1

IMPORTANT INSTRUCTIONS

There are three parts in the question paper A, B, C consisting of Chemistry, Physics

and Mathematics having 30 questions in each part of equal weightage.

Each question is allotted 4 (four) marks for each correct response.

Candidates will be awarded marks as stated above in instruction for correct response

of each question. 1/4 (one fourth) marks will be deducted for indicating incorrect

response of each question.

No deduction from the total score will be made if no response is indicated for an item

in the answer sheet.

There is only one correct response for each question.

Filling up more than one response in each question will be treated as wrong response

and marks for wrong response will be deducted accordingly as per instruction

Maximum time is 3 hrs.

Maximum number of marks is 360

JEE-MAIN

MODEL GRAND TEST

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PHYSICS

1. A cylindrical metallic shell has inner radius r1=2cm and outer radius r2=4cm and it has a

length L=50cm. The inner and outer surfaces are maintained at T1 = 00C and T2=1000C. the

thermal conductivity of metal is 69.3Wm–1 K–1. The rate of flow of heat from outer to inner

surface

1) 3140 J sec–1 2) 4130 J sec–1 3) 1430 J sec–1 4) 5130 J sec–1

2. A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of

the light at a distance of 1m from the diode is

1) 1.73V/m 2) 2.45 V/m 3) 5.48 V/m 4) 7.75 V/m

3. A carpet of mass M made of inexetensible material is rolled along its length in the form of a

cylinder of radius R and is kept on a rough floor. The carpet starts unrolling without sliding

on the floor when a negligibly small push is given to it. The horizontal velocity of the axis

of the cylindrical part of the carpet when its radius reduces to 2

R

1) gR 2) 7

3gR 3)

8

3gR 4)

14

3gR

4. One mole of a monoatomic ideal gas fallows a process 1 2 . The specific heat of the

process is 13

6

R. The value of x on p-axis is

P

x

3p0

v0 5v0

2

v

1

1) 4Po 2) 6Po 3) 5Po 4) 18Po

5. A particle execute in S.H.M whose displacement equation is Y = 5 sin 10 / 3t . The

minimum time after which the K.E of the particle will be maximum is

1) A

B

C 2)

sec

60

3)

sec

90

4) sec

120

6. 2

sin 12

xdt ta

aat t

the value of x is

1) 1 2) –1 3) 0 4) 2

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7. Three identical particles are joined together by a thread as shown in the figure. All the three

particles are moving in a horizontal plane. If the velocity of the outermost particle is 0v ,

then the ratio of tensions in the three sections of the string (T1:T2:T3=?) is

T1 T2 T3

A B CO

1) 6:5:3 2) 3:5:6 3) 3:4:5 4) none of these

8. A particle moves from rest at A on the surface of a smooth circular cylinder of radius r as

shown. At B it leaves the cylinder. The equation relating and is

A

B

r

1) 3 sin = 2cos 2) 2 sin = 3cos 3) 3 sin = 2cos 4) 3 sin = 3cos

9. A stationary body explodes in to four identical fragments such that three of them fly

mutually perpendicular to each other, each with same KE( oE ). The energy of explosion will

be

1) 6E0 2) 3E0 3) 4E0 4) 2E0

10. A wire is suspended vertically from a rigid support. When loaded with a steel weight in air,

the wire extends by 16cm. when the weight is completely immersed in water, the extension

is reduced to 14cm. The relative density of the material of the weight is

1) 2 g/cm3 2) 6 g/cm3 3) 8 g/cm3 4) 16 g/cm3

11. A large open tank has two holes in the wall. One is a square hole of side L at a depth Y from

the top and the other one is a circular hole of radius R at a depth 4Y from the top. When the

tank is completely filled with water, the quantities of water flowing out per second from

both holes are the same. Then R is equal to

1) 2

L

2) 2 L 3) L 4)

2

L

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12. From a solid sphere of mass M and radius R, a spherical portion of radius 2

R removed, as

shown in the figure. Taking gravitational potential V=0 at r= , the potential at the centre of

the cavity thus formed is (G = gravitational constant)

1) 2GM

R

2)

2

GM

R

3)

GM

R

4)

2

3

GM

R

13. A ball of weight w is thrown upwards with a velocity u. If air exerts an average resisting

force F, the speed with which the ball returns back to the thrower is

1) w

uw F

2) w

uw F

3) w F

uw F

4)

w Fu

w F

14. A particle moves in a straight line with retardation proportional to its displacement. Its loss

of kinetic energy for any displacement x is proportional to

1) x2 2) ex 3) x 4) nl x

15. A sample of gas has N molecules with velocity ,2 ,3 .......N . The ratio of the root mean

square to the average speed is

1) 3

8

2)

2 2 1

3 1

N

N

3) 1 2 1

3

N N 4)

2 1

3 2 1

N

N

16. A particle of charge – q and mass “m” moves In a circle of Radius r around an Infinetly long

line charge of linear charge density + then time period will be 1

4K

o

1) 22

mT r

k q

2) 2

mT r

K q

3) 1 2

2

k qT

r m

4)

1

2 2

mT

r k q

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17. A combination of capacitors is setup as shown in figure the magnitude of the electric field

due to a point charge ‘Q’ (having charge equal to the sum of the charges on the 4 f and

