

d

TT

Wd = Ww + Fb – T

Fb = Buoyant Force

T = Tension in String

Ans. (B) 70. mL = 540 × 80 = 54 × 8 × 100 cal.

Fb

Fb

Fb

Fb

T

Fb

Fb



®




amount of ice water at 0°C Heat = 540 × 80 cal.

water (80°) water at 0°C heat available = 540 × 80 cal.

So the temperature of mixture = 0°C

CHEMSITRY

71. Higher the Ksp, more soluble is that

compound in H2O.

Ksp of FeS (11 × 10–20) is highest.

So it is more soluble and has maximum

solubility in H2O.

72. For the equation

B2H6(g) 3O2(g) B2O2(g) 3H2O(g)

sEqs. (i) + 3 (ii) + 3 (iii) – (iv)

H = – 1273 + 3(–286) + 3(44) – 36

= – 1273 – 858 + 132 – 36

= – 2035 kJ/mol

73. v = 2.188 × 106 n

Z m/s

Following the above relation, option (D)

represents the incorrect graph between v

and n.

74. Heat of combustion

moleculeofStability

1 No. of

‘C’ atoms in molecule. 75. Order of Acidic strength stability of

conjugate base.



FULL SYLLABUS TEST-2


(KVPY) 2023 | STREAM (SA)_XI

®

D Time: 3 Hours Max. Marks : 100.

Ñi;k bu funsZ'kksa dks /;ku ls i<+saA vkidks 5 feuV fo'ks"k :i ls bl dke ds fy, fn;s x;s gSaA

INSTRUCTIONS/ funsZ'k :

1. iz'u iqfLrdk esa fn, x, LFkku fu/kkZfjr LFkku ij rRdky viuk uke vkSj jksy uEcj dkys@uhys ck¡y ikbUV iSu ls Hkfj;sA isfUly dk mi;ksx izfrcaf/kr gSA

2. bl iz'u&i=k ds nks Hkkx A ,oa B ¼ nksuksa Hkkxksa dsoy cgqfodYi iz'u gSa½ gSA izR;sd Hkkx ds pkj [k.M gSaaA Hkkx&A ¼izR;sd [k.M esa 15 iz'u½ o Hkkx& B ¼izR;sd [k.M esa 5 iz'u½ gSaA

3. bl iz'u&i=k ds nks Hkkx A ,oa B gSaA lgh mRrj ds fy;s nkuksa Hkkxksaa esa fo"k;kuqlkj vadksa dk forj.k fn;k x;k gSA a

vadu ;kstuk a:

Hkkx&A :

xf.kr

iz'u la[;k 1 ls 15 rd izR;sd lgh mRrj ds fy, (1) vad o xyr mRrj ds fy, – 0.25 vad fn;s tk;sxsasA

HkkSfrd foKku iz'u la[;k 16 ls 30 rd izR;sd lgh mRrj ds fy, (1) vad o xyr mRrj ds fy, – 0.25 vad fn;s tk;sxsasA

jlk;u foKku iz'u la[;k 31 ls 45 rd izR;sd lgh mRrj ds fy, (1) vad o xyr mRrj ds fy, – 0.25 vad fn;s tk;sxsasA

tho foKku iz'u la[;k 46 ls 60 rd izR;sd lgh mRrj ds fy, (1) vad o xyr mRrj ds fy, – 0.25 vad fn;s tk;sxsasA

Hkkx& B :

xf.kr

iz'u la[;k 61 ls 65 rd izR;sd lgh mRrj ds fy, (2) vaad o xyr mRrj ds fy, – 0.5 vad fn;s tk;sxsasA

HkkSfrdh foKku


jlk;u foKku


tho foKku


4. gj iz'u esa dsoy lgh mÙkj okys cqycqys(BUBBLES) dks dkyk djus ij vad iznku fd, tk;saxsA vU; lHkh fLFkfr;ksa esa vad ugha iznku fd;k tk;sxkA

5. vuqefr i=k ds vfrfjDr] dksjs dkxt] fDyi cksMZ (CLIP BOARD)] ykWx rkfydk] LykbM:y] dSYdqysVj] dSejk] lsyQksu] istj vkSj fdlh izdkj ds bysDVªkfud midj.k ijh{kk d{k esa vuqefr ugha gSA

6. dPps dke ds fy;s [kkyh txg bl iqfLrdk esa gh gSA dPps dke ds fy, dksbZ vfrfjDr dkxt ugha fn;k tk;sxkA

7. iz'u i=k iw.kZ gks tkus ij mRrj iqfLrdk dks ijh{kk gkWy esa mifLFkr ijh{kd dks lkSai nsaA fo|kFkhZ dks iz'u i=k ys tkus dh vuqefr gSA

8. ORS dks u dkVs u gh QkMs u gh xUnk ugha djsa rFkk dksbZ Hkh fu'kku ;k lQsnh ORS ij ugha yxk;sA

Resonance Eduventures Limited REGISTERED & CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005

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ijh{kkFkhZ dk uke :

eSaus lHkh funsZ'kksa dks i<+ fy;k gS vkSj eSa mudk vo'; ikyu d:¡xk@d:¡xhA

...................................... ijh{kkFkhZ ds gLrk{kj

jksy uEcj:

ijh{kkFkhZ }kjk Hkjh xbZ lkjh tkudkjh dks eSusa tk¡p fy;k gSA

...................................... ijh{kd ds gLrk{kj



®




(KVPY-STREAM-SA)_XI FULL SYLLABUS TEST-2

PART- A

One Mark Questions

xf.kr

1. ik¡p vadks dh ,d la[;k abcde, dks 9 ls xq.kk djus ij 5 vadksa dh la[;k edcba nsrk gS] bl la[;k esa vadksa dk ;ksx

gS% (A) 18 (B) 27 (C) 36 (D) 45

2. eku fy;k fd x o y nks /kukRed okLrfod la[;k,a bl izdkj gSa fd x + y = 1 rc y

1

x

1 dk U;wure eku gS

(A) 2 (B) 2

5 (C) 3 (D) 4

3. eku ysa fd p,q,r /kukUed ifjes; la[;k,a bl izdkj gSa fd rqp Hkh ifjes; gSa] rc

(A) r,q,p vifjes; gSa (B) qr,pr,pq ifjes; gSa] ijUrq r,q,p vifjes; gS

(C) r,q,p ifjes; gaS (D) qr,pr,pq vifjes; gSa

4. eku ys fd x, y okLrfod la[;k,sab l izdkj gS fd x > 2y > 0 ,oa 2log(x – 2y) = logx + logy

rc y

x ds laHkkfor eku gS &

(A) 1 dsoy (B) 1 ,oa 4 gS (C) dsoy 4 gS (D) dsoy 8 gS

5. ekuk fd p1(x) = x3 – 2020x2 + b1x + c1 o p2(x) = x3 – 2021x2 + b2x + c2 nks cgqin gSa] ftlds o nks mHk;fu"V ewy gSa] eku ys fd q1(x) o q2(x) cgqin ,sls gSa fd p1(x)q1(x) + p2(x)q2 (x) = x2 – 3x + 2. rc lgh rRled gS%

(A) p1(3) + p2(1) + 4028 = 0 (B) p1(3) + p2(1) + 4026 = 0 (C) p1(2) + p2(1) + 4028 = 0 (D) p1(1) + p2(2) + 4028 = 0

