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PERCT1090815C0-1

PART � A

SECTION - I Straight Objective Type

This section contains 20 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 prism having refractive index 2 and

refracting angle 30º, has one of the refracting

surfaces polished. A beam of light incident

on the other refracting surface will retrace its

path if the angle of incidence is:

(1) 0º (2) 30º

(3) 45º (4) 60º

2. The radius of curvature of a convex spherical

mirror is 1.2 m. How far away from the mirror

is an object of height 1.2 cm if the distance

between its virtual image and the mirror is

0.35 m? What is the height of the image?

[Apply formula for paraxial rays]

(1) 84 cm, 0.5 cm (2) 84 cm, 0.25 cm

(3) 84 cm, 1.2 cm (4) None of these

3. The cross section of a glass prism has the

form of an equilateral triangle. A ray is

incident onto one of the faces perpendicular

to it. The angle between the ray that leaves

the prism and the base of the prisme is, (the

refractive index of glass is µ = 1.5)

(1) 60º (2) 90º

(3) 30º (4) 45º

4. A concave mirror forms an image of the sun

at a distance of 6 cm from it.

(1) the radius of curvature of this mirror is 6 cm

(2) to use it as a shaving mirror, it can be held

at a distance of 8 - 10 cm from the face

(3) if an object is kept at a distance of 24 cm

from it, the image formed will be of the

same size as the object

(4) all the above alternatives are incorrect.

5. In the figure shown. A particle �P� moves with

velocity 10 m/s towards the intersection point

�O� of the plane mirror kept at right angle to

each other. 1 and 2 are the images formed

due to direct reflection from m1 and m2

respectively. In the position shown,find

difference in speed of image I1 and I2 .

(1) 20 m/s

(2) 10 m/s

(3) 0 m/s

(4) None of these

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PERCT1090815C0-2

6. Which of the following statements are

correct for spherical mirrors.

(1) a concave mirror forms only virtual

images for any position of real object

(2) a convex mirror forms only virtual images

for any position of a real object

(3) a concave mirror forms only a virtual

diminished image of an object placed

between its pole and the focus

(4) a convex mirror forms a virtual enlarged

image of an object if it lies between its

pole and the focus.

7. For the refraction of light through a prism

kept in air

(1) For every angle of deviation there are

two angles of incidence.

(2) The light travelling inside an isosceles

prism is necessarily parallel to the base

when prism is set for minimum deviation.

(3) There are two angles of incidence for

maximum deviation.

(4) Angle of minimum deviation will decrease

if refractive index of prism is increased

keeping the outside medium unchanged.

8. In the figure shown the radius of curvature of

the left & right surface of the concave lens

are 10 cm & 15 cm respectively. The radius

of curvature of the mirror is 15 cm.

(1) equivalent focal length of the combination

is �16 cm

(2) equivalent focal length of the combination

is +36 cm

(3) the system behaves like a concave

mirror

(4) the system behaves like a convex mirror.

9. A flat mirror M is arranged parallel to a wall

W at a distance L from it as shown in the

figure. The light produced by a point source

S kept on the wall is reflected by the mirror

and produces a light patch on the wall. The

mirror moves with velocity v towards the wall.

L S

wall w

V

M

(1) The patch of light will move with the

speed v on the wall.

(2) The patch of light will not move on the

wall.

(3) As the mirror comes closer the patch of

light will become larger and shift away

from the wall with speed larger than v.

(4) None of these







PERCT1090815C0-3

10. A concave mirror of focal length 2 cm is

placed on a glass slab as shown in the

figure. Then the image of object O formed

due to reflection at mirror and then refraction

by the slab:

(1) will be virtual and will be at 2 cm from the

pole of the concave mirror

(2) will be virtual and formed on the pole of

the mirror

(3) will be virtual and on the object itself

(4) none of these

11. A flint glass prism and a crown glass prism

are to be combined in such a way that the

deviation of the mean ray is zero. The

refractive index of flint and crown glasses for

the mean ray are 1.6 and 1.9 respectively. If

the refracting angle of the flint prism is 6°,

what would be the refracting angle of crown

prism?

(1) 2º

(2) 4º

(3) 6º

(4) 8º

12. A plano convex lens of refractive index

1.5 and radius of curvature 30 cm is silvered

at the curved surface. Now this lens has

been used to from the image of an object.

The distance from this lens an object be

placed in order to have a real image of the

size of the object is 10x cm, then x is :

(1) 2 (2) 4

(3) 6 (4) 8

13. White light travelling in air is refracted by

water then which statement is incorrect :

(1) It is possible that dispersion does not

take place.

(2) Dispersion necessarily takes place.

(3) Red colour has highest speed in water

(4) If light is dispersed than violet colour

undergoes maximum deviation.

14. A prism of refractive index 2 and apex

angle A is shown. Light is incident from PQ

side at angle of incidence i (0 < i 90º).

Then which option is incorrect :

(1) If A = 40º then light incident at all angles

will be refracted from surface PR.

(2) If A = 80º then light incident at some

angles will be refracted and at some

other angles light will be reflected.

(3) If A = 92º then light incident at all angles

will be reflected from the face PR.

(4) Whatever is the value of A, light

definitely emerge from the surface PR.







PERCT1090815C0-4

15. A small wooden rod of length 5mm is fixed at

the bottom of the container filled with water

(water = 43

). It is making an angle 37º with

vertical. If an observer (in air) observes the

rod paraxially, then length of rod is

appearing. (in mm)

(1) 3 3 (2) 3 2

(3) 5 2 (4) 4 2

16. Light of wavelength 4800 Å is incident at

small angle on a prism of apex angle 4º. The

prism has the refractive index for violet and

red rays respectively nv = 1.62 & nr = 1.6.

The angle of dispersion produced by the

prism in this light is:

(1) 0.8º (2) 0.08º

(3) 0.06º (4) None of these

17. In the figure shown find the total magnification

after two successive reflections first on M1

and then on M2. (Assume paraxial rays only)

(1) + 6 (2) � 6

(3) + 3 (4) +2

18. Light travelling in air falls at an incidence

angle of 2° on one refracting surface of a

prism of refractive index 1.5 and angle of

refraction 4º. The medium on the other side

is water (n = 4/3). Find the deviation

produced by the prism.

(1) 1º (2) 2º

(3) 3º (4) 4º

19. Dlameter of a planoconvex lens of focal

length 37 cm is 6 cm. It�s thickness at the

centre is 5 mm. The speed of light in the

material of the lens is

(1) 108 m/s

(2) 2.4 108 m/s

(3) 12 108 m/s

(4) 3 108 m/s

20. The power of a convex lens [ = 1.5]

(1) will decrease on immersing in water

(2) will increase in immersing in water

(3) is less for violet rays compared to red

rays

(4) None of these







PERCT1090815C0-5

SECTION - II Straight Objective Type

This section contains 10 multiple choice questions. Each question has 4 choices (1), (2), (3), (4) for its answer, out of which ONLY ONE is correct.

21. A stationary person A throws a ball as shown with speed 20 m/sec at an angle of 45° with

horizontal. At the same instant, another person B, at a distance of 10 m from A starts running with constant velocity and catches the ball at point C. The velocity of ball and velocity of person B always lie in same vertical plane; also the vertical level of point of projection and point of catching the ball from ground is same. Then the speed v of person B will be : (g = 10 m/s2, Neglect air friction)

V

BA

u

C

10m

(1) 15 2 m/sec (2) 7.5 2 m/sec

(3) 6 2 m/sec (4) 2.5 2 m/sec 22. A stone is projected horizontally with speed v

from a height h above ground. A horizontal wind is blowing in direction opposite to velocity of projection and gives the stone a constant horizontal acceleration f (in direction opposite to initial velocity). As a result the stone falls on ground at a point vertically below the point of projection. Then the value of height h in terms of f, g, v is (g is acceleration due to gravity)

(1) 2

2

gv2f

(2) 2

2

gvf

(3) 2

2

2gvf

(4) 2

2

2gvf

23. In projectile motion of a particle under gravity on an inclined plane (Assuming ground surface to be horizontal)

(1) Horizontal velocity is constant (2) Vertical velocity is constant (3) Velocity parallel to inclined plane is constant (4) Velocity perpendicular to inclined plane

is constant

24. A particle is moving along straight line and its motion is represented by given velocity-time graph. Based on the given graph, select the correct alternatives.

(1) Motion from A to B is case of retardation (2) Motion from B to C is case of varying

position (3) Magnitude of velocity at point C is lesser

than at point A (4) Particle changes its direction of motion

once between A to C.

25. Rain is falling down on ground making certain angle �� with vertical. A man starts walking on

level ground. Mark the correct statement. (1) If the man increases his speed of

walking, then it is possible that the angle made by rain with vertical (as observed by man) goes on decreasing.

(2) If the man goes on increasing his speed, then with respect to him, it is sure that the angle made by rain with horizontal will go on decreasing.

(3) Keeping his own velocity constant on horizontal, if the man observes rain to be falling vertically downwards but with variable magnitude then it is sure that velocity of rain with respect to ground is constant.

(4) If the man goes on decreasing his speed, then with respect to him, it is sure that the angle made by rain with horizontal will go on decreasing.







PERCT1090815C0-6

26. A man is standing on a road and observes that rain is falling at angle 45º with the

vertical. The man starts running on the road with constant acceleration 0.5 m/s2. After a certain time from the start of the motion, it appears to him that rain is still falling at angle

45º with the vertical, with speed 2 2 m/s . Motion of the man is in the same vertical plane in which the rain is falling. Then which of the following statement(s) are true.

(1) It is not possible (2) Speed of the rain relative to the ground

is 2 m/s. (3) Speed of the man when he finds rain to be

falling at angle 45º with the vertical, is 6m/s. (4) The man has travelled a distance 16m

on the road by the time he again finds rain to be falling at angle 45°.

27. Two stones are projected from level ground.

Trajectories of two stones are shown in figure. Both stones have same maximum heights above level ground as shown. Let T1 and T2 be their time of flights and u1 and u2 be their speeds of projection respectively (neglect air resistance). Then

y

x

1 2

(1) T2 > T1 (2) T1 = T2 (3) u1 > u2 (4) u1 = u2

28. A particle is moving rectilinearly so that its acceleration is given as a = 3t2+1 m/s2.Its initial velocity is zero.

(1) The velocity of the particle at t=1 sec will be 4m/s.

(2) The displacement of the particle in 1 sec will be 2m.

(3) The particle will continue to move in positive direction.

(4) The particle will come back to its starting point after some time.

29. As shown in figure, a particle P is fixed at

certain point in horizontal plane and another

particle Q is moving around P in circular path

(with centre O) of radius �r� with constant

speed �u�. �P� observes the motion of �Q�. Pick

out correct statement :

(1) Velocity of approach between P and Q

will be variable.


will be always positive.


will be always constant.


will be always zero.

30. A ball is thrown vertically upwards in air. If

the air resistance cannot be neglected

(assume it to be directly proportional to

velocity), then choose the correct options.

(Assume that resistance force is less than

weight of ball and g is the acceleration due to

gravity)

(1) at the highest point acceleration of the

ball is g.

(2) in upward motion (when ball is moving

upwards) acceleration of the ball is more

(in magnitude) than g.

(3) in downward motion (when ball is moving

downwards) acceleration of the ball is

less (in magnitude) than g.

(4) All of these right







CERCT1090815C0-7

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]

SECTION - I

Straight Objective Type

This section contains 20 multiple choice

questions. Each question has 4 choices (1),

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

ONLY ONE is correct.

31. For elements belonging to the same period,

first ionization energy is maximum for :

(1) halogen

(2) inert gas

(3) alkaline earth metal

(4) alkali metal

32. The angular momentum of electron in 'd'

orbital is equal to :

(1) 2 (2) 2 3

(3) 0 (4) 6

33. Number of electrons having + m value

equal to zero in 26Fe may be

(1) 13 (2) 12

(3) 7 (4) 12

34. The compressibility factor of a gas is less

than unity at 273 K and Vm = 22.4 L

therefore:

(1) P > 1 atm (2) P = 1 atm

(3) P < 1 atm (4) P = 2atm

35. Assertion : One molal aqueous solution of

glucose contains 180 g of glucose in 1 kg

water.

Reason : Solution containing one mole of

solute in 1000 g of solvent is called one

molal solution.

(1) If both assertion and reason are true and

reason is the correct explanation of

assertion.

(2) If both assertion and reason are true but

reason is not the correct explanation of

assertion.

(3) If Assertion is true but reason is false.

(4) If both assertion and reason are false.

36. 1021 molecules are removed from 200 mg of

CO2. The mole of CO2 left are :

(1) 2.88 × 10�3 (2) 28.2 × 10

�3

(3) 288 × 10�3

(4) 28.8 × 103







CERCT1090815C0-8

37. The molarity of H2SO4 solution, which has a

density 1.84 g/cc at 35°C and contains 98%

by weight is :

(1) 1.84 M (2) 18.4 M

(3) 20.6 M (4) 24.5 M

38. The weight of NaCl decomposed by

4.9 gram of H2SO4, if 6 gram of sodium

hydrogen sulphate and 1.825 gram of HCl

were produced in the reaction.