9 f capacitors) at a point distant 30m from it would equal to

2 f

3 f4 f

9 f

8v

1) 240 N/C 2) 360 N/C 3) 420 N/C 4) 480 N/C

18. A thin glass rod is bent into a semicircle of radius R. A charge +Q is uniformly distributed

along the upper half and a chage –Q is uniformly distributed along the lower half as shown

in figure find electric field “E” at P the centre of semicircle

+

_ P

+

+

+

__

__

1) 2 2

Q

oR 2)

Q

oR 3)

2

Q

oR 4)

2 2

Q o

R

19. The length of a given cylindrical wire is increased by 100%. Due to the consequent decrease

in diameter the change in the resistance of the wire will be

1) 300% 2) 200% 3) 100% 4) 50%

20. A green light is incident from the water to the air-water interface at the critical angle

select the correct statement

1) the spectrum of visible light whose frequency is more than that of green light will come

out to the air medium.

2) the entire spectrum of visible light will come out of the water at various angles to the

normal.

3) The entire spectrum of visible light will come out of the water at angle of 900 to the

normal.

4) The spectrum of visible light whose frequency is less than that of green light will come

out to the air medium.

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21. A compound microscope has a magnifying power 30. The focal length of its eycpiece is

50m. Assuming the final image to be at the least distance of distant vision (25cm) calculate

the magnification produced by objective.

1) 6 2) –5 3) –4 4) 5

22. In young’s double slit experiment intensity at point is 1

4 of the maximum Intensity. Angular

position of this point is.

1) 1cos3d

2) 1sin

3d

3) 1tan

4d

4) 1cot

3d

23. A non conducting disc having uniform positive charge Q is rotating about its axis with

uniform angular velocity the magnetic field at the center of the disc is

(a) directed out word (b) having magnitude 0

4

Q

R

(c) directed Inword (d) having magnitude 0

2

Q

R

1) (a) only correct 2) (a) and (d) correct

3) b only correct 4) None of the above

24. Shown In the figure is a circular loop of radius r and resistance “R” a variable magnetic field

of induction. B=Boe–t is established inside the coil if the key K is closed the electrical power

developed right after closing the switch is equal to

× RB

k

× ×

× × ×

× × ×

1) 2 2

oB r

R

2)

310oB r

R 3)

2 2 4

5

oB r R 4)

2 2 4

oB r

R

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25. According to Bohr’s theory the expressions for the kinetic and potential energy of an

electron revolving in an orbit is given respectively by

1) 2 2

8 4o o

e eand

r r

2)

2 28 4o ore eand

r r

3) 2 2

8 4o o

e eand

r r

4)

2 2

8 4o o

e eand

r r

26. In a photo electric experiment with light of wavelength the fastest electron has a speed

“ v ” if the exciting wave length is changed to 3

4

the speed of the fastest emitted electron

will be

1) 3

4v 2)

4

3v 3) les than

3

4v 4) greater than

4

3v

27. A fusion reactor uses deuteron as fuel. When 2 deuterons are fused 24 Mev of

energy is released mass of deuteron required to run the reactor of 160 MW

power for 1 hour is

1) 4 gm 2) 2 gm 3) 1 gm 4) 8 gm

28. In the circuit shown in fig the ac source given a voltage V=20cos2000t neglecting source

resistance the volt meter and ammeter reading will be

~6

4

v

5mH

A

50 f

1) 0V, 0.97A 2) 1.68V, 0.47A 3) 0V, 1.4A 4) 5.64V, 1.4A

29. In the following circuit the equivalent resistance between A and B is

2

4 6

8 12

A B10V 2V

1) 20

3 2) 10 3) 16 4) 20

30. The minimum force required to move a body up an inclined plane of inclination 300 is found

to be thrice the minimum force required to prevent it from sliding down the plane. The co-

efficient of friction between the body and the plane is

1) 1

3 2)

1

2 3 3)

1

3 3 4)

1

4 3

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CHEMISTRY

31. Both geometrical and optical isomerisms are shown by

1) [Co(en)2Cl2]+ 2) [Co(NH3)5Cl]2+ 3) [Co(NH3)4Cl2]+ 4) [Cr(ox)3]3-

32. For which of the following parameters the structural isomers C2H5OH and CH3OCH3 would

be expected to have the same values? (Assume ideal behaviour)

1) Boiling points 2) vapour pressure at the same temperature

3) Heat of vaporization

4) Gaseous densities at the same temperature and pressure

33. If the bond length and dipolemoment of a diatomic molecule are 1.25A0 and 1.0D

respectively, what is the percentage of ionic character of the bond ?