6. eku ysa fd ABCD ,d prqHkqZt bl izdkj gS fd] prqHkqZt ds Hkhrj ,d fcUnq E gS tks AE = BE = CE = DE dk larq"V djrk gS] eku ysa fd DAB, ABC,BCD ,d lekUrj Js<h (AP) gS] rc leqPp; {DAB, ABC,

BCD} dk ek/; gS (A)

6

(B)

4

(C)

3

(D)

2

7. 20212020 dks 20202 ls Hkkx nsus ij cps 'ks"k r dk eku fdlds chp esa gksxk (A) 0 ,oa 5 (B) 10 ,oa 15 (C) 20 ,oa 100 (D) 107 ,oa 120

8. ;fn 1 ls ysdj 2021 rd ds iw.kkZadksa dks 123…91011…20202021 dh rjg ,dy iw.kkZad la[;k ds :Ik esa fy[kk tk,] rks ifj.kkeh la[;k esa ck,Î¡ ls fxuus ij 2021oka vad gksxk

(A) 0 (B) 1 (C) 6 (D) 9



®




9. fdlh f=kHkqt ABC esa, BC ij ,d fcUnq D izdkj pquk tkrk gS fd BD : DC = 2:5 eku ysa fd ifjo`Ùk ABC ij ,d fcUnq P bl izdkj gS fd PDB = BAC rc PD : PC gS

(A) 5:2 (B) 2 : 5 (C) 2 : 7 (D) 7:2

10. fdlh okLrfod x, ds fy, x, ls NksVk ;k x ds cjkcj lcls mPpre iw.kkZad [x] gS

16

16

15

15

14

14

13

13

12

122018

2020

2018

2020

2018

2020

2018

2020

2018

2020

gS (A) 80 (B) 85 (C) 90 (D) 95

11. eku M = 230 – 215 + 1 ,oa M2 dks vk/kkj 2 ij O;Dr fd;k tkrk gS] M2 ds vk/kkj 2 ds bl fu:i.k esa fdrus 1 dh la[;k gS&

(A) 29 (B) 30 (C) 59 (D) 60

12. fdlh f=kHkqt ABC esa 'kh"kZyEc AD ,oa ek/; AE, A dks rhu leku Hkkxksa esa ckaVrk gS] ;fn BC = 28 rks AB + AC

dk fudVre iw.kkZad gS

(A) 38 (B) 37 (C) 36 (D) 33

13. v{kjksa a1, a2, a3, a4, a5 ds Øep;ksa dh la[;k D;k gksxh ftlesa igyk v{kj a1 ck,¡ ls igyk LFkku rFkk nwljk v{kj a2 ck;sa ls nwljk LFkku xzg.k ugha djrk

(A) 96 (B) 78 (C) 60 (D) 42

14. m iqLrdsa dkys vkoj.k esa vkSj n iwLrdsa uhys vkoj.k esa gS vkSj lHkh iwLrdsa fHkUu gS] dqy (m + n) iwLrdksa dks vkyekjh esa fdrus <ax ls ltk;k tk ldrk gS ftlls fd dkys vkoj.k okyh lHkh iqLrdsa lkFk&lkFk jgs

(A) m!n! (B) m!(n + 1)! (C) (n + 1)! (D) (m + n)!

15. eku fy;k fd fdlh okLrfod la[;k [x] ds fy,] x ls NksVk ;k mlds cjkcj egÙke iw.kkZad [x] gS] rc lehdj.k [x2] = x + 1 ds fdrus gy gSa (A) nks gy (B) ,d gy

(C) dksbZ gy ugha (D) nks ls vf/kd gy HkkSfrd foKku

16. R f=kT;k ds ,d fpdus tM+or~ xksys esa fp=kkuqlkj fLFkr ?k"kZ.kghu rkj AB gSA rkj dh yEckbZ xksys dh lrg ds vuqfn'k lek;ksftr gks ldrh gSA ,d cgqr NksVk xksykdkj eksrh bl rkj ds vuqfn'k n'kkZ;s vuqlkj fQlyrk gSaA (;fn 0 < /2) dks.k ds Qyu ds :i esa eksrh dks fQlydj A ls B vkus esa fy;k x;k le; fdl vkjs[k ds vuqlkj gksxk&

O

A

C

B

(A)

t

(B)

t

(C)

t

(D)

t

17. ,d xsan dks tehu ls {kSfrt ls 30° dks.k ij Qsadk tkrk gS tks ç{ksi.k fcUnq ls 20 m nwj fLFkr bZekjr dh vf/kdre lrg ij ykSVrh gSA vf/kdre lrg ls ç{ksi.k fcUnq ls 5m Å¡ph gSA rks xsan dh çkjfEHkd pky gksxh (g = 10 m/s2) :

(A) 10 m/s (B) 20 m/s (C) 25 m/s (D) 30 m/s



®




18. ,d rSjkd unh dks ikj djrs gq, A fcUnq ls lh/ks B ij igq¡prk gSA rSjkd dh pky ¼ikuh ds lkis{k½ rFkk unh izokg dh pky ,d leku gSA fcUnqor js[kk AN unh izokg ds yEcor~ gSA fcUnq B ij igq¡pus ds fy, rSjkd dk unh ds lkis{k osx js[kk AN ds lkFk dks.k cukrk gS rks dk eku gksxk &

(A) 60° (B) 30° (C) 45° (D) 0°

19. m1 = 8 kg rFkk m2 = 7 kg nzO;eku ds nks xqVdk gYdh rkj ls ?k"kZ.kjfgr f?kjuh ls gksdj tqM+k gqvk gSA nzO;eku m1

urry ij fLFkjkoLFkk esa gS rFkk m2 Å/okZ/kj >qy jgk gSA urdks.k dk eku 30º gSA vr% m1 ij dk;Zjr~ ?k"kZ.k cy dk eku gksxk

(A) 30 N lrg ds vuqfn'k Åij dh vksj (B) 30 N lrg ds vuqfn'k uhps dh vksj

(C) 40 N lrg ds vuqfn'k Åij dh vksj (D) 40 N lrg ds vuqfn'k uhps dh vksj

20. lh/kh js[kk esa xfr'khy d.k dh xfrt ÅtkZ le; ds lkFk K = 4t2 ds vuqlkj cnyrh gSA d.k ij dk;Zjr cy gS (A) fu;r gS (B) c<+ jgk gS (C) ?kV jgk gS (D) igys c<+ jgk gS rFkk ckn esa ?kV jgk gS

21. nks leku nzO;eku ds CykWd A rFkk B vfoLrkfjr o nzO;eku jfgr jLlh ds }kjk vkn'kZ f?kjuh ls tksM+s tkrs gSa izkjEHk esa CykWd B dks fp=kkuqlkj j[kk tkrk gSA vc CykWd B dks NksM+k tkrk gSA CykWd A lrg ij fQlyrk gqvk f?kjuh ls tA le; esa Vdjkrk gSA CykWd B >qyrk gqvk lrg ls tB le; esa Vdjkrk gSA ¼ekuk lrg ?k"kZ.k jfgr gSA½

(A) tA = tB (B) tA < tB (C) tA > tB

(D) lwpuk vi;kZIr gS blfy, tA rFkk tB esa laca/k iznf'kZr ugha fd;k tk ldrkA

22. nks ,dleku pdfr;k¡ Å/okZ/kj v{k ij n'kkZ;s vuqlkj voLFkk esa uhps okyh pdrh o ls ?kwe jgh gS rFkk bldh ?kw.kZu

xfrt ÅtkZ ko gSA Åij okyh pdrh izkjEHk esa fojkekoLFkk esa gSA Åij okyh pdrh dks fxjus ds fy;s NksM+k x;k gSA