(1) 6.921 g (2) 4.65 g

(3) 2.925 g (4) 1.4 g

39. Which is correct order for ionic radii ?

(1) N3� < O2� < F� < Na+ < Mg2+

(2) N3� > O2� > F� > Na+ > Mg2+

(3) N3� < O2� < Mg2+ < F� < Na+

(4) F� < O2� < N3� < Na+ < Mg2+

40. Which of the following represents the correct

order of increasing electron gain enthalpy

with negative sign for the elements O, S, F

and Cl ?

(1) Cl < F < O < S

(2) O < S < F < Cl

(3) F < S < O < Cl

(4) S < O < Cl < F

41. What is the value of electron gain enthalpy of

Na+ if IE1 of Na = 5.1 eV ?

(1) �5.1 eV (2) �10.2 eV

(3) +2.55 eV (4) +10.2 eV

42. Reason of lanthanoid contraction is :

(1) Negligible screening effect of 'f' orbitals

(2) Increasing nuclear charge

(3) Decreasing nuclear charge

(4) Decreasing screening effect

43. CH2=CH�CH=CH�CH3 O3

Zn, H O2 Products.

Which product is not obtained in the above

reaction ?

(1) CH2=O

(2) CH3�CH=O

(3) OHC�CHO

(4) CH3�COOH

44. Test to differentiate between ethanol

(CH3CH2OH) and phenol (Ph�OH) is/are :

(1) Litmus test

(2) Neutral FeCl3

(3) Sodium metal test

(4) All of these







CERCT1090815C0-9

45. Which is the position isomers of

Br

(1)

Br

(2)

Br

(3)

Br

(4)

Br

46. How many tertiary alcohols is/are possible

with molecular formula C5H12O ?

(1) 1 (2) 2

(3) 3 (4) 4

47. A compound (P) on reaction with "Q" in basic

medium (KOH) gives a bad smelling

compound (CH3CH2NC). Compound Q can

be prepare by reaction of acetone with

calciumhypochlorite (Ca(OCl)2].

P and Q can

(1) CH3�CH2�NH2 & CHCl3

(2) CH3�CH2�NO2 & CH3Cl

(3) CH3�CH2�NH�CH3 & COCl2

(4) (CH3�CH2) 3N & Cl2

48. Product

No. of monochloro structure isomers in

products is:

(1) 3 (2) 4

(3) 6 (4) 7

49. Acetaldehyde and benzaldehyde can be

differenitated by :

(a) Fehling test (b) Iodoform test

(c) Tollen�s reagent (d) 2,4-DNP test

(1) a & b (2) a & c

(3) b & c (4) c & d

50. Which statement is correct for inductive

effect ?

(1) I effect increases with distance.

(2) �I order is O

N+

O�

OC

O�<

(3) In compound CH3�CN complete positive

charge is developed at CH3 due to

strong �I of cyanide group

(4) In aldehydes and ketones partial +ve

charge is developed at carbonyl carbon

atom.







CERCT1090815C0-10

SECTION - II


This section contains 10 multiple choice

questions. Each question has 4 four choices (1),

(2), (3), (4) out of which ONLY ONE is correct.

51. Consider the following statements :

I. The radius of an anion is larger than that

of the parent atom.

II. The ionization energy generally

increases with increasing atomic

number in a period.

III. The electronegativity of an element is the

tendency of an isolated atom to attract

an electron.

Which of the above statements is/are

correct?

(1) I alone (2) II alone

(3) I and II (4) II and III

52. In the vander Waals equation, �a� signifies :

(1) intermolecular attraction

(2) intramolecular attraction

(3) attraction between molecules and wall of

container

(4) volume of molecules

53. X ml of H2 gas effuse through a hole in a

container in 5 seconds. The time taken for

the effusion of the same volume of the gas

specified below under identical conditions is :

(1) 10 seconds : He

(2) 20 seconds : O2

(3) 25 seconds : CO

(4) 55 seconds : CO2

54. 2 gram of H2 gas and 2 gram of He gas are

filled in a closed container. Pressure of

Hydrogen gas, in term of total pressure is :

(1) 23

(2) 32

(3) 13

(4) 12

55. Positive deviation from ideal behaviour takes

place because of

(1) Molecular interaction between atoms

and PV

1nRT

(2) Molecular interaction between atoms and

PV

1nRT

(3) Finite size of atoms and PV

1nRT

(4) Finite size of atoms and PV

1nRT







CERCT1090815C0-11

56. Magnetic moment 2.84 B.M. is given by

(At. nos, Ni =28, Ti= 22, Cr =24, Co = 27 )

(1) Ti3+

(2) Cr2+

(3) Co2+

(4) Ni2+

57. Resonance is not possible in :

(1)

(2)

(3) CH2=CH�Cl

(4)

O

58. Which is the functional isomers of 2-Butanol?

(1) Butan-2-one

(2) Dimethyl ether

(3) 1-Butanol

(4) Diethyl ether

59. Which common name is wrong ?

(1) COCl

Benzoyl chloride

(2)

C�CH3 || O

Acetophenone

(3) CHO

Benzaldehyde

(4) CH2�OH

Phenol

60. Which is the correct structure for

Ethylacetate ?

(1) CH3�CH2�COOCH3

(2) CH3�CH2�COOC2H5

(3) CH3COOC2H5

(4) CH3�CH2�CO�CH2�CH3



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MERCT1090815C0-12


PART � C

SECTION - I



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

its answer, out of which ONLY ONE is correct.

61. If a � b + c < 0 and equation ax2 + bx + c = 0

does not have any real root then expression

a + 3b + 9c is

(1) Always negative

(2) always positive

(3) may be positive

(4) may be positive or negative

62. Let y = 2

2

x 3x 1

x x 1

x R, then

(1) y 5

1,3

(2) y 1 5

,3 3

(3) y 1

, 33

(4) y [� 1, 3]

63. Sign of a, b, c respectively for graph

f(x) = ax2 + bx + c.

(1) a < 0, b < 0, c < 0

(2) a > 0, b > 0, c > 0

(3) a > 0, b < 0, c > 0

(4) a 0, b < 0, c 0

64. The equation of straight line which passes

through the point (2, 1) and make an angle

4

with the straight line 2x + 3y + 4 = 0

(1) y + 5x = 11

(2) x � 5y � 7 = 0

(3) 5y � x = 13

(4) x � y = 1

65. A(0, 1), B(2, 1), C(1, 0), D(�1, 0) are vertices

of

(1) square

(2) rectangle

(3) parallelogram

(4) rhombus





MERCT1090815C0-13


66. The area of triangle formed by the mid points

of sides of triangle whose vertices are (0, 0),

(6, 0), (0 , 8)

(1) 3

(2) 6

(3) 12

(4) 24

67. If f(x) = 13sin x 4cosx � 2

then range of the

expression f(x) is

(1) 2

,3

(2) 2 2

� ,7 3

(3)

71

,��

,

31

(4) (�, 0) (0, )

68. Negation of (p ~ q) is

(1) (~ p q)

(2) (p ~ q)

(3) (p q)

(4) (~ p ~ q)

69. If the equation |x � k| + |x + k| = x2 has three

real solution then number of integral values

of k is/are

(1) 2

(2) 1

(3) 0

(4) infinite

70. Let 0,4

and t1 = (tan )tan ,

t2 = (tan )cot , t3 = (cot )tan and t4 = (cot )cot ,

then

(1) t1 > t2 > t3 > t4

(2) t2 < t1 < t3 < t4

(3) t3 > t1 > t2 > t4

(4) t2 > t3 > t1 > t4

71. Number of real roots of equation

(x � 1)2 + (x � 2)2 + (x � 3)2 = 0

(1) 1

(2) 2

(3) 3

(4) 0





MERCT1090815C0-14


72. Number of non positive integers satisfying

the inequation x x(3 � 4 ) n(x 2)

(x � 5)

0 is/are

(1) 1

(2) 2

(3) 0

(4) infinite

73. Sum of all real roots of the equation

|x|2 � 3|x| + 2 = 0 is

(1) 3

(2) 6

(3) 4

(4) 0

74. If x and y are the real number satisfying the

equation 12 sinx + 5 cos x = 2y2 � 8y + 21,

then the value of 12 cot xy2

is

(1) 2

(2) 3

(3) 4

(4) 5

75. If ylnx = x2lny & x,y N and n(x2y3) = 3 then

number of ordered pairs of (x, y) satisfying

the equation is/are -

(1) 0

(2) 1

(3) 2

(4) infinite

76. The equivalent statement of p q is

(1) (p q) (p q)

(2) (p q) (q p)

(3) (~ p q) (p ~ q)

(4) (~ p q) (p ~ q)

77. Number of acute angles such that cos

cos2 cos4 = 18

are

(1) 1

(2) 2

(3) 3

(4) 4





MERCT1090815C0-15


78. Let f(x) = x3 + x + 1. Suppose P(x) is a cubic

polynomial such that P(0) = �1 and the roots

of P(x) = 0 are the squares of the roots of

f(x) = 0, find the value of P(4) � 90 .

(1) 8

(2) 7

(3) 9

(4) 6

79. The expression logm m m m

mlog ........... m n radical rign

where m 2, m N; n N when simplified

is :

(1) independent of m

(2) independent of m & n

(3) dependent on m & n

(4) positive

80. If a, b, p, q are non zero real numbers then

two equations 2a2x2 � 2abx + b2 = 0 and

p2x2 + 2pqx + q2 have

(1) no common root

(2) one common root of 2a2 + b2 = p2 + q2

(3) two common roots if 3pq = 2ab

(4) two common roots if 3qb = 2ap

SECTION - II



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

answer, out of which ONLY ONE is correct.

81. The complete solution set of inequality

||x � 1| �1| > 2 is

(1) (�, �2] [4, )

(2) (�2, 4)

(3) (�, � 2) (4, )

(4) (�, �1) (2, )

82. If 5 lies between the roots of equation

x2 � px + 4 = 0 then number of positive

integral values of p which do not satisfying

the given condition will be -

(1) 0

(2) 1

(3) 6

(4) 5





MERCT1090815C0-16


83. If log0.5log5(x2 � 4) > log0.51 then set of value

of x will be -

(1) (�3, � 5 ) ( 5 , 3)

(2) (�, � 3) (3, )

(3) ( 5 , 3 5 )

(4) (�3, � 2) (2, 3)

84. Number of real solutions of the equation

xlog 2x + x2 = 3x is

(1) 0

(2) 1

(3) 2

(4) 4

85. If be root of 4x2 � 16x + = 0, where

R such that 1 < < 2 & 2 < < 3, then

number of integral values of is :

(1) 5

(2) 6

(3) 2

(4) 3

86. The number of integers less than 15

satisfying the inequation

|x2 � x| � |x2 �1| < |x � 1| is/are

(1) 0

(2) 13

(3) 14

(4) 15

87. Orthocentre of the triangle having vertices

A(1, 1), B(3, 0), C 2, 3 , is

(1) ( 2 , 0)

(2) (�1, 1)

(3) (1, 1)

(4) (3, 0)

88. If sin = p, |p| 1, then the quadratic

equation whose roots are tan 2

and cot 2

is

(1) px2 + 2x + p = 0

(2) px2 � x + p = 0

(3) px2 � 2x + p = 0

(4) px2 � x + 2p = 0





MERCT1090815C0-17


89. The number of solutions of equation

sinx = |x| are

(1) 2

(2) 1

(3) 0

(4) 4

90. If x2 � x � 1 = 0, then the value of x3 � 2x + 3

is

(1) 0

(2) 1

(3) 2

(4) 4



dPps dk;Z ds fy, LFkku




PERCT1090815C0-1

Hkkx � A

[k.M- I

lh/ks oLrqfu"B izdkj

bl [k.M esa 20 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. 30º viorZd dks.k o 2 viorZukad okys ,d

fizTe dk dksbZ ,d viorZd i"B ikWfy'k fd;k x;k

gSA nwljs viorZd i"B ij vkifrr ,d izdk'k iqat

blds iFk dks iqu% vuqjsf[kr (retrace) djsxk ;fn

vkiru dks.k gS %

(1) 0º (2) 30º

(3) 45º (4) 60º

2. ,d mÙky xksyh; niZ.k dh oØrk f=kT;k 1.2 ehVj

gSA 1.2 lseh- Å ¡pkbZ dh ,d oLrq niZ.k ls fdruh

nwjh ij fLFkr gS ;fn blds vkHkklh çfrfcEc o

niZ.k ds chp dh nwjh 0.35 ehVj gS ? çfrfcEc dh

Å ¡pkbZ D;k gS ? [v{k ds utnhd fdj.kksa ds fy,

lw=k yxkus ij ]