1) 10.66 2) 12.33 3) 16.66 4) 19.33

34. Which one of the following reactions represents the oxidizing property of H2O2 ?

1) 4 2 4 2 22KMnO 3H SO 5H O 2 4 4 2 2K SO 2MnSO 8H O 5O

2) 3 6 2 22K ]Fe(CN) ] 2KOH H O 4 2 262K Fe CN 2H O O

3) 2 2 2 2 2PbO H O PbO H O O

4) 2 4 2 2 2 4 2 22KI H SO H O K SO I 2H O

35. 100ml of a sample of water requires 0.98mg of K2 Cr2 O7 (M.W. = 294) in presence of the

H2SO4 for the oxidation of dissolved organic matter in it. The C.O.D of the water sample is

1) 78.4 ppm 2) 1.6 ppm 3) 3.2 ppm 4) 6.4 ppm

36. The mass of 80% pure calcium carbonate required to prepare 11.2 L of CO2 at STP is

1) 50g 2) 62.5g 3) 40g 4) 75g

37. N2 gas is present in one litre flask at a pressure of 7.6 × 10-10 mm of Hg. The number of N2

gas molecules in the flask at O0C is

1) 2.68 × 109 2) 2.68 × 1010 3) 1.34 × 1028 4) 2.68 × 1022

38. What would be the heat released when 350 cm3 of 0.20M H2SO4 is mixed with 650 cm3 of

0.10 M NaOH.

1) 37.0KJ 2) 3.72KJ 3) 3.17KJ 4) 0.317KJ

39. The reaction quotient (Q) for the reaction involved for manufacture of NH3 by Haber process

is given by

2

3

3

2 2

NHQ=

N H. The reaction will proceed form right to left if

1) Q = 0 2) Q = K 3) Q < Kc 4) Q > Kc

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40. The precipitate of CaF2 (Ksp = 1.7 x 10-10) is obtained when equal volumes of the following

are mixed ?

1) 10-4M Ca2+ + 10-4 F- 2) 10-2M Ca2+ + 10-3M F-

3) 10-5M Ca2+ + 10-3M F- 4) 10-3M Ca2+ + 10-5M F-

41. 2 2 3 2CH CH HCl CH CH Cl X . What is ‘X’ ?

1) Al2O3 2) Anhy. AlCl3 3) NaCl 4) MgCl2

42. Aluminium oxide may be electrolysed at 10000C to furnish aluminium metal (At. Mass =

27 amu; 1 Farady = 96,500 Coulombs). The cathode reaction is 3 3 Al e Al . To

prepare 5.12 kg of aluminium metal by this method we require

1) 15.49 10 C of electricity 2)

45.49 10 C of electricity

3) 71.83 10 C of electricity 4)

75.49 10 C of electricity

43. An ionic compound has a unit cell consisting of A ions at the corners of a cube and B ions

on the centres of the faces of the cube. The empirical formula for this compound would be

1) A3B 2) AB3 3) A2B 4) AB

44. (I): HCHO; (II): CH3CHO; (III): CCl3CHO; (IV): CH3COCH3 (V): CCl3COCH3 and

(VI): C6H5CHO

Which of the above compounds undergo aldol condensation ?

1) Only, II, III, IV and VI 2) Only II, IV and V

3) Only II and III 4) All except I

45. The correct order of increasing basic nature for the bases NH3, CH3NH2 and (CH3)2 NH in

aqueous solutions

1) CH3 NH2 < NH3 < (CH3)2 NH 2) NH3 < CH3 NH2 < (CH3)2NH

3) CH3NH2 << NH3 4) (CH3)2 NH < NH3 < CH3 NH2

46. Which of the following hormone is produced by testis ?

1) Progesterone 2) Estradiol 3) Testosterone 4) Estrone

47. Incorrect statement about PHBV is

1) it is co-polymer of 3-hydroxybutanoic acid and 3 – hydroxypentanoic acid

2) it has ester linkage

3) excess of hydroxy pentanoic acid makes the polymer more tougher

4) it undergoes degradation by bacteria

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6 5 3 6 4C H OH +CHCl + NaOH C H

+ONa

CHO

3 2CH -CH == CH

X

Y2 61. B H

2 2 2. H O / O H

2 4d il . H S O

-

48. The trans-alkenes are formed by the reduction of alkynes with

1) 2 4H Pd / BaSO in quinoline 2)

3) 4) Sn HCl

49.

Here, the products X and Y are

X Y

1) n-propyl alcohol n-propyl alcohol 2) iso- propyl alcohol, iso-propyl alcohol

3) n- propyl alcohol iso- propyl alcohol 4) iso-propyl alcohol, n-propyl alcohol

50. .

The electrophile involved in the above reaction is

1) Dichloro carbene (: CCl2) 2) Trichloro methyl anion

3CCl

-

3) Formyl cation (C+ HO) 4) Dichloro methyl cation (C+ HCl2)

51. Oxidation state of +1 for phosphorous is found in

1) H3PO3 2) H3PO4 3) H3PO2 4) H4P2O7

52. A and B are two gases. ‘A’ is identified with a glass rod dipped in NH3 and the gas ‘B’ is

identified with a glass rod dipped in HCl. ‘A’ and ‘B’ are respectively

1) HCl, NO2 2) HCl, NH3 3) NH3, HCl 4) NH3, SO2

53. The hybridization present in IF3 is

1) sp3d 2) sp3 3) sp3d2 4) sp3d3

54. When 1 lit of air is burnt with a mixture calcium carbide and anhydrous calcium chloride, the

reduction in volume of air is about

1) 10 ml 2) 990 ml 3) 100 ml 4) 900 ml

55. The metal extracted by leaching with a cyanide is

1) Mg 2) Ag 3) Cu 4) Na

3Na / liq.NH

4NaBH

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56. For the non – stoichiometric reaction , the following kinetic data were obtained in three

separate experiments, all at 298 K

Initial Concentration

(A)

Initial Concentration

(B)

Rate of formation of C

(mol L-1 s-1)