;g uhps okyh pdrh ij fpid tkrh gSA VDdj ds ckn fudk; dh ?kw.kZu xfrt ÅtkZ esa ifjorZu gksxk:

(A) ko/2 (B) (1/2) k

o (C) (1/4) k

o (D) k

o/4

23. fp=k esa pkj izxkeh rjaxs A, B, C rFkk D iznf'kZr gSa fp=k ls ;g fu"d"kZ fudky ldrs gSa fd rjax A ds lkis{k

(A) rjax C, /2 dyk dks.k ls vkxs rFkk rjax B, /2 dyk dks.k ls ihNs gS (B) rjax C, /2 dyk dks.k ls ihNs rFkk rjax B, /2 dyk dks.k ls vkxs gS (C) rjax C, dyk dks.k ls vkxs rFkk rjax B, dyk dks.k ls ihNs gS (D) rjax C, dyk dks.k ls ihNs rFkk rjax B, dyk dks.k ls vkxs gS



®




24. fuEu fn;s x;s esa fdldh dqy vkSlr xfrt ÅtkZ 300K ij egRre gksxhA (A) H

2 dk 1 kg

(B) He dk 1 kg

(C) H2 dk ½ kg + He dk ½ kg (D) H

2 dk 1

4 kg + He dk 3

4 kg

25. ,d vkn'kZ xSl fn;s x;s izØe PV2 = C, (P1, V1, T1) ls (P2, V2, T2) rd vuqlj.k djrh gS (C fu;rkad gS). rc

(A) ;fn P1 > P2 rc T2 > T1 (B) ;fn V2 > V1 rc T2 < T1

(C) ;fn V2 > V1 rc T2 > T1 (D) ;fn P1 > P2 rc V1 > V2

26. fp=kkuqlkj 6 lseh- f=kT;k dh v)Zo`Ùkkdkj pdrh esa pdrh ds dsUnz C ls 8 cm nwjh ij 2 cm f=kT;k dk fNnz dkVrs gSaA bl fudk; ds nzO;eku dsUnz dh fcUnq C ls nwjh gSA

(A) 4 cm (B) 8 cm (C) 6 cm (D) 12 cm

27. R;kSgkj ds ekSle esa /ofu dh rhozrk 100 xquk c<+ tkrh gSA blls ifj.kke fudkyk tk ldrk gS fd Msflcy esa

c<+krjh gS % (A) 20 to 120 dB (B) 70 to 72 dB (C) 100 to 10000 dB (D) 80 to 100 dB

28. ,d vkneh ,d cM+h nhokj ds lkeus fojke ij [kM+k gSA ,d 400 Hz vko`fÙk dk /ofu L=kksr nhokj vkSj mlds chp esa j[kk gSA vc L=kksr nhokj dh rjQ 1 m/s dh pky ls xfr djrk gSA izfr lsd.M+ lqus x;s LiUnksa dh la[;k gksxh (gok esa /ofu dh pky 345 m/s gSA)

(A) 0.8 (B) 0.58 (C) 1.16 (D) 2.32

29. nks nzO;ekuksa 1kg. rFkk 2kg dh ‘'kh"kkZfHkeq[k VDdj ds nkSjku laihMu dh egRre ÅtkZ 100

3twy gSA ;fn VDdj ds

igys nzO;eku leku fn'kk esa xfreku gSa ] rc mudk VDdj ls igys lkehI; osx gS &

(A) 10 m/sec. (B) 5 m/sec. (C) 20 m/sec. (D) 10 2 m/sec.

30. ljy vkorZ xfr djrs gq;s d, d.k ds fy,] ;fn foLFkkiu dk vk;ke ‘a’ gS rFkk osx dk vk;ke ‘v’ gS] rks Roj.k dk vk;ke gksxk&

(A) va (B) 2v

a (C)

2v

2a (D)

v

a

jlk;u foKku 31. Na+, Mg2+, Al3+ rFkk Si4+ vk;u lebysDVªkWfud gS] bu vk;uks fd vk;fud f=kT;k ds eku dk Øe gksxk& (A) Na+ > Mg2+ > Al3+ > Si4+ (B) Na+ < Mg2+ < Al3+ < Si4+

(C) Na+ > Mg2+ > Al3+ < Si4+ (D) Na+ < Mg2+ > Al3+ > Si4+

32. C, S rFkk CS2 ds ngu fd Å"ek –393.3 kJ, –293.7 kJ rFkk –1108.76 kJ gSA rc CS2 ds fuekZ.k fd Å"ek D;k gksxh \

(A) –128.06 kJ (B) +970 kJ (C) +1108.7 kJ (D) +12 kJ

33. csfdax ikmMj] jkWdlkWYV] okf'kax lksMk rFkk LysDM ykbe ds tyh; foy;u ds pH ds eku dk c<+rk gqvk lgh Øe gS&

(A) csfadax lksMk < jkWdlkWYV < okf'kax lksMk < LysDM ykbe

(B) jkWdlkWYV < csfadax lksMk < okf'kax lksMk < LysDM ykbe

(C) LysDM ykbe < okf'kax lksMk < jkWdlkWYV < csfadax lksMk (D) okf'kax lksMk < csfadax lksMk < jkWdlkWYV < LysDM ykbe



®




34. uksjey NaOH ds 100 mL foy;u ds mnklhuhdj.k ds fy, vko';d vkWDtsfyd vEy dh ek=kk gS&

(A) 6.3 g (B) 126 g (C) 530 g (D) 63 g

35. fdldk e/m vuqikr mPpre gS \

(A) He2+ (B) H+ (C) He+ (D) D+

36. fuEu esa ls dkSulk ;kSfxd leryh; rFkk v/kqzoh; gS \

(A) XeO4 (B) SF4 (C) XeF4 (D) CF4

37. vkWDlhdkjd tks fd vipk;d ds leku O;ogkj ugha dj ldrk gSA og gS&

(A) SO2 (B) NO2 (C) CO2 (D) C1O2

38. IykWd fu;rkad dk eku 6.63 × 10–34 Js gSA izdk'k dk osx 3.0 × 108 m s–1 gSA 8 × 1015 s–1 vko`fÙk ds izdk'k ds ,d DokaVe fd rjax}S/;Z dk fudVre eku usuksehVj esa D;k gS\

(A) 3 × 107 (B) 2 × 10–25 (C) 5 × 10–18 (D) 4 × 101

39. leku {kerk okys ik=kksa esa] leku rki ij] ,d ik=k esa H2 ds 44 g rFkk vU; ik=k esa CO2 ds 44 g Hkjh x;h gSA ;fn f}rh; ik=k esa dkcZuMkbZvkWDlkbM dk nkc 1atm gS rc izFke ik=k esa gkbMªkstu dk nkc gksxk&

(A) 1 atm (B) 10 atm (C) 22 atm (D) 44 atm

40. N2+ dk ca/k Øe gS&

(A) 1.5 (B) 3.0 (C) 2.5 (D) 2.0

41. dkSuls v.kqvksa esa] v.kqvks esa js[kkafdr ijek.kqvks dk ladj.k sp-ladj.k gS&

