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Page No.1

Sub: PHYSICS JEE (MAIN) GRAND TEST - 1 1. In an experiment of simple pendulum, time period measured was 50 S for 20 vibrations when the

length of the simple pendulum was taken 100 cm. If the least count of stop watch is 0.1 S and that of meter scale is 0.1 cm, the maximum possible percentage error in the measurement of acceleration due to gravity ‘g’ is

A) 0.2 % B) 0.3 % C) 0.4 % D) 0.5 % 2. A particle is projected with a certain velocity at an angle α above the horizon from the foot of an

inclined plane of inclination 030 . If the particle strikes the plane normally, then α is equal to

A) ( )0 130 tan 2 3−+ B) 045 C) 0 1 330 tan

2−

+

D) 060

3. From what minimum height ‘h’ must the system be released when a spring is unstreched so that after perfectly inelastic collision (e = 0) with ground, ‘B’ may be lifted off the ground (spring constant = k)

A) 4

mg

k B)

4mg

k C)

2

mg

k D)

3

mg

k

4. A standing wave having 4 nodes and 3 antinodes is formed between two atoms having a distance 0

1.8A between them. The wavelength of the standing wave is

A) 0

1.2A B) 0

2.4A C) 0

3.6A D)0

4.8A 5. An ideal battery of emf E, an inductor of inductance ‘L’ and resistor of resistance ‘R’ are connected in

series. Maximum energy stored in the magnetic field is

A) 2

24

LE

R B)

2

2

2LE

R C)

2

2

LE

R D)

2

22

LE

R

6. The electric field associated with a wave in air has an amplitude of 121Vm− . Then the amplitude of oscillations of magnetic field is

A) 25 10 T−× B) 82.5 10 T−× C) 87 10 T−× D) 73.5 10 T−×

7. 5 g of water of 030 C and 5 g of ice at 020 C− are mixed together in a calorimeter. The final weight of ice in the mixture will be

A) 1.25 g B) 2.5 g C) 3.75 g D) 0 g 8. Given that mass of the earth is M and its radius is ‘R’. A body is dropped from a height equal to the

radius of the earth above the surface of the earth. When it reaches the ground its velocity will be

A) 1/2

2GM

R

B) 1/2

GM

R

C) GM

R D)

2GM

R

9. The time constant of a L-R circuit is 15 s. When a resistance of 20Ω is connected in series in a previous circuit then time constant becomes 3 seconds, then the self inductance of the circuit is

A) 75 H B) 50 H C) 25 H D) 100 H

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Page No.2

10. A plot of binding energy per nucleon bE , against the nuclear mass A is given : P,Q,R,S,T,U

corresponding to different nuclei. consider four reactions (i) P Q R E+ → + (ii) R P Q E→ + + (iii) S T U E+ → + iv) U S T E→ + +

Where E is the energy released, in which reaction is ‘E’ positive A) (i) & (iv) B) (i) & (iii) C) (ii) & (iv) D) (ii ) & (iii) 11. Two capacitors 1C & 2C are charged to 120 V and 200 V respectively. It is found that by connecting

them together the potential on each one can be made zero. Then, A) 1 25 3C C= B) 1 23 5C C= C) 1 23 5 0C C+ = D) 1 29 5C C=

12. A thin semi-circular ring of radius ‘r’ has a positive charge ‘q’ distributed uniformly over it. The net field ‘E’ at the centre ‘O’ is

A) ^

2 204

qj

rπ ∈ B)

^

2 204

qj

rπ−∈

C) ^

2 202

qj

rπ−∈

D) ^

2 202

qj

rπ ∈

13. A gas is found today 2P V = constant. The initial temperature and volume are 100 K & 0V . If the gas

expands to volume 03V . The final temperature of gas in Kelvin is

A) 100 3 B) 100 / 3 C) 300 D) 300 14. Two particles ‘A’ & ‘B’ , each of mass ‘m’ and moving with velocity ‘V’, hit the ends of a rigid of

mass ‘M’ and length ‘l’ simultaneously and stick to the bar. The bar is kept on a smooth horizontal plane (as shown). The angular speed of the system (bar + particle) after the collision ( M = 6m)

A) V

l B) zero C)

3V

l D)

2V

l

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Page No.3

15. To double the covering range of a T.V. transmitter tower, its height should be made

A) 2 times B) 4 times C) 1

2 times D)

1

4 times

16. Magnification produced by an Astronomical telescope for normal adjustment is 10 and length of telescope is 1.1 m. The magnification when the image is formed at least distance of distinct vision

(D = 25 cm) is A) 6 B) 14 C) 16 D) 18

17. In the following circuit if a) input P = 1 & 0Q = then output is 1Z

b) 1& 1P Q= = then output is 2Z . Which of the following is correct?