0.1 M 0.1 M 31.2 10

0.1 M 0.2 M 31.2 10

0.2 M 0.1 M 32.4 10

The rate law for the formation of C is

1)

dck A B

dt

2)

2dck A B

dt

3)

2dck A B

dt

4)

dck A

dt

57. Chemically heroin is

1) morphine monoacetate 2) morphine dibenzoate

3) morphine diacetate 4) morphine monobenzoate

58. Arrange Ce3+, La3+, Pm3+ and Yb3+ in

increasing order of their ionic radii

1) Yb3+ < Pm3+ < Ce3+ < La3+ 2) Ce3+ < Yb3+ < Pm3+ < La3+

3) Yb3+ < Pm3+ < La3+ < Ce3+ 4) Le3+ < Ce3+ < Yb3+ < Pm3+

59. [Co2(CO)8] displays

1) one Co – Co bond, four terminal CO and four bridging CO

2) one Co – Co bond, six terminal CO and two bridging CO

3) no Co – Co bond, four terminal CO and four bridging CO

4) no Co – Co bond, six terminal CO and two bridging CO

60. The electrons identified by quantum numbers n and 1 are

I) n = 4, I = 1 II) n = 4, I = 0 III) n = 3, I = 2 IV) n = 3, I = 1

Can be placed in order of increasing energy as

1) IV < II < III < I 2) II < IV < I < III 3) I < III < II < IV 4) III < I < IV < II

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MATHEMATICS

61. The false statement in the following is

1) p ~ p is a contradiction 2) p q ~ q ~ p is a contradiction

3) ~ ~ p p is a tautology 4) p ~ p is a tautology

62. Let f :R Z be defined such that f(x) = [x] for all x R and 0 a 1 , 1 b 0 then value of

f a 1 f b

f b f a 1

1) 0 2) 1 3) 2 4) 3

63. ABCD is a square plot. The angle of elevation of the top of a pole standing at D from A or C

is 300 and that from B is , then tan is equal to

1) 1

6 2)

3

2 3) 6 4)

2

3

64. If the system of linear equations x + ky + 3z = 0, 3x + ky – 2z = 0, 2x + 4y – 3z = 0, has a non

– zero solution (x, y, z) then 2

x z

yis equal to

1) 30 2) – 10 3) 10 4) – 30

65. Centre of the arc represented by z 3i

Argz 2i 4 4

is

1) 1

5 9i2

2) 1

9i 52

3) 1

9i 52

4) 1

9 5i2

66. If the mean deviation about the median of the numbers a, 2a, 3a, …….. 50a, is 50 then |a|

equals

1) 4 2) 5 3) 2 4) 3

67. If the sum of the first ten terms of the series 2 2 2 2

23 2 1 41 2 3 4 4 .....5 5 5 5

is 16

5m

then m is equal to

1) 102 2) 101 3) 100 4) 99

68. A biased coin with probability p, 0 < p < 1 of heads is tossed until a head appears for the

first time. If the probability that the number of tosses required is even is 2

5, then p is equal

to

1) 2

3 2)

1

2 3)

1

3 4)

1

4

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69. Let A 1,2,3,4 and R be a relation in A given by R 1,1 2,2 3,3 4,4 1,2 3,1 1,3

Then R is

1) reflexive and transitive only 2) transitive and symmetric only

3) equivalence 4) reflexive only

70. OABC is a regular tetrahedron of unit edge then its volume is

1) 1

3 2)

1

72 3)

1

3 2 4)

1

6 3

71. 0 0

1,

n n

n nn nr rr r

rs t

C C

then n

n

t

s

1) 1

4n 2)

1

3n 3)

1

2n 4) n

72. The number of solutions of sin 3x=cos 2x, in the interval ,2

is

1) 1 2) 2 3) 3 4) 4

73. If sin-1x + sin-1y = π/ 2 then 4 4

2 2 2 2

1+x + y

x -x y +yis equal to

1) 1 2) 2 3) 1 / 2 4) 5

74. Find the number of n digit numbers which contain the digits 2 and 7, but not the digits 0, 1,

8, 9.

1) 6n – 5n 2) 6n – 5n – 4n 3) 6n – 5n – 5n + 4n 4) 6n + 4n

75. Number of real roots of (6–x)4 + (8–x)4 = 16 is

1) 1 2) 2 3) 3 4) 4

76. Three positive numbers form an increasing G.P. If the middle term in this G.P is tripled, the

new numbers are in A.P then the common ratio of G.P is

1) 3 2 2 2) 2 2 3 3) 2 2 3 4) 2 3 3

77. If dy

y tan x sec xdx

and y(0) = 0. Then y2

1) 0 2) 1 3) 3

2 4)

1

2

78. The area of the region 2 2 2x, y | y x, x y 2 is

1) 1

4 3

2)

1

4 3

3)

1

4 6

4)

1

2 3

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79. If a, b > 0 and

2 2

0

log bx πdx =

2ax +a. Then the value of ab is

1) 1 2) 1

e 3) e 4) 1

2

80.