(u) CH2CHCH3

(w) CH3CH2+

(y) CH3CN

(v) CH2CCHCl

(x) H — C C — H

(z) (CH3)2CNNH2

(A) x rFkk z (B) x, y rFkk z (C) u, w rFkk x (D) v, x rFkk y

42. ;kSfxd dk IUPAC uke D;k gS \

(A) 2-Dyksjks- 2-C;wVsu (B) 3-Dyksjks-1- C;wVsu

(C) 3-esfFky-3-Dyksjksizksisu-1 (D) 3-Dyksjks-3-esfFky-1-izksisu

43. lkbDyksgsDlsu fd B.Ms {kkjh; KMnO4 ds lkFk vfHkfØ;k ls fufeZr gksrk gS&

(A) foi{k-gsDlsuMkbZvkWy (B) gsDlkMkbZdhVksu (C) lei{k-gsDlkMkbZvkWy (D) isUVkMkbZdhVksu

44. uhps n'kkZ;s x;s] inkFkZ ds dqlhZ vfHkfoU;kl esa esfFky lewgksa ds e/; lgh lEcU/k dh O;k[;k fuEu esa dkSu djrk gS&

(A) foi{k (B) izfr (C) xkSlst (D) xzflr



®




45. 2 2 2

||O

OHC CH CH C CH COOH dk IUPAC uke gS & (A) 1-QkWfeZy-3-vkWDlksisUVsukWbd vEy (B) 5-QkWfeZy-3-vkWDlksisUVsukWbd vEy

(C) 3-vkWDlks-5-QkWfeZyisUVsukWbd vEy (D) 3-vkWDlks-1-QkWfeZyisUVsukWbd vEy

tho foKku

46. ukWu dkbusVksdksj lw{eufydkvksa dk dk;Z gSA

i. xq.klw=kksa dks bDosVj ij O;ofLFkr djusa esa lgk;rk djuk ii. ,ukQst ds nkSjku dksf'kdk dks fn?khZr djuk iii. ,ukQst ds nkSjku v/kZ-xq.klw=kksa (chromatids) dks i`Fkd djuk dkSulk/ dkSuls dFku lgh gS/ gSa?

(A) dsoy i (B) dsoy ii (C) dsoy iii (D) i rFkk iii

47. Loiks"kh jlku'kys"kh ftok.kq cgqr ls vdkfcuhd inkFkZ tSls ukbVsªVl, ukbVªkbVl vkSj veksuhyk dks (A) DNA izrhfyihdj.k (B) RNA la'ys"k.k (C) ATP fuekZ.k (D) izksVhu fuekZ.k 48. fxuh fix esa Ropk dk 'osr jax (b) dkys jax (B) dh rqyuk esa vaizHkkoh gSA dkys] le;qXeuth fxuh fix ds v.Mk'k;

dks 'osr vksojsDVksekbTM eknk eas izR;kjksfir fd;k tkrk gSA ;fn ;g 'osr eknk] 'osr uj ds lkFk tuu djrh gS rks larkus

(A) thu iz:i BB ;qDr dkyh (B) thuiz:Ik bb ;qDr 'osr

(C) thu iz:I Bb ;k bb ;qDr dkyh ;k 'osr (D) tuiz:Ik Bb ;qDr dkyh 49. eDds esa js'ksnkj ewysa fdlls fodflr gksrh gSA (A) fupyh iOkZlaf/k;ksa ls (B) •Åijh ioZlaf/k;ksa ls (C) Åijh ioksaZ ls (D) mijksDr dksbZ ugha 50. ,d gheksfQfyd iq:"k dh iq=kh lkekU; y{k.k iz:Ik dh gSaA mldk fookg ,d ,sls iq:"k ls gksrk gS tks gheksfQfyd

ugha gSaA ;fn bl naifr ds pkj iq=k gSa rks bl ckr dh D;k laHkkouk gS fd pkjksa iq=k gheksfQfyd mRiUu gksa?

(A) 1/4 (B) 1/32 (C) 1/16 (D) 1/8

51. dkSulk thok.kq dksf'kdk dk y{k.k gS\

(A) dsfUnzdk dh mifLFkfr (B) dksf'kdknzO;h izokg xfr dh vuqifLFkfr

(C) ,dylw=kh] o`Ùkkdkj DNA dh mifLFkfr (D) lsyqykst dh cuh dksf'kdk fHkfÙk dh mifLFkfr

52. ;g /;ku nsus ;ksX; egRo dh ckr gS fd euq"; dh dksf'kdk ds vkSlru vof/k 24 ?kaVs dh dksf'kdk pØ esa dksf'kdk foHkktu flQZ yxHkx--------------esa iw.kZ gksrk gS%&

(A) pkj ?kaVs (B) 90 feuV (C) ,d ?kaVs (D) 10 ?kaVs 53. e`nk ls tkbye esa ty ds çokg dk iFk gS& (A) esVktkbyu izksVkstkbyu cYdqV e`nk ewy jkse (B) cYdqV ewyjkse varlLRopk ifjjEHk izksVkstkbye esVktkbye (C) e`nk ewyjkse cYdqV varLRopk ifjjEe izksVkstkbye esVktkbye (D) ifjjEHk e`nk ewyjkse cYdqV varLRopk izksVkstkbye esVktkbye



®




54. fuEUk esa ls dkSulk dFku lR; gS\ (A) lHkh d'ks:dh tjk;qt gSA (B) lHkh fNidfy;ka fo"kSyh gksrh gSaA

(C) lHkh ,fEQfc;k vius v.Ms ty esa nsrs gSA (D) lHkh ,oht mM+us ;ksX; gksrs gSa 55. mHk;pjksa rFkk ljhl`iksZa esa ck;k¡ vkfyan fxYl/QsaQM+ksa/Ropk ls vkWDlhtfur jDr xzg.k djrk gSA rFkk nk;k¡ vkfyan

'kjhj ds fofHkUu Hkkxksa ls deoxygenated jDr xzg.k djrk gSA ;s nksuksa ,d gh fuy; eas fefJr gks tkrs gSa tks fefJr jDr dks iEi dj nsrk gSA bl izdkj dk ifjlapj.k dgykrk gSA

(A) viw.kZ bdgjk ifjlapj.k (B) iw.kZ ,dgjk ifjlapj.k (C) viw.kZ nksgjk ifjlapj.k (D) iw.kZ nksgjk ifjalpj.k 56. QSQMs fuEu ds vkoj.k ls <ds gksrs gS (A) 'ys"e dyk (B) Iy;woy dyk (C) ifjdkfMZ;e (D) isfjVksfuj;e

57. usÝksu dk dkSulk Hkkx o`ô ds oYdqVh; {ks=k esa fLFkr ugha gksrk gS&

(A) eSYih/kh nsgk.kq (B) PCT

(C) DCT (D) gsuys dk ywi

58. fuEufyf[kr esa fdldks jlk;fud ;qXekuqca/k ls lEc) ugh fd;k tk ldrk gS\

(A) iwoZ flusfIVd f>Yyh (B) i'p flusfIVd f>Yyh (C) raf=kdk lapkjh (D) vkosx dk f}fn'kh; lapj.k

59. izksVhu gkWeksZuksa ds fy, xzkgh mifLFkr gksrs gSa&

(A) dksf'kdk lrg ij (B) dsUnzd ij

(C) vUr%iznzO;h tkfydk (D) dksf'kdk nzO;

