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PEBPT2160815C0-1

PART-A

SECTION - I

Straight Objective Type

This section contains 28 multiple choice questions.

Each question has 4 choices (1), (2), (3) and (4) for

its answer, out of which ONLY ONE is correct.

1. A car travels from A to B at a speed of

20 km h�1 and returns at a speed of

30 km h�1. The average speed of the car for

the whole journey is :

(1) 5 km h�1

(2) 24 km h�1

(3) 25 km h�1

(4) 50 km h�1

2. A stone is thrown vertically upward with an

initial speed u from the top of a tower,

reaches the ground with a speed 3u. The

height of the tower is:

(1) 23u

g

(2) 24u

g

(3) 26u

g

(4) 29u

g

3. A body starts from rest and is uniformly

acclerated for 30 s. The distance travelled in

the first 10 s is x1, next 10 s is x2 and the last

10 s is x3. Then x1 : x2 : x3 is the same as

(1) 1 : 2 : 4

(2) 1 : 2 : 5

(3) 1 : 3 : 5

(4) 1 : 3 : 9

4. Two balls of equal masses are thrown

upward, along the same vertical line at an

interval of 2 seconds, with the same initial

velocity of 40 m/s. Then these collide at a

height of (Take g = 10 m/s2)

(1) 120 m

(2) 75 m

(3) 200 m

(4) 45 m

5. A balloon is moving upwards with velocity

10 ms�1. It releases a stone which comes

down to the ground in 11 s. The height of the

balloon from the ground at the moment when

the stone was dropped is :

(1) 495 m

(2) 592 m

(3) 460 m

(4) 500 m

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PEBPT2160815C0-2

6. The acceleration time plot for a particle

(starting from rest) moving on a straight line

is shown in figure. For given time interval,

choose the incorrect option :

(1) The particle has zero average acceleration

(2) The particle has never turned around.

(3) The particle has zero displacement

(4) The average speed in the interval 0 to 10s

is the same as the average speed in the

interval 10s to 20s.

7. Mark the incorrect statement for a particle

going on a straight line (x�position coordinate,

v�velocity, a�acceleration) :

(1) If the v and a have opposite sign, the

object is slowing down.

(2) If the x and v have opposite sign, the

particle is moving towards the origin.

(3) If the v is zero at an instant, the a should

also be zero at that instant.

(4) If the v is zero for a time interval, then a

is zero at every instant within the time

interval.

8. A particle is initially at rest, It is subjected to

a linear acceleration a, as shown in the

figure. The maximum speed attained by the

particle is

(1) 605 m/s (2) 110 m/s

(3) 55 m/s (4) 550 m/s

9. The coordinates of a moving particle at any

time t are given by x = t3 and y = t3. The

speed of the particle at time t is given by :

(1) 2 2

(2) 3t2 2 2

(3) t2 2 2

(4) 2 2

10. A particle is moving eastwards with a velocity

of 5 ms�1. In 10 second the velocity changes

to 5 ms�1 northwards. The average acceleration

in this time is :

(1) 1

2 ms�1 towards north-west

(2) 12

ms�2 towards north

(3) zero

(4) 12

ms�2 towards north-west.







PEBPT2160815C0-3

11. During projectile motion, acceleration of a

particle at the highest point of its trajectory is

(1) g

(2) zero

(3) less than g

(4) dependent upon projection velocity

12. The velocity of projection of a projectile is

(6 �i + 8 �j ) ms�1.. The horizontal range of the

projectile is : (g = 10 m/sec2)

(1) 4.9 m

(2) 9.6 m

(3) 19.6 m

(4) 14 m

13. A body is projected horizontally from the top

of a tower with initial velocity 18 ms�1. It hits

the ground at angle 45º. What is the vertical

component of velocity when it strikes the

ground?

(1) 18 2 ms�1

(2) 18 ms�1

(3) 9 2 ms�1

(4) 9 ms�1

14. A ball is horizontally projected with a speed v

from the top of a plane inclined at an angle

45º with the horizontal. How far from the

point of projection will the ball strike the

plane?

(1) 2v

g

(2) 22v

g

(3) 22v

g

(4) 22 2v

g

15. On an inclined plane of inclination 30º, a ball

is thrown at an angle of 60º with the

horizontal from the foot of the incline with a

velocity of 10 3 ms�1. If g = 10 ms�2, then

the time in which ball will hit the inclined

plane is -

(1) 1 sec.

(2) 6 sec.

(3) 2 sec.

(4) 4 sec.







PEBPT2160815C0-4

16. A particle moves in the xy plane with only an

x-component of acceleration of 2 ms�2. The

particle starts from the origin at t = 0 with an

initial velocity having an x-component of

8 ms�1 and y-component of �15 ms�1.

Velocity of particle after time t is :

(1) [(8 + 2t) �i � 15 �j ] m s�1

(2) zero

(3) 2t �i + 15 �j

(4) directed along z-axis.

17. A stone projected at an angle of 60º from the

ground level strikes at an angle of 30º on the

roof of a building of height �h�. Then the

speed of projection of the stone is :

(1) 2gh

(2) 6gh

(3) 3gh

(4) gh

18. A particle is projected at 60° to the horizontal

with a kinetic energy K. The kinetic energy at

the highest point is :

(1) K

(2) Zero

(3) K/4

(4) K/2

19. Shown in the figure are the position time

graph for two children going home from the

school. Which of the following statements

about their relative motion is true after both

of them started moving ?

Their relative velocity : (consider 1-D motion)

(1) first increases and then decreases

(2) first decreases and then increases

(3) is zero

(4) is non zero constant.

20. A boat which can move with a speed of

5 m/s relative to water crosses a river of

width 480 m flowing with a constant speed of

4 m/s. What is the time taken by the boat to

cross the river along the shortest path.

(1) 80 s

(2) 160 s

(3) 240 s

(4) 320 s







PEBPT2160815C0-5

21. A man walks in rain with a velocity of

5 kmh�1. The rain drops strike at him at an

angle of 45° with the horizontal. Velocity of

the rain if it is falling vertically downward �

(1) 5 kmh�1 (2) 4 kmh�1

(3) 3 kmh�1 (4) 1 kmh�1

22. Two particles A and B move with velocities v1

and v2 respectively along the x & y axis. The

initial separation between them is �d� as

shown in the figure. Find the least distance

between them during their motion.

(1) 21

2 21 2

d.v

v v (2)

22

2 21 2

d.v

v v

(3) 1

2 21 2

d.v

v v (4) 2

2 21 2

d.v

v v

23. A body is thrown up in a lift with a velocity u

relative to the lift and the time of flight is

found to be � t �. The acceleration with which

the lift is moving up is :

(1) u gt

t

(2) 2u gt

t

(3) u gt

t

(4) 2u gt

t

24. Which figure represents the correct F.B.D. of rod of mass m as shown in figure :

(1) (2)

(3) (4) None of these

25. Two forces of 6N and 3N are acting on the

two blocks of 2kg and 1kg kept on

frictionless floor. What is the force exerted on

2kg block by 1kg block ?

(1) 1N (2) 2N

(3) 4N (4) 5N

26. A mass M is suspended by a rope from a rigid support at A as shown in figure. Another rope is tied at the end B, and it is pulled horizontally with a force F. If the rope AB makes an angle with the vertical in equilibrium, then the tension in the string AB is :

/////////////////////////////

M

B

A

F

(1) F sin (2) F/sin

(3) F cos (4) F/cos







PEBPT2160815C0-6

27. A body of mass 8 kg is hanging from another

body of mass 12 kg. The combination is

being pulled by a string with an acceleration

of 2.2 m s�2. The tension T1 and T2 will be

respectively : (use g = 9.8m/s2 )

(1) 200 N, 80 N

(2) 220 N, 90 N

(3) 240 N, 96 N

(4) 260 N, 96 N

28. A block is dragged on smooth plane with the

help of a rope which moves with velocity v.

The horizontal velocity of the block is :

//////

/////

/////

//////

/////

/

m

//////////////////

V

(1) v

(2) v

sin

(3) v sin

(4) v

cos

SECTION - II

Reasoning Type

This section contains 2 reasoning type questions.

Each question has 4 choices (1), (2), (3) and (4), out

of which ONLY ONE is correct.

29. STATEMENT-1 : A particle having negative

acceleration will slow down. STATEMENT-2 : Direction of the

acceleration is not dependent upon direction of the velocity.

(1) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1

(2) Statement-1 is true, statement-2 is true and statement-2 is not correct explanation for statement-1

(3) Statement-1 is true, statement-2 is false (4) Statement-1 is false, statement-2 is true 30. STATEMENT-1 : Two stones are

simultaneously projected from level ground from same point with same speeds but different angles with horizontal. Both stones move in same vertical plane. Then the two stones may collide in mid air.

STATEMENT-2 : For two stones projected simultaneously from same point with same speed at different angles with horizontal, their trajectories may intersect at some point.

(1) Statement-1 is true, statement-2 is true

and statement-2 is correct explanation

for statement-1

(2) Statement-1 is true, statement-2 is true

and statement-2 is not correct

explanation for statement-1

(3) Statement-1 is true, statement-2 is false

(4) Statement-1 is false, statement-2 is true







CEBPT2160815C0-7

PART � B

Atomic masses : [H = 1, D = 2, Li = 7, C = 12,

N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27,

Si = 28, P = 31, S = 32, Cl = 35.5, K = 39, Ca = 40,

Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65,

As = 75, Br = 80, Ag = 108, I = 127, Ba = 137,

Hg = 200, Pb = 207]


This section contains 30 multiple choice

questions. Each question has 4 choices (1),

(2), (3) and (4) for its answer, out of which

ONLY ONE is correct.

31. The ratio of the difference in energy between

the first and second Bohr orbit to that

between the second and third Bohr orbit in a

H-like species is

(1) 12

(2) 13

(3) 49

(4) 275

32. If the de-Broglie wavelength of an

electron revolving in 2nd orbit of

H-atom is x, then radius of that orbit is

given by :

(1) x

(2) 2x

(3) x

2

(4) Cannot be determined

33. The uncertainty in position and velocity of a

particle are 0.5 Å and 5.27 × 10�24 m/s

respectively. Then, the approximate mass of

the particle is :

(1) 0.1 Kg

(2) 0.2 Kg

(3) 0.3 Kg

(4) 0.4 Kg

34. The wavenumber of the spectral line of

shortest wavelength of Balmer series of He+

ion is : (R = Rydberg�s constant)

(1) R (2) 3R

(3) 4R (4) 4R/9







CEBPT2160815C0-8

35. The value of azimuthal quantum number of

an electron present in the orbital designated

with quantum numbers as n = 4, m = �3 may

be :

(1) 0 (2) 1

(3) 2 (4) 3

36. In a sample of H-atom electrons make

transition from 5th excited state to ground

state, producing all possible types of

photons, then number of lines in infrared

region are

(1) 4 (2) 5

(3) 6 (4) 3

37. Which of the following set of quantum

numbers is correct for an electron in 4f

orbital ?

(1) n = 4, l =3, m = +4, s = +1/2

(2) n = 4, l = 4, m = �4, s = �1/2

(3) n = 4, l = 3, m = +1, s = +1/2

(4) n = 3, l=2, m =�2, s = +1/2

38. The wave motion of an electron in a Bohr's

orbit of Hydrogen atom is as shown in

diagram. The orbit number is :

(1) 2 (2) 3

(3) 4 (4) 6

39. One atom of an element x weigh

6.643 × 10�23 g. Number of moles of atom in

20 kg is :

(1) 4

(2) 40

(3) 100

(4) 500

40. For reaction P + 2Q 3R, if reaction start

with 0.1 mole of Q then find out the mole of

R produed.

(1) 0.2

(2) 0.3

(3) 1.5

(4) 0.15







CEBPT2160815C0-9

41. 2K + 2 + 22 HNO3 2HO3 + 2KO3 +

22NO2 + 10H2O

If 3 mole of K & 2 moles 2 are reacted with

excess of HNO3. Volume of NO2 gas evolved

at NTP is

(1) 739.2 Lt

(2) 1075.2 Lt

(3) 44.8 Lt

(4) 67.2 Lt

42. A solution containing 0.1 mol of a metal

chloride MClx requires 500 ml of 0.8 M

AgNO3 solution for complete precipitation.

The value of x is

(1) 1

(2) 2

(3) 4

(4) 3

43. The atomic mass of an element is 27. if

valency is 3, the vapour density of the

volatile chloride will be:

(1) 66.75

(2) 6.675

(3) 667.5

(4) 81

44. A hydrocarbon contains 80% of carbon, then

the hydrocarbon is :

(1) CH4

(2) C2H5

(3) C2H6

(4) C2H2

45. 20 gm. CaCO3 on decomposition gives CO2

at STP

(1) 4.48 litre

(2) 22.4 litre

(3) 2.24 litre

(4) None of these

46. How many alcohols give immediate turbidity

with lucas reagent having molecular formula

(C5H12O)

(1) 1

(2) 2

(3) 3

(4) 4







CEBPT2160815C0-10

47. Which of the following compound can give

test with Tollen's reagent, and yellow

precipitate with iodine in NaOH.