6

44 7 7 7

cos xdx

sin x cos x sin x is equal to

1) 3

7 71

1+tan x +c21

2) 3

7 7-1

1+cot x +c3

3) 7

7 3-1

1+cot x +c3

4) 7

7 3-1

1+cot x +c21

81. A hyperbola having the transverse axis of length 2sin is confocal with the ellipse

3x2 + 4y2 = 12 then its equation is

1) 2 2 2 2x cosec y sec 1 2) 2 2 2 2x sec y cosec 1

3) 2 2 2 2x sin y cos 1 4) 2 2 2 2x cos y sin 1

82. The distance of the point (1, 0, 2) from the point of Intersection of the lines x-2 y+1 z-2

= =3 4 12

and the plane x – y + z = 16 is

1) 2 14 2) 8 3) 3 21 4) 13

83. A tangent at a point on the circle x2 + y2 = a2 intersects a concentric circle ‘C’ at two points

P and Q. The tangents to the circle ‘C’ at P and Q meet at a point on the circle x2 + y2 = b2

then the equation of circle ‘C’ is

1) x2 + y2 = (a – b)2 2) x2 + y2 = ab 3) x2 + y2 = (a + b)2 4) x2 + y2 = a2 + b2

84. Consider a family of straight lines (x + y) + λ (2x – y + 1) = 0. The equation of the

straight line belonging to this family that is farthest from (1, -3)

1) 6x + 15y + 7 = 0 2) 6x – 15y – 1 = 0

3) 15y – 6x – 7 = 0 4) 15x – 6y + 7 = 0

85. The value of parameter ‘a’ so that the line (3 – a) x + ay + (a2 – 1) = 0 is normal to the

Curve xy = 1 may lie in the Interval.

1) , 0 3, 2) (1, 3) 3) (-3, 3) 4) (2, 3)

86. If 3

-1

2

3-xf x =Tan

x+3x

then value of ' 1f is

1) 3

10

2)

7

10 3) 1 4)

3

10

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87. If

1f x

x 1 x 2

and

2

1g x

x . Then points of discontinuity of f(g(x)) are

1) 1

1, 0, 1,2

2) 1 1

, 1, 0, 1,2 2

3) 0, 1 4)

10, 1,

2

88. Direction ratios of normal to plane passing through the points (-1, 1, 1), (1, -1, 1) and

perpendicular to the plane x + 2y + 2z = 5 are

1) (2, -1,0) 2) (4, -1,-1) 3) (2, 2, -3) 4) (2, -1, 3)

89. Tangents to the parabola y = x2 + 6 at a point (1, 7) touches the circle

x2 + y2 + 16x + 12y + c = 0 at a point Q then the coordinates of Q are

1) (-6, -11) 2) (-9, -13) 3) (-10, -15) 4) (-6, -7)

90. If the function 2

ax bf x

x 1

has a local

Maximum value 1 at x = 3. Then the value of a – b is equal to

1) 10 2) 13 3) 16 4) 4

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16

JEE MAIN MODEL GRAND TEST

KEY SHEET PHYSICS

1)

1 2) 2 3) 4 4) 2 5) 1 6) 3 7) 1 8) 3 9) 1 10) 3

11) 1 12) 3 13) 4 14) 1 15) 2 16) 1 17) 3 18) 1 19) 1 20) 4

21) 2 22) 2 23) 2 24) 4 25) 3 26) 4 27) 3 28) 4 29) 3 30) 2

CHESMISTRY

31) 1 32) 4 33) 3 34) 4 35) 2 36) 2 37) 2 38) 2 39) 4 40) 2

41) 2 42) 4 43) 2 44) 2 45) 2 46) 3 47) 3 48) 3 49) 4 50) 1

51) 3 52) 2 53) 1 54) 2 55) 2 56) 4 57) 3 58) 1 59) 2 60) 1

MATHEMATICS

61) 2 62) 1 63) 1 64) 3 65) 2 66) 1 67) 2 68) 3 69) 4 70) 2

71) 3 72) 1 73) 2 74) 3 75) 2 76) 1 77) 2 78) 4 79) 3 80) 2

81) 1 82) 4 83) 2 84) 3 85) 1 86) 2 87) 2 88) 3 89) 4 90) 3

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17

JEE MAIN MODEL GRAND TEST

SOLUTIONS

1. Consider an element of thicken dr at a distance r from axis of the shell. Let dT is the

temp. diff between the inner and into surface of the element surface are of element =

2 rL

2Q K rL dTH outer

t dr

2 2

1 1

2r T

r T

dr KLdT

r H

22 1

1

2log

r KLT T

r H

4 2 3.14 69.3 5

log 100 02 H

2.303 0.3010=2 3.14 69.3 5 100

H

2 3.14 69.3 5 100

2.303 0.3010H

= 3140Jsec–1

2. 24

av

PI U c

r

2

0 0

1

2avU E

2

0 02

1

4 2

PE c

r

0 2

0

22.45 /

4

PE V m

r c

3. Loss in G.P.E = gain in Rotational k.E. + gain in translational k.E.

in MgR–M1g R1 = 1 2 1 21 1

2 2cmI w M V

212

1 1

2 4 4

R R l MR M R l

2

2

1 1 2

1

1 1 1

4 2 2 2 2 4

cmcm

M R V MMgR g M R V

R

2 27 1

8 2 8 8

M MMgR Vcm V cm

14

3

gRVcm

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4. C=13

6

R for monoatomc gas Cv=

3

2

R

dQ=du+dw

CdT=CvdT+dw

(C–Cv)= 2 1

dw

T T

dw = 2 1

13 3

6 2

R RT T

2 2 1 1

4

6

Rdw PV PV

0 0 0

25 3

3R x V PV

0 0 0

102

3

xV PV

Area under the graph

dw =1

2(3P0+x)(5V0–V0)=2V0(3P0+x) =6P0V0+2V0x

6P0V0+2V0X=10

3

xV0–2P0V0

18P0V0+6V0x=10V0x–+P0V0

24P0V6=4V0x, x=6P0

5. 5sin 103

y t

210W

T

sec

5T

1

sec6 6 2 30

Tt

6. The quantity 1t

a is dimensions i.e., [a]=[t]

22at t t

0 0 0

22

dt tor M L T

tat t

i.e., ax should also be dimensionless or x=0.