60. LoLFk nkar o vfLFk;ksa ds fuekZ.k ds fy, vko';d nks rRo gSa&

(A) vk;ju o dSfY'k;e (B) QkWLQksjl o vk;ju

(C) dSfY'k;e o QkLQksjl (D) CO2 o H2O

PART- B

Two Mark Questions

xf.kr

61. ,d le:i n'k Hkqt ij fopkj djsa ftlds 'kh"kZ ,dkbZ o`Ùk ij gSa] ,d 'kh"kZ fu/kkZfjr dj vU; 9 'kh"kZ ij lh/kh js[kk [khpsa mUgs L1, L2, ….., L9 uke nsa] muds yEckbZ dks 1, 2, ,,,,,9 ls fu:fir djsa] rc 1, 2, ,,,,,9 dk xq.kuQy gS&

(A) 10 (B) 10 3 (C) 3

50 (D) 20



®




62. eku ysa fd ABC ,d f=kHkqt gS ftlesa AB = 4, BC = 5 vkSj CA = 6 gS] AB, BC ,oa CA ij Øe'k% fcanq D, E, F pqusa

ftlls fd AD = 2, BE = 3, CF = 4 rc {ks=kQyDEF

{ks=kQyABC

gS%

(A) 4

1 (B)

15

3 (C)

15

4 (D)

30

7

63. [0,] ds fy, eku ysa f() = sin(cos ,oa g() = cos(sin) eku ysa a = fmax

0, b =

fmin

0 ,

c = gmax

0vkSj d =

gmin

0 a, b, c, d }kjk larq"V lgh vlerk,sag S &

(A) b < d < c < a (B) d < b < a < c (C) b < d < a < c (D) b < a < d < c

64. h Å¡pkbZ okys yEc o`Ùkh; 'kadq ds vkÑfr ds ,d cksry esa dqN ty j[kk gS] blds vk/kkj dks lery lrg ij j[kus

ij 'kh"kZ ls ty lrg dh Å¡pkbZ bdkbZ gS] 'kh"kZ dks uhps dh vksj j[kus ij vk/kkj ls ty ds ry dh Å¡pkbZ 4

a

bZdkbZ gS] rc a

h vuqikr gS%

(A) 4

851 (B)

8

851 (C)

4

651 (D)

8

651

65. fuEufyf[kr nks dFkuksa ij fopkj djsa% I. ;fn n ,d la;qDr la[;k (composite number) gS rks n, (n —1)! dks foHkkftr djrk gS] II. ,slh vaur izkÑfrd la[;k,a n bl izdkj gSa fd n3 + 2n2 + n, n! dks foHkkftr djrk gS] rc

(A) I ,oa II lgh gS (B) I ,oa II xyr gS (C) I lgha gSa ,oa II xyr gS (D) I xyr gSa ,oa II lgha gS

HkkSfrd foKku

66. ,d eksVj lkbfdy ftl {k.k pyuk izkjEHk djrh gS] ,d dkj mlls vkxs fudy tkrh gSA nksuksa ,d gh fn'kk esa tk jgs gSaA ;fn eksVjlkbfdy rFkk dkj ds fy;s pky le; vkjs[k Øe'k% OAB rFkk CD ls fu:fir dh tkrh gSA rc

C

A B

D

60

40

20

O 27 18 12

speed

(m/s)

s.

(A) t = 18 s ij eksVjlkbfdy vkSj dkj ds e/; dh nwjh 180 m gSA

(B) t = 18 s ij eksVjlkbfdy vkSj dkj ds e/; dh nwjh 720 m gSA

(C) eksVjlkbfdy vkSj dkj ds e/; dh nwjh lnSo fu;r jgrh gSA (D) buesa ls dksbZ ugha

67. v0 osx ls xfr dj jgh m nzO;eku dh ,d xksyh fp=kkuqlkj [kqjnjh {kSfrt lrg ij fojke ij j[kh M nzO;eku o R

f=kT;k dh ,dleku o`Ùkkdkj pdrh ds 'kh"kZ ls Bhd Li'kZ djrh gqbZ fudyrh gSA (xksyh pdrh ds lkFk lEidZ ds nkSjku pdrh ij fQlyrh ugha gS);g ekfu, fd pdrh fcuk fQlys yksVuh xfr djrh gS] pdrh dks Li'kZ djus ds Bhd i'pkr~ xksyh dk osx Kkr djksA

(A) M3m8

mu4

(B)

M3m8

mu8

(C)

M3m4

mu8

(D)

M3m4

mu4



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68. nks yksydks dh yEckbZ dk vUrj 22 cm gSA nksuksa yksyd ,d gh LFkku ij bl izdkj ls nkSyu djrs gS fd ,d yksyd 30 nkSyu ,oa nwljk yksyd 36 nkSyu leku esa djrs gSA yksydks dh yEckbZ;k¡ (cm esa) gksxh

(A) 72 and rFkk 50 (B) 60 and rFkk 38 (C) 50 and rFkk 28 (D) 80 and rFkk 58

69. fp=kkuqlkj pkj ,d leku chdj esa leku ek=kk esa ikuh j[kk gqvk gSA chdj 'a' esa dsoy ikuh gSA ,d lhls (lead)

dh xsan dks ,d /kkxs ls Åij ls ck¡/k dj chdj 'b' esa iwjh rjg Mwck;k x;k gSA leku vkdkj dh ,d IykfLVd dh xsan (eku yhft, Vscy Vsful dh xsan] TT) dks ,d /kkxs ds }kjk cak/kdj chdj 'c' esa iwjh rjg Mqck;k x; gSA& bl ifjfLFkfr esa /kkxs dks ,d ckgj j[ks ,d vk/kkj (stand) ls ck¡/kk x;k gSA leku vkdkj dh Vscy Vsful dh ,d nwljh xsan dks ,d /kkxs ls ck¡/k dh chdj 'd' esa iwjh rjg Mqck;k tkrk gS& bl ifjfLFkfr es /kkxs ds nwljs f'kjs dks chdj ds fupys ry ls ck¡/kk tkrk gSA bu pkjksa chdkjksa dks (fcuk vk/kkj ds) ,d Hkkj ekid rqyk ij j[kk tkrk gSA ;g rqyk chdj a,b,c ,oa d dk Hkkj Øe'k% Wa, Wb, Wc ,oa Wd ekirk gSA (/kkxs ,oa vk/kkj ds vk;ru vkSj nzO;eku ux.; gS)

a

b

lead

c

TT

d

TT

(A) Wa = Wb = Wc = Wd (B) Wb = Wc > Wd > Wa

(C) Wb = Wc > Wa > Wd (D) Wb > Wc > Wd > Wa

70. 0ºC ds 540 g cQZ dks 80ºC ds 540 g ikuh esa feyk;k tkrk gSA feJ.k dk vfUre rki gksxk –

(A) 0ºC (B) 40ºC (C) 80ºC (D) 100°C

jlk;u foKku

71. fuEu esa ls dkSulk /kkrq lYQkbM ty esa lokZf/kd foy;srk j[krk gS \

(A) CdS (Ksp = 36 × 10–30) (B) FeS (Ksp= 11 × 10–20)

(C) HgS (Ksp = 32 × 10–54) (D) ZnS (Ksp = 11 × 10–22)