(1) 84 cm, 0.5 cm (2) 84 cm, 0.25 cm

(3) 84 cm, 1.2 cm (4) buesa ls dksbZ ughaA

3. dk¡p ds ,d fçTe dk vuqçLFk dkV leckgq f=kHkqt

:i dk gSA fdlh ,d Qyd ij blds yEcor~ ,d

fdj.k vkifrr gksrh gSA fizTe ds vk/kkj vkSj

fizTe ls fuxZr fdj.k ds chp dk dks.k gksxkA

¼dk¡p dk viorZukad µ = 1.5 gSA½

(1) 60º (2) 90º

(3) 30º (4) 45º

4. ,d vory niZ.k lw;Z dk çfrfcEc cukrk gS tksfd

blls 6 cm nwj gS &

(1) bl niZ.k dh oØrk f=kT;k 6 cm gSA

(2) bldks 'ksfoax ds fy, iz;qDr djus ij bldks

psgjs ls 8 - 10 cm nwjh ij j[krs gSA

(3) ;fn bldks oLrq ls 24 cm nwj j[krs gS rks

çfrfcEc oLrq ds vkdkj dk cusxkA

(4) mijksDr lHkh fodYi xyr gSA

5. fp=kkuqlkj ,d d.k �P� nks yEcor~ j[ks lery

niZ.k ds izfrPNsnu fcUnq dh rjQ 10 m/s ds osx

ls xfr dj jgk gSA 1 rFkk 2 niZ.k m1 rFkk m2 ls

lh/ks ijkorZu ls cuk d.k dk izfrfcEc gSA fn[kkbZ

xbZ fLFkfr ds fy, 1 rFkk 2 dh pky esa vUrj

Kkr dhft,A

(1) 20 m/s

(2) 10 m/s

(3) 0 m/s

(4) buesa ls dksbZ ugha







PERCT1090815C0-2

6. fuEu esa ls dkSulk dFku xksyh; niZ.k ds fy, lgh

gS &

(1) ,d vory niZ.k okLrfod oLrq dh fdlh Hkh

fLFkfr ds fy, dsoy vkHkklh çfrfcEc cukrk

gSA

(2) ,d mÙky niZ.k okLrfod oLrq dh fdlh Hkh

fLFkfr ds fy, dsoy vkHkklh çfrfcEc cukrk

gSA

(3) ,d vory niZ.k] /kzqo o Qksdl ds chp j[kh

oLrq dk dsoy vkHkklh o NksVk çfrfcEc cukrk

gSA

(4) ;fn oLrq /kzqo rFkk Qksdl ds e/; gks rks

mÙky niZ.k vkHkklh rFkk vkof/kZr çfrfcEc

cukrk gSA

7. gok esa j[ks ,d fçTe ls çdk'k ds viorZu ds fy,

(1) çR;sd fopyu dks.k ds fy, nks vkiru dks.k

gksrs gSaA

(2) tc fçTe dks U;wure fopyu ds fy,

O;ofLFkr fd;k tkrk gS rc lef}ckgq fçTe ds

vUnj xqtjus okyk çdk'k vko';d :i ls

vk/kkj ds lekukUrj gksrk gSA

(3) vf/kdre fopyu ds fy, nks vkiru dks.k

gksrs gSaA

(4) U;wure fopyu dks.k ?kVsxk ;fn ckg~; ek/;e

dks vifjofrZr j[krs gq, fçTe dk viorZukad

c<+k;k tkrk gSA

8. fp=k esa iznf'kZr ,d vory ySUl ds cka;s o nka;s

i"B dh oØrk f=kT;k Øe'k% 10 lseh- o 15 lseh- gSA

niZ.k dh oØrk f=kT;k 15 lseh- gS %

(1) la;kstu dh rqY; Qksdl nwjh �16 lseh- gSA

(2) la;kstu dh rqY; Qksdl nwjh +36 lseh- gSA

(3) fudk; ,d vory niZ.k dh Hkkafr O;ogkj

djrk gSA

(4) fudk; ,d mÙky niZ.k dh Hkkafr O;ogkj

djrk gSA

9. ,d nhokj W ls L nwjh ij nhokj ds lekUrj ,d

lery niZ.k M fp=kkuqlkj O;ofLFkr fd;k tkrk

gSA nhokj ij fLFkr ,d fcUnq lzksr S }kjk mRiUu

çdk'k niZ.k ls ijkofrZr gksrk gS rFkk nhokj ij

,d çdkf'kd {ks=k cukrk gSA ;fn niZ.k v osx ls

nhokj dh vksj xfr djrk gS] rc

L S

nhokj w

V

M

(1) çdkf'kd {ks=k] nhokj ij v pky ls xfr djsxkA

(2) çdkf'kd {ks=k] nhokj ij xfr ugha djsxkA

(3) niZ.k tSls&tSls ikl vkrk gS çdkf'kd {ks=k

cM+k gksrk tkrk gS vkSj v dh rqyuk es vf/kd

pky ls nhokj ij ¼nwj dh vksj½ foLFkkfir

gksrk gSA

(4) bueas ls dksbZ ughaA







PERCT1090815C0-3

10. fp=kkuqlkj ,d 2 lseh Qksdl nwjh dk vory niZ.k

,d dk¡p dh ifV~Vdk ij j[kk gS] rks oLrq 'O' dk

izfrfcEc niZ.k ls ijkorZu ,oa rRi'pkr~ ifV~Vdk ls

viorZu }kjk cusxk&

(1) vory niZ.k ds /kzqo ls 2 lseh- nwj rFkk

vkHkklh gksxkA

(2) niZ.k ds ?kzqo ij rFkk vkHkklh gksxkA

(3) oLrq ij gh rFkk vkHkklh gksxkA

(4) buesa ls dksbZ ugh

11. ,d f¶yaV dk¡p fçTe o ,d Økmu dk¡p fçTe dks

,d nwljs ls bl çdkj tksMk tkrk gS fd ek/;

fdj.k dk fopyu 'kwU; gksrk gSA ek/; fdj.k ds

fy, f¶yaV dk¡p o Økmu dk¡p dk viorZukad

Øe'k%1.6 o 1.9 gSA ;fn f¶yaV fçTe dk viorZd

dks.k 6°, gS rks Økmu fçTe dk viorZd dks.k D;k

gksxk ?

(1) 2º (2) 4º

(3) 6º (4) 8º

12. 30 lseh oØrk f=kT;k rFkk 1.5 viorZukad ds fdlh

leryksÙky ySal ds ofØr i"B dks ikWfy'k fd;k

x;k gSA vc bl ySal dk mi;ksx fdlh oLrq dk

izfrfcEc cukus esa fd;k tkrk gSA bl ySal ls oLrq

dh nwjh ftlls fd oLrq dk mlh eki dk okLrfod

izfrfcEc cus] 10x cm gS] rks x gksxkA

(1) 2 (2) 4

(3) 6 (4) 8

13. ok;q esa tk jgk lQsn izdk'k ty }kjk viofrZr

gksrk gS rks fuEu ls dkSulk dFku vlR; gS &

(1) ;g lEHko gS fd fo{ksi.k u gks

(2) fo{ksi.k vo'; gksxk

(3) ty esa yky jax dh pky lokZf/kd gS

(4) ;fn izdk'k fo{ksfir gksrk gS rks cSaxuh jax dk

fopyu lokZf/kd gksrk gSA

14. 2 viorZukad o 'kh"kZ dks.k A dk ,d fizTe fp=k

esa iznf'kZr gSA izdk'k Hkqtk PQ ls vkiru dks.k

i (0 < i 90º) ij vkifrr gS rks fuEu esa ls

dkSulk fodYi xyr gS &

(1) ;fn A = 40º gS rks lHkh dks.kksa ij vkifrr

izdk'k lrg PR ls viofrZr gksxkA

(2) ;fn A = 80º gS rks dqN dks.k ij vkifrr

izdk'k viofrZr gksxk rFkk dqN vU; dks.k ij

izdk'k ijkofrZr gksxkA

(3) ;fn A = 92º gS rks lHkh dks.kksa ij vkifrr

izdk'k lrg PR ls ijkofrZr gksxkA

(4) A ds eku dqN Hkh gks izdk'k lrg PR ls

fuf'pr :i ls mRlftZr gksxkA







PERCT1090815C0-4

15. 5 mm yEckbZ dh NksVh dk"B NM+, ty

(ty = 43

) ls Hkjs ik=k dh ryh esa tM+or~ gSA ;g

Å /oZ ls 37º dks.k cuk jgh gSA ;fn ,d çs{kd

(ok;q esa) v{kh; fdj.kksa ds çs{k.k }kjk NM+ dh

yEckbZ D;k ikrk gSA (feeh0 esa)

(1) 3 3 (2) 3 2

(3) 5 2 (4) 4 2

16. 4800 Å rjaxnS/;Z dk çdk'k vYi dks.k ij

4º 'kh"kZ dks.k okys fçTe ij vkifrr gksrk gSA

cSaxuh fdj.k ds fy, fçTe dk viorZukad

nv = 1.62 rFkk yky fdj.k ds fy, nr = 1.6 gSA

bl çdk'k esa fçTe }kjk mRiUu fo{ksi.k dks.k gS :

(1) 0.8º (2) 0.08º

(3) 0.06º (4) buesa ls dksbZ ugha

17. n'kkZ;s x;s fp=k esa igys M1 o fQj M2 ij gksus

okys nks mÙkjksÙkj ijkorZuksa ds ckn dqy vko/kZu

Kkr dhft,A (dsoy lek+{kh; fdj.ksa ekusa)

(1) + 6 (2) � 6

(3) + 3 (4) +2

18. 1.5 viorZukad rFkk fizTe dks.k 4º okys fizTe ds

viorZd ry ij gok ls xqtjrh gqbZ fdj.k 2° ds

vkiru dks.k ij vkifrr gksrh gSA nwljs rjQ dk

ek/;e ikuh (n = 4/3) gSA fizTe }kjk mRiUu

fopyu Kkr djksA

(1) 1º

(2) 2º

(3) 3º

(4) 4º

19. Qksdl nwjh 37 cm ds leryksÙky ySal dk O;kl 6

cm gSA e/; esa bldh eksVkbZ 5 mm gSA ySal inkFkZ

esas izdk'k pky gksxh

(1) 108 m/s

(2) 2.4 108 m/s

(3) 12 108 m/s

(4) 3 108 m/s

20. mÙky ySUl [ = 1.5] dh 'kfDr gS

(1) ty esa Mqcksus ij ?kVsxhA

(2) ty esa Mqcksus ij c<+sxhA

(3) yky fdj.kksa dh rqyuk esa cSaxuh fdj.kksa ds fy,

de gksrh gSA

(4) buesa ls dksbZ ugha







PERCT1090815C0-5

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

bl [k.M esa 10 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4

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

21. fp=kkuqlkj ,d fLFkj O;fä A {kSfrt ls 45° ds dks.k ij 20 m/sec ds osx ls ,d xsan dks Qsdrk gSA blh {k.k A ls 10 m ehVj dh nwjh ij fLFkr nwljk O;fä B fu;r osx ls xfr djuk izkjEHk djrk gS vkSj fcUnq C ij xsan dks idM+ ysrk gSA xsan rFkk O;fä B nksuksa ds osx leku Å /okZ/kj ry esa gS rFkk iz{ksi.k fcUnq ,oa O;fä B }kjk xsan dks idM+s tkus okys fcUnq C dh /kjkry ls Å ¡pkbZ leku gS rks O;fä B dh pky gksxh &

(g = 10 m/s2, ok;q dk ?k"kZ.k ux.; eku ysus ij)

V

BA

u

C

10m (1) 15 2 m/sec (2) 7.5 2 m/sec

(3) 6 2 m/sec (4) 2.5 2 m/sec 22. ,d iRFkj dks v pky ls tehu ls h Å ¡pkbZ Å ij ls

{kSfrt fn'kk esa iz{ksfir fd;k tkrk gSA {kSfrt gok] iz{ksi.k osx ds Bhd foifjr fn'kk esa cg jgh gS rFkk gok iRFkj dks fu;r {kSfrt Roj.k f iRFkj dks iznku djrh gS ftldh fn'kk izkjfEHkd osx ds foijhr fn'kk esa gSA ftlds ifj.kkeLo:i iRFkj iz{ksi.k fcUnq ds Bhd Å /okZ/kj uhps tehu ij fxjrk gSA rks Å ¡pkbZ h dk eku f, g, v ds inksa esa gksxk (g xq:Ro ds dkj.k Roj.k gS)

(1) 2

2

gv2f

(2) 2

2

gvf

(3) 2

2

2gvf

(4) 2

2

2gvf

23. ur ry ij xq:Ro ds vUrxZr iz{ksI; xfr esa (/kjkry lrg dks {kSfrt ekfu;s)

(1) {kSfrt osx fu;r gS (2) m/okZ/kj osx fu;r gS (3) ur ry ds lekUrj osx fu;r gS (4) ur ry ds yEcor osx fu;r gS

24. ,d d.k ljy js[kk ds vuqfn'k xfr dj jgk gS rFkk bldh xfr fn;s x;s osx≤ xzkQ ls iznf'kZr dh tkrh gS] fn;s x;s xzkQ ij vk/kkfjr lR; fodYiksa dk p;u dhft,A