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19

VoV2V1

T3T1 T2A B C

W

7. 00 2

23 , 2

3

vv l v l and 0

13

vv l

2

3 3T m l

2 2

2 3 22 5T T m l T m l

2 2

1 2 1 6T T m l T m l

T3:T2:T1=3:5:6

T1:T2:T3=6:5:3

8. By law of conservation of mechanical energy (K+U)A=(K+U)B

0+mgR cos = –mv+mgR sin

2 2 cos sinBv gR

At B since particle leaves contact with the surface. Hence normal reaction equals zero

2

sin Bmvmg

R

sin 2 cos sinm

mg gRR

3sin 2cos

9. Let three of the fragments move along X, Y and Z axes. Therefore their velocities can

be given as

21 ,v vi v v j and 3v vk

Where v=speed of each of the three fragments. Since, in explosion no net external force

is involved, the net of momentum of the system remains conserved just before and after

the explosion.

ifp p

1 2 3 4 0mv mv mv mv

(p1=0 because the body was stationary), putting the values of 1 2,v v and 3v we obtain,

4v v i j k

Therefore , 4 3v v

The energy of explosion KE SYSTEM

f iE KE KE

2 2 2 2

1 2 3 4

1 1 1 10

2 2 2 2mv mv mv mv

Putting 41 2 3

3

vv v v v and putting 2

0

1

2mv E

We obtain E = 6E0

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10. ; ,Fl

Y Y la l

and a are constants

F l

In the first case, V g=16

In the second case, (V g–V 1 g)=14

Dividing, 1 14 7

16 8

Vg

Vg

1 7

8or

8 7or 38 /or g cm

11. cir sq

vol vol

t t

cir sq

AV AV

2 22 (4 ) 2 ( )r g y L g y

1

2R

12. Potential due to the total sphere (with out removing the body) at the centre of cavity

2

2

1 2 2

3 3 3 12

2 2 2 2 2 8

R

GM r GM GMV

R R R R R

12 1 11

8 8

GM GM

R R

(for remaining)

Potential due to removed portion at its centre

2

8

4

2

MG

GMV

R R

Total potential = 11

8 4

GM GM GM

R R R

13.

W Fa

m

1 W F

am

2 2 2u as 2 2 2 '2 0 2u as a s

20 2W F

u sm

2

22

W F mu

m W F

2

2

mus

W F

W Fu

W F

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21

4

2

9

2q

1q

+

8v

14. Kinetic energy is given by KE= 21

2m

.d d dx

x given xdt dx dt

dvv x

dx

2 2

2

0 0

;2 2

xx

d xdx d xdx or x

As KE 2 , loss in KE x2;

15. 2 2 1 2 1

6rms

N N N

N N

11

2av

N Nand

N N

2 2 1

3 1

rms

av

N

N

16. 2

2

Mq

r or

2 2

2 4

q q q K

om om m

22

2

r mT r

q k

17.

4 128 24

4 12eqq C V C

1

924 18

9 3q C c

Q=q+q1=(24+18)=42 c

9 6

22

1 9 10 42 10420 /

4 0 30

QE N C

r

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22

dq

d

dE

p

d

dq

18. The charge density of each half

2

2

Q Q

The charge on the element dq=2Q

d

Electric field at P due to dq is dE=2

1

4

dq

o R

The resultant field of the two elements is 2dE sin

The result an field due to semicircle /2

0

2 sinE d E

/2

2 2 2

0

1 22 sin

4 0

qd QE

R oR

19. If l is the original length. Then the New length is 2 l if “a” is Initial area of cross

section. Then the new cross sectional area is “2

a” since the volume of wire remains

constant.

1 2 4

2

l lR

a a

R1=4R

The increase in resistance

R1–R=4R–R=3R

The % Increase in resistance

3

100 300%R

R

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20. For critical angle we can write sin1

.

The refractive index of medium inversely proportional to wave length of light “ ”

is lesser for larger wave length. (i.e., lesser frequency).