A) 1 0Z = B) 1 1Z = C) 1 1Z = D) 1 0Z =

2 0Z = 2 1Z = 2 0Z = 2 1Z = 18. Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the

minimum separation between two objects that human eye can resolve at 500 nm wavelength is A) 300 mµ B) 100 mµ C) 30 mµ D) 1 mµ 19. In the given circuit the impedance of the circuit will be

A) 30Ω B) 20 Ω C) 90 Ω D) 60 Ω 20. Consider a uniformly charged solid non conducting sphere of radius ‘R’ and charge ‘Q’. Consider

three spherical regions centered at the centre of the uniformly charged solid non conducting sphere, given below

Region (1) 1 0R r≥ ≥

Region (2) 2 1R r R≥ ≥

Region (3) 2r R∞ ≥ ≥

If electrostatic potential energy stored in each of the three regions is equal then 2

1

R

R=

A) 4 B) 1 C) 3 D) 2

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Page No.4

21. In the arrangement shown density of block ‘B’ is 2ρ and density of liquid is ‘ρ ’. The system is

released from self sothat the block ‘B’ moves up when in liquid and moves down when out of liquid with the same acceleration. Then the mass of the block A is

A) 2 m B) 7

6

m C)

5

3

m D)

9

4

m

22. Electromagnets are made of soft iron, because soft iron has A) Low susceptibility and low retentivity B) High susceptibility and low retentivity C) High susceptibility and high retentivity D) Low permeability and high retentivity 23. A stationary hydrogen atom of mass M emits a photon corresponding to the first line of lyman series.

If ‘R’ is the Rydberg’s constant, the velocity of that atom acquires is ( h = Planck’s constant)

A) 3

4

Rh

M B)

4

Rh

M C)

2

Rh

M D)

Rh

M

24. How many photons of a radiation of wavelength 75 10 mλ −= × must fall per second on a blackened

plate in order to produce a force of 56.62 10 N−× ?

A) 225 10× B) 223 10× C) 181.67 10× D) 182 10× 25. In a Young’s double slit experiment, first maxima is observed at a fixed point ‘P’ on the screen. Now

the screen is continuously moved away from the plane of the slits. The ratio of intensity at point ‘P’ to the intensity at point ‘O’ (Centre of the screen)

A) remains constant B) keeps on decreasing C) first decreases and then increases D) first decreases and then becomes constant 26. An electron is projected at an angle θ with a uniform magnetic field. If the pitch of the helical path is

equal to its radius, then the angle of projection is A) ( )1tan π− B) ( )1tan 2π− C) ( )1cot π− D) ( )1cot 2π−

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Page No.5

27. A wire carrying a 10A current is bent to pass through various sides of a cube of side 10 cm as shown

in figure. A magnetic field ^ ^ ^

2 3B i j k T = − +

is present in the region. Then the net torque on the loop

is ( in Nm)

A) ^ ^

0.2 0.4i k− B) ^ ^

0.1 0.4i k− C) ^ ^

0.1 0.4i k− − D) ^ ^

0.1 0.4i k− + 28. One of the refracting surface of a prism of angle 030 is silvered. A ray of light incident at an angle of

060 retraces its path. The refractive index of the material of prism is

A) 3 B) 2 C) 3

2 D) 2

29. The potential energy of a particle of mass 1 kg in motion along the x-axis is given by ( )4 1 cos2U x J= − . Where ‘x’ in metre . The time period of small oscillations (in second) is

A) 2π B) 3

π C)

2

π D) 2π

30. An ideal monoatomic gas undergoes a cyclic process ABCA as shown in the figure. The rate of heat absorbed during AB to the work done on the gas during BC is

A) 5

ln 2 B)

5

3 C)

5

4ln 2 D)

5

6

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Page No.6

PHYSICS KEY :

1) D 2) C 3) B 4) A 5) D 6) C 7) C 8) B 9) A 10) A

11) B 12) C 13) A 14) A 15) B 16) B 17) B 18) C 19) D 20) D

21) D 22) B 23) A 24) A 25) C 26) B 27) A 28) A 29) C 30) C

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Page No.7

PHYSICS JEE MAINS GRAND TEST-1 SOLUTIONS

1) 22

4l

gT

π=

2 0.1 0.1100 100 100 100 2 100 0.5%

100 50

g l T

g l T

∆ ∆ ∆× = × × × = × + × × =

2)

( )2 sin 30

cos30

UT

g

α −=

0x x xV U a T= = −

x xU a T=

( ) ( )2 sin 30cos 30 sin30

cos30

UU g

g

αα

−− = ×

( ) cot 30tan 30

2α − =

1 330 tan

2α −

= +

3) 2U gh=

To lift the block ‘B’, the elonagation ‘x’ must be 2 /mg k work-energy theorem

2 21 1 10

2 2 2mU mg x kx

−− = −

Substitute 2mg

xk

=

2

2 8 42

mg mgU gh h

k k⇒ = = ⇒ =

4) 4 nodes and 3 antinodes

031.8 1.2

2A

λ λ= ⇒ =

5) 2 2

2max max 2

1 1

2 2 2

E LEU Li L

R R = = =

6) 80 00 8

0

217 10

3 10

E EC B T

B C−= ⇒ = = = ×

×

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Page No.8

7) water iceQ Q=

water ice icems ms xLθ θ∆ = ∆ +

5 1 30 5 0.5 20 80x× × = × × × × 1.25x = g of ice will melt Mass ice remained = 5-1.25 = 3.75 g 8) Total initial energy = T.E final