(1) A ls B rd dh xfr eanu dh fLFkfr gSA (2) B ls C rd dh xfr fLFkfr ifjorZu dh fLFkfr gSA (3) fcUnq C ij osx dk ifjek.k A ij osx ds

ifjek.k ls de gSA (4) A ls C rd xfr esa ,d ckj d.k dh fn'kk

ifjofrZr gksrh gSA

25. ckfj'k Å /okZ/kj ds lkFk �� dks.k cukrs gq, /kjkry ij fxj jgh gSA ,d O;fDr /kjkry ij pyuk izkjEHk djrk gSA lR; dFkuksa dk p;u dhft;saA

(1) ;fn O;fDr vius pyus dh pky dks c<+krk gS] rks ;g laHko gS fd Å/okZ/kj ds lkFk ckfj'k ds }kjk cuk;k x;k dks.k ¼O;fDr ds lkis{k½ ?kVrk tk;sxkA

(2) ;fn O;fDr viuh pky dks c<+krk tkrk gS rks mlds lkis{k ;g fuf'pr gS fd ckfj'k }kjk {kSfrt ds lkFk cuk;k x;k dks.k ?kVrk tk;sxkA

(3) mlds Lo;a ds osx dks {kSfrt lrg ij fu;r j[krs gq, ;fn O;fDr o"kkZ dks Å /okZ/kj uhps dh vksj fxjrs gq, ns[krk gS] fdUrq ifjofrZr ifjek.k ds lkFk] rks ;g fuf'pr gS fd /kjkry ds lkis{k ckfj'k dk osx fu;r gSA

(4) ;fn O;fDr viuh pky dks ?kVrk tkrk gS rks mlds lkis{k ;g fuf'pr gS fd ckfj'k }kjk {kSfrt ds lkFk cuk;k x;k dks.k ?kVrk tk;sxkA







PERCT1090815C0-6

26. ,d vkneh ,d lM+d ij [kM+k gS rFkk m/okZ/kj ds

lkFk 45º ds dks.k ij ckfj'k dks fxjrh gqbZ ikrk

gSA vc vkneh lM+d ij fu;r Roj.k 0.5 m/s2 ls

nkSM+uk izkjEHk djrk gSA vkneh dh xfr izkjEHk

djus ds dqN nsj ckn og vkneh ckfj'k dks fQj ls

m/okZ/kj ds lkFk 45º dk dks.k cukrs gq, 2 2 m/s

dh pky ls fxjrh gqbZ ikrk gS ckfj'k ds fxjus dk

m/okZ/kj ry gh vkneh dh xfr dk ry gSA rks

fuEu esa ls dkSuls dFku lR; gS &

(1) ;g laHko ugha gSA

(2) tehu ds lkis{k ckfj'k dh pky 2 m/s gSA

(3) vkneh dh pky tc og ckfj'k dks m/okZ/kj ls

45º dks.k ij ikrk gS rc 6m/s gSA

(4) vkneh dks tc nqckjk ckfj'k 45° dks.k ij

feyrh gS rks bl le; vUrjky esa r; nwjh

16m gSA

27. nks iRFkjksa dks tehu dh lrg ls iz{ksfir fd;k

tkrk gSA nksuksa iRFkjksa ds iz{ksI;&iFk fp=k esa fn[kk;s

x;s gSaA fn[kk;s fp=kkuqlkj nksuksa iRFkjksa dh

vf/kdre Å ¡pkbZ;k¡ leku gSA ekuk fd muds

mM~M;u dky Øe'k% T1 rFkk T2 gS vkSj muds iz{ksi

osx Øe'k% u1 rFkk u2 gSA ¼gok ds izfrjks/k dks

ux.; ekfu;s½ rks � y

x

1 2

(1) T2 > T1 (2) T1 = T2

(3) u1 > u2 (4) u1 = u2

28. ,d foeh; esa xfr djrs gq, d.k dk Roj.k a = 3t2+1 m/s2 gSA bldk çkjfEHkd osx 'kwU; gS rks

(1) t = 1 ij d.k dk osx 4m/s osx gksxkA (2) 1 sec esa d.k dk foLFkkiu 2m gksxkA (3) d.k ljy js[kk esa /kukRed fn'kk esa xfr djrk

jgsxkA (4) dqN le; ckn d.k vius çkjfEHkd fcUnq ij

vk,xkA 29. fp=kkuqlkj ,d d.k P {kSfrt ry esa fuf'pr fcUnq

ij fLFkj gS rFkk vU; d.k Q, 'r' f=kT;k ds oÙkkdkj iFk (ftldk dsUnz O gS) esa �u� fu;r pky ls �P� ds pkjksa vksj xfr dj jgk gSA 'P', �Q� dh xfr dks ns[krk gSA lgh dFku dk p;u dhft;sA

(1) fcUnq P o Q ds e/; igq¡pus dk osx

ifjorZu'khy gksxkA (2) fcUnq P o Q ds e/; igq¡pus dk osx lnSo

/kukRed gksxkA (3) fcUnq P o Q ds e/; igq¡pus dk osx lnSo

fu;r gksxkA (4) fcUnq P o Q ds e/; igq¡pus dk osx lnSo 'kwU;

gksxkA

30. ,d xsan Å /okZ/kj Åij dh rjQ gok esa Qsadk x;k gSA ;fn ok;q dk çfrjks/k ux.; ugha gS rFkk bls xsan ds osx ds lekuqikrh ekuk tk;s rks fuEu esa ls dkSuls dFku lR; gSA (;g ekfu, fd çfrjks/kh cy xsan ds Hkkj ls vf/kd ugha gS rFkk g xq:Roh; Roj.k gS)

(1) mPpre fcUnq ij xsan dk Roj.k g gSA (2) xsan ds Å ij dh rjQ dh xfr esa Roj.k dk

ifjek.k g ls vf/kd gksxkA (3) xsan ds uhps dh rjQ dh xfr esa Roj.k dk

ifjek.k g ls de gksxkA (4) lHkh lR; gSA






CERCT1090815C0-7


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]

[k.M - I


bl [k.M esa 20 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. leku vkorZ ls lEcfU/kr rRoksa ds fy,] izFke

vk;uu Å tkZ fuEu ds fy, vf/kdre gksrh gS %

(1) gSykstu

(2) vfØ; xSl

(3) {kkjh; enk /kkrq

(4) {kkj /kkrq

32. 'd' d{kd esa bysDVªkWu dk dks.kh; laosx fuEu ds

cjkcj gksrk gS:

(1) 2 (2) 2 3

(3) 0 (4) 6

33. 26Fe esa mu bysDVªksuksa dh la[;k D;k gksxh ftuds

+ m dk eku 'kwU; ds cjkcj gksxk-

(1) 13 (2) 12

(3) 7 (4) 12

34. 273 K o Vm = 22.4 L ij ,d xSl dh lEihfM~;rk

xq.kkad ,d ls de gS] vr% &

(1) P > 1 atm (2) P = 1 atm

(3) P < 1 atm (4) P = 2atm

35. dFku : Xywdksl dk ,d eksyy tyh; foy;u 1

kg esa 180 g Xywdksl j[krk gSA

dkj.k : 1000 g foyk;d esa foys; ds ,d eksy

;qDr foy;u] 1 eksyy foy;u dgykrk gSA

(1) ;fn dFku rFkk dkj.k nksuks lgh gSa rFkk dkj.k

dFku dh lgh O;k[;k djrk gSA

(2) ;fn dFku rFkk dkj.k nksuks lgh gSa ysfdu

dkj.k dFku dh lgh O;k[;k ugha djrk gSA

(3) ;fn dFku lgh gS rFkk dkj.k xyr gSA

(4) ;fn dFku rFkk dkj.k nksuksa xyr gSaA

36. 200 mg CO2 ls 1021 v.kq ?kVk;s tkrs gSaA 'ks"k cps

CO2 ds eksy fuEu gSa %

(1) 2.88 × 10�3 (2) 28.2 × 10

�3

(3) 288 × 10�3

(4) 28.8 × 103






CERCT1090815C0-8


37. 35°C ij H2SO4 foy;u dh eksyjrk D;k gksxh

ftldk ?kuRo 1.84 g/cc gS rFkk tks Hkkj dk 98%

;qDr gS %

(1) 1.84 M (2) 18.4 M

(3) 20.6 M (4) 24.5 M

38. 4.9 xzke H2SO4 }kjk fo[kf.Mr NaCl dk Hkkj D;k

gksxk] ;fn vfHkfØ;k esa 6 xzke lksfM;e gkbMªkstu

lYQsV o 1.825 xzke HCl mRikfnr gksrs gksa %

(1) 6.921 g (2) 4.65 g

(3) 2.925 g (4) 1.4 g

39. vk;fud f=kT;kvksa dk lgh Øe dkSulk gS \

(1) N3� < O2� < F� < Na+ < Mg2+

(2) N3� > O2� > F� > Na+ > Mg2+

(3) N3� < O2� < Mg2+ < F� < Na+

(4) F� < O2� < N3� < Na+ < Mg2+

40. fuEu esa dkSulk Øe O, S, F rFkk Cl rRoksa ds fy,

_ .kkRed fpUg ds lkFk bysDVªkWu xzg.k ,UFkSYih ds

c<+rs gq, lgh Øe dks iznf'kZr djrk gS \

(1) Cl < F < O < S

(2) O < S < F < Cl

(3) F < S < O < Cl

(4) S < O < Cl < F

41. ;fn Na dh IE1 = 5.1 eV gS] rks Na+ dh bysDVªkWu

xzg.k ,UFkSYih dk eku D;k gksxk \

(1) �5.1 eV (2) �10.2 eV

(3) +2.55 eV (4) +10.2 eV

42. ySUFkSuksbM ladqpu dk dkj.k gS %

(1) 'f' d{kdksa dk ux.; vkoj.k izHkko

(2) ukfHkdh; vkos'k esa of)

(3) ukfHkdh; vkos'k eas deh

(4) vkoj.k izHkko esa deh


Zn, H O2 mRikn

fuEu esa ls dkSulk mRikn mijksDr vfHkfØ;k esa

izkIr ugha gksrk gS ?

(1) CH2=O

(2) CH3�CH=O

(3) OHC�CHO

(4) CH3�COOH

44. ,FksukWy (CH3CH2OH) rFkk fQukWy (Ph�OH) ds

e/; vUrj djus ds fy, iz;qDr ijh{k.k gS@gSa&

(1) fyVel ijh{k.k

(2) mnklhu FeCl3

(3) lksfM;e /kkrq ijh{k.k

(4) ;s lHkh






CERCT1090815C0-9


45. ;kSfxd

Br

dk fLFkfr leko;oh

dkSulk gS \

Br

(1)

Br

(2)

Br

(3)

Br

(4)

Br

46. v.kqlw=k C5H12O ds fdrus rRkh;d ,YdksgkWy

lEHko gS ? (1) 1 (2) 2 (3) 3 (4) 4 47. ,d ;kSfxd (P) {kkjh; ek/;e (KOH) dh mifLFkfr

esa ;kSfxd "Q" ds lkFk vfHkfØ;k djus ij ,d

nqxZa/k xa/k okyk ;kSfxd (CH3CH2NC) nsrk gSA

;kSfxd Q dSfYl;egkbiksDyksjkbV (Ca(OCl)2] ds

lkFk ,lhVksu dh vfHkfØ;k djkus ij curk gSA

vr% ;kSfxd P rFkk Q gks ldrs gSa & (1) CH3�CH2�NH2 & CHCl3

(2) CH3�CH2�NO2 & CH3Cl (3) CH3�CH2�NH�CH3 & COCl2

(4) (CH3�CH2) 3N & Cl2

48. mRikn

izkIr mRikn esa eksuksDyksjks lajpuk leko;fo;ksa dh

la[;k gS&

(1) 3 (2) 4

(3) 6 (4) 7

49. ,lhVsfYMgkbM rFkk csUtsfYMgkbM fuEu }kjk

foHksfnr gks ldrs gS&

(a) Qsgfyax ijh{k.k (b) vk;ksMksQkWeZ ijh{k.k

(c) VkWysu ijh{k.k (d) 2,4-DNP ijh{k.k

(1) a & b (2) a & c

(3) b & c (4) c & d

50. izsjf.kd izHkko ds lUnHkZ esa dkSulk dFku lR; gS \

(1) nwjh c<+us ds lkFk I izHkko Hkh c<+rk gSA

(2) �I dk Øe O

N+

O�

OC

O�<

(3) ;kSfxd CH3�CN esa CH3 lewg ij lk;ukbM

lewg ds izcy �I izHkko ds dkj.k iw.kZ /kukos'k

mRiUu gksrk gSA

(4) ,fYMgkbM ,oa dhVksu ds dkcksZfuy dkcZu

ijek.kq ij vkaf'kd /kukos'k +ve mRiUu gksrk

gSA






CERCT1090815C0-10


________________________________________________________________

[k.M - II


bl [k.M esa 10 iz'u gSaA izR;sd iz'u ds 4 fodYi

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

______________________________________________________________________

51. fuEufyf[kr dFkuksa dk voyksdu dhft, %

I. ,d _ .kk;u dh f=kT;k iSrd ijek.kq ls

vf/kd gksrh gSA

II. vkorZ esa ijek.kq Øekad c<+us ds lkFk

lkekU;r% vk;uu Å tkZ c<+rh gSA

III. ,d rRo dh fo|qr_ .krk] ,d foyfxr ijek.kq

dh ,d bysDVªkWu dks vkd f"kZr djus dh çofr

gksrh gSA

fuEu esa ls dkSulk@dkSuls dFku lgh gS@gSa %

(1) dsoy I (2) dsoy II

(3) I o II (4) II o III

52. ok.Mj okWy lehdj.k esa] �a� fuEu ds fy, lkFkZd

gS :