(1) CH2=O

(2) CH3�CH=O

(3) CH3�CH2�CH=O

(4)

48. Which is incorrect match with respect to the

reagent used for lab test ?

(1) Carbohydrates �Naphthol (Molish

reagent)

(2) Nitro ethane Zn, NH4Cl and AgNO3

(Muliken Barker test)

(3) Phenol Anhydrous ZnCl2 + Conc.

HCl (Lucas Reagent)

(4) Benzoic acid NaHCO3

49. How many structural isomers of C5H10 give

bromine water test ?

(1) 1 (2) 3

(3) 5 (4) 10

50. How many structural isomeric ketones

having molecular formula (C5H10O) give

iodoform test ?

(1) 1

(2) 2

(3) 3

(4) 4

51. On oxidative ozonolysis of 3-Methylhex-3-

ene, two products A & B are formed.

A gives CO2 gas with sodium bicarbonate,

but B can not. The structures of A & B are

respectively :

(1) & CH3�CH2�COOH

(2) CH3�CH2�COOH & CH3�CH2�CH=O

(3) CH3�CH2�COOH &

(4) CH3�CH2�CH2�COOH &







CEBPT2160815C0-11

52. How many monochloro structure isomers will

produce when 3-Methylpentane reacts with

chlorine in presence of sunlight ?

(1) 2

(2) 4

(3) 6

(4) 3

53. Which of the following hydrocarbon can give

only acetone and CO2 on ozonolysis in

presence of Zinc ?

(1) CH3�CH=C=CH�CH3

(2) CH3�CH=CH�CH=C(CH3)2

(3) (CH3)2C=C=CH2

(4)

54. Which is incorrect order for atomic radii?

(1) Mg < Ca < Sr < Ba

(2) B < Al < Ga < In

(3) F < O < N < C

(4) F < Cl < Br < I

55. Which statement is correct for the ionization

energy of second period elements (Be, B, C,

N, O, F)

(1) Order of Ist and IInd ionization energy is

same for the above elements.

(2) Ist ionization energy is highest for Be

among the given elements.

(3) Ionization energy of N is higher than

Oxygen & Fluorine

(4) First Ionization energy of Be is higher

than Boron but lower than Carbon

56. The set representing the correct order of first

ionization potential is :

(1) K > Na > Li

(2) Be > Mg > Ca

(3) B > C > N

(4) Ge > Si > C

57. Element with atomic number 38, belongs to

(1) II A group and 5th period

(2) II A group and 2nd period

(3) V A group and 2nd period

(4) III A group and 5th period







CEBPT2160815C0-12

58. Which gas is released in the given reaction ?

(1) H2 gas

(2) SO2 gas

(3) CO2 gas

(4) NO2 gas

59. Which is the position isomers of

(1)

(2)

(3)

(4)

60. Which is the chain isomers of

(1)

(2)

(3)

(4)



Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005

Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PTC024029

MEBPT2160815C0-13


PART � C

SECTION - I


This section contains 27 multiple choice questions.

Each question has 4 choices (1), (2), (3) and (4) for

its answer, out of which ONLY ONE is correct.

61. If a1, a2, .......... an are positive number in

A.P. then 21

n1

aa

aa

+

32

n1

aa

aa

+........ +

n1�n

n1

aa

aa

is equal to:

(1) n + 1

(2) n � 1

(3) n

(4) None of these

62. log(0.3)(x � 1) < log(0.09)(x � 1) then x lies in the

interval

(1) (2, )

(2) (�2, �1) (2, )

(3) (1, 2)

(4) (�, 1) (2, 8)

63. If c�b2log

= a�c3log

= b�a5log

then 2a 3b 5c is

equal to :

(1) 1

(2) 10

(3) 15

(4) None of these

64. The value of x for which |x2 + 3x| + x2 � 2 0 is

(1) R � [0, 1)

(2) R

(3) R �

21

,32

�

(4) R � (� 2 , 2 )

65. If P(x) = ax2 + bx + c & Q(x) = � ax2 + dx + c,

where ac 0, then P(x).Q(x) = 0 has at least

(1) Two real roots

(2) Four imaginary roots

(3) Four real roots

(4) can not say





MEBPT2160815C0-14


66. If roots of the equation

(a2 + b2)x2 � 2(ac + bd)x + c2 + d2 = 0 are

equal then :

(1) ad + bc = 0

(2) ab = dc

(3) ac = bd

(4) ad = bc

67. If a,b,c,d and x are distinct real numbers such

that (a2+b2+c2)x2 � 2(ab+bc+cd)x + b2+c2+d2 0

then a, b, c, d :

(1) are in A.P.

(2) are in G.P.

(3) are in H.P.

(4) satisfy ab = cd

68. Solution of

1x2�x2)32( + 1�x2�x2

)3�2( =4

2 3are

(1) 1 ± 3 , 1

(2) 1 ± 2 , 1

(3) 1 ± 3 , 2

(4) 1 ± 2 , ± 1

69. If a1, a2, a3 ....... a2k are in A.P., then

21a � 2

2a + 23a � 2

4a + ........ � 2k2a =

(1) 1�k2

k( 2

1a � 2k2a )

(2) 1�k

k2 ( 2

k2a �21a )

(3) 1k

k

( 21a + 2

k2a )

(4) None of these

70. If ratio of sum of m terms and n terms of an

A.P. be m2 : n2, then the ratio of its mth and

nth terms will be

(1) (2m + 1) : (2n + 1)

(2) m : n

(3) (2m � 1) : (2n � 1)

(4) none of these

71. If sum of the A.M. & G.M. of two positive

distinct numbers is equal to the difference

between the numbers then numbers are in

ratio :

(1) 1 : 3

(2) 1 : 6

(3) 9 : 1

(4) 1 : 12





MEBPT2160815C0-15


72. Suppose x1, x2 be the roots of ax2 + bx + c = 0

and x3, x4 be the roots of px2 + qx + r = 0

and x1, x2, x3, x4 are in A.P. then common

difference of this A.P. is

(1) 21

pq

�ab

(2) 31

pq

�ab

(3) 41

pq

�ab

(4) None of these

73. Sum of positive roots of the equation

(x2 � 12x + 35)(x2 + 10x + 24) = 504 is

(1) 9

(2) 10

(3) 11

(4) 12

74. Solution of the equation x x�1x5 (64) = 2000

is (where x 2 and x is an integer)

(1) divisible by 2

(2) divisible by 3

(3) divisible by 5

(4) divisible by 6

75. If the equations x2 + ax + b = 0 and

x2 + bx + a = 0 have a common root, then the

value of a + b is

(1) 1

(2) 0

(3) � 1

(4) none of these

76. Roots of the quadratic equation

2 2x 4x 3 x 6x 8 0, R will

be

(1) always real

(2) real only when is positive

(3) real only when is negative

(4) always imaginary

77. The solution set of the inequality

(x + 1)2 > (x + 3) is

(1) {x : �3 < x < �1}

(2) {x : x > �1}

(3) {x : �3 x �2}

(4) {x : x > 1 or x < �2}





MEBPT2160815C0-16


78. The number of values of k for which

equation |x + 1| + |x � 1| = k has infinite

solution is

(1) greater than 2

(2) 2

(3) less than 2

(4) none of these

79. If |x| � |x�2| = 2 then number of prime

solution of this equation less than 10 are

(1) 4

(2) 3

(3) 5

(4) none of these

80. Common solution of the equation

2(x �3x 2)| x � 2 | = 1 and inequality x(x � 2) 0

is

(1) 1

(2) 2

(3) 3

(4) 1, 2

81. Number of positive integers which satisfy the

inequality 2(x � 2)(�x 1)(x 1)

(x 1)

0 are

(1) two

(2) one

(3) three

(4) infinite

82. Sum of the all possible solutions of

||x � 1| � 3| = 2 is

(1) 0

(2) 2

(3) 4

(4) 6

83. If {x} = 2.1 then complete set of values of x

is (where {.} denotes fractional part function).

(1)

(2) 0.1

(3) 2

(4) 0.9





MEBPT2160815C0-17


84. 3 3 3(171) � (123) � (48)

9 19 123 16 3 =

(1) 8

(2) 2

(3) �3

(4) 3

85. If Hn = 1 + 21

+ 31

+ ........+ n1

then value of

1 +23

+ 35

+ ........+ n

1�n2 is :

(1) 2n � Hn

(2) 2n + Hn

(3) Hn � 2n

(4) Hn + n

86. If positive numbers a, b, c, d are in HP then

(1) ab cd

(2) ac bd

(3) ad bc

(4) none of these

87. The value of 91/3 . 91/9. 91/27 ......upto , is

(1) 1

(2) 3

(3) 9

(4) None of these

SECTION - II

Reasoning Type

This section contains 3 reasoning type questions.

Each question has 4 choices (1), (2), (3) and (4), out

of which ONLY ONE is correct.

88. Let a, b, c, d, e are positive numbers

STATEMENT -1 : If a, b, c, d, e are in A.P.,

then bcde, acde, abde, abcd and abce are in

H.P.

STATEMENT -2 : If a

a�cb ,

bb�ac

,

cc�ba

are in A.P., then a, b, c are in H.P.

(1) Statement-1 is True, Statement-2 is True;

Statement-2 is a correct explanation for

Statement-1.


Statement-2 is NOT a correct

explanation for Statement-1

(3) Statement-1 is True, Statement-2 is False

(4) Statement-1 is False, Statement-2 is True





MEBPT2160815C0-18


89. STATEMENT -1 :

Equation x2(a) + x (a2 � 3a + 1) � a2 + 2a � 1

is an identity in 'a' if x = �1.

STATEMENT-2 : If all the coefficients of

equation are zero then equation is called an

identity.


Statement-2 is a correct explanation for

Statement-1.


Statement-2 is NOT a correct


(3) Statement-1 is True, Statement-2 is False

(4) Statement-1 is False, Statement-2 is True

90. STATEMENT- 1 : 12| 1� 3 | � 4 2 3

is a rational number.

STATEMENT - 2 : |x| = x x 0

�x x 0

(1) Statement -1 is True, Statement -2 is

True ; Statement -2 is a correct

explanation for Statement -1

(2) Statement-1 is True, Statement-2 is True

; Statement-2 is NOT a correct


(3) Statement -1 is True, Statement -2 is

False

(4) Statement -1 is False, Statement -2 is

True



dPps dk;Z ds fy, LFkku




PEBPT2160815C0-1

Hkkx-A

[k.M- I

lh/ks oLrqfu"B izdkj

bl [k.M esa 28 cgq&fodYih iz'u gSaA izR;sd iz'u ds

4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA

1. ,d dkj A ls B rd 20 fdeh-@?k.Vk ls xfr

djrh gS vkSj 30 fdeh-@?k.Vk ls okil ykSVdj

vk tkrh gSA iwjh ;k=kk ds nkSjku dkj dh vkSlr

pky D;k gksxh \

(1) 5 fdeh-@?k.Vk

(2) 24 fdeh-@?k.Vk

(3) 25 fdeh-@?k.Vk

(4) 50 fdeh-@?k.Vk

2. fdlh iRFkj dks feukj ls m/okZ/kj Å ij dh vksj

u çkjfEHkd osx ls Qsadus ij ;g tehu ij 3u

osx ls igq¡prk gS rks feukj dh Å ¡pkbZ gS:

(1) 23u

g (2)

24ug

(3) 26u

g (4)

29ug

3. ,d oLrq fojkekoLFkk ls çkjEHk gksdj 30 lSd.M

rd ,d leku Rojfr gksrh gSA ;fn igys

10 lSd.M esa r; nwjh x1] vxys 10 lSd.M eas

x2 rFkk vfUre 10 lSd.M esa x3 gS rks x1 : x2 : x3 dk

vuqikr gksxk &

(1) 1 : 2 : 4 (2) 1 : 2 : 5

(3) 1 : 3 : 5 (4) 1 : 3 : 9

4. leku nzO;eku dh nks xsanksa dks leku Å /okZ/kj js[kk

esa 2 lSd.M ds vUrjky esa ,d leku çkjfEHkd osx

40 eh-/lS- ls Å ij dh vksj Qsadk tkrk gS rks ;s

fdl Å ¡pkbZ ij Vdjk,xhA (g = 10 m/s2)