72. MkbZcksjsu] ,d vPNk jkWdsV b±/ku gS] ;g fuEu vfHkfØ;k lehdj.k ds vuqlkj ngu vfHkfØ;k n'kkZrk gSA B2H6(g) + 3O2(g) B2O3(s) + 3H2O(g)

MkbZcksjsu ds ngu ds fy, ,UFkSYih ifjorZu dh x.kuk dhft,A fn;k gS&

(i) 3B(s) +2

3O2(g) B2O3(s) ; H = – 1273 kJ per mol

(ii) H2(g) +2

1O2(g) H2O (l) ; H = – 286 kJ per mol

(iii) H2O(l) H2O (g) ; H = 44 kJ per mol

(iv) 2B(s) + 3H2(g) B2H2 (g) ; H = 36 kJ per mol

(A) + 2035 kJ per mol (B) – 2035 kJ per mol

(C) + 2167 kJ per mol (D) – 2167 kJ per mol



®




73. ,d d{kk esa ,d bysDVªkWu ds osx rFkk Z, n

1 rFkk n ds e/; vkjs[k ds fy, xyr vkjs[k dk p;u dhft,&

(A) (B)

(C) (D)

74. fuEu inkFkksZ dks buds ngu fd Å"ek ds ?kVrs Øe esa fyf[k;s ¼vf/kdre ls U;wure rd½

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

75. fuEu dks vEyh;rk ds c<+rs Øe esa O;ofLFkr dhft,&

(A) III < I < II (B) I < III < II (C) III < II < I (D) II < I < III

tho foKku

76. i`"B/kkjh ifÙk;ksa ds fy, dkSuls dFku lR; gS \ (a) laogu iwyks dk vkdkj] f'kjkvksa ds vkdkj ij fuHkZj djrk gSA

(b) laogu iwy ds pkjks vksj eksVh fHkfÙk okyh dksf'kdkvksa dh ijr gksrh gS ftls iwykPNn ¼caMy vkPNn½ dgrs gSA

(c) i.kZ e/;ksrd foHksfnr ugha gksrk gSA

(A) a and b (B) b and c (C) a, b and c (D) a and c

77. fuEu dFkuksa dk v/;;u dhft, rFkk lR; dFku okys fodYi dk p;u dhft,A

(i) S-izkoLFkk ds i'pkr DNA dh la[;k nqxuh ijUrq xq.klw=k dh la[;k leku jgrh gS vFkkZr 2nA

(ii) ,d dksf'kdk dk G0 izkoLFkk esa mikip;h :i ls lØh; gksrh gS rFkk tho dh vko';drkuqlkj foHkktu esa izos'k dj ldrh gSA

(iii) dksf'kdk foHkktu ,d of̀) izØe gS tks fofHkUu izkoLFkkvksa ds e/; foHksnu lhek dks n'kkZrk gSA

(iv) izksQst dh lekfIr ij GB, ER, U;wfDyvksyl rFkk dsUnzh; vkoj.k foyqIr gks tkrs gSA

(A) i, ii o iv (B) i, ii o iii (C) ii, iii o iv (D) lHkh



®




78. vR;f/kd ty gkfu ifÙk;ksa esa eykfu dk dkj.k gS tks jksdk tk ldrk gSA

(A) ikni dks rst izdk'k esa j[k dj (B) iÙkh dh lrg ij cslrhu yxkdj

(C) e`nk esa vf/kd ek=kk esa moZjd feyk dj (D) ikniksa ij ,Ydksgu dk fNMdko dj

79. xyr dFku dk p;u dhft,&

(A) Je dk;ksZa dk dqN foHkktu] la?k iksfjQsjk ds lnL;ksa dh dksf'kdkvksa esa feyrk gSA

(B) Je ¼dk;ksZa½ dk foHkktu] iksfjQsjk tarqvksa dh dksf'kdkvksa esa iw.kZr% vuqifLFkr gksrk gSA

(C) [kqyk ifjlapj.k ra=k V~;wfudsV~l] gsehdkWMsZV~l] vkFkzksZiksM~l o ukWu&lhQsyksiksM~ eksyLd tarqvksa esa feyrk gSA

(D) can ifjlapj.k ra=k esa jDr o 'kjhj dksf'kdkvksa ds chp inkFkksZa dk fofue;] jDr dsf'kdkvksa ds Lrj ij gksrk gSA

80. fuEufyf[kr dFkuksa esa ls dkSulk xyr gS \

(A) dy'k ¼xksCysV½ dksf'kdk,¡ vka=k ds E;wdkslk ¼'ysf"edk½ esa fLFkr gksrh gSa rFkk E;wdl ¼'ys"ek½ dk L=kko djrh gSaA

(B) vEytu ¼vkWfDlfUVd½ dksf'kdk,¡ vkek'k; ds E;wdkslk ¼'ysf"edk½ esa fLFkr gksrh gSa rFkk gkbMªksDyksfjd vEy dk L=kko djrh gSaA

(C) xqPNdks"Bd ¼,sflul½ vXU;k'k; eas fLFkr gksrs gSa vkSj dkcksZDlhisfIVMst dk L=kko djrs gSaA

(D) czwuj xzafFk;k¡ vkek'k; ds lcE;wdkslk ¼v/k%'ysf"edk½ esa fLFkr gksrh gSa rFkk isfIlukstu dk L=kko djrh gSA



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FULL SYLLABUS TEST-2 (KFST-2) KISHORE VAIGYANIK PROTSAHAN YOJANA


ANSWER KEY

PART- A

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

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

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

22. (B) 23. (B) 24. (A) 25. (B) 26. (B) 27. (D) 28. (D)

29. (A) 30. (B)

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

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

45. (B)

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

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

60. (C)

PART- B

61. (A) 62. (C) 63. (C) 64. (B) 65. (D) 66. (A) 67. (B)

68. (A) 69. (B) 70. (A)

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

76. (A) 77. (A) 78. (B) 79. (B) 80. (D)



®




HINTS & SOLUTIONS

PART- A

xf.kr

1. fn;k x;k gS, 9 × abcde = edcba

(10–a) abcde = edcba

a b c d e o - a b c d e _________ e d c b a a = 1, e = 9, b = 0, c = 9, d = 8

abcde = 10989

sum of digits vadksa dk ;ksx = 27

2. x+y = 1, x, y > 0

ekuk x = sin2& y = cos2

vr% y

1

x

1 = cosec + sec2

= 2+ tan2 + cot2 4

3. p,q,r /kukRed ifjes; gSA

vkSj rqp ifjes;

blfy,, p + q + r +

2 rp2qr2pq ifjes;

rpqrpq2 ifjes;

4. 2 log (x – 2y) = log x + log y

log (x – 2y)2 = log xy

(x – 2y)2 = xy x2 + 4y2 – 5xy = 0

2

2

y

x + 4 – 5

y

x = 0

2

y

x

–

y

x5 + 4 = 0

1

y

x

4

y

x = 0

y

x = 1, 4

x > 2y y

x = 4

5. P1(x) = x3 – 2020x2 + bi x + c1 P2(x) = x3 – 2021 x2 + b2x + c2 P1(x) q1(x) + P2(x) q2(x) = x2 – 3x + 2

P1(x) q1 (x) + P2 (x) = 0 ds , mHk;fu"B ewy gSA

P1(x) q(x) + P2(x) q2(x) nks xq.kuQy gSA

(x – ), (x – )

x2 – 3x + 2 = 0 ds xq.kuQy x – , – = 1 vkSj = 2

P1(x) = (x – 1) (x – 2) (x – 2017)

vkSj P2(x) = (x – 1) (x – 2) (x – 2018)

P1(3) = (2) (1) (–2014) = – 4028 vkSj P2(1) = 0

P1(3) + P2(1) + 4028 = 0

6.