(1) vUrjkf.od vkd"kZ.k

(2) vUr%vkf.od vkd"kZ.k

(3) v.kqvksa o ik=k dh nhokj ds chp vkd"kZ.k

(4) v.kqvksa dk vk;ru

53. X ml, H2 xSl ,d ik=k esa fLFkr ,d fNnz esa ls 5

lSd.M ds fy, fulfjr gksrh gSA leku ifjLFkfr;ksa

esa fuEu esas ls dkSulh xSl ds leku vk;ru ds

fulj.k ds fy, fdruk le; yxsxk\

(1) 10 lSd.M : He

(2) 20 lSd.M : O2

(3) 25 lSd.M : CO

(4) 55 lSd.M : CO2

(5) 35 lSd.M : Cl2

54. ,d cUn ik=k esa 2 xzke H2 xSl o 2 xzke He xSl

Hkjh tkrh gSA gkbMªkstu xSl dk nkc] dqy nkc ds

in esa gksxk &

(1) 23

(2) 32

(3) 13

(4) 12

55. vkn'kZ O;ogkj ls èkukRed fopyu gksrk gS ftldk

dkj.k gS%

(1) ijek.kqvksa ds chp vkf.od vUrZfØ;k rFkk

PV

1nRT

(2) ijek.kqvksa ds chp vkf.od vUrZfØ;k rFkk

PV

1nRT

(3) ijek.kqvksa dk fuf'pr vkdkj rFkk PV

1nRT

(4) ijek.kqvksa dk fuf'pr vkdkj rFkk PV

1nRT






CERCT1090815C0-11


56. fuEu esa ls fdldk pqacdh; vk?kw.kZ 2.84 B.M. gS\

(i-la- Ni =28, Ti= 22, Cr =24, Co = 27 )

(1) Ti3+

(2) Cr2+

(3) Co2+

(4) Ni2+

57. fuEu esa ls fdlesa vuqukn lEHko ugha gS \

(1)

(2)

(3) CH2=CH�Cl

(4)

O

58. fuEu esa ls dkSu 2-C;wVsukWy dk fØ;kRed leko;oh

gS ?

(1) C;wVsu-2-vkWu

(2) MkbZesfFky bZFkj

(3) 1-C;wVsukWy

(4) MkbZ,fFky bZFkj

59. dkSulk lkekU; uke xyr gS \

(1) COCl

csatks;y DyksjkbM

(2)

C�CH3 || O

,lhVksQhuksu

(3) CHO

csUtsfYMgkbM

(4) CH2�OH

fQuksy

60. ,fFky,lhVsV ds fy, fuEu esa ls lgh lajpuk gS \

(1) CH3�CH2�COOCH3

(2) CH3�CH2�COOC2H5

(3) CH3COOC2H5

(4) CH3�CH2�CO�CH2�CH3





MERCT1090815C0-12

(dPps dk;Z ds fy, LFkku )

PART � C

[k.M - I


bl [k.M esa 20 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 a � b + c < 0 vkSj lehdj.k ax2 + bx + c = 0

dksbZ okLrfod ewy ugha j[krk gS rc O;atd

a + 3b + 9c gS&

(1) lnSo _ .kkRed gSA

(2) lnSo /kukRed gSA

(3) /kukRed gks ldrk gSA

(4) /kukRed ;k _ .kkRed gks ldrk gS

62. ekuk y = 2

2

x 3x 1

x x 1

x R, rks

(1) y 5

1,3

(2) y 1 5

,3 3

(3) y 1

, 33

(4) y [� 1, 3]

63. vkjs[k f(x) = ax2 + bx + c ds fy, a, b, c dk

fpUg Øe'k% gS&

(1) a < 0, b < 0, c < 0

(2) a > 0, b > 0, c > 0

(3) a > 0, b < 0, c > 0

(4) a 0, b < 0, c 0

64. ljy js[kk dk lehdj.k gksxk tks fcUnq (2, 1) ls

xqtjrk gS rFkk ljy js[kk 2x + 3y + 4 = 0 ds

lkFk 4

dks.k cukrk gS&

(1) y + 5x = 11

(2) x � 5y � 7 = 0

(3) 5y � x = 13

(4) x � y = 1

65. A(0, 1), B(2, 1), C(1, 0), D(�1, 0) 'kh"kZ gS&

(1) oxZ

(2) vk;r

(3) lekUrj prqHkZqt

(4) leprqHkZqt





MERCT1090815C0-13


66. 'kh"kZ (0, 0), (6, 0), (0 , 8) ls cus f=kHkqt dh

Hkqtkvksa ds e/; fcUnqvksa ls cus f=kHkqt dk {ks=kQy

gS&

(1) 3

(2) 6

(3) 12

(4) 24

67. ;fn f(x) = 1

3sinx 4cosx � 2 gks] rks O;atd

f(x) dk ifjlj gS&

(1) 2

,3

(2) 2 2

� ,7 3

(3)

71

,��

,

31

(4) (�, 0) (0, )

68. (p ~ q) dk udkjkRed gS&

(1) (~ p q)

(2) (p ~ q)

(3) (p q)

(4) (~ p ~ q)

69. ;fn lehdj.k |x � k| + |x + k| = x2 ds rhu

okLrfod gy j[krk gS rc k ds iw.kk±d ekuksa dh

la[;k gS&

(1) 2

(2) 1

(3) 0

(4) vuUr

70. ekuk 0,4

vkSj t1 = (tan ) t a n ,

t2 = (tan )co t , t3 = (cot ) t a n vkSj

t4 = (cot ) co t gks] rks

(1) t1 > t2 > t3 > t4

(2) t2 < t1 < t3 < t4

(3) t3 > t1 > t2 > t4

(4) t2 > t3 > t1 > t4

71. lehdj.k (x � 1)2 + (x � 2)2 + (x � 3)2 = 0

ds okLrfod ewyksa dh la[;k gS&

(1) 1

(2) 2

(3) 3

(4) 0





MERCT1090815C0-14


72. vlfedk x x(3 � 4 ) n(x 2)

(x � 5)

0 dks larq"V djus

okys v/kukRed iw.kk±dkas dh la[;k gS&

(1) 1

(2) 2

(3) 0

(4) vuUr

73. lehdj.k |x|2 � 3|x| + 2 = 0 ds lHkh okLrfod

ewyksa dk ;ksxQy gS&

(1) 3

(2) 6

(3) 4

(4) 0

74. ;fn x vkSj y okLrfod la[;k,a gS tks lehdj.k

12 sinx + 5 cos x = 2y2 � 8y + 21 dks larq"V

djrh gS] rks 12 cot xy2

dk eku gS&

(1) 2

(2) 3

(3) 4

(4) 5

75. ;fn ylnx = x2lny vkSj x,y N vkSj n(x2y3) = 3

gks] rks lehdj.k dks larq"V djus okys (x, y) ds

Øfer ;qXeksa dh la[;k gS&

(1) 0

(2) 1

(3) 2

(4) vuUr

76. p q dk rqY; dFku gS&

(1) (p q) (p q)

(2) (p q) (q p)

(3) (~ p q) (p ~ q)

(4) (~ p q) (p ~ q)

77. U;wu dks.kksa dh la[;k tcfd

cos cos2 cos4 = 18

gS&

(1) 1

(2) 2

(3) 3

(4) 4





MERCT1090815C0-15


78. ekukfd f(x) = x3 + x + 1. ekukfd P(x) ,d

f=k?kkrh cgqin bl izdkj gS fd P(0) = �1 rFkk

P(x) = 0 ds ewy] f(x) = 0 ds ewyksa ds oxZ gS] rks

P(4) � 90 dk eku gS&

(1) 8

(2) 7

(3) 9

(4) 6

79. O;atd logm m m m

mlog ........... m n d j.kh fpUg

tgk¡ m 2,

m N; n N tc ljyhd r :i gS&

(1) m ls Lora=k

(2) m vkSj n ls Lora=k

(3) m vkSj n ij fuHkZj

(4) /kukRed

80. ;fn a, b, p, q v'kwU; okLrfod la[;k,a gS] rc nks

lehdj.ksa 2a2x2 � 2abx + b2 = 0 vkSj

p2x2 + 2pqx + q2 j[krk gS] rc

(1) dksbZ ewy mHk;fu"B ugha gSA

(2) ,d mHk;fu"B ewy 2a2 + b2 = p2 + q2

(3) nks mHk;fu"B ewy ;fn 3pq = 2ab

(4) nks mHk;fU"B ewy ;fn 3qb = 2ap

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

bl [k.M esa 10 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3), (4) gSa] ftuesa ls flQZ ,d lgh gSA

81. vlfedk ||x � 1| �1| > 2 ds lEiw.kZ gy leqPp;

gS&

(1) (�, �2] [4, )

(2) (�2, 4)

(3) (�, � 2) (4, )

(4) (�, �1) (2, )

(5) (�, �1) (3, )

82. ;fn lehdj.k x2 � px + 4 = 0 ds ewyksa ds e/; 5

gS rc p ds /kukRed iw.kk±d ekuksa dh la[;k tks fn,

x, izfrcU/k dks larq"V ugha djrk gS&

(1) 0

(2) 1

(3) 6

(4) 5

(5) vuUr





MERCT1090815C0-16


83. ;fn log0.5log5(x2 � 4) > log0.51 rc x vUrjky esa

fLFkr gS&

(1) (�3, � 5 ) ( 5 , 3)

(2) (�, � 3) (3, )

(3) ( 5 , 3 5 )

(4) (�3, � 2) (2, 3)

(5)

84. lehdj.k xlog 2x + x2 = 3x ds okLrfod gyksa dh

la[;k gS &

(1) 0

(2) 1

(3) 2

(4) 4

(5) vuUr

85. ;fn lehdj.k 4x2 � 16x + = 0, ds ewy gS]

tgk¡ R bl çdkj 1 < < 2 rFkk 2 < < 3,

rc ds iw.kk±d ekuksa dh la[;k gS&

(1) 5

(2) 6

(3) 2

(4) 3

(5) 4

86. vlfedk |x2 � x| � |x2 �1| < |x � 1| dks larq"V

djus okys 15 ls de iw.kk±dksa dh la[;k gS&

(1) 0

(2) 13

(3) 14

(4) 15

(5) vuUr

87. 'kh"kZ A(1, 1), B(3, 0), C 2, 3 ls cus f=kHkqt dk

yEcdsUæ gS&

(1) ( 2 , 0)

(2) (�1, 1)

(3) (1, 1)

(4) (3, 0)

(5) (2, 3)

88. ;fn sin = p, |p| 1 gS] rks og f}?kkr

lehd j.k ft ld s ewy tan2

vkSj cot2

gS]

gS&

(1) px2 + 2x + p = 0

(2) px2 � x + p = 0

(3) px2 � 2x + p = 0

(4) px2 � x + 2p = 0

(5) x2 � 2x + 1 = 0





MERCT1090815C0-17


89. lehdj.k sinx = |x| ds gyksa dh la[;k gS&

(1) 2

(2) 1

(3) 0

(4) 4

(5) vuUr

90. ;fn x2 � x � 1 = 0, rks x3 � 2x + 3 dk eku gS&

(1) 0

(2) 1

(3) 2

(4) 4

(5) 3





SOLERCT1090815-1

CUMULATIVE TEST-1 (CT-1)

TARGET : JEE (MAIN)-2016

HINTS & SOLUTIONS ¼ladsr ,oa gy½

PART-A PHYSICS

1. A prism having refractive...........

30º viorZd dks.k o 2 ........... Sol.

sini

2sin30º

sin i =1 1

22 2

i = 45º .

2. The radius of curvature of a........... ,d mÙky xksyh; niZ.k dh oØrk........... Sol.

m = 2

1

h �v

h u

h2 = �v

uh1 =

�0.35 1.2cm

�0.84

= 0.5 cm.

3. The cross section of a glass........... dk¡p ds ,d fçTe dk vuqçLFk........... Sol.

Here vr% ic = sin�1 1

1.5 = sin�1

23

< 60º

So, T..R. takes place at second surface

vr% T..R. f}rh; lrg ij gksxkA

+ 120º = 180º = 60º

= 90º

5. In the figure shown. A particle...........

fp=kkuqlkj ,d d.k �P� nks yEcor~...........

Sol.

The image will move as shown in the figure. It is

very clear from the figure that the v2 � v1 = 0

izfrfcEc fp=kkuqlkj xfr djsaxsA vr% iz'u esa iwNk x;k

v2 � v1 = 0

6. Which of the following...........

fuEu esa ls dkSulk dFku...........