(1) 120 m (2) 75 m

(3) 200 m (4) 45 m

5. ,d xqCckjk 10 eh0@lS0 osx ls tehu ls Å ij dh

vksj tk jgk gSA blls ,d iRFkj fxjk;k tkrk gS]

tks tehu ij 11 lSd.M esa igq¡prk gSA tc iRFkj

fxjk;k x;k Fkk] rc xqCckjs dh tehu ls Å pk¡bZ D;k

Fkh\

(1) 495 m (2) 592 m

(3) 460 m (4) 500 m



mailto:fdeh-@?k.Vk

mailto:fdeh-@?k.Vk

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mailto:fdeh-@?k.Vk

mailto:fdeh-@?k.Vk





PEBPT2160815C0-2

6. lh/kh js[kk esa fLFkjkoLFkk ls xfreku d.k dk

Roj.k le; vkjs[k n'kkZ;k x;k gSA fn;s x;s

le;kUrjky ds fy, vlR; dFku gSA

(1) d.k dk vkSlr Roj.k 'kwU; gSA

(2) d.k dHkh Hkh okil ugha eqM+rk gSA

(3) d.k dk foLFkkiu 'kwU; gksxkA

(4) d.k dh 0 ls 10 lSd.M eas vkSlr pky ogh gksxh

tks 10 lSd.M ls 20 lSd.M ds nkSjku gSA

7. ,d d.k lh/kh js[kk esa xfr dj jgk gSA bl d.k ds

fy, xyr dFku igpkfu;s&(x�fLFkfr funsZ'kkad ,

v�osx, a�Roj.k gS) :

(1) vxj v vkSj a foijhr fn'kk esa gSa] rks d.k

dh pky de gks jgh gSA

(2) vxj x vkSj v foijhr fpUg ds gaS] rks d.k

ewy fcUnq dh vksj tk jgk gSA

(3) vxj fdlh {k.k v 'kwU; gS] rks ml {k.k ij a

Hkh 'kwU; gksxkA

(4) vxj fdlh le;kUrjky esa v 'kwU; gS] rks ml

le;kUrjky esa fdlh Hkh {k.k a Hkh 'kwU;

gksxkA

8. fojkekoLFkk ls çkjEHk gqvk ,d d.k] fu;r js[kh;

Roj.k a ls fp=kkuqlkj xfreku gSA d.k }kjk çkIr

vf/kdre pky gksxh&

(1) 605 m/s (2) 110 m/s

(3) 55 m/s (4) 550 m/s

9. xfr'khy d.k ds fdlh le; t ij funZs'kkad x = t3

o y = t3A le; t ij d.k dh pky gksxhA

(1) 2 2 (2) 3t2 2 2

(3) t2 2 2 (4) 2 2

10. ,d d.k 5 ms�1 pky ls iwoZ fn'kk esa xfr dj jgk

gSA 10 lSd.M ckn ;g mÙrj fn'kk esa 5 ms�1 pky

ls py jgk gksrk gSA rks bl le; esa d.k dk vkSlr

Roj.k D;k gksxkA

(1) 1

2 ms�1 mÙrj&if'pe dh vksj

(2) 12

ms�2 mÙrj dh vksj

(3) 'kwU;

(4) 12

ms�2 mÙrj&if'pe dh vksj







PEBPT2160815C0-3

11. iz{ksI; xfr esa] iFk ds mPpre fcUnq ij d.k dk

Roj.k gksrk gS :

(1) g

(2) 'kwU;

(3) g ls de

(4) iz{ksi.k osx ij fuHkZj djrk gSA

12. ,d iz{ksI; dk iz{ksi.k osx (6 �i + 8 �j ) ms�1 gSA rks

iz{ksI; dh {kSfrt ijkl gksxh & (g = 10 m/sec2)

(1) 4.9 m

(2) 9.6 m

(3) 19.6 m

(4) 14 m

13. ,d oLrq ,d Å ¡ph bekjr ls 18 ms�1 ds

izkjfEHkd osx ls {kSfrt fn'kk esa iz{ksfir dh tkrh

gS ;g tehu ij 45º dks.k ij Vdjkrh gS rks

oLrq }kjk tehu ij Vdjkrs le; osx dk

Å /okZ/kj ?kVd gksxkA

(1) 18 2 ms�1

(2) 18 ms�1

(3) 9 2 ms�1

(4) 9 ms�1

14. ,d xsn dks 45º dks.k ij >qds urry ds Å ijh

fljs ls {kSfrt fn'kk esas v pky ls iz{ksfir fd;k

tkrk gSA rks iz{ksfir fcUnq ls xsan urry ij fdruh

nwj Vdjk,xh &

(1) 2v

g

(2) 22 v

g

(3) 22v

g

(4) 22 2 v

g

15. 30º mUu;u dks.k okys ,d urry ds vk/kkj ls

{kSfrt ls 60º ds dks.k ij ,d xsan dks

10 3 ms�1 ds osx ls Qsadrs gSaA ;fn g=10 ms�2

gS rks fdrus le; ckn xsan okil ur ry ls

Vdjk,xhA

(1) 1 sec.

(2) 6 sec.

(3) 2 sec.

(4) 4 sec.







PEBPT2160815C0-4

16. x-y ry esa xfr'khy d.k] dsoy Roj.k dk

x�?kVd 2 ms�2 j[krk gSA d.k t = 0 ij ewy

fcUnq ls xfr izkjEHk djrk gSA izkjfEHkd osx dk

x-?kVd 8 ms�1 rFkk y-?kVd �15 ms�1 gSA rks

t le; i'pkr d.k dk osx gksxk &

(1) [(8 + 2t) �i � 15 �j ] m s�1

(2) 'kwU;

(3) 2t + 15 �j

(4) z-v{k ds vuqfn'k funsZf'kr

17. ,d iRFkj tehu ls 60º ds dks.k ij iz{ksfir

fd;k tkrk gS ,oa h Å ¡pkbZ dh ,d bekjr dh

Nr ij 30º ds dks.k ij Vdjkrk gSA rks iRFkj

dh iz{ksi.k pky gSA

(1) 2gh

(2) 6gh

(3) 3gh

(4) gh

18. ,d d.k dks {kSfrt ls 60° dk dks.k cukrs gq,

xfrt Å tkZ K ls iz{ksfir fd;k tkrk gSA mPpre

fcUnq ij xfrt Å tkZ gksxh %

(1) K (2) 'kwU;

(3) K/4 (4) K/2

19. ?kj ls Ldwy tkrs gq, cPpksa ds fLFkfr le; xzkQ

fp=k esa çnf'kZr gSA rks nksuks ds xfr izkjEHk djus ds

i'pkr~ buds lkis{k xfr ds lEcU/k esa dkSuls dFku

lR; gS? budk lkis{k osx% ¼ljy js[kh; xfr ekfu;s½

(1) igys c<+sxk ckn esa ?kVsxkA

(2) igys ?kVsxk ckn esa c<+sxkA

(3) 'kwU; gksxkA

(4) v'kwU; fu;rakd gksxkA

20. ikuh ds lkis{k 5 m/s dh pky ls xfreku ,d ukoa

4 m/s dh fu;r pky ls cgrh gqbZ 480 m pkSM+h

unh dks ikj djrh gS rks U;wure nwjh ls unh dks

ikj djus esa yxk le; Kkr djksA

(1) 80 s

(2) 160 s

(3) 240 s

(4) 320 s







PEBPT2160815C0-5

21. ,d vkneh 5 kmh�1 osx ls xfr'khy gSA vkneh

dks c"kkZ dh cwans {kSfrt ls 45° ds dks.k ij vkrh

çrhr gksrh gS rks m/okZ/kj uhps fxjrh cwanksa dk osx

gksxk \

(1) 5 kmh�1 (2) 4 kmh�1

(3) 3 kmh�1 (4) 1 kmh�1

22. nks d.k A rFkk B Øe'k% v1 rFkk v2 osx ls x rFkk

y v{k esa xfreku gSA nksuksa d.kksa ds e/; izkjfEHkd

nwjh d fp=k esa n'kkZ;s vuqlkj gS rks xfr ds nkSjku

nksuksa ds e/; U;wure nwjh Kkr djksA

(1) 21

2 21 2

d.v

v v (2)

22

2 21 2

d.v

v v

(3) 1

2 21 2

d.v

v v (4) 2

2 21 2

d.v

v v

23. ,d oLrq dks Å ij dh rjQ xfr'khy fy¶V esa

fy¶V ds lkis{k Å ij dh rjQ u osx ls QSadk

tkrk gS vkSj oLrq dk mM~M;u dky � t � çkIr

gksrk gSA rks Å ij dh rjQ xfr'khy fy¶V dk

Roj.k gksxk %

(1) u gt

t

(2) 2u gt

t

(3) u gt

t

(4) 2u gt

t

24. fp=k esa iznf'kZr m nzO;eku dh NM+ dk eqDr oLrq

js[kkfp=k fuEu esa ls dkSuls fp=k }kjk fn;k tkrk gSA

(1)

(2)

(3)

(4) buesa ls dksbZ ugha

25. fp=kkuqlkj ?k"kZ.k jfgr lrg ij j[ks 2 kg o 1 kg ds

nks CykWdks ij nks cy 6N vkSj 3 N dk;Zjr gS] rks

1 kg ds CykWd }kjk 2kg ds CykWd ij vkjksfir cy

D;k gksxk \

(1) 1N (2) 2N

(3) 4N (4) 5N







PEBPT2160815C0-6

26. fp=kkuqlkj ,d M nzO;eku dks jLlh dh lgk;rk

ls n< vk/kkj ij fcUnq A ls yVdk;k x;k gSA ,d nwljh jLlh fcUnq B ij ca/kh gS vkSj bldks {kSfrt fn'kk esa cy F ls [khapk tkrk gSA ;fn m/okZ/kj lkE;koLFkk eas jLlh AB Å /okZ/kj ls

dks.k cukrh gks rks jLlh AB esa ruko gksxk & /////////////////////////////

M

B

A

F

(1) F sin (2) F/sin (3) F cos (4) F/cos

27. ,d 8 kg nzO;eku dh oLrq ,d nwljh 12 kg dh oLrq ls yVdh gqbZ gSA bl la;kstu dks jLlh dh lgk;rk ls Å ij dh rjQ 2.2 m s�2 ds Roj.k ls [khapk tkrk gS rks ruko T1 o T2 Øe'k% gksaxs : (fn;k gS g = 9.8m/s2 )

(1) 200 N, 80 N (2) 220 N, 90 N (3) 240 N, 96 N (4) 260 N, 96 N 28. ,d CykWd dks fpdus ry ij v osx ls xfr'khy

jLlh }kjk fp=kkuqlkj [khapk tkrk gSA CykWd dk {kSfrt osx gksxk&

//////

/////

/////

//////

/////

/

m

//////////////////

V

(1) v (2) v

sin

(3) v sin (4) v

cos

[k.M - II

dkj.k&izdkj

bl [k.M esa 2 dkj.k ds ç'u gSA çR;sd ç'u ds 4 fodYi

(1), (2), (3) rFkk (4) gS] ftlesa ls flQZ ,d lgh gSA

29. oDrO;-1 % ,d d.k ftldk Roj.k _ .kkRed gS]

eafnr gksxkA

oDrO;-2 % Roj.k dh fn'kk osx dh fn'kk ij fuHkZj

ugha djrhA

(1) oDrO;&1 lR; gS] oDrO;&2 lR; gS ;

oDrO;&2, oDrO;&1 dk lgh Li"Vhdj.k gSA


oDrO;&2, oDrO;&1 dk lgh Li"Vhdj.k ugha

gSA

(3) oDrO;&1 lR; gS] oDrO;&2 vlR; gSA

(4) oDrO;&1 vlR; gS] oDrO;&2 lR; gSA 30. oDrO;-1 : nks iRFkj tehu ls ,d lkFk ,d gh

fcUnq ls ,d gh pky ls ysfdu {kSfrt ls

fHkUu&fHkUu dks.kksa ij iz{ksfir fd;s tkrs gSaA nksuksa

iRFkj ,d gh Å /okZ/kj ry esa xfr djrs gSaA rks

nksuksa iRFkj chp gok esa Vdjk ldrs gSaA

oDrO;-2 : ,d gh fcUnq ls ,d lkFk leku pky ls

{kSfrt ls fHkUu&fHkUu dks.kksa ij iz{ksfir nks iRFkj ds

iz{ksI; iFk chp gok esa fdlh fcUnq ij dkV ldrs

gSaA


oDrO;&2, oDrO;&1 dk lgh Li"Vhdj.k gSA


oDrO;&2, oDrO;&1 dk lgh Li"Vhdj.k ugha

gSA

(3) oDrO;&1 lR; gS] oDrO;&2 vlR; gSA

(4) oDrO;&1 vlR; gS] oDrO;&2 lR; gSA



d Pps d k;Z d s fy, LFkku




CEBPT2160815C0-7

PART � B

Atomic masses : [H = 1, D = 2, Li = 7, C = 12,

N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27,

Si = 28, P = 31, S = 32, Cl = 35.5, K = 39, Ca = 40,

Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65,

As = 75, Br = 80, Ag = 108, I = 127, Ba = 137,

Hg = 200, Pb = 207]

lh/ks oLrqfu"B izdkj

bl [k.M esa 30 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4

fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d

lgh gSA

31. fdlh H ds leku Lih'kht esa çFke o f}rh; cksj

d{k rFkk f}rh; o rrh; cksj d{k ds Å tkZ vUrj

dk vuqikr gksxk µ

(1) 12

(2) 13

(3) 49

(4) 275

32. ;fn H-ijek.kq dh 2nd d{kk esa ifjØe.k djus

okys ,d bysDVªkWu dh Mh&czksXyh rjax}S/;Z x gS] rks

bl d{kk dh f=kT;k D;k gksxh :

