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Sub: PHYSICS JEE(Main) GRAND TEST - 2

1. A stone is projected from a balloon which is ascending with a velocity 13ms− . The velocity of the

stone w.r.to balloon is 12ms− at an angle of 030 with the horizontal. The velocity of the stone with

respect to ground