•

D C

B A

y

w

x

z E

x + y + z + w = 360º

+ + + = 180º

+ + + = 2( + ) using A.P.

+ = + = 180º – ( + )

+ = + = 90º

blfy, ekf/;dk ds dks.k =

3

BCDABCDAB

= 3

= 3

)º90(2º90 = 90º =

2

7. (2021)2020 = (1 + 2020)2020 = 1 + 2020C1. 2020 + 2020C0. (2020)2 +...... = 1 + (2020)2 + (2020)2 I

blfy, 'ks"kQy = 1

8. 710,709........101100

610

99........121110

90

9......321

18302909

189

189 + 1830 = 2019

709 rd dqy vadksa dh la[;k = 2019

710 dk vxyk vad gSA blfy, 2021 ds vad 1

gSA

9. ekuk PCB =

A

B C

P

D

DPC = –

BPC = DPB = – ( – ) = DPB & PCB

DPB = PCB



®




vkSj DPB = PBC

blfy, DPB = PBC

rc BC

PB

PB

DB

PC

DP

vc BC

PB

PB

DB

PC

PD2

blfy, 7

2

PC

PD

10. ?

16

16

15

1

14

14

13

13

12

122018

2020

2018

2020

2018

2020

2018

2020

2018

2020

22020 +1 = 22 (22018) + 1 = 22 (22018+1)–3 = 4(22018+1)–3

1212

12 2

2018

2020

…..(1)

blfy, 1313

13 2

2018

2020

1414

14 2

2018

2020

1515

15 2

2018

2020

& 1616

16 2

2018

2020

;ksxQy = (22 + 32 +42 + 52 + 62)–5

=

66

1376

= 91–6 = 85

11. M = 230 – 215 + 1 M2

= 260 – 246 + 231 + 230 – 216 + 1 = 246 (214 – 1) + 231 + 216 (214 – 1) +1

246 112

)12(22

12

)12( 141631

14

246 ( 1 + 2 + 22 + .....+ 213) + 231 + 1 216 (1 + 2 + 22 +.....+213) +1

14

)2+........ + 2 + (2 594746

+ 231 +

0291716

214

)2+........ + 2 + (2

There will be 30 times digit 1 12.

A

B C D E

c b

14 7 7

BC = 28 AB + AC = c + b

tan2 =

21, tan =

7

2

49–1

27

=

21 37 so = 30º

vc c + b =

)31(143

22114

º60sin

21

º30sin

7

13. O O O O O

a2 4 = 24

O O O O O a1

3C1 3C1. 3 = 54

dqy = 24 +54 = 78

14. dqy iqLrdsa = m + n

dqy iqLrdsa = n + 1

(consider m books at 1) dqy O;oLFkk = m.1n

15. [x2] = x + 1

x = [x2] – 1

Li"V;k x

[ [x2] vkSj nksuksa iw.kk±d gS)

x = x2 – 1

x2 – x – 1 = 0

x2 – x – 1 = 0

x = 2

51dksbZ gy ugha

HkkSfrd foKku

16. tSlk fd s = 1

2 at2

2R cos = 1

2 g cos t2

‘t’ dks.k ‘’ ij fuHkZj ugha djrk gSA

17. y = x tan – 2

2 2

gx

2u cos

5 = 20 tan30° – 2 2

10 400

2u cos 30

u 20 m/s

18. D;ksafd rSjkd o unh dk osx ,d leku gSA vr% AB dh ifj.kkeh fn'kk bl rjg gksxhA



®




vr% 90

2

= 60 = 30°

19.

30° m2

80×sin30° m1

T = 70N 70N

= 40 N

f = 30 N

20. 2 21mv 4.t

2

1 dvm 2 v

2 dt = 4 × 2t

m. dv

dt =

8t

v =

2

8t8m

8t

m

21.

Tcos

Tsin

A

T

T sin < T

tA < tB

22. I = 2I'

' = 2

k = ki – kf =

2

2 ik1 1(2I) I

2 2 2 2

23. fp=k esa 'C' ml fLFkfr ij 'A' ds t = /2 le; ckn igqprk gSs] tgkW 'A'igys gh igqWp pqdk gSa rFkk

'A' ml fLFkfr ij t = /2 le; ckn igWqprk gSa tgkW 'B' igys igqp pqdk gSA

vr% (B).

24. dqy xfrt ÅtkZ = U = 2

nRT

H2 ds ft, LokU=kR; dksfV lokZf/kd ,oa eksayks dh la[;k lcls T;knk gksxhA vr% bldh xfrt ÅtkZ lokZf/kd gksxhA

26. C dks ewy fcUnq ysa vkSj x o y–v{k fp=k esa fn[kk;k x;k gSA

y–v{k ds ifjr% leferh ds dkj.k

xcm = 0

ycm = 1 1 2 2

1 2

m y m y

m m

=

22

22

(6 ) 4(6 ) –[ (2) (8)]2 3

(6 )(2)

2

(m Area)

= 2

2

8(18 4)

(18 4)

= 8 cm.

27. /ofu rhozrk = 10 log10

0I

I,

I = 100 I0 so

log10 0

0100

i

i

= 20

vr% Lrj 20 dB ls c<+ tkrk gSA

28. f = 2 2

c c cf2u 2uf.f f

c u c u cc u

f = 2.1.400 800

345 345 = 2.32

29. U = 1 2

1 2

1 m m

2 (m m ) (V1 – V2)2 =

100

3

(V1 – V2)2 × 2m .m

2 (m 2m) =

100

3

Putting m = 1 kg : (V1 – V2) = 10 m/sec. AlternateSolution: When deformation is maximum

both the particles are moving with same velocity. So applying momentum conservation.

m1v1 + m2v2 = m1v 1’ + m2v1’

v1’ = 1 1 2 2

1 2

m v m v

m m

Applying energy conservation:

1

2 m1v1

2 + 1

2m2v2

2 = 1

2 (m1 + m2) (v1´ )2

+ Udeformation

U deformation= 1

2 1 2

1 2

m m

m m

× (v1 – v2)2 = 100

3

v1 – v2 = 10m/sec. 30. vf/kdre osx v = a

vf/kdre Roj.k f = 2 a f = 2v

a

jlk;u foKku

31. Na+ Mg2+ Al3+ Si4+

izksVksu 11 12 13 14

bysDVªkWu 10 10 10 10

lebysDVªksfud /kuk;uksa dk vkdkj] ukfHkdh; vkos'k ds

ifjek.k esa o`f) ds lkFk ?kVrk gSA



®




vkdkj dk ?kVrk Øe gS %

Na+ > Mg2+ > Al3+ > Si4+

32. C + 2S CS2

H = H – HR

= –1108.76 – [–393.3 – 2 ´ (293.7)] = –1108.76 + 393.3 + 587.4

= –128.06 kJ 33. ;kSfxd esa vEyh; xq.k vf/kd gksus ij pH dk eku de

gksxkA

jkWdlkWYV (NaCl) < csfadax lksMk (NaHCO3) < okf'kax

lksMk (Na2CO3) < LysDM ykbe (Ca(OH)2).