(1) x

(2) 2x

(3) x

2

(4) Kkr ugha fd;k tk ldrk

33. ;fn fdlh ,d d.k dh fLFkfr rFkk osx esa

vfuf'pr~rk Øe'k% 0.5 Å rFkk 5.27 × 10�24 m/s

gSa] rks d.k dk nzO;eku yxHkx fdruk gksxk\

(1) 0.1 Kg (2) 0.2 Kg

(3) 0.3 Kg (4) 0.4 Kg

34. He+ vk;u dh ckej Js.kh esa izkIr U;wure rjax}S/;Z

dh LiSDVªeh js[kk dh rjaxla[;k D;k gksxh\

(R = fjMcxZ fu;rkad )

(1) R (2) 3R

(3) 4R (4) 4R/9







CEBPT2160815C0-8

35. DokaVe la[;k n = 4, m = �3 okys ,d d{kd esa

mifLFkr ,d bysDVªkWu ds fy, f}xa'kh DokaVe

la[;k dk eku D;k gks ldrk gS %

(1) 0

(2) 1

(3) 2

(4) 3

36. H-ijek.kq ds uewus esa bysDVªkWu 5th mÙksftr voLFkk

ls ewy voLFkk rd laØe.k dj lHkh laHko QksVksu

mRikfnr djrs gSa] rks vojDr {ks=k esa js[kkvksa dh

la[;k fuEu gSa &

(1) 4

(2) 5

(3) 6

(4) 3

37. 4f d{kd esa ,d by sDVªkWu ds fy, DokaVe la[;k

dk dkSulk leqPp; lgh gS ?

(1) n = 4, l =3, m = +4, s = +1/2

(2) n = 4, l = 4, m = �4, s = �1/2

(3) n = 4, l = 3, m = +1, s = +1/2

(4) n = 3, l=2, m =�2, s = +1/2

38. fuEu js[kkfp=k esa gkbMªkstu ijek.kq dh ,d cksgj

d{kk esa ,d bysDVªkWu dh rjax xfr n'kkZ;h x;h gSA

d{kk la[;k fuEu gS%

(1) 2

(2) 3

(3) 4

(4) 6

39. rRo x ds ,d ijek.kq dk Hkkj 6.643 × 10�23 xzke

gSA 20 kg esa ijek.kq ds eksyksa dh la[;k fuEu gS %

(1) 4

(2) 40

(3) 100

(4) 500

40. vfHkfØ;k P + 2Q 3R, ds fy, ;fn vfHkfØ;k

Q ds 0.1 eksy ysdj izkjEHk dh xbZ rc fufeZr

R ds eksy Kkr dhft,A

(1) 0.2 (2) 0.3

(3) 1.5 (4) 0.15







CEBPT2160815C0-9

41. 2K + 2 + 22 HNO3 2HO3 + 2KO3 +

22NO2 + 10H2O

;fn K ds 3 eksy rFkk 2 ds 2 eksy HNO3 ds

vkf/kD; ls fØ;k djrs gS] rks ekud rki o nkc

(NTP) ij NO2 dk izkIr vk;ru Kkr djksA

(1) 739.2 Lt

(2) 1075.2 Lt

(3) 44.8 Lt

(4) 67.2 Lt

42. /kkrq DyksjkbM MClx ds 0.1 eksy ;qDr ,d foy;u

ds iw.kZ :i ls vo{ksi.k ds fy, 0.8 M AgNO3

foy;u ds 500 ml vko';d gSA x dk eku fuEu gS%

(1) 1

(2) 2

(3) 4

(4) 3

43. ,d rRo dk ijek.kq Hkkj 27 gSA ;fn la;kstdrk 3

gS] rks ok"i'khy DyksjkbM dk ok"i ?kuRo gksxk :

(1) 66.75

(2) 6.675

(3) 667.5

(4) 81

44. ;fn ,d gkbMªksdkcZu esa 80% dkcZu gks] rks

gkbMªksdkcZu gksxk :

(1) CH4

(2) C2H5

(3) C2H6

(4) C2H2

45. 20 xzke CaCO3 dks fo?kfVr djus ij fdruh CO

2

STP ij çkIr gksrh gS\

(1) 4.48 yhVj

(2) 22.4 yhVj

(3) 2.24 yhVj


46. (C5H

12O) v.kqlw=k okys fdrus ,YdksgkWy Y;wdkWl

vfHkdeZd ds lkFk rqjUr xanykiu nsrs gSa \

(1) 1

(2) 2

(3) 3

(4) 4







CEBPT2160815C0-10

47. fuEu esa ls dkSulk ;kSfxd VkWysu vfHkdeZd ds

lkFk ijh{k.k nsrk gS rFkk NaOH esa vk;ksMhu ds

lkFk ihyk vo{ksi nsrk gS \

(1) CH2=O

(2) CH3�CH=O

(3) CH3�CH

2�CH=O

(4)

48. iz;ksx'kkyk ijh{k.k ds fy, fuEu esa ls dkSu mlds

vfHkdeZd ds lkFk lgh lqesfyr ugha gSA

(1) dkcksZgkbMªsV �us¶Fkksy (eksfy'k vfHkdeZd )

(2) ukbVªks,Fksu Zn, NH4Cl rFkk AgNO

3

(eqfydu cdZj ijh{k.k)

(3) fQuksy futZy ZnCl2 + Conc. HCl

(Y;wdkWl vfHkdeZd )

(4) csUtksbd vEy NaHCO3

49. C5H

10 ds fdrus lajpukRed leko;oh czksehu ty

ijh{k.k nsrs gSa ?

(1) 1 (2) 3

(3) 5 (4) 10

50. (C5H

10O) v.kqlw=k okys fdrus lajpukRed

leko;oh dhVksu vk;ksMksQkWeZ ijh{k.k nsrs gSa?

(1) 1 (2) 2

(3) 3 (4) 4

51. 3-esfFkygsDl-3-bZu ds vkWDlhdkjh vkstksuhvi?kVu

ij nks mRikn A o B curs gSaA mRikn A lksfM;e

ckbdkcksZusV ds lkFk CO2 xSl nsrk gS fdUrq mRikn

B ughaA A rFkk B dh lajpuk,a Øe'k% gS :

(1) & CH3�CH

2�COOH

(2) CH3�CH

2�COOH & CH

3�CH

2�CH=O

(3) CH3�CH

2�COOH &

(4) CH3�CH

2�CH

2�COOH &







CEBPT2160815C0-11

52. tc 3-esfFkyisUVsu dh vfHkfØ;k lw;Z ds izdk'k dh

mifLFkfr esa Dyksjhu ds lkFk dh tkrh gS rks fdrus

eksuksDyksjks lajpuk leko;oh izkIr gksrs gSa\

(1) 2 (2) 4

(3) 6 (4) 3

53. dkSulk gkbMªksdkcZu ftad dh mifLFkfr esa

vkstksuhdj.k djus ij dsoy ,lhVksu ,oa CO2 nsrk

gS \

(1) CH3�CH=C=CH�CH

3

(2) CH3�CH=CH�CH=C(CH

3)

2

(3) (CH3)2C=C=CH

2

(4)

54. fuEu esa ls dkSulk ijekf.od f=kT;kvksa dk xyr

Øe gS \

(1) Mg < Ca < Sr < Ba

(2) B < Al < Ga < In

(3) F < O < N < C

(4) F < Cl < Br < I

55. f}rh; vkorZ rRoksa (Be, B, C, N, O, F) dh

vk;uu Å tkZ ds lEcU/k esa lgh dFku gS \

(1) mijksDr rRoksa ds fy, izFke rFkk f}rh;

vk;uu Å tkZvksa dk Øe leku gSA

(2) fn;s x;s rRoksa esa ls Be dh izFke vk;uu Å tkZ

lcls vf/kd gksrh gSA

(3) N dh vk;uu Å tkZ vkWDlhtu o ¶yksjhu ls

vf/kd gksrh gSA

(4) Be dh izFke vk;uu Å tkZ cksjkWu ls vf/kd

fdUrq dkcZu ls de gksrh gSA

56. fuEu lewg esa izFke vk;uu foHko dk lgh Øe

fuEu gS %

(1) K > Na > Li

(2) Be > Mg > Ca

(3) B > C > N

(4) Ge > Si > C







CEBPT2160815C0-12

57. vkorZ lkj.kh esa ijek.kq Øekad 38 okyk rRo

mifLFkr gksxkA

(1) II A lewg rFkk 5th vkorZ esa

(2) II A lewg rFkk 2nd vkorZ esa

(3) V A lewg rFkk 2nd vkorZ esa

(4) III A lewg rFkk 5th vkorZ esa

58. nh xbZ vfHkfØ;k esa dkSulh xSl eqDr gksrh gS \

(1) H2 gas

(2) SO2 gas

(3) CO2 gas

(4) NO2 gas

59. ;kSfxd dk fLFkfr leko;oh

dkSulk gS

(1)

(2)

(3)

(4)

60. ;kSfxd dk Ja[kyk leko;oh

dkSulk gS \

(1)

(2)

(3)

(4)





MEBPT2160815C0-13

(dPps dk;Z ds fy, LFkku )

PART � C

[k.M - I lh/ks oLrqfu"B izdkj

bl [k.M esa 27 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4

fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA

61. ;fn a1, a2, .......... an /kukRed la[;k,¡ lekUrj

Js.kh esa gS rc 21

n1

aa

aa

+

32

n1

aa

aa

+

.......... + n1�n

n1

aa

aa

cjkcj gS -

(1) n + 1

(2) n � 1

(3) n


62. log(0.3)(x � 1) < log(0.09)(x � 1) gks x vUrjky esa

fLFkr gS -

(1) (2, )

(2) (�2, �1) (2, )

(3) (1, 2)

(4) (�, 1) (2, 8)

63. ;fn c�b2log

= a�c3log

= b�a5log

rc 2a 3b 5c cjkcj

gS -

(1) 1

(2) 10

(3) 15


64. |x2 + 3x| + x2 � 2 0 ds fy, x dk eku gS -

(1) R � [0, 1)

(2) R

(3) R �

21

,32

�

(4) R � (� 2 , 2 )

65. ;fn P(x) = ax2 + bx + c vkSj Q(x) = � ax2 + dx + c

tgk¡ ac 0 rc P(x).Q(x) = 0 de ls de -

(1) nks okLrfod ewy j[krh gS

(2) pkj dkYifud ewy j[krk gS

(3) pkj okLrfod ewy j[krk gS

(4) dqN dgk ugh tk ldrk





MEBPT2160815C0-14


66. ;fn lehdj.k

(a2 + b2)x2 � 2(ac + bd)x + c2 + d2 = 0 ls ewy

cjkcj gS rc -

(1) ad + bc = 0

(2) ab = dc

(3) ac = bd

(4) ad = bc

67. ;fn a, b, c, d vkSj x fofHkUu okLrfod la[;k,¡ bl

izdkj gS fd (a2 + b2 + c2)x2 � 2(ab + bc + cd)x

+ b2 + c2 + d2 0 rc a, b, c, d -

(1) lekUrj Js.kh es

(2) xq.kksÙkj Js.kh esa

(3) gjkRed Js.kh esa

(4) ab = cd dks larq"B djrk gSA

68. 1x2�x2)32( + 1�x2�x2

)3�2( =4

2 3 dk

gy gS -

(1) 1 ± 3 , 1

(2) 1 ± 2 , 1

(3) 1 ± 3 , 2

(4) 1 ± 2 , ± 1

69. a1, a2, a3 ....... a2k lekUrj Js.kh esa gS] rc

21a � 2

2a + 23a � 2

4a + ........ � 2k2a =

(1) 1�k2

k( 2

1a � 2k2a )

(2) 1�k

k2 ( 2

k2a �21a )

(3) 1k

k

( 21a + 2

k2a )

(4) buesa ls dksbZ ugh

70. ;fn lekUrj Js.kh ds m inksa vkSj n inksa ds

;ksxQyksa dk vuqikr m2 : n2 gS rc blds m osa

vkSj n osa inks dk vuqikr gksxk -

(1) (2m + 1) : (2n + 1)