So will be more for leser frequency of light Hence “d” is correct

21. In case of compound microscope M=mome ------(1)

And incase of final image at leant distance of distant vision

me= 1e

D

f

----(2)

From (1) and (2) M = mo 1e

D

f

M= –30 D=25cm fe=5cm

–30 = mo25

15

= mo=30

56

22. I=Imax cos2

2

2mm

II cos

4 2

axax

11 1cos cos

2 2 2 2

2

2 3 3

2x

2 2sin

3d

sinx d

1sin sin3 3d d

23. Consider a ring of radius x and thikness dx.

Equivalent current in this ring

= 2

charge on ring

= 2

22

Qxdx

R

“dB” due to ring = 0

2

2

2 2

xQdx

x R

2 2

02 2 2

R

o o oQ Q QB dx R

R R R

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24

24. 2 t

o

d dB de A r B e

dt dt dt

2 t

oe r B e

20 oAt e r B

Power developed

222 2 4 2

o or Be r B

PR R R

25. 2 2

0

1

4

Ke ePE

r r

2

0

1 1

2 8

eKE PE

r

26. 2

max

1

2o

hcmV w

Case I 21(1)

2o

hcmv w

Case II 2

11

2o

hcm v w

2

11 4(2)

2 3o

hcm v w

From (I) and (II) 21 24

3 3

wv v

1 4

3v v

27.

E MP NA

t m

6 196 2324 10 1.6 10

160 10 6 103600 4

m

1m gm

28. 22

L cZ R X X

R=10 XL=L=2000 5 10–3=10

Xc= 6

1 110

2000 50 10c

Thus z=10

00

20 22 I 2 1.414

10 2 2

orms

V II A

z

Vrms=4 Irms=5.64V

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29. According to the given figure A is at lowest potential with reference to B. Hence both

the diodes are in reverse bias. Hence equivalent circuit can be re drawn as follows.

Equivalent resistance between A and B is R=8+2+6=16

30. F1=mg sin 300+ 0cos30mg

F2=+ 0cos30mg =mg sin 300

F1=3F2

3 3 3 3

2 2 2 2

mg mg mgmg

34.

2mg mg

1

2 3

31. 2 2[Co(en) Cl ] shows optical isomerism as well as geometrical isomerism

32. Same density of vapours at identical pressure

33. % ionic character observed

calculated100

observed 100e d

18

10 8

10100

4.8 10 1.25 10

= 16.66

34. Iodide is oxidized to iodine

35. Calculate using the formula,

COD = 62 2 7wtofK Cr O 810

49 wtofwater

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3 2CaCO CaO CO

W? 11.2 Litre

22.4 Litre100 g

36.

Weight of CaCO3 required to produce 11.2 Litres of CO2 = 50 g

80 g of CaCO3 is present in 100 g of the sample

Therefore, 50g of CaCO3 is present in 100 50

62.5g80

37. PV = nRT

107.6 10

1 n 0.0821 273760

n = number of moles

Number of molecules = n x 6 x 1023

38. Number of gram equivalents of Acid =

350 0.2 2

0.141000

Number of gram equivalents of Base = 650 0.1

0.0651000

One gram equivalent of H+ & OH- produces 57.3 KJ

Therefore, 0.065 gm equivalents of H+ & OH- produces = 57.3 x 0.065 = 3.71 KJ

39. Concept of reaction quotient and equilibrium constant

40. Kw = [Ca2+] [F-]2; w SPK K

41. Addition of HCl is catalysed by AlCl3

42. 27 gms – 3 x 96500 coulombs

5.12 x 103gms - ?

33 96500 5.12 10

27

43. Number of A atoms = 8 x 1/8 = 1

Number of B atoms = 6 x 1/2 = 3

Composition is AB3

44. Presence of – hydrogen atom usual

45. Usual strength is in the order : 0 032 1 NH

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46.

2x + 3(93 – x) – 2 x 100 = 0

47. Polymer is tougher if hyroxybutanoic acid is more

48. Reagent is sodium metal in liquid ammonia

49. –OH goes to carbon 2 in the presence of dilute sulphuric acid

50. Electrophile is Cl2C :

51. 2.3032.303 5 5

log log10 4 30 p

5 5

3log log4 p

35 5

P 2.564 p

52. A is acid and B is base. In both cases NH4Cl fumes are formed

53. Central atom is Iodine

Iodine has 3 bond pairs and 2 lone pairs

54. air contains 1% of noble gases.

100ml air ------ 1 ml

1000 --------- ? = 10 ml

Reduction in volume

= 1000 – 10 = 990 ml.

55. 0 20.06E E log Cu

n

20.060.34 log10

2

= 0.34 – 0.06 = + 0.28

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56

n2 2

1 1

r c

r c

Does not depend on B

With respective A

n3

3

2.4 10 0.2

0.11.2 10

n2 2

Therefore, n = 1

57. Morphine diacetate

58. Order of ionic radii of lanthanides

59. Bond is present between Co atoms. Two ‘ carbonyls’ are bridging

60. Energy of orbital = n + l

More energy, if ‘n’ is more.

61. By truth table

62. f(a) = 0 f(b) = - 1

1 1

01 1

63.

AD a BD 2a

0 htan 30

a

a 3 h

h h

tanBD 2 a

h 1

tan62 3 h

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64. Eliminating y from first two equations, we get

2x – 5z = 0 2x = 5z

x = 5λ , z = 2λ

2x + 4y – 3z = 0 y = - λ

2 2

xz 5 . 210

y

.

65.

z 3i

argz 2i 4 4

z 3i

argz 4 2i 4

C is the centre BCA / 2

i /21

1

z 3ie i

z 4 2i

z1 – 3i = i z1 + 4i + 2

(1 – i) z1 = 2 + 7i 1

2 7iz

1 i

1

1z 9i 5

2 .