34. NaOH ds rqY;kadks dh la[;k = vkWDtsfyd vEy ds

rqY;kadks dh la[;k

rc 631000

1100 eku;vEy dk nzOvkWDtsfyd

W= 6.3 g

35. He2+ ds fy, e/m = 2/4 H+ ds fy, e/m = 1/1

He+ ds fy, e/m = 1/4 D+ ds fy, e/m = 1/2

H+ ds fy, e/m dk eku mPpre gSA

36.

37. CO2 vf/kdre vkWDlhdj.k voLFkk +4 j[krk gSA vr% ;g

vksj vf/kd mPp vkWDlhdj.k voLFkk iznf'kZr ugha dj

ldrkA

38.

39.

40. N2+ = 7 + 7 – 1 = 13 bysDVªkWu

bysDVªkWfud foU;kl : 1s2 *1s2 2s2

*2s22px2 2py

2 2pz1

ca/k Øe = 2

1 [ca/kh bysDVªkWuks fd la[;k – izfrca/kh

bysDVªkWuksa dh la[;k]

=2

1(9 – 4) =

2

1× 5 = 2.5

41.

42. ;g 4C ijek.kqvksa dh Ja`[kyk j[krk gS rFkk f}ca/k dks Cl

ijek.kq ls vf/kd izkf;drk nh tkrh gSA

43. lkbDyksgsDlsu + KMnO4 (B.Mk rFkk {kkjh;)

lei{k-gsDlkMkbZvkWyA

44.

45. 2 2 2

||O

OHC CH CH C CH COOH 12345

5-QkWfeZy-3-vkWDlksisUVsukWbd vEy

PART- B

xf.kr

61.

10

28l

9l10A

1l1A 2A

3A

4A

3l

2l

II

O

cos)1)(1(2

–11

5

21

22

5

cos2–221

5

cos–12



®




10

sin4 2

1 = 2sin10

= 9

2 = 2 sin 810

2

10

2cos2

10

3sin2 73

,

10

5sin2,

10cos2

10

4sin2 564

1.2.3 ………..9 =

2222

10cos2.

10

2cos2.

10

2sin2

10sin2

.10

5sin2

= 2.10

4sin2.

10

2sin2

22

= 32. 2

22

4

5210

4

52–1032

5

2sin.

5sin

= 10 62.

A

C B E

F D

3 2

2

2 6

4

5

2

4

ekuk ABC dk {ks=kQy =

= 2

1× 6 × 4sinA = 12sinA

rFkk, = 2

1 × 4 × 5sinB = 10sinB

rFkk, = 2

1× 5 × 6sinC = 15sinC

ekuk ADF dk {ks=kQy = 1 = 2

1 × 2 × 2sinA

= 2sinA

ekuk BDE dk {ks=kQy = 2 = 2

1 × 2 ×

3sinB = 3sinB

ekuk ECF dk {ks=kQy = 3 = 2

1 × 4 ×

2sinC = 4sinC

321 –––

ABCofArea

DEFofArea

321 –––1

= 1 – Asin12

Asin2 –

Bsin10

Bsin3 –

Csin15

Csin4

= 1 – 6

1 –

10

3 –

15

4 =

15

4 Ans.

63. cos [-1,1]

sin (cos) [–sin1, sin1]

a = sin1, b = –sin1

sin [0,1] [0, ]

cos(sin) [cos1, cos0]

cos(sin) [cos1,1]

b < d < a < c

64.

B C

A

mr

a r1

h

a

r

h

h

a

h

arr1

V of water = a.h

ra

3

1hr

3

12

222

4

a

m

h

4

ahr

rr

r

4

ah

h1

1

V of water =

4

ah.

h

4

ahr

3

12

22

2

2

2

2

32

h

4

ah

.r3

1

h

ahr

3

1

3

33

4

ahah

2

3

3

4

1

a

h1

a

h

64

1

16

1.t3

4

1.t3t1t 233

–64 = 48t2 + 12t – 1

48t2 – 12t – 63 = 0

96

1224012

96

1209614412t



®




= 96

851212

8

851

a

ht

65. n3 + 2n2 + n = n(n +1)2

So, 2)1n(n

!n

=

2)1n(

)!1–n(

if (n +1) is prime number, then (n –1)! will not divisible by n + 1

Ex. = 7 If (n +1) is a composite number, then

(n –1)! may be divisible by (n +1). Ex. 24, 25, 27, .................................... So, statement I is false and statement II is

true.

HkkSfrd foKku

66. t = 18 lsd.M ij eksVjlkbfdy vkSj dkj ds e/; dh nwjh xzkQ ds {ks=kQy ds vUrj ds cjkcj gksxk

18x40 – ½(18x60) = 180 m

27 lsd.M ij izR;sd vkjs[k ds vUrxZr {ks=kQy 27x40 gS =1080 m

nksuksa ds e/; lkis{k nwjh 'kwU; gksxh tc nksuksa ds vUrxZr {ks=kQy leku gksaxs

1/2 × 18 × 60 + 9 × 60 – 27 × 40 = 0

67. fcUnq O ds lkis{k fudk; (xksyh + pdrh) ds dks.kh; laosx lja{k.k ls

mv02R = mv(2R) +

0

v = 2R ; 0 =

2MR2

3

v = M3m8

mu8

68. T1 = 12

g

T2 = 22

g

1

t30

T 2

1

T 5

T 6

2

t

T = 36 6T

2 = 5T

1

T1

2 = 88 36 6 2

100 11 10

=

6 2

5

6 2

5 = 12

g

36 2

25

= 4 × 10 × 1

10 Ans. (A)

69.

a Wa = Ww = ikuh dk Hkkj

b

lead

Wb = Ww + Fb ;

Fb = mRIykod cy

c

TT

Wc = Wb + Fb ;

Fb = mRIykod cy

d

TT

Wd = Ww + Fb – T

Fb = mRIykod cy

T = rkj esa ruko

Ans. (B) 70. mL = 540 × 80 = 54 × 8 × 100 cal.

cQZ dh ek=kk 0°C ij ikuh Å"ek = 540 × 80 cal.

ikuh (80°) 0°C ij ikuh miyC/k Å"ek = 540 × 80 cal.

vr% feJ.k dk rki = 0°C

Fb

Fb

Fb

Fb

T

Fb

Fb



®




jlk;u foKku

71. Ksp dk mPp eku j[kus okyk ;kSfxd H2O esa vf/kd

foys; gksrk gSA

FeS dk Ksp (11 × 10–20) vf/kdre gSA

blfy, ;g vf/kd foys; gSA ;g H2O vf/kdre foys;rk

n'kkZrk gSA

72. lehdj.k ds fy,

B2H6(g) 3O2(g) B2O2(g) 3H2O(g)

sEqs. (i) + 3 (ii) + 3 (iii) – (iv)

H = – 1273 + 3(–286) + 3(44) – 36

= – 1273 – 858 + 132 – 36

= – 2035 kJ/mol

73. v = 2.188 × 106 n

Z m/s

mijksDr lEcU/k ds vuqlkj fodYi (D) v rFkk n ds e/; vkjs[k dk lgh fu:i.k ugha djrk gSA

74. ngu dh Å"ek Rokkf;v.kq dk LF

1v.kq esa ‘C’

ijek.kqvksa dh la[;kA

75. vEyh; lkeF;Zrk dk Øe la;qXeh {kkj dk LFkkf;RoA



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