(2) m : n

(3) (2m � 1) : (2n � 1)


71. ;fn nks fofHkUu /kukRed la[;kvksa ds lekUrj ek/;

vkSj xq.kksÙkj ek/; dk ;ksxQy la[;kvksa ds vUrj

ds cjkcj gS rc la[;kvksa dk vuqikr gS -

(1) 1 : 3

(2) 1 : 6

(3) 9 : 1

(4) 1 : 12





MEBPT2160815C0-15


72. ekuk fd lehdj.k ax2 + bx + c = 0 ds ewy x1

vkSj x2 vkSj lehdj.k px2 + qx + r = 0 ds ewy

x3, x4 gS rFkk x1, x2, x3, x4 lekUrj Js.kh esa gS rc

bl lekUrj Js.kh dk lkoZvUrj gS -

(1) 21

pq

�ab

(2) 31

pq

�ab

(3) 41

pq

�ab


73. lehdj.k (x2 � 12x + 35)(x2 + 10x + 24) = 504

ds /kukRed ewyksa dk ;ksxQy gS -

(1) 9

(2) 10

(3) 11

(4) 12

74. lehdj.k x x�1x5 (64) = 2000 dk gy gS

(tgk¡ x 2 ,oa x iw.kk±d gS)

(1) 2 ls foHkkftr

(2) 3 ls foHkkftr

(3) 5 ls foHkkftr

(4) 6 ls foHkkftr

75. ;fn x2 + ax + b = 0 rFkk x2 + bx + a = 0

dk ,d mHk;fu"B ewy gS] rks a + b dk eku gS&

(1) 1

(2) 0

(3) � 1


76. f}?kkr lehdj.k

2 2x 4x 3 x 6x 8 0, R ds

ewy gksxsa&

(1) lnSo okLrfod

(2) okLrfod dsoy tc /kukRed gSA

(3) okLrfod dsoy tc _ .kkRed gSA

(4) lnSo dkYifud

77. vlfedk (x + 1)2 > (x + 3) dk gy leqPp; gS&

(1) {x : �3 < x < �1}

(2) {x : x > �1}

(3) {x : �3 x �2}

(4) {x : x > 1 ;k x < �2}





MEBPT2160815C0-16


78. lehdj.k |x + 1| + |x � 1| = k ds vuUr gy gksus

ds fy, k ds ekuksa dh la[;k gS&

(1) 2 ls vf/kd

(2) 2

(3) 2 ls de


79. ;fn |x| � |x �2| = 2, rc bl lehdj.k ds] 10 ls

NksVs vHkkT; gyksa dh la[;k gS&

(1) 4

(2) 3

(3) 5


80. lehdj.k 2(x �3x 2)| x � 2 |

= 1 ,oa vlfedk

x(x � 2) 0 dk mHk;fu"B gy gS&

(1) 1

(2) 2

(3) 3

(4) 1, 2

81. vlfedk 2(x � 2)(�x 1)(x 1)

(x 1)

0 dks larq"V

djus okys /kukRed iw.kk±dksa dh la[;k gS&

(1) nks

(2) ,d

(3) rhu

(4) vuUr

82. lehd j.k ||x � 1| � 3| = 2 d s lHkh

laHkkfor gy ksa d k ;ksxQ y gS&

(1) 0

(2) 2

(3) 4

(4) 6

83. ;fn {x} = 2.1 gS rks x ds lEi.wkZ ekuksa dk leqPp;

gS (tgk¡ {.} fHkUukRed HkkxQyu dks O;Dr djrk

gSA)

(1)

(2) 0.1

(3) 2

(4) 0.9





MEBPT2160815C0-17


84. 3 3 3(171) � (123) � (48)

9 19 123 16 3 =

(1) 8

(2) 2

(3) �3

(4) 3

85. ;fn Hn = 1 + 21

+ 31

+ ........+ n1

gks] rks

1 +23

+ 35

+ ........+ n

1�n2dk eku gS &

(1) 2n � Hn

(2) 2n + Hn

(3) Hn � 2n

(4) Hn + n

86. ;fn /kukRed la[;k,sa a, b, c, d gjkRed Js<+h esa

gks] rks &

(1) ab cd

(2) ac bd

(3) ad bc


87. 91/3 . 91/9. 91/27 ......rd] dk eku gS &

(1) 1

(2) 3

(3) 9


[k.M- II

dkj.k çdkj

bl [k.M esa 3 dkj.k çdkj ds ç'u gSA çR;sd ç'u ds 4

fodYi (1), (2), (3),rFkk (4) gS, ftuesa ls flQZ ,d lgh gSA

88. ekuk fd a, b, c, d, e /kukRed la[;k,¡ gSA

oDrO;-1 : ;fn a, b, c, d, e lekUrj Js.kh esa gS rc

bcde, acde, abde, abcd vkSj abce gjkRed Js.kh

esa gSA

oDrO;-2 : ;fn a

a�cb ,

bb�ac

,

cc�ba lekUrj Js.kh esa gS rc a, b, c gjkRed

Js.kh esa gSA

(1) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2,

oDrO;-1 dk lgh Li"Vhdj.k gSA


oDrO;-1 dk lgh Li"Vhdj.k ugha gSA

(3) oDrO;-1 lR; gS] oDrO;-2 vlR; gSA

(4) oDrO;-1 vlR; gS] oDrO;-2 lR; gSA





MEBPT2160815C0-18


89. oDrO; -1 :

lehdj.k x2(a) + x (a2 � 3a + 1) � a2 + 2a � 1,

'a' esa ,d loZ lfedk gS ;fn x = �1 gSA

oDrO; -2 : ;fn fdlh lehdj.k ds lHkh xq.kkad

'kwU; gksa rks lehdj.k loZlfedk dgykrh gSA







90. oDrO;-1 : 12| 1� 3 | � 4 2 3 ,d

vifjes; la[;k gS&

oDrO;-2 : |x| = x x 0

�x x 0











SOLEBPT2160815-1

PART TEST-2 (PT-2)

TARGET : JEE (MAIN)-2017

HINTS & SOLUTIONS ¼ladsr ,oa gy½

PART-A PHYSICS

1. A car travels from A ........................

,d dkj A ls B rd ........................ Sol. Suppose ekuk AB = x km Average speed

= Total distance cov ered

Total time taken

vkSlr pky = r; d h xb Zd qy nwjh

fy;k x;k dqy le;

= 2x

x x20 30

= 2

1 120 30

= 20 6020 30

= 24 km/h = 24 kmh�1

2. A stone is thrown ........................ fdlh iRFkj dks feukj ........................ Sol. I mothod � Let downward direction is taken

as +ve. initial vel is �ve = � u (say) From the equation; v2 � u2 = 2as we get

(3u)2 � (�u)2 = 2hg izFke fof/k � ekuk uhps dh vksj nwjh;ksa dks /kukRed

fy;k tkrk gS izkjfEHkd osx _ .kkRed gS = �u (sayekuk)

lehdj.k ls ; v2 � u2 = 2as ge izkIr djrs gS

(3u)2 � (�u)2 = 2hg

h = 24u

g

The stone is thrown vertically upward with an

initial velocity u from the top of a tower it

reaches the highest point and returns back

and reaches the top of tower with the same

velocity u vertically downward.

Now, from the equation, V2 = u2 + 2gh

feukj ds 'kh"kZ ls ,d iRFkj m/oZ Å ij dh vksj

izkjfEHkd osx u ls Qsadk x;k gS ;g mPpre fcUnq rd igq¡prk gS vkSj okil ykSVrk gS o feukj ds 'kh"kZ ij m/oZ uhps dh vksj leku pky ls igq¡prk gS

lehdj.k ls V2 = u2 + 2gh (3u)2 = u2 + 2 gh 2gh = 9u2 � u2

h = 28u

2g h =

24ug

3. A body starts from ........................ ,d oLrq fojkekoLFkk ls ........................ Sol. u = 0, Let acceleration ekuk Roj.k = a

Total time dqy le; t = 30 s X1 = distance travelled in the first 10 s.

X1 = igys 10 s esa r; dh xbZ nwjh

Using , S = ut + 12

at2, we get

S = ut + 12

at2 dk mi;ksx djds ge izkIr

djrs gSA

X1 = 0 + 12

a (10)2 , i.e., X1 = 50 a

Similarly, blh izdkj X2 = distance travelled in the next 10 s

X2 = vxys 10 s esa r; dh xbZ nwjh

So blfy,, X2 = (0 + 10a ) 10 + 12

a (10)2

So blfy,, X2 = 100 a + 50 a

or ;k, X2 = 150 a

and, X3 = distance travelled in the last 10 s

o, X3 = vfUre 10 s esa r; dh xbZ nwjh

Soblfy,, X3 = (10 a + 10 a) 10 + 12

a (10)2

or ;k, X3 = 200a + 50a

or ;k, X3 = 250a

Hence vr%, X1 : X2 : X3 = 50a : 150 a : 250a

= 1 : 3 : 5

DATE : 16-08-2015 | CLASS-XI





SOLEBPT2160815-2

4. Two balls of equal ........................ leku nzO;eku dh nks xsanksa ........................ Sol.

f}rh; xsan

t=3S

u = 40 m/s , g = 10 m/s2 Let t be time taken by the first ball to reach

the highest point. ekuk izFke xsan }kjk mPpre fcUnq rd igq¡pus esa

fy;k x;k le; t gS V = u � gt ; 0 = 40 � 10 t ; t = 4 s From figure second ball will collide with first

ball after 3 second, there fore the height of collision point

= height gained by the second ball in 3 sec vr% Vdjkus okys fcUnq dh Å ¡pkbZ = f}rh; xsan }kjk 3 sec esa izkIr Å ¡pkbZ

= 40 (3) � 12

(10) (3)2 = 120 � 45 = 75 m 5. A balloon is moving ........................ ,d xqCckjk 10 eh0@lS0 ........................ Sol.

A

B

10 m/s

t = 11 secH

As pwafd s = ut + 12

at2

� H = 10 × 11 � 5 × (11)2 � H = 110 � 605 H = 495 m Aliter : oSdfYid fof/k

10ms�1

10ms�1

u=�10ms�1

t=11s

At the time of release, velocity of stone will

be same as that of balloon, hence eqDr djus ds le; iRFkj dk osx xqCckjs ds osx ds

rqY; gksxk vr% u = � 10 ms�1 , t = 11 s xqCckjs dh Å ¡pkbZ gksxh

h = ut + 12

gt2

= (� 10)×11 + 12

(10)(11)2 =�110 + 605 = 495 m

6. The acceleration time ........................ lh/kh js[kk esa fLFkjkoLFkk ........................ Sol.

aAvg vkSlr Roj.k = vt

= 020

= 0

From 0 to 20 time interval velocity of particle doesn't change it's direction.

Area under v�t curve is not zero. As the magnitude of area under v � t graph

from t = 0 to 10 is same as from t = 10 to 20, hence the average speed in both the intervals will be same.

0 ls 20 lsd.M vUrjky esa d.k dk osx viuh fn'kk ifjofrZr ugh djrkA

v�t vkjs[k ds vUrxZr {ks=kQy 'kwU; ugh gSA t = 0 ls 10 lsd.M rFkk t = 10 ls 20 lsd.M ds

fy, v � t vkjs[k ds vUrxZr {ks=kQy leku gSA vr% bu nksuks le; vUrjkyksa esa vkSlr pky leku gksxhA

7. Mark the incorrect statement ........................ ,d d.k lh/kh js[kk esa xfr dj ........................

Sol. If the velocity (u) and acceleration (a) have

opposite directions, then velocity (v) will decrease, therefore the object is slowing down.

If the position (x) and velocity (u) have opposite sign the position (x) reduces to become zero. hence the particle is moving towards the origin.

If a v 0

speed will increase. If velocity V = 0 , t1 < t < t2





SOLEBPT2160815-3

Hence; acceleration a = Vt

= 0 ;

t1 < t < t2 Therefore if the velocity is zero for a time

interval, the acceleration is zero at any instant within the time interval.

[acc, a = dvdt

v = u + at ]

Now , v = 0 a = 0 a = � u/t acceleration may not be zero when vel. 'V' = 0, 'c' is incorrect.

;fn osx a v 0

(u) rFkk Roj.k (a) foifjr fn'kkvksa esa gS rks vfUre osx (v) ?kVsxk vr% d.k /khek gks tk;sxkA

;fn fLFkfr (x) rFkk osx (v) foifjr fn'kkvksa esa gS rks fLFkfr ?kVdj 'kwU; gks tk;sxh vr% d.k ewy fcUnq dh vksj xfr dj jgk gSA

;fn rks pky c<sxh . ;fn osx V = 0 , t1 < t < t2

vr% Roj.k a = = 0 ; t1 < t t2

blfy, ;fn fdlh le; vUrjky esa osx 'kwU; gS rks ml le; vUrjky esa fdlh Hkh {k.k ij Roj.k 'kwU; gSA

[ a = dvdt

v = u + at ]

vc , v = 0 a = 0 a = � u/t Roj.k 'kwU; ugh Hkh gks ldrk gSA ;fn osx 'V' = 0, 'c'

xyr gSA 8. A particle is initially at ........................ fojkekoLFkk ls çkjEHk gqvk ........................