Sol. (1) No, when object is between infinite and

focus ,image is real.

ugh, tc oLrq vuUr rFkk Qksdl ds e/; gS] izfrfcEc

okLrfod gS

(3) when object is between pole and focus,

image is magnified.

tc oLrq /kzqo rFkk Qksdl ds e/; gS] izfrfcEc vkof/kZr

gksxk

(4) when object is between pole and focus

image formed by convex mirror is real.

tc oLrq /kzqo rFkk Qksdl ds e/; gS] mÙky niZ.k }kjk

cuk izfrfcEc okLrfod gksxk

DATE : 09-08-2015 |CLASS-XII/XIII





SOLERCT1090815-2

7. For the refraction of light...........

gok esa j[ks ,d fçTe ls çdk'k...........

(1) For every angle of deviation there are two

angles of incidence.

çR;sd fopyu dks.k ds fy, nks vkiru dks.k gksrs gSaA

(2) The light travelling inside an isosceles prism

is necessarily parallel to the base when prism is

set for minimum deviation.

tc fçTe dks U;wure fopyu ds fy, O;ofLFkr fd;k

tkrk gS rc lef}ckgq fçTe ds vUnj xqtjus okyk

çdk'k vko';d :i ls vk/kkj ds lekukUrj gksrk gSA

(3*) There are two angles of incidence for

maximum deviation.

vf/kdre fopyu ds fy, nks vkiru dks.k gksrs gSaA

(4) Angle of minimum deviation will decrease if

refractive index of prism is increased keeping

the outside medium unchanged.

U;wure fopyu dks.k ?kVsxk ;fn ckg~; ek/;e dks

vifjofrZr j[krs gq, fçTe dk viorZukad c<+k;k tkrk

gSA

Sol. (1) is not true for minimum deviation.

U;wure fopyu ds fy, A lgh ugh gS

(2) is true only if refracting side are equal.

;fn dsoy viofrZr lrg leku gS rks B lgh gS

(3) Two angles for maximum deviation are 90º

and imin.

vf/kdre fopyu ds fy, nks dks.k 90º rFkk imin. gSA

(4) min. = ( � 1) A.

8. In the figure shown the radius...........

fp=k esa iznf'kZr ,d vory ySUl ...........

Sol.

1

1f

= 3

12

1 110 15

= � 1

12 ;

2

1f

= 4

13

215

=245

; m

1f

= � 2

15

1 2

1 1 1f f f

eq m

1 1 2f f f

= � 1

18

feq = � 18 cm

So, the combination behaves as a concave

mirror

vr% la;kstu vory niZ.k dh Hkkafr O;ogkj djrk gSA

9. A flat mirror M is arranged...........

,d nhokj W ls L nwjh ij nhokj...........

Sol.

Here ;gka, sp = PA and rFkk SQ = QB

so, position of A and B doesn't depend on

separation of mirror from the wall so, the patch

AB will not move on the wall.

vr% A rFkk B dh fLFkfr nhokj ls niZ.k dh nwjh ij

fuHkZj ugh djrh gS] vr% izdkf'kd Hkkx (patch) AB

nhokj ij xfr ugh djsxkA

SA and SB are constant

SA rFkk SB fu;r gS

So, AB = constant.

vr% AB = fu;r gS

11. A flint glass prism and a crown...........

,d f¶yaV dk¡p fçTe o ,d Økmu...........

Sol. Since deviation of mean ray is zero. pqafd ek/;

fdj.k dk fopyu 'kwU; gSA

A (m1 � 1) = A2 (m2 � 1)

6º (1.6 � 1) = A2 (1.9 � 1)





SOLERCT1090815-3

A2 = 6º 0.6

0.9

= 4º

12. A plano convex lens of refractive...........

30 lseh oØrk f=kT;k rFkk 1.5...........

Sol. To get real image of the size of the object,

object should be placed at the centre of

curvature of equivalent mirror.

oLrq ds vkdkj dk okLrfod izfrfcEc izkIr djus ds

fy, oLrq lerqY; niZ.k ds od rk dsUnz ij j[kh gksuh

pkfg,A

m

1 1 2F f f

fm = � 15cm

and rFkk 2

1 1 2f 15 60

F2 = � 10 cm

Hence, object should be placed at 20cm from

the combination

vr% oLrq la;kstu ls 20cm nwjh ij j[kh gksuh

pkfg,A

13. White light travelling in air is...........

ok;q esa tk jgk lQsn izdk'k ty ...........

Sol. (A) It is true when i = 0.

(C) By cauchy's formula n = a + 2

b

, medium

has lowest 'n' for red colour (out of all the

colours in white light)

V = C/n V is max. for red.

(D) = | i � r |

1 × sin i = n sin r

for violet, n is max r is min. in max.

Sol. (A) ;g i = 0 ds fy, lR; gSA

(C) dksfPp ds lw=k n = a + 2

b

, ls yky jax ds fy,

ek/;e dk n U;wure gSA ('osr izdk'k ds lHkh jaxksa esa ls)

V = C/n

yky ds fy, V vf/kdre gSA

(D) = | i � r |

1 × sin i = n sin r

cSaxuh ds fy, n vf/kdre gS r U;wure gS

vf/kdre gSA

14. A prism of refractive index 2 ...........

2 viorZukad o 'kh"kZ dks.k A...........

Sol. sin C = 1

2 c = 45º

for A > 2C no. refraction

A > 2C ds fy, viorZu ugha gksxkA

C < A < 2C some refraction, some reflection

C < A < 2C dqN viorZu] dqN ijkorZu

A < C all refraction.

A < C lHkh viorZu

16. Light of wavelength 4800 Å...........

4800 Å rjaxnS/;Z dk çdk'k vYi...........

Sol. Dispersion will not occur for a monochromatic

light.

,d of.kZ; izdk'k ds fy, fo{ksi.k ugha gksxkA

17. In the figure shown find the...........

n'kkZ;s x;s fp=k esa igys M1 o...........

Sol. For M1 ds fy, :

v1 =uf (�30) (�20)

u � f (�30) � (�20)

= � 60

M = � 1v

u = � 2.

For M2 ds fy, : u = + 20. f = 30

1v

+ 1

20 =

130

v = �60





SOLERCT1090815-4

m2 = � 6020

= 3 m = m1 × m2 = � 6

18. Light travelling in air falls...........

1.5 viorZukad rFkk fizTe dks.k...........

Sol.

we can use sin x = x

(where x is in radians) ge sin x = x dks mi;ksx

dj ldrs gS (tgk¡ x jsfM;u esa gSAis in radians)

sin2° 32

= sin r1°

r1 =43

; r2 = A � r1 = 83

32

sin r2 =43

sine e =98

r2 = 3°

= i + e � A = 2° + 3° � 4° = 1°

19. Dlameter of a planoconvex lens...........

Qksdl nwjh 37 cm ds leryksÙky ...........

Sol. R2 = 9 + (R� 0.5)2

R = 9.25 cm

1

37 = ( � 1) (

1

+1R

)

= 1.25 = cv

v = 83 10

1.25

= 2.4 × 108 m/s.

20. The power of a convex...........

mÙky ySUl [ = 1.5] ...........

Sol. 1f

= rd 1 1 2

1 1R R

on immersing rd decrease hence P decrease

hence (A)

Sol. 1f

= rd 1 1 2

1 1R R

ty esa Mqcksus ij rd de gksrk gS vr% P de gksxh

vr% (A)

21. A stationary person A throws...........

fp=kkuqlkj ,d fLFkj O;fä A {kSfrt ...........

Sol. Stetp (1) Range of projectile

(1) iz{ksI; dh ijkl

2u sin2

g

= 40 m

AC = 40 cm, BC = 30 m

time of flight (mM~M;udky )

2usin

g

= 2 20 sin45

10

=40 110 2

= 4

2= 2 2 sec.

In this same time, the person should catch

the ball.

vr% mM~M;u dky ds cjkcj le; esa O;fä B xsan dks

idM+ ysxkA

V × 2 2 = 30 V = 30 15

2 2 2 m/sec.

= 15 2

2= 7.5 2 m/sec





SOLERCT1090815-5

22. A stone is projected horizontally...........

,d iRFkj dks v pky ls tehu...........

Sol. Time taken to reach the ground is given by h =

12

gt2 .... (1)

Since horizontal displacement in time t is zero

tehu ij igq¡pus dk le; fn;k tk;sxk h =12

gt2

.... (1)

le; t ds inksa esa {kSfrt foLFkkiu 'kwU; gSA

t = 2vf

.... (2)

lehdj.k (1) o (2) ls

h = 2

2

2gvf

23. In projectile motion of a...........

ur ry ij xq:Ro ds vUrxZr...........

Sol. Components of acceleration of particle parallel

and perpendicular to the inclined plane is non

zero. Hence velocity of particle is varying with

time along and perpendicular to inclined

surface.

ur ry ds yEcor~ rFkk lekUrj d.k ds Roj.k dk

?kVd v'kwU; gS] vr% d.k dk osx le; ds lkFk bu

nksuksa fn'kkvksa esa ifjofrZr gksxkA

24. A particle is moving along...........

,d d.k ljy js[kk ds vuqfn'k...........

Sol. From A to B speed increases.

The particle always moves in negative direction.

A ls B rd pky c<+rh gSA

d.k lnSo _ .kkRed fn'kk esa xfr djrk gSA

25. Rain is falling down on ground...........

ckfj'k Å /okZ/kj ds lkFk �� dks.k...........

Sol. If the man increases his speed of walking, then

it is possible that the angle made by rain with

vertical (as observed by man) goes on

decreasing.

;fn O;fDr vius pyus dh pky dks c<+krk gS] rks ;g

laHko gS fd Å /okZ/kj ds lkFk ckfj'k ds }kjk cuk;k

x;k dks.k ¼O;fDr ds lkis{k½ ?kVrk tk;sxkA

26. A man is standing on a road...........

,d vkneh ,d lM+d ij [kM+k...........

Sol. rg rm mgV V V

rg rg mgV V V

45°45°

Initial Final

VVmg

Vrg Vrm

Vrm cos 45° = Vrg cos45°

Vrm = 2 2 m/s = Vrg

Vrm cos 45° = Vmg � Vrg cos45°

m g

1V 2 2

2 +

12 2

2 = 4 m/s

VrgVrm

�Vmg

45° 45°

using v2 = u2 + 2as for the motion of man,

v2 = u2 + 2as vkneh dh xfr ds fy,]

s = 16 m.





SOLERCT1090815-6

27. Two stones are projected...........

nks iRFkjksa dks tehu dh lrg ls...........

Sol. Since maximum heights are same, their time of

flight should be same

T1 = T2

Also, vertical components of initial velocity are

same.

Since range of 2 is greater than range of 1.

Horizontal component of velocity of 2 >

horizontal component of velocity of 1.

Hence, u2 > u1 .

gy pwafd vf/kdre Å pk¡bZ;ka leku gS] blfy, muds

mM~M;udky leku gksaxsA

T1 = T2

çkjfEHkd osx ds Å /okZ/kj ?kVd Hkh leku gSA

pwafd 2 dh ijkl , 1 dh ijkl ls vf/kd gSA

blfy, 2 ds osx dk {kSfrt ?kVd > 1 ds osx dk

{kSfrt ?kVd

blfy, ] u2 > u1

28. A particle is moving rectilinearly...........

,d foeh; esa xfr djrs gq, d.k...........

Sol. a = 3t2 + 1

dvdt

= 3 t2 + 1

v 12

0 0

dv (3t 1)dt

v = 13

0t t = 2 m/s. v = t3 + t

s 1

3

0 0

ds (t t)dt S =1 14 2 = 0.75





SOLERCT1090815-7

PART- B

CHEMISTRY

31. For elements belonging to the same ............

leku vkorZ ls lEcfU/kr rRoksa ds fy, ----------------------

Sol. Inert gases have stable noble gas configuration.

So, they have maximum first ionization energy

in a period. vfØ; xSlsa LFkk;h mRÑ"V xSl foU;kl

j[krh gSaA blfy, ;s ,d vkorZ esa mudk izFke vk;uu

Å tkZ vf/kdre gksrk gSaA

32. The angular momentum of electron ................

'd' d{kd esa bysDVªkWu dk dks.kh; laosx ---------------------

Sol. Angular momentum = h

( 1)2

=

( 1)

For d orbital = 2

Angular momentum = 2(2 1) = 6

gy% dks.kh; laosx = h

( 1)2

= ( 1)

d-d{kd ds fy, = 2

dks.kh; laosx = 2(2 1) = 6

33. Number of electrons having + m .................

26Fe esa mu bysDVªksuksa dh la[;k D;k gksxh ------------------

Sol. 26Fe � 1s2 , 2s2 2p6 , 3s2 3p6, 3d6, 4s2

+ m = 0

= 0, m = 0 i.e. s-subshell 8e�

= 1, m = �1 i.e. one orbital of p. 4e�

= 2, m = �2 i.e. one of d-orbitals

1e� or 2e�

hence there are 13 or 14 electron as in d-orbital

it may be one or two electron having m = �2.

gy- 26Fe � 1s2 , 2s2 2p6 , 3s2 3p6, 3d6, 4s2

+ m = 0

= 0, m = 0 i.e. s- midks'k 8e�

= 1, m = � 1 i.e. p. dk ,d d{kd

4e�

= 2, m = �2 i.e. d- dk ,d d{kd

1e� or 2e�

pwfd ;gk¡ d- d{kd esa 13 ;k 14 bysDVªkWu gS bles

,d ;k nks bysDVªkWu m = �2. j[ksxkA

34. The compressibility factor of a gas ...............

273 K o Vm = 22.4 L ij ,d xSl ----------------

Sol. m real real

m ideal

(PV ) P 22.4Z 1

(PV ) 1 22.4

Preal < 1 atm.