21

2

GMm GMmV

R R R

− −= ++

GM

RV =

9) 1 15L

Rτ ==

2 320

L

Rτ =

+=

75L H⇒ = 10) 1st reaction is fusion and 4th reaction is fission

11) 1 1 2 2

1 2

0C V C V

C CV

−+

= =

1 1 2 2 0V C VC − =

1 2120 0200CC × − =×

1 253 CC =

12)

q

rπλ =

q

ddq θπ

×=

2

0

^1sin sin

4

dqE dE j

rθ θ

π = = − ∈

∫ ∫

= 2 2 2

0 00

^^1sin

4 2

q d qj j

r r

θ θπ π π

− = − ∈ ∈ ∫

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Page No.9

13) 2VP = constant but V nRTP =

2

constantnRT

VV

=

2 VT α

VT α

3V V→

3 100 3T KT =→

14) Conservation of angular momentum 1 2LL = given M = 6m

2 2

l lmv IWmv + =

2 26

12 12

Ml mlI ==

( )22

2

2

6/ 2

2

ml lm l m w

lmvl

+ +

=

/V lw w V l= ⇒ =

15) ( ) 2 ET Range R hTD =

2T Dh α

2D D→ 4T Thh →

16) 0 1.1ef mL f + ==

0 10e

f

fm ==

0.1ef =

0 1f =

0near point

0.11 10 1 14

25e

e

ff

f Dm = + = + =

17) .P Q QZ +=

1 1 0 . 0Z += = 0.1 0 1= =

2 1 1 . 1 0.0 0 1Z + = = ==

18) 1.22d

D a

λθ ==

9

2 2

1.22 500 10

25 10 0.25 2 10

d −

− −

× ×=× × ×

5 303 10 m md µ− == ×

19) L Cii i +=

1 1 1 1

6060 30 60L C

v v Vz

z x x z= ⇒ = = ⇒ = Ω∼ ∼

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Page No.10

20) 23

5

kQ

RE = ,

2 2

,2 10out in

kQ kQE

R RE ==

Given 2

1 2 3 5

KQE E

RE = = =

2 2

2

5

2 5 2

KQ kQ RR

R R= ⇒ =

2 2

1

5 5

2 5 4

KQ kQ RR

R R= ⇒ =

2

1

2R

R=

21) in liquid in airaa = ↓↑ But 3 3

2 2B B l

mV g g mgf ρ ρ

ρ= = × =

( ) ( )A B B B A

A B A B

m g m g f m mg

m m m m

− − −=

+ +

33

323 3

A

A A

m g mg mgm mA

gm m m m

− − − =+ +

3

22A Am m mm − = −

9

42 A

mm =

22) Conceptual 23) Law of conservation of linear momentum

h

MVλ

=

2 2

1 1

1 2hRMV −

=

3

4

hR

MV =

24) dp nh

dt tF

λ==

5 7

2234

6.62 10 5 105 10

6.62 10

n F

t h

λ − −

−

× × ×= = = ××

25) ( )2max

2cos / 2 , &

Y L

dI I

π λφ φ ββ

= ==

First maxima ( )/ 2 1cos φ =→

As ‘L’ cosβ φ φ↑ ↑ ↓ is -ve, max

I

I↓

As cosφ is + ve max

I

I↑

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Page No.11

26) sinmv

Bqr

θ= Pitch 2

cos cosm

BqP V T V

πθ θ= = ×

Given P r= ( )1 2tan πθ −⇒ =

27) M Bτ = ×

1

2A = base x height

iA Bτ = ×

4 21

^110 10 10

2A j m− = × × ×

^

0.1 0.05 0.05

2 3 1

i j k

τ =−

4 22

^110 10 10

2A k m− = × × ×

^ ^

0.2 0.4i k Nmτ = −

( )4 23

^10 10 10A i m−= × ×

28) 2 0r = ⇒ 1 2 Ar r+ =

⇒ 1 300r + = ⇒ 01 30r =

1

1

sin sin 60 3 / 23

sin sin30 1/ 2gi

rµ = = = =

29) Given ( )1 cos 24 xU −=

2dUmw x

dxF

− = −=

2

2

48sin 2 1 x

Tx

π ×− = −

For small oscillations sin 2 2x x≃ 2

Tπ

⇒ = second

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Page No.12

30) ( )0 0

52

2AB p

RnC dT n T T

isobaric

dQ = = × −

00 0

0

22 ln 2 ln 2BCBC

VdW nR T nR T

VdQ = = × =

5

4ln 2AB

BC

dQ

W=