66. a, 2a, ----- 50 a

Median = 25a 26a

25.5a2

M.D = 50.

i

1x Median 50

n

i

1x Median 50

50

| a | (25) (25) = 50 x 50

| a | = 4

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67.

2 2 2 228 12 16 244 ....

5 5 5 5

2

2 2 2 2 2 242 3 4 5 6 ....... 11

5

216

11 125

16

50525

68. 3 5 2qp q p q p .......

5

2

qp 2

51 q

2

p 1 p 2P 3p 1 0

51 1 p

1

P3

69. Reflexive only

70. Consider 2

a.a a .b a.c 1 1/ 2 1/ 2

a b c b .a b.b b.c 1/ 2 1 1/ 2

1/ 2 1/ 2 1c.a c.b c.c

= 1

2

1a b c

2

volume = 1 1 1

a b c6 6 2 72

72. cos2x sin3x

cos 3x cos2x2

2x 2n 3x2

5x 2n2

2n

x5 10

73. x = 1, y = 0.

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74. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

2, 3, 4, 5, 6, 7

Req. no. of ways = 6n –[ (digit containing 2 but not 7) + (digit containing 7 but not 2) –

(not containing the digits 2 & 7)]

= 6n – [5n + 5n – 4n ]

= 6n – 5n – 5n + 4n.

75. 7 – x = t

(t – 1)4 + (t + 1)4 = 16

t4 + 6t2 + 1 = 8

t4 + 6t2 – 7 = 0

(t2 + 7) (t2 – 1) = 0 t = 1

7 – x = 1, 7 – x = - 1

X = 6. X = 8

76. 2a, ar, ar G .P

2a, 3ar, ar A .P

2(3ar) = a + ar2

r2 – 6r + 1= 0

r = 3 + 2 2

77. dy

yTan x sec xdx

Tan x dx

I.F e sec x

Solution y sec x = 2sec x dx Tanx C

If x 0 y 0 c 0

y sin x, y / 2 1

78.

y2 = x, x2 + y2 = 2 meet at x = 1

1

2 2

0

A 2 2 y y dy

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1

A2 3

79.

2 2

0

log bxI dx

2ax a

Put x = a tanθ

/2

2

2 2

0

log b a TanI a sec d

a sec

/2

0

1I log ab log Tan d

a

I log ab 02a 2a

log ab = 1

ab = e

80.

6

44 7 7 7

cos xdx

sin x cos x sin x

6

44 4 7 7

cos xdx

sin x cos x 1 Tan x

822

4 44 7 77 7

sec x Tan xsec xdx dx

Tan x 1 Tan x 1 Tan x

Put 1 + (Tan x)-7 = t.

90. Length of transverse axis is 2a = 2sinθ

a =sinθ also for the ellipse 3x2 + 4y2 = 12

1

e =2

Foci = (1, 0)

as the hyperbola is confocal with the ellipse

then focus of the hyperbola is (1, 0) 1ae 1

1e cosec

22 2 1 2b a e 1 cos

Equation of hyperbola is x2

2 2 2 2x cosec y sec 1

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82. y +1 z -2x-2

= = = λ3 4 12

P= 3 2, 4 1,12 2

Plies on x – y + z = 16 1

P = (5, 3, 14)

Given Q = (1, 0, 2)

PQ 16 9 144 13

83. 1) x2 + y2 = a2

2) x2 + y2 = r2 (Req – circle)

Chord of contact of R (h, K)

Wrt (2) is S1 = 0

2xh yk r PQ

Which is tangent to x2 + y2 = a2

2

2 2

ra

h K

Since R lies on x2 + y2 = b2

2 2 2h K b

2

2ra r ab

b

2 22 x y ab

90. (x + y) + λ (2x – y + 1) = 0 1 2L +λL 0

Given family is concurrent at 1 1

P ,3 3

Given point Q = (1, -3)

Slope of 5

PQ2

Req line is 1 2 1

y x3 5 3

15y 6x 7 0

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90. Slop of the line (3 – a)x + ay + a2 – 1 = 0 is

a 3

a

and 2

dy 1xy 1

dx x

Slope of normal = x2

2 a 3x 0 a a 3 0

aa , 0 3,

86.

3 31 1

2

3 x 3 xf x Tan Tan

x 1 3xx 3x

2

1

3x

xTan1 3x

1 1 23Tan Tan x

x

1

2 4

2

1 3 2xf x

9 x 1 x1

x

1 3 7f 1 1

10 10

87.

4

2 2

2 2

1 xf g x

1 1 1 x 1 2x1 2

x x

F(g(x)) is discontinuous at 1

x 1, x2

And also g is discontinuous at x = 0.

88.

x 1 y 1 z 1

2 2 0 0

1 2 2

2x + 2y – 3z + 3 = 0

Dr’s = (2, 2, -3).

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89. Y = x2 + 6 ----- (1)

Tangent at (1, 7) w.rt(1) is 2x – y + 5 = 0 ---(2)

(2) touches the circle at a point Q

Q is the Foot of the r from centre

16 6 5h 8 K 6

2 1 5

Q (h, K) = (-6, - 7).

90. 2

ax bf x

x 1

2

1

22

x 1 a ax b 2xf x

x 1

1f 3 0 5a 3b 0 1

And f(3) = 1 3a + b = 8 (2)

Solve (1) and (2)

a = 6, b = -10

a – b = 16

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