35. Assertion : One molal aqueous ................

dFku : Xywdksl dk ,d eksyy tyh; .................

Sol. Molality = Number of moles of solute

weight of solvent (in kg)

If number of moles of solute = 1

weight of solvent = 1 kg

then, molality = 1, i.e., one molal

Glucose (C6H12O6),

Molecular weight = 180

Number of moles = 180180

= 1

Weight of water = 1 kg

Hence, molality of the solution is one.





SOLERCT1090815-8

gy eksyyrk =

(kg )foys; d seksyksad hl[;k

foyk;d dkHkkj esa

;fn foys; eksyksa dh la[;k = 1

foyk;d dk Hkkj = 1 kg

rc, eksyyrk = 1, vFkkZr~ ,d eksyy~

Xywdksl (C6H12O6),

v.kqlw=k Hkkj = 180

eksyksa dh la[;k = 180180

= 1

ty dk Hkkj = 1 kg

vr%, ,d foys; dh eksyyrkA

36. 1021 molecules are removed from 200 mg .......

200 mg CO2 ls 1021 v.kq ?kVk;s tkrs gSaA -----------------

Sol. 6.023 × 1023 molecules = 1 mole of CO2 = 44 g

1021 molecules of CO2 = 21

23

44 10

6.023 10

g = 7.31

× 10�2 g = 73.1 mg

Mass of CO2 left = 200 � 73.1 = 126.9 mg.

No. of moles of CO2 left = �3126.9 10

44

= 2.88

× 10�3 moles.

Sol. 6.023 × 1023 v.kq = CO2 ds 1 eksy = 44 g

CO2 ds 1021 v.kq = 21

23

44 10

6.023 10

g = 7.31 × 10

�2 g

= 73.1 mg

'ks"k CO2 dk nzO;eku = 200 � 73.1 = 126.9 mg.

'ks"k CO2 ds eksyksa dh la[;k = �3126.9 10

44

= 2.88

× 10�3 eksy

37. The molarity of H2SO4 solution ...............

35°C ij H2SO4 foy;u dh eksyjrk ----------------------

Sol. 98% by mass means-98 g of H2SO4 is present

in 100 g of acid.

M = given mass 1000

mol. mass V

Given, mass = 98 g

Mol. mass = 98 g

V = mass

density =

1001.84

[ density of H2SO4 is 1.84 g/cc]

Putting the values :

M = 98 1000 1.84

98 100

= 18.4 M

Sol. 98% ww

dk vFkZ 100 g vEy esa 98 g H2SO4

mifLFkr gksrk gSA

M = 1000

V

fn;k x;k nzO;eku

v.kqHkkj

fn;k x;k nzO;eku = 98 g

v.kqHkkj = 98 g

V = nzO; eku

?kuRo =

1001.84

[ H2SO4 dk ?kuRo 1.84 g/cc gSA]

eku j[kus ij : M=98 1000 1.84

98 100

=18.4 M

38. The weight of NaCl decomposed by ...............

4.9 xzke H2SO4 }kjk fo[kf.Mr NaCl dk Hkkj ------------

Sol. The reaction takes place as follows

NaCl + H2SO4 NaHSO4 + HCl

x 4.9 g 6 g 1.825 g

According to law of conservation of mass "mass

is neither created, nor destroyed during any

chemical change".

mass of reactants = mass of products

x + 4.9 = 6 + 1.825

x = 2.925 g





SOLERCT1090815-9

Sol. vfHkfØ;k fuEu izdkj ls gksrh gSA

NaCl + H2SO4 NaHSO4 + HCl

x 4.9 g 6 g 1.825 g

nzO;eku ds laj{k.k ds fu;ekuqlkj ^fdlh jklk;fud

ifjorZu ds nkSjku u rks nzO;eku dks cuk;k tk ldrk

gS vkSj uk gh u"V fd;k tk ldrk gSA

vfHkdkjdksa dk nzO;eku = mRiknksa dk nzO;eku

x + 4.9 = 6 + 1.825

x = 2.925 g

39. Which is correct order for ionic ..............

vk;fud f=kT;kvksa dk lgh Øe -----------------------

Sol. Size of anion charge on the ion.

Size of cation 1

chargeonthe ion.

Size of anion > Size of cation.

Sol. _ .kk;u dk vkdkj vk;u ij vkos'k

/kuk;u dk vkdkj 1

vk;u ij vkos'k

_ .kk;u dk vkdkj > /kuk;u dk vkdkj

40. Which of the following represents the................

fuEu esa dkSulk Øe O, S, F rFkk Cl rRoksa ----------------------

Sol. Electron gain enthalpy, generally, increases in a

period from left to right and decreases in a

group on moving downwards. However,

members of III period have samewhat higher

electron gain enthalpy as compared to the

coressponding members of second period,

because of their small size.

O and S belong to VI A (16) group and Cl and F

belong to VII A (17) group. Thus, the electron

gain enthalpy of Cl and F is higher as compared

to O and S.

Cl and F > O and S

Between Cl and F, Cl has higher electron gain

enthalpy as in F, the incoming electron

experiences a greater force of repulsion

because of small size of F atom. Similar is true

in case of O and S ie, the electron gain enthalpy

of S is higher as compared to O due to its small

size. Thus, the correct order of electron gain

enthalpy of given elements is

O < S < F< Cl

Sol. bysDVªkWu xzg.k ,UFkSYih vkorZ esa ck;as ls nk;sa tkus ij

c<+rh gS rFkk oxZ ds vuqfn'k tkus ij ?kVrh gSA fQj

Hkh III vkorZ ds rRo dh bysDVªkWu xzg.k ,UFkSYih NksVs

vkdkj ds f}rh; vkorZ ds rRoksa dh vis{kk vf/kd gksrh

gSA

O rFkk S VI A (16) oxZ rFkk Cl o F VII A (17) oxZ

ls lfEcfU/kr gSA bl izdkj Cl rFkk F dh bysDVªkWu

xzg.k ,UFkSYih] O rFkk S dh vis{kk mPp gksrh gSA

Cl rFkk F > O rFkk S

Cl rFkk F esa Cl dh bysDVªkWu xzg.k ,UFkSYih F ls

vf/kd gksrh gS D;ksafd F ds NksVs vkdkj ds dkj.k vkus

okys bysDVªkWu ij vf/kd izfrd"kZ.k cy yxrk gSA blh

izdkj O rFkk S dh ifjfLFkfr esa Hkh lR; gSA O ds

NksVs vkdkj dh rqyuk esa S dh bysDVªkWu xzg.k ,UFkSYih

mPp gksrh gSA bl izdkj rRoksa dh bysDVªkWu xzg.k

,UFkSYih dk Øe

O < S < F< Cl gSA

41. What is the value of electron gain ................

;fn Na dh IE1 = 5.1 eV gS] rks Na+ ---------------------

Sol. IE1 of Na = � Electron given enthalpy of

Na+ = � 5.1 Volt.

Sol. Na dh IE1 = � Na+ dh bysDVªkWu izkfIr ,UFkSYih

= � 5.1 Volt.





SOLERCT1090815-10

42. Reason of lanthanoid contraction ..............

ySUFkSuksbM ladqpu dk dkj.k -------------------

Sol. Poor screening effect of f-orbital. (f-d{kdksa dk

nqcZy vkoj.k izHkko)


Zn, H O2 ...............

CH2=CH�CH=CH�CH3 O3

Zn, H O2 mRikn ---------------------

Sol. CH2=CH�CH=CH�CH3 O3

Zn, H O2 CH2=O + CH3�

CH=O + OHC�CHO

44. Test to differentiate between ethanol .............

,FksukWy (CH3CH2OH) rFkk fQukWy (Ph�OH) ------------

Sol. Ethanol can not give neutral FeCl3 test but

phenol gives this test.

,FksukWy mnklhu FeCl3 ijh{k.k ugha ns ldrk ysfdu

fQukWy ;g ijh{k.k nsrk gSA

45. Which is the position isomers ...................

;kSfxd

Br

dk fLFkfr leko;oh ------------------

-------

Sol. IUPAC name of

Br

1

2 3 4 5

6 7 is 4-Bromo-3,5-

dimethyl heptane and IUPAC name of

Br

1

2 3 4 5 6 7 is 4-Bromo-3,6-

dimethylheptane. So, position of methyl group is

different hence position isomers.

Sol.

Br

1

2 3 4 5

6 7 dk IUPAC uke 4-czkseks-3,5-

MkbZesfFky gsIVsu rFkk

Br

1

2 3 4 5 6 7 dk IUPAC

uke 4-czkseks-3,6-MkbZesfFky gsIVsu gSA vr% esfFky lewg

dh fLFkfr fHkUu gS] blfy, nksuksa ,d nwljs ds fLFkfr

leko;oh gSA

46. How many tertiary alcohols is/are ...........

v.kqlw=k C5H12O ds fdrus rRkh;d ,YdksgkWy --------------

Sol.

OH

(Only one tertiary alcohol with C5H12O)

OH (C5H12O v.kqlw=k dk dsoy ,d gh rrh;d

,YdksgkWy lEHko gSA)

47. A compound (P) on reaction with "Q" ...........

,d ;kSfxd (P) {kkjh; ek/;e (KOH) dh ------------------

Sol. CH3�CH2�NH2 + CHCl3 + KOH CH3CH2NC

CH3�CO�CH3 + Ca(OCl)2 CHCl3 +

(CH3COO)2Ca

48. Product (mRikn)

No. of monochloro structure ..................

izkIr mRikn esa eksuksDyksjks lajpuk ---------------------

Sol. Alkyl group of this molecule has four types of

replaceable hydrogen atoms.

bl v.kq dk ,fYdy lewg izfrLFkkfir gksus ;ksX;

(replaceable) gkbMªkstu ijek.kqvksa ds pkj izdkj

j[krk gSA





SOLERCT1090815-11

49. Acetaldehyde and benzaldehyde ..............

,lhVsfYMgkbM rFkk csUtsfYMgkbM ----------------------

Sol. Benzaldehyde can not give Fehling test and

Iodoform test but Acetaldehyde gives.

csUtsfYMgkbM Qsgfyax ijh{k.k rFkk vk;ksMksQkWeZ ijh{k.k

ugha ns ldrk gS ysfdu ,lhVsfYMgkbM nsrk gSA

50. Which statement is correct for inductive ...........

izsjf.kd izHkko ds lUnHkZ esa dkSulk dFku -----------------

Sol. Partial charges is developed due to I effect.

vkaf'kd vkos'k I izHkko ds dkj.k mRiUu gksrk gSA

52. In the vander Waals equation, �a� ................

ok.Mj okWy lehdj.k esa] �a� fuEu ---------------------

Sol. In vander Waal�s equation, a signifies the

intermolecular force of attraction.

ok.Mj okWy lehdj.k esa] a vkd"kZ.k ds vUrjkf.od cy

dks crkrk gSA

53. X ml of H2 gas effuse through a hole ................

X ml, H2 xSl ,d ik=k esa fLFkr ,d fNnz -----------------

Sol. For effusion of same volume, 1

2

t

t = 1

2

M

M

1

1

t

M = 2

2

t

M

This is clearly seen from the options that the

ratio of t

M is same for H2 and O2.

5 20 5

2 32 2

gy- leku vk;ru ds fulj.k ds fy,] 1

2

t

t = 1

2

M

M

1

1

t

M = 2

2

t

M

fodYiksa ls ;g Li"V gS fd H2 o O2 ds fy, t

M

vuqikr leku gS 5 20 5

2 32 2

54. 2 gram of H2 gas and 2 gram of He.........

,d cUn ik=k esa 2 xzke H2 xSl o 2 xzke He -------------

Sol. 2HP =

2HX × PT

56. Magnetic moment 2.84 B.M. ..................

fuEu esa ls fdldk pqacdh; vk?kw.kZ 2.84 B.M. .......

Sol. Magnetic moment = n n 2

2.84 Bohr magneton, means 2 unpaired

electrons are present in ion.