Sol. Area under acceleration-time graph gives the

change in velocity.

Roj.k≤ vkjs[k dk {ks=kQy] osx esa ifjorZu

nsrk gSA

Hence, vr% vmax = 12

× 10 × 11 = 55 m/s

Therefore, the correct option is (3)

vr% (3) lgh gSA

9. The coordinates of a ........................ xfr'khy d.k ds fdlh ........................ Sol. x = t3, y = t3

Vx = dxdt

= 3t2

Vy = dydt

= 3t2

Resultant velocity ifj.kkeh osx

V = 2 2x yv v

= 2 4 2 49 t 9 t

= 3t2 2 2a 10. A particle is moving........................ ,d d.k 5 ms�1 pky ........................ Sol. 1

�v �5i

2�v 5 j

2� �v 5 j 5i

v

= 5 2

a = v 5 2 1t 10 2

ms�2

W E

vN

1v 1v

2v

S

For direction, fn'kk ds fy,

tan = �55

= � 1

Average acceleration is 1

2 ms�2

towards north-west.

vkSlr Roj.k 1

2 ms�2 mÙkj if'pe dh

vksj gSA 11. During projectile motion........................ iz{ksI; xfr esa iFk ........................ Sol. Gravitational acceleration is constant near

the surface of the earth. iFoh dh lrg ds ikl xq:Roh; Roj.k fu;r jgrk

gSA





SOLEBPT2160815-4

12. The velocity of projection........................ ,d iz{ksI; dk iz{ksi.k........................ Sol. xu

= 6 �l + 8 �j

x�u 6l

uy = 8 �j

R = x y2u u

g =

2 6 810

= 9.6

13. A body is projected ........................ ,d oLrq ,d Å ¡ph ........................

Sol. tan45° = y

x

v

v

vy = vx = 18m/s Ans. 14. A ball is horizontally ........................ ,d xsn dks 45º dks.k........................

Sol. R = 2

2

vsin(2 ) � sin

gcos

Putting ;gka = 45º & = � 3

4 4

R =

2

2

v 3sin 2 � sin

4 4 41g

2

sx = 2 2v 2 1 v

�2 � 2 2g g2

(�ve sign indicates that the displacement is in �ve x direction)

(�ve fpUg ;g n'kkZrk gS fd foLFkkiu _ .kkRed x

fn'kk esa gksxk )

Range ijkl = 2v

2 2g

Ans "(4)"

Alternate II method

�4

& 4

R = 2

2

usin (2 ) � sin

g cos

= 2

2

usin � sin �

4 41g

2

R =22 2 u

g (along +ve x die.)

(+ve x fn'kk esa ) III Method

ux = u cos , T = 2 usingcos

, ax = g sin

; ay = � g cos

sx = ux t +12

axt2

= (u cos ) 2

2 u sin 1 2 u sin(gsin )

g cos 2 gcos

Let ;fn = = 45º So blfy,, sx

=2 2

2

u 2 1 1 1 2.2 ug

g 2 g2 2

= 22 u

[1 1]g2

sx = 2u

2 2g

Ans "(4)"

15. On an inclined plane ........................ 30º mUu;u dks.k okys ........................ Sol. u = 10m/s

Time of flight on the incline plane ur ry ij mM~M;u dky

60o

30o

u

T= 2u sing cos

given fn;k gS =30o & =30o & u =

10 3 m/s

T =o

o

2 10 3 sin30

10 cos 30

so vr% T= 2 sec .





SOLEBPT2160815-5

16. A particle moves in ........................ x-y ry esa xfr'khy d.k........................ Sol. ax = 2 m/s2 ; ay = 0 ux = 8 m/s uy = � 15 m/s.

V

= Vx�i + Vy

�j

Vy = uy + ay t Vy = � 15 m/s Vx = ux + ax t Vx = 8 + 2 t

V = [(8 + 2 t) �i � 15 �j ] m/s. Ans.

17. A stone projected at ........................ ,d iRFkj tehu ls 60º ........................ Sol. Let initial and final speeds of stone be u and v. ekuk iRFkj dh izkjfEHkd o vfUre pky u rFkk v gSA v2 = u2 � 2gh .........(1) and vkSj v cos 30° = u cos 60° ..........(2) solving 1 and 2 we get (1) o (2) dks gy djus ij u = 3gh

18. A particle is projected at 60º ........................ ,d d.k dks {kSfrt ls ........................

Sol. 2H

1mv

2 = K, 21

m(vcos60)2

= 21 v

m2 4

= 21 1mv

4 2

= K4

19. Shown in the figure are ........................ ?kj ls Ldwy tkrs gq, ........................ Sol. 1 = slope of C1 line = constant 2 = slope of C2 line = constant 1 � 2 0 but constant

1 = C1 js[kk dh <ky = fu;rkad 2 = C2 js[kk dh <ky = fu;rkad

1 � 2 0 ysfdu fu;rkad 20. A boat which can move ........................ ikuh ds lkis{k 5 m/s dh ........................

Sol. Vb = 2 25 4 = 3 m/s

t = 4803

= 160 s

21. A man walks in rain with ........................ ,d vkneh 5 kmh�1 osx ls ........................

Sol. r yV j

m�5i

r m y� �V � V (�5) i j

tan = 1 = y

5

so y = 5 km/hr

22. Two particles A and ........................ nks d.k A rFkk B Øe'k% ........................ Sol.

tan = 1

2

v

v

rmin = d sin

= 1

2 21 2

vd .

v v

23. A body is thrown up ........................ ,d oLrq dks Å ij dh ........................

Sol. We have, ge tkurs gSa] Srel = urelt + 12

arel t2

0 = ut � 12

(a + g) t2

a = 2ut� g =

2u gtt

24. Which figure represents........................ fp=k esa iznf'kZr m nzO;eku ........................ Sol. Force exerted by string is always along the

string and of pull type. When there is a contact between a point

and a surface the normal reaction is perpendicular to the surface and of push type.

Mksjh }kjk vkjksfir fd;k x;k cy ges'kk Mksjh ds vuqfn'k rFkk f[kapko ds :i esa gksrk gSA tc ,d fcUnq rFkk ,d lrg ds lEidZ gksrk gS rks vfHkyEc izfrfØ;k lrg ds yEcor~ rFkk ncko ds :i esa gksrk gSA





SOLEBPT2160815-6

25. Two forces of 6N and........................ fp=kkuqlkj ?k"kZ.k jfgr........................

Sol.

Both blocks are constrained to move with same acceleration.

nksuks CYkkWd leku Roj.k ls xfr djus ds fy, cfU/kr gSA

6 � N = 2a [Newtons II law for 2 kg block] [2 kg CykWd ds fy, U;wVu dk f}rh; fu;e] N � 3 = 1a [Newtons II law for 1 kg block]

[1 kg CYkkWd ds fy, U;wVu dk f}rh; fu;e] N = 4 Newton 26. A mass M is suspended ........................ fp=kkuqlkj ,d M nzO;eku ........................ Sol.

Point A is mass less so net force on it most

be zero otherwise it will have acceleration. fcUnq A nzO;ekujfgr gS blfy;s bl ij dqy cy

'kwU; gksxk vU;Fkk bldk Roj.k gksxkA F � Tsin = 0 [Equilibrium of A in

horizontal direction] [{kSfrt fn'kk esa A dh lkE;koLFkk]

T = F

sin

27. A body of mass 8 kg ........................ ,d 8 kg nzO;eku dh ........................ Sol.

T2 � 8g = 8a [Newton�s II law for 8 kg

block] [8 kg CykWd ds fy, U;wVu dk f}rh; fu;e] T2 = 8 × 2.2 + 8 × 9.8 = 96 N T1 � 12 g � T2 = 12 a [Newton�s II law for 12 kg

block] [12 kg CykWd ds fy, U;wVu dk f}rh; fu;e] T1 = 12 × 2.2 + 12 × 9.8 + 96 T1 = 240 N 28. A block is dragged on ........................ ,d CykWd dks fpdus ry ........................ Sol.

The length of string AB is constant. Mksjh AB dh yEckbZ fu;r gSA speed A and B along the string are same

u sin = V Mksjh ds vuqfn'k A rFkk B dh pky leku gS u

sin = V

u sin = V u = V

sin

29. STATEMENT-1 : A particle having ............... oDrO; 1 % ,d d.k ftldk ........................ Sol. Assertion is false , because direction of

velocity is not specified .It is not necessary that when acceleration is negative positive slow down .As when both velocity and acceleration are in opposite directions, then particle will increase its speed.

dFku vlR; gS D;ksfd osx dh fn'kk ugh crkbZ xbZ gSA ;g t:jh ugh gS fd tc Roj.k _ .kkRed gks rks d.k /khek gksA D;ksfd tc osx o Roj.k nksuks foijhr fn'kk es gksa rc d.k dh pky c<sxhA

30. STATEMENT-1 : Two stones........................ dFku-1 : nks iRFkj tehu ........................ Sol. Both the stones cannot meet (collide)

because their horizontal component of velocities are different. Hence statement I is false.

nksuks iRFkj vkil esa Vdjk ugha ldrs gS D;ksafd muds osxksa ds {kSfrt ?kVd vyx&vyx gSA vr% dFku-I vlR; gSA





SOLEBPT2160815-7

PART- B

CHEMISTRY

31. The ratio of the difference in energy ..................

fdlh H ds leku Lih'kht esa çFke ------------------------

Sol. Use : E1 � E2 / E2 � E3

32. If the de-Broglie wavelength ......... ... .. .

;fn H-ijek.kq dh 2nd d{kk esa ifjØe.k -------------------

Sol. 2r = n

Here (;gk¡), n = 2. 2r = 2x r = x

.

33. The uncertainty in position and ................

;fn fdlh ,d d.k dh fLFkfr rFkk osx ------------------

Sol. x × (mv) = h4

= 0.527 × 10�34

0.5 × 10�10 × m × 5.27 × 10�24 = 0.527 × 10�34

m = 0.2 Kg.

34. The wavenumber of the spectral line ............

He+ vk;u dh ckej Js.kh esa izkIr U;wure -----------------

Sol. Z = 2 n1 = 2 n2 =

= R (2)2 2 2

1 1

2

= R

35. The value of azimuthal quantum ..............

DokaVe la[;k n = 4, m = �3 okys ,d d{kd --------------

Sol. n = 4, = 0, 1, 2, 3

For m = �3, the only possible value of = 3

gy . n = 4, = 0, 1, 2, 3

m = �3 ds fy, dk laHkkfor eku = 3 gSA

36. In a sample of H-atom electrons ...............

H-ijek.kq ds uewus esa bysDVªkWu 5th mÙksftr ----------------

Sol. 65432n = 1

maximum number of lines produced = 6 (6 1)

2

= 15

out of these 15 lines,

5 lines belong to ultra violet region and 4 lines

are in visible region and rest are in infrared

region

Sol. 65432n = 1

mRikfnr js[kkvksa dh vf/kdre la[;k=6 (6 1)

2

= 15

15 js[kkvksa esa ls]

5 js[kk,¡ ijkcSaxuh {ks=k ls] 4 js[kk,¡ n'; {ks=k ls rFkk

'ks"k js[kk,¡ vojDr Js.kh ls gSaA

37. Which of the following set of quantum ...............

4f d{kd esa ,d by sDVªkWu ds fy, DokaVe --------------------

Sol. For 4 orbital electrons, n = 4

= 3 (because ) m = + 3, + 2, + 1, 0, �1, �2,

�3 s = + 1/2.

gy- 4 d{kh; bysDVªkWu ds fy;s] n = 4

= 3 (D;ksafd ) m = + 3, + 2, + 1, 0, �1, �2,

�3 s = + 1/2.

39. One atom of an element x weigh ..................

rRo x ds ,d ijek.kq dk Hkkj ...................

Sol. Atomic weight of an element

x = 6.643 × 10�23 × NA = 40

Number of moles of x = 20 1000

40

= 500

gy- ,d rRo dk ijek.kq Hkkj

x = 6.643 × 10�23 × NA = 40

x ds eksyksa dh la[;k = 20 1000

40

= 500

40. For reaction P + 2Q 3R, if reaction .................

vfHkfØ;k P + 2Q 3R, ds fy, ...............

Sol. P + 2Q 3R 0.1/2 = Rmole / 3

R = 0.15 mole

41. 2K + 2 + 22 HNO3 2HO3 + 2KO3 +

22NO2 + 10H2O

If 3 mole of K & 2 moles 2 ..................

;fn K ds 3 eksy rFkk 2 ds 2 eksy HNO3 --------------





SOLEBPT2160815-8

Sol. KI is limiting reagent

3 mole of KI will give 33 mole of NO2

according to stoichiometry.

Volume of NO2 at STP = 33 22.04

= 739.2 Lt

Sol. KI lhekUr vfHkdeZd gSaA

vfHkfØ;k dh jllehdj.kferh lsµ 3 eksy KI, }kjk

33 eksy NO2 nsrs gSA

Volume of NO2 at STP = 33 22.04

= 739.2 Lt

42. A solution containing 0.1 mol ................

/kkrq DyksjkbM MClx ds 0.1 eksy ;qDr -----------------

Sol. MClx + AgNO3 AgCl + MNO3

POAC on Ag

500

1000 × 0.8 = 1 × mole of AgCl

mole of AgCl = 0.4 ...........(1)

POAC on Cl

0.1 × x = 1 × mole of AgCl = 0.4........(1)

mole of AgCl = 0.1 x ....................(2)

put eq (2) in eq (1)

0.1 x = 0.4

x = 4

gy % MClx + AgNO3 AgCl + MNO3

Ag ij POAC

500

1000 × 0.8 = 1 × AgCl ds eksy

AgCl ds eksy = 0.4 ...........(1)

Cl ij POAC

0.1 × x = 1 × AgCl ds eksy = 0.4.......(1)

AgCl ds eksy = 0.1 x ....................(2)

leh- (1) esa (2) dks j[kus ij

0.1 x = 0.4

x = 4

43. The atomic mass of an element is 27 ...........

,d rRo dk ijek.kq Hkkj 27 -------------------

Sol. It may be AlCl3

V.D = 3AlClM.Wt

2

V.D = 133.5

2 = 66.75

gy % ;g AlCl3 gks ldrk gS\

ok-?k- = 3AlClM.Wt

2

ok- ?k- = 133.5

2 = 66.75

44. A hydrocarbon contains 80% .............

;fn ,d gkbMªksdkcZu esa 80% --------------------- Sol. ratiosimplest ratio

C 80 /12 6.6680 6.66 / 6.66 1 1

3H 20 / 6.66 320 20 /1 20

Hence hydrocarbon = n × CH3 (n = 1, 2, 3.....)

i.e. CH3, C2H6.................

gy %

C 80 /12 6.6680 16.66 / 6.66 1 1

3H 20 / 6.66 320 20 /1 20

lk/kkj.kre vuqikr vuqikr

vr % gkbMªkssdkcZu = n × CH3 (n = 1, 2, 3........)

i.e. CH3, C2H6.................

45. 20 gm. CaCO3 on decomposition ...............

20 xzke CaCO3 dks fo?kfVr -----------------

Sol. The decomposition of CaCO3 takes place as

below

3 256 gm 22.4 lit100 gm

CaCO CaO CO

100 gm CaCO3 at STP produce = 22.4 lit CO

2

20 gm CaCO3 at STP produce =

22.420

100

= 4.48 lit CO2

gy- 3 256 gm 22.4 lit100 gm

CaCO CaO CO

100 xzke CaCO3 nsrk gS 22.4 yhVj CO

2

20 xzke CaCO3 nsxk CO

2 =

22.420

100

= 4.48 yhVj

46. How many alcohols give immediate ............

(C5H

12O) v.kqlw=k okys fdrus ,YdksgkWy -------------

Sol.





SOLEBPT2160815-9

47. Which of the following compound ...............

fuEu esa ls dkSulk ;kSfxd VkWysu ---------------------

Sol. Aldehyde and Ketones having �C�CH3

O

group always gives +ve tollen's test.

,YMhgkbM rFkk dhVksu tks fd �C�CH3

O

lewg j[krs gS lnSo VkWysu ifj{k.k +ve nsrs gSA

48. Which is incorrect match with respect .............

iz;ksx'kkyk ijh{k.k ds fy, fuEu esa ls dkSu ----------

Sol. It is fact.

rF; gSA

49. How many structural isomers of C5H

10 ...........

C5H

10 ds fdrus lajpukRed leko;oh ----------------

Sol. C5H10

D.U. = 1

Unsaturated compound which having >C=C<

and �CC� always shown bromine water test

positive.

vlarIr ;kSfxd tks fd >C=C< rFkk �CC� j[krs

gS lnSo czksehu ty ifj{k.k /kukRed nsrs gSA

C�C�C�C=C

C�C�C=C�C

C=C�C�C

C

C�C=C�C

C

C�C�C=C

C 50. How many structural isomeric ketones ...............

(C5H

10O) v.kqlw=k okys fdrus lajpukRed --------------

Sol. ,

51. On oxidative ozonolysis of 3-Methylhex .............

3-esfFkygsDl -3-bZu ds vkWDlhdkjh -------------------

Sol.

CH3�CH2�C=CH�CH2�CH3

CH3

(i) O H O3 2�H O2 2

CH3�CH2�C=O

CH3

+ O=CH�CH2�CH3

CH3�CH2�C=O

CH3

+H2O2

+ CH3�CH2�COOH

52. How many monochloro structure ............

tc 3-esfFkyisUVsu dh vfHkfØ;k lw;Z ------------------

Sol.

53. Which of the following hydrocarbon ................

dkSulk gkbMªksdkcZu ftad dh mifLFkfr ---------------------

Sol. (1) CH3�CH=C=CH�CH3 (i) O H O3 2

(ii) Zn

CH3�CH=O + CO2

Acetaldehyde

,flVsfYMgkbM

(2) CH3�CH=CH�CH=C(CH3)2 (i) O H O3 2

(ii) Zn

CH3CH=O + CHO

CHO + (CH3)2C=O

Acetaldehyde glyoxal acetone

,flVsfYMgkbM XykbvkWDtsy ,flVksu

(3) (CH3)2C=C=CH2 (i) O H O3 2

(ii) Zn

(CH3)2C=O +

CO2 + H�CHO

Acetone Formaldehyd

,flVksu QkeZsfYMgkbM

(4) CH3�C=C=C�CH3

CH3 CH3

(i) O H O3 2(ii) Zn

CH3�C=O + CO2

CH3 Acetone

,flVksu

54. Which is incorrect order ..............





SOLEBPT2160815-10

fuEu esa ls dkSulk ijekf.od f=kT;kvksa --------------------

Sol. Refer notes At. radii = B < Al ~� Ga < In

uksV~l ns[ksaA ijekf.od f=kT;k = B < Al ~� Ga < In

55. Which statement is correct for the ..............

f}rh; vkorZ rRoksa (Be, B, C, N, O, F) --------------

Sol. First ionization energy of Be(2s2) in higher than

Boron (3s22p1) but lower than carbon (2s22p2)

due to full field electronic configuration of

Be(2s2).

Be(2s2) izFke vk;uu Å tkZ B(3s22p1) ls vf/kd

gksrh gS ysfdu dkcZu (2s22p2) dh rqyuk esa de

gksrh gSA

D;ksafd Be(2s2) iw.kZiwfjr bysDVªksfud foU;kl

j[krk gSA

56. The set representing the correct.................

fuEu lewg esa izFke vk;uu foHko ---------------------

Sol. Down the group the effective nuclear charge

remains almost constant. But down the group

with increasing atomic number, the number of

shells increase and thereby atomic size

increases. As a result, the distance of valence

shell electron from nucleus increases, attraction

between them decreases and therefore

ionization energy decreases.

gy- oxZ esa uhps tkus ij izHkkoh ukfHkdh; vkos'k yxHkx

fLFkj jgrk gS] ijUrq oxZ esa uhps tkus ij ijek.kq

Øekad c<+us ij dks'kksa dh la[;k esa of) gksrh gS

vkSj ijek.kq dk vkdkj c<+rk gSA bl dkj.k]

ukfHkd ls la;ksth dks'k bysDVªkWu dh nwjh c<+rh gS

vkSj muds chp vkd"kZ.k ?kVrk gS] ftlds

ifj.kkeLo:i vk;uu Å tkZ ?kVrh gSA

57. Element with atomic number 38 ..............

vkorZ lkj.kh esa ijek.kq Øekad 38 ----------------------

Sol. Refer Periodic table.

vkoZr lkj.kh esa ns[ksaA

58. Which gas is released ................

nh xbZ vfHkfØ;k esa dkSulh xSl -------------------

Sol. SO3H NO2

OH

Na

SO3Na NO2

ONa

+ H2

59. Which is the position isomers ............

;kSfxd dk fLFkfr ----------------

Sol. (1) Identical

(2) Position

(3) Chain

(4) Homologous

(1) le:i

(2) fLFkrh

(3) Ja[kyk

(4) ltkrh;

60. Which is the chain isomers ..............

;kSfxd dk Ja[kyk --------------------

Sol. Only 1st is chain isomer while 2nd and 3rd are

positional isomers. 4th is functional isomer.

dsoy 1st Ja[kyk leko;oh gS tcfd 2nd rFkk 3rd

fLFkfr leko;oh gSA 4th fØ;kRed leko;oh gSA

PART- C





SOLEBPT2160815-11

MATHEMATICS 61. If a1, a2, .......... an are.................... ;fn a1, a2, .......... an ....................

Sol. 21

n1

aa

aa

+

32

n1

aa

aa

+...... +

n1�n

n1

aa

aa

= 1a + na

1�nn

1�nn

n1�n23

23

3212

12

21 a�a

a�a

aa

1.....

a�a

a�a

aa

1

a�a

a�a

aa

1

= 1a + nad

]a�a.......a�aa�a[ 1�nn2312

= 1 n n 1[ a a ] [ a � a ]

d

=

d

a�a 1n

[as. a2 � a1 = a3 � a2 = a4 � a3 = .....d]

= d

d)1�n( = n � 1

62. log(0.3)(x � 1) < log(0.09) ....................

log(0.3)(x � 1) < log(0.09)(x � 1) gks x .................

Sol. log0.3(x � 1) < 2)3.0(log (x �1)

log0.3(x � 1) < 21

log0.3(x � 1)

log0.3(x � 1)2 < log0.3(x �1) (x � 1)2 > (x � 1) x2 � 3x + 2 > 0 (x � 1)(x � 2) > 0 xt(�, 1) (2, ) but ysfdu x � 1 > 0 x > 1 so blfy, x (2, )

63. If c�b2log

= a�c3log

= b�a5log

....................

;fn c�b2log

= a�c3log

= b�a5log

rc ....................

Sol. Let ekuk c�b2log

= a�c3log

= b�a5log

= k

log 2 = k(b � c) log 3 = k(c � a) log 5 = k(a � b) let 2a 3b 5c = x taking log both sides nksuksa rjQ log ysus ij a log 2 + b log 3 + c log 5 = log x a k(b � c) + bk(c � a) + ck(a � b) = log x k[ab � ac + bc � ab + ac � bc] = logx logx = 0 x = 1 2a 3b 5c = 1

64. The value of x, |x2 + 3x| ....................

x, |x2 + 3x| + x2 � 2 0 ds ....................

Sol. case fLFkfr- CasefLFkfr-

x2 + 3x 0 x2 + 3x 0

x (�, � 3] [0, ) x [�3, 0]

2x2 + 3x � 2 0 �x2 � 3x + x2 � 2 0

(2x � 1)(x + 2) 0 3x + 2 0

x (�, � 2]

,

21

x � 32

x (, �3]

,

21

x

32

�,3�

x

32

�,�

,

21

x R �

21

,32

�

65. If P(x) = ax2 + bx + c & ....................

;fn P(x) = ax2 + bx + c vkSj .................... Sol. P(x) = ax2 + bx + c D1 b2 � 4ac Q(x) = � ax2 + dx + c D2 d2 + 4ac D1 + D2 = b2 � 4ac + d2 + 4ac D1 + D2 = b2 + d2 0 D1 + D2 0 So at least one of D1 and D2 0 blfy, de ls de ,d D1 vkSj D2 0 So at least two real roots blfy, de ls de nks okLrfod ewy gSA 66. If roots of the equation.................... ;fn lehdj.k (a2 + b2)x2 ....................

Sol. D 4(ac + bd)2 � 4(a2 + b2)(c2 + d2) = 0 (ac + bd)2 = (a2 + b2)(c2 + d2) a2c2 + b2d2 + 2abcd = a2c2 + b2c2 + a2d2 + b2d2 b2c2 + a2d2 � 2abcd = 0 (bc � ad)2 = 0 bc = ad 67. If a, b, c, d and x are .................... ;fn a, b, c, d vkSj x fofHkUu .................... Sol. (a2 + b2 + c2)x2 � (ab + bc + cd)x + b2 + c2 + d2 0 (ax � b)2 + (bx � c)2 + (cx � d)2 0 ax � b = 0 bx � c = 0 cx � d = 0

x = ab

= bc

= cd

= k(common ratio lkoZvuqikr)

so a, b, c, d are in G.P.

blfy, a, b, c, d xq.kksÙkj Js.kh esa gSA

68. Solution of 1x2�x2)32( ....................





SOLEBPT2160815-12

1x2�x2)32( + 1�x2�x2

)3�2( ....................

Sol. 1x2�x2)32( + 1�x2�x2

)3�2( = 3�2

4

1x2�x2)32( (2 � 3 ) + x2�x2

)3�2( = 4

x2�x2)32( (2 + 3 )(2 � 3 ) + x2�x2

)3�2(

= 4

x2�x2)32( + x2�x2

)3�2( = 4

let ekuk x2�x2)32( = y then rc

x2�x2)3�2( =

y1

y + y1

= 4 y = 2 ± 3

y2 � 4y + 1 = 0

x2�x2)32( = 2 + 3 or 2 � 3

x2 � 2x = ± 1 x2 � 2x + 1 = 0 (x � 1)2 = 0 &

x2 � 2x � 1 = 0 x = 2

82 = 1 ± 2

so x = 1 ± 2 , 1

69. If a1, a2, a3 ....... a2k........................

a1, a2, a3 ....... a2k lekUrj Js.kh esa........... Sol. a2 = a1 + d a2k = a2k�1 + d S = 2

1a � 22a + 2

3a � 24a + .......... 2

k2a

= ( 21a � 2

2a ) + ( 23a � 2

4a ) + ........ ( 21�ka � 2

k2a ) S = � d(a1 + a2) + (a3 + a4) + ....... (a2k�1 + a2k) S = � d[a1 + a2 + a3 + a4 + .......... a2k]

S = � d

2k2

[a1 + a2k]

S = � k

1�k2a�a 1k2 [a1 + a2k] a2k = a1 + (2k � 1)d

S = 1�k2

k( 2

1a � 2k2a ) d =

)1�k2(

a�a k2

70. If ratio of sum of m .................... ;fn lekUrj Js.kh ds m ....................

Sol. Sm = 2m

[2a + (m �1)d]

Sn = 2n

[2a + (n � 1)d]

n

m

SS

= ]d)1�n(a2[

2n

]d)1�m(a2[2m

= 2

2

n

m

d)1�n(a2d)1�m(a2

=

nm

; d

21�n

a

d2

1�ma

= nm

d)1�n(ad)1�m(a

=

1�n21�m2

n

m

TT

= 1�n21�m2

71. If sum of the A.M. & G.M. .................... ;fn nks fofHkUu /kukRed la[;k,¡ ....................

Sol. 2

ba + ab = a � b

a + b + 2 ab = 2a � 2b (a � 3b)2 = 4ab a2 + 9b2 � 10ab = 0 (a � 9b)(a � b) = 0 a = 9b as a � b 0

ba

= 19

72. Suppose x1, x2 be the .................... ekuk fd lehdj.k ax2 + bx + c = 0............

Sol. x1 + x2 = � ab

x1 = a

x3 + x4 = � pq

x2 = a + d

x3 = a + 2d x4 = a + 3d (x3 + x4) � (x1 + x2)

(a + 3d + a + 2d) � (a + a + d) = ab�

pq

4d = ab

� pq

d = 41

pq

�ab

73. Sum of positive roots .................... lehdj.k (x2 � 12x + 35) .................... Sol. (x2 � 12x + 35)(x2 + 10x + 24) = 504 (x � 5)(x � 7) (x + 6)(x + 4) = 504 (x � 7)(x + 6)(x � 5)(x + 4) = 504 (x2 � x � 42)(x2 � x � 20) = 504 x2 � x = t (t � 42)(t � 20) = 504 x2 � x = 6 t2 � 62t + 840 � 504 = 0 x2 � x � 6 = 0 t2 � 62t + 336 = 0 (x � 3)(x + 2) = 0 t2 � 56t � 6t + 336 = 0 x = 3 & � 2 (t � 6)(t � 56) = 0 x2 � x = 56 t = 6 or 56 x2 � x � 56 = 0 (x � 8)(x + 7) = 0 x = � 78 x1 + x2 = 8 + 3 = 1





SOLEBPT2160815-13

74. Solution of the equation ....................

lehdj.k x x�1x5 (64) = 2000 ....................

Sol. x�1

x x5 .(64) = 53. 16

5x. 3(x�1)

x(4) = 53. 42

x = 3

75. If the equations ....................

;fn x2 + ax + b = 0 ....................

Sol. Let be a common root of

x2 + ax + b = 0 and x2 + bx + a = 0. Then,

ekuk x2 + ax + b = 0 vkSj x2 + bx + a = 0 dk

mHk;fu"B ewy gS] rks

2 + a + b = 0 andvkSj 2 + b + a = 0

= 1

Putting = 1, either of the two, we get

a + b = � 1.

= 1 j[kus ij, a + b = � 1.

76. Roots of the quadratic ....................

f}?kkr lehdj.k ....................

Sol. 2 2x 4x 3 x 6x 8 0 ,

2x 1 2x 2 3 3 8 0

Discriminant. foospd

2

D 4 2 3 4 1 3 8

2D 4 1

If R then D > 0 ;fn R rc D > 0

so root of given quadratic always real blfy, nh

xbZ f}?kkr lehdj.k ds ewy lnSo okLrfod gksxsaA

Answer A.

77. The solution set of ....................

vlfedk (x + 1)2 > (x + 3) ....................

Sol. (x + 1)2 > x + 3 or ;k x2 + x � 2 > 0

or ;k (x + 2)(x � 1) > 0

x < � 2 or ;k x > 1

78. The number of values....................

lehdj.k |x + 1| + |x � 1| = k ....................

Sol. |x + 1| + |x � 1| =

�2x , x �1

2 , �1 x 1

2x , x 1

for infinite solutionvuUr gy ds fy, k = 2

79. If |x| � |x�2| = 2 then ....................

;fn |x| � |x �2| = 2, rc 10 ....................

Sol. |x| = 2 + |x�2|

|a+b| = |a| + |b|

a.b 0

2(x � 2) 0 x 2

4 such solutions 4 gy gSA

80. Common solution of ....................

lehdj.k 2(x �3x 2)| x � 2 |

= 1 ,oa ....................

Sol. 2(x �3x 2)(x � 2)

= 1

CasefLFkfr-I x2 � 3x + 2 = 0 x = 1 or x = 2

(RejectedvLohdk;Z)

CasefLFkfr-II x � 2 = ±1 x = 2 ± 1

x = 3,1

andvkSj x(x � 2) 0 x [0, 2]

common solution mHk;fu"B gy x = 1





SOLEBPT2160815-14

81. Number of positive integers ....................

vlfedk 2(x � 2)(�x 1)(x 1)

(x 1)

....................

Sol. 2(x � 2)(�x 1)(x 1)

(x 1)

0

2(x � 2)(x �1)(x 1)

(x 1)

0

x (�, � 1) [1, 2]

82. Sum of the all possible....................

lehdj.k ||x � 1| � 3| = 2 ds ....................

Sol. |x � 1| � 3 = 2 or ;k |x � 1| � 3 = � 2

|x � 1| = 5 or ;k |x � 1| = 1

x � 1 = ± 5 or ;k x � 1 = ±1

x = 6, � 4 or ;k x = 2, 0

sum ;ksxQy = 4

83. If {x} = 2.1 then complete....................

;fn {x} = 2.1 gS rks x ds ....................

Sol. {x} = 2.1 not possible lEHko ugha

as pwfd 0 {x} < 1

so blfy, x

84. 3 3 3(171) � (123) � (48)

9 19 123 16 3 ....................

Sol. Let a = 171, b = �123, c = � 48

then a + b + c = 0

a3 + b3 + c3 = 3abc

3 3 3(171) � (123) � (48)

9 19 123 16 3

= 3 171 (�123)(�48)

9 19 123 16 3

= 3

Hindi. ekuk a = 171, b = �123, c = � 48

rc a + b + c = 0

a3 + b3 + c3 = 3abc

3 3 3(171) � (123) � (48)

9 19 123 16 3

= 3 171 (�123)(�48)

9 19 123 16 3

= 3

85. If Hn = 1 + 21

+ 31

+ ........+ ....................

;fn Hn = 1 + 21

+ 31

+ ........+ ....................

Sol. Hn = 1 + 12

+ 13

+ ...... + 1n

1 + 32

+ 53

+ ....... + 2n 1

n

Here ;gk¡ Tn = 2n 1

n

Tn = 2 � 1n

T1 = 2 � 11

; T2 = 2 � 12

T3 = 2 � 13

Tn = 2 � 1n

Sn = T1 + T2 + T3 + ....... Tn

Sn = 2n � 1 1 1

1 .....2 3 n

Sn = 2n � Hn

86. If positive numbers .................... ;fn /kukRed la[;k,sa .................... Sol. HM between a and c = b

and GM = ac Also HM between b and d = c

and GM = bd But GM HM

ac b and bd c

ac bd bc ad bc





SOLEBPT2160815-15

Hindi. a ,oa c ds e/; gjkRed ek/; = b

rFkk xq.kksÙkj ek/; = ac

blh izdkj b ,oa d dk gjkRed ek/; = c

,oa xq.kksÙkj ek/; = bd

ysfdu GM HM

ac b ,oa bd c

ac bd bc

ad bc

87. The value of 91/3 . 91/9. 91/27....................

91/3 . 91/9. 91/27 ......rd ....................

Sol. We have to find value of 91/3 . 91/9 . 91/27 .... = x

(say)

We can see that powers are in G.P. and |r| < 1

x =

13

11

39

= 91/2 = 3

Hindi. gesa 91/3 . 91/9 . 91/27 ......... = x (ekuk) dk eku

Kkr djuk gSA

?kkrs xq.kksÙkj Js<+h esa gS rFkk |r| < 1

x =

13

11

39

= 91/2 = 3

88. Let a, b, c, d, e ..................... Statement-1 : If a, b, c, d, e .................... ekuk fd a, b, c, d, e /kukRed .................... oDrO;-1 : ;fn a, b, c, d, e .................... Sol. a, b, c, d, e A.P.

a1

, b1

, c1

, d1

, e1

H.P.

bcde, acde, abde, abce, abcd multiplying by abcde ls xq.kk djus ij

bcde, acde, abde, abce, abcd are in H.P. gjkRed Js.kh esa gSA

Statement-1 is false dFku-1 vlR; gSA Statement-2 dFku-2

a

a�cb ,

bb�ac

, a b c

c

A.P.

a

a2�acb ,

bb2�bac

, a b c 2c

c

A.P.

a

acb � 2,

bbac � 2,

ccba � 2 A.P.

a

acb ,

bbac

, c

cba A.P.

a1

, b1

, c1

A.P. statement-2 is true dFku 2 lR; gSA

a, b , c H.P. 89. Statement 1 : Equation x2(a) + x .................... oDrO; -1 : lehdj.k x2(a) + x .................... Sol. Obvious Li"Vr%

90. Statement- 1 : 12| 1� 3 | � 4 2 3 .............

oDrO;-1 : 12| 1� 3 | � 4 2 3 ....................

Sol. 12| 1� 3 | � 4 2 3 = 3 � 1 � 3 1 = 2





SOLEBPT2160815-16

PART TEST-2 (PT-2)

TARGET : JEE (MAIN)-2017

ANSWER KEY

CODE-0

PHYSICS

1. (2) 2. (2) 3. (3) 4. (2) 5. (1) 6. (3) 7. (3)

8. (3) 9. (2) 10. (1) 11. (1) 12. (2) 13. (2) 14. (4)

15. (3) 16. (1) 17. (3) 18. (3) 19. (4) 20. (2) 21. (1)

22. (3) 23. (2) 24. (3) 25. (3) 26. (2) 27. (3) 28. (2)

29. (4) 30. (4)

CHEMISTRY

31. (4) 32. (1) 33. (2) 34. (1) 35. (4) 36. (3) 37. (3)

38. (2) 39. (4) 40. (4) 41. (1) 42. (3) 43. (1) 44. (3)

45. (1) 46. (1) 47. (2) 48. (3) 49. (3) 50. (2) 51. (3)

52. (2) 53. (4) 54. (2) 55. (4) 56. (2) 57. (1) 58. (1)

59. (2) 60. (1)

MATHEMATICS

61. (2) 62. (1) 63. (1) 64. (3) 65. (1) 66. (4) 67. (2)

68. (2) 69. (1) 70. (3) 71. (3) 72. (3) 73. (3) 74. (2)

75. (3) 76. (1) 77. (4) 78. (3) 79. (1) 80. (1) 81. (1)

82. (3) 83. (1) 84. (4) 85. (1) 86. (3) 87. (2) 88. (4)

89. (4) 90. (1)

DATE : 16-08-2015 | CLASS-XI |COURSE : ALL INDIA TEST SERIES (VIKALP)



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