Ni+2 = 4s0 3d8

gy% pqEcdh; vk/kw.kZ = n n 2

2.84 B.M. vFkkZr vk;u esa 2 v;qfXer bysDVªkWu

mifLFkr gSA

Ni+2 = 4s0 3d8





SOLERCT1090815-12

57. Resonance is not possible ..............

fuEu esa ls fdlesa vuqukn lEHko ---------------------

Sol. In the conjugation is absent.

esa la;qXeu vuqifLFkr gSA

58. Which is the functional isomers ................

fuEu esa ls dkSu 2-C;wVsukWy dk fØ;kRed -------------------

Sol. 2-Butanol is i.e. CH3�CH�CH2�CH3

OH

C4H10O

Diethyl ether is CH3�CH2�O�CH2�CH3

i.e. C4H10O

2-C;wVsukWy vFkkZr~ CH3�CH�CH2�CH3

OH

C4H10O gSA

MkbZ,fFky bZFkj CH3�CH2�O�CH2�CH3

vFkkZr~ C4H10O gSA

59. Which common name is wrong ..................

dkSulk lkekU; uke xyr -----------------

Sol. CH2�OH

Benzyl alcohol (csfUty

,Ydksgy )

60. Which is the correct structure .............

,fFky,lhVsV ds fy, fuEu ------------------

Sol. CH3COOC2H5 (Ethylacetate) (,fFky,lhVsV)





SOLERCT1090815-13

PART- C

MATHEMATICS 61. If a � b + c < 0 and ....................

;fn a � b + c < 0 vkSj ....................

Sol. (1) ax2 + bx + c = f(x)

imaginary root dkYifud ewy

means vFkZ D < 0 & f(�1) = a � b + c < 0

f(x) < 0 x R

62. Let y = 2

2

x 3x 1

x x 1

x R, ....................

ekuk y = 2

2

x 3x 1

x x 1

x R, ....................

Sol. y(x2 + x + 1) = x2 + 3x + 1

x2(y � 1) + x (y � 3) + y � 1 = 0

sincepawfd x R Sovr%, D 0 (if y 1)

(y � 3)2 � 4(y � 1)2 0

� 1 y 53

, also y can be 1 rFkk y = 1 Hkh

laHko gks ldrk gSA

63. Sign of a, b, c respectively ....................

vkjs[k f(x) = ax2 + bx + c ....................

Sol. parabola upward Å ijh ijoy; a > 0

f(0) > 0 c > 0

x(coordinate) of vertex > 0

'kh"kZ dk x- funsZ'kkad > 0

�b2a

> 0 b < 0

64. The equation of straight ....................

ljy js[kk dk lehdj.k ....................

Sol. Let slope of straight line = m

ekuk ljy js[kk dh ço.krk = m

line js[kk 2x + 3y + 4 = 0

slope ço.krk m1 = � 23

tan 45° = 1

1

m �m

1 mm m =

15

or m = � 5

then equation of line may be rc js[kk dk lehdj.k

(y � 1) = 15

(x � 2) or (y � 1) = � 5(x � 2)

5y � x = 3 or ;k y + 5x = 11

65. A(0, 1), B(2, 1), C(1, 0) ....................

A(0, 1), B(2, 1), C(1, 0) ....................

Sol. Mid point of A & C coinside with mid point of B & D

A vkSj C dk e/; fcUnq] B vkSj D dk e/; fcUnq laikrh gSA

(0, 1)

B (2, 1)

C (1, 0)

A

D (�1,0)

66. The area of triangle ....................

'kh"kZ (0, 0), (6, 0), (0 , 8) ....................

Sol.

C(0, 8)

B(6, 0) A

Area of triangle as formed by mid point = 14

Area of triangle = 14

16 8

2

= 6

e/; fcUnqvksa ls cuk f=kHkqt dk {ks=kQy = 14

f=kHkqt

dk {ks=kQy = 14

16 8

2

= 6

67. If f(x) = 1

3sinx 4cosx � 2....................

;fn f(x) = 1

3sinx 4cosx � 2 ....................

Sol. 3sin x + 4 cosx [�5 , 5]

3 sinx + cos x � 2 [� 7, 3]

2�xcos4xsin3

1

71

,��

,

31





SOLERCT1090815-14

68. Negation of (p ~ q) ....................

(p ~ q) dk udkjkRed ....................

Sol. Negation of ~ (p q) will be (p q)

~ (p q) dk udkjkRed (p q) gksxkA

69. If the equation |x � k| + ....................

;fn lehdj.k |x � k| + ....................

Sol.

k = 0

70. Let 0,4

and ....................

ekuk 0,4

vkSj ....................

Sol. 0,4

tan in 0,4

and 0 < tan < 1

cot in 0,4

and cot > 1

Let tan = 1 � 1 and cot = 1 + 2

where 1 and 2 are very small and

posit ive, then

t1 = 111(1 )

, t2 = 211(1 )

,

t3 = 112(1 )

, t4 = 2(1 )2(1 )

t4 > t3 > t1 > t2

OR

tan in 0,4

and 0 < tan < 1

cot in 0,4

and cot > 1

think only above and conculude result.

Hindi 0,4

0,4

essa tan vkSj 0 < tan < 1

0,4

ess cot vkSj cot > 1

ekuk tan = 1 � 1 vkSj cot = 1 + 2 t gk¡

1 vkSj 2 cgqr NksVs gS rFkk /kukRed gSa] rks

t1 = 111(1 )

, t2 = 211(1 )

, t3 =

112(1 )

, t4 = 2(1 )2(1 )

t4 > t3 > t1 > t2

OR

tan in a 0,4

esa o/kZeku vkSj

0 < tan < 1

cot in 0,4

and cot > 1

mijksDr O;at d ls Li"V : i ls ge gy Kkr

d j ld rs gSA

71. Number of real roots ....................

lehdj.k (x � 1)2 + (x � 2)2 ....................

Sol. (x � 1)2 + (x � 2)2 + (x � 3)2 = 0

only possible is dsoy laHko (x � 1) = 0 & rFkk x �

2 = 0 & x � 3 = 0

so number solutions = 0

blfy, gyksa dh la[;k = 0

72. Number of non positive ....................

vlfedk x x(3 � 4 ) n(x 2)

(x � 5)

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

Sol. x x(4 � 3 ) n(x 2)

(x � 5)

0

(+) (�)

(+)

�1 0 5 (�)

x [�1, 0] (5, )

73. Sum of all real roots ....................

lehdj.k |x|2 � 3|x| + 2 = 0 ....................

Sol. |x|2 � 3|x| + 2 = 0

(|x| � 2)(|x| � 1)

x = ± 2 & x = ± 1





SOLERCT1090815-15

74. If x and y are the real....................

;fn x vkSj y okLrfod ....................

Sol. 12 sinx + 5 cos x 13 and 2y2 � 8y + 21 13

Equality holds if 12 sinx + 5 cos x = 13

and y = 2

lerk gksxh ;fn 12 sinx + 5 cos x = 13 vkSj y = 2

cot x = 5

12 and vkSj y = 2

Hence vr% 12.cot xy2

= 12.cot2x2

= 12 cot x = 12. 5

12= 5

75. If ylnx = x2lny & x,y N ....................

;fn ylnx = x2lny vkSj x,y N ....................

Sol. (lnx) lny = (2lny)lnx

lnx = 0 or ;k lny = 0

x = 1 or ;k y = 1

if ;fn x = 1 3lny = 3

(x, y) = (1, e)

if ;fn y = 1 2lnx = 3

76. The equivalent statement ....................

p q dk rqY; dFku ....................

Sol. p q p q & vkSj q p

(~ p q) (p ~ q)

77. Number of acute angles ....................

U;wu dks.kksa dh la[;k....................

Sol. sin88sin

=

18

8 = n + (�1)n

n = 2m 7 = 2m

= 2m

7

: m = 1, = 27

n = (2m + 1) 8 = (2m + 1) �

= (2m + 1) 9

m = 0, 1 = ,9 3

Number of acute angles are 3

U;wudks.kksa dh la[;k 3 gSA

78. Let f(x) = x3 + x + 1 ....................

ekukfd f(x) = x3 + x + 1....................

Sol. Let f(x) = (x � a)(x � b)(x � c), ...........(i)

hence the roots of P(x) = 0 are a2, b2, c2

P(x) = k(x � a2)(x � b2)(x � c2) for some k

...........(ii)

put x = 0 in (ii)

P(0) = � ka2b2c2 = � 1 (given)

ka2b2c2 = 1

but abc = � 1 k = 1

now,P(x2) = (x2 � a2)(x2 � b2)(x2 � c2)

= (x � a)(x � b)(x � c)(x + a)(x + b)(x + c)

= � f(x) . f(� x)

put x = 2, P(4) = � f(2) . f(�2) = � (11)(�9) = 99

Hindi ekukfd f(x) = (x � a)(x � b)(x � c),

tgk¡ f(x) = 0 ds ewy a, b, c gS ...........(i)

vr% P(x) = 0 ds ewy a2, b2, c2 gksaxs

k ds fdlh okLrfod eku ds fy,

P(x) = k(x � a2)(x � b2)(x � c2) .......(ii)

(ii) esa x = 0 j[kus ij

P(0) = � ka2b2c2 = � 1 (fn;k x;k gS)

ka2b2c2 = 1

ijUrq abc = � 1 k = 1

vc] P(x2) = (x2 � a2)(x2 � b2)(x2 � c2)

= (x � a)(x � b)(x � c)(x + a)(x + b)(x + c)

= � f(x) . f(� x)

x = 2 j[kus ij P(4) = � f(2) . f(�2)

= � (11)(�9) = 99

79. The expression logm ....................

O;atd logm m m m

mlog ........... m n dj.kh fpUg

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

Sol. logm logm

1mnm

= logm n

1

m

= � n

80. If a, b, p, q are non....................

;fn a, b, p, q v'kwU; ....................

Sol. D < 0 and D2 = 0 no common root





SOLERCT1090815-16

81. The complete solution....................

vlfedk ||x � 1| �1| > 2 .................... Sol. ||x � 1|�1| > 2

|x �1|�1 < � 2 |x�1|<�1

no solution (x�1) < � 3

|x �1| > 3

x < � 2

(x � 1) > 3

x > 4 or Solution is gy x (�, � 2) (4, )

82. If 5 lies between the ....................

;fn lehdj.k x2 � px + 4 = 0....................

Sol. f(x) = x2 � px + 4 f(5) < 0

� 5p + 29 < 0 p > 295

83. If log0.5log5(x2 � 4) ....................

;fn log0.5log5(x2 � 4) ....................

Sol. log0.5 log5(x2 � 4) > (log 0.5 1)

0 < log5(x2 � 4) < 1

log5 1 < log5(x2 � 4) < log55

1 < x2 � 4 < 9

= x (�3, � 5 ) ( 5 , 3) 84. Number of real ....................

lehdj.k xlog 2x + x2 = 3x ....................

Sol. 0 < x and x 1 the equation is x2 � 3x + 2 = 0 i.e. x = 1, 2 x = 2 is the only solution

Hindi 0 < x rFkk x 1

lehdj.k x2 � 3x + 2 = 0 gSA

vFkkZr~ x = 1, 2

dsoy x = 2 gy gSA 85. If be root of ....................

;fn lehdj.k .................... Sol. f(x) = 4x2 � 16x +

f(1) . f(2) < 0 & f(2). f(3) < 0 (12, 16) 86. The number of integers ....................

vlfedk |x2 � x| � |x2 �1| .................... Sol. |x2 � 1| + |x �1| > |x2 � x| |a| + |b| > |a � b| ab > 0 (x2 � 1)(x � 1) > 0 (x + 1)(x � 1)2 > 0 x (�1, ) � {1}

87. Orthocentre of the ....................

'kh"kZ A(1, 1), B(3, 0), C....................

Sol. Line A B & AC slopes

AB vkSj AC dh ço.krk

are mAB = � 12

& rFkk mAC = 2

mAB . mAC = � 1

A = orthocentre yEcdsUæ gSA 88. If sin = p, |p| 1, then....................

;fn sin = p, |p| 1 gS] ....................

Sol. Sum of roots ewy ksa d k ;ksx

= tan2

+ 1

tan2

=

2tan 12

tan2

= 2

2tan 12

2tan2

=2

sin=

2p

Product of roots ewy ksa d k xq.kuQ y

= tan2

.cot 2

= 1

equation islehd j.k x2 � 2p

x + 1 = 0 gSA

89. The number of solutions ....................

lehdj.k sinx = |x| ds .................... Sol.

90. If x2 � x � 1 = 0, then ....................

;fn x2 � x � 1 = 0, rks .................... Sol. x3 � 2 x + 3 = (x2 � x � 1) (x + 1) + 4 if x2 � x � 1 = 0, then x3 � 2 x + 3 = 4 Sol. x3 � 2 x + 3 = (x2 � x � 1) (x + 1) + 4

;fn x2 � x � 1 = 0,

rks x3 � 2 x + 3 = 4





SOLERCT1090815-17

CUMULATIVE TEST-1 (CT-1)

TARGET : JEE (MAIN)-2016

ANSWER KEY

CODE-0

PHYSICS

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

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

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

22. (4) 23. (1) 24. (2) 25. (1) 26. (4) 27. (2) 28. (3)

29. (1) 30. (4)

CHEMISTRY

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

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

45. (2) 46. (1) 47. (1) 48. (2) 49. (1) 50. (4) 51. (3)

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

59. (4) 60. (3)

MATHEMATICS

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

68. (3) 69. (2) 70. (2) 71. (4) 72. (2) 73. (4) 74. (4)

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

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

89. (1) 90. (4)

DATE : 09-08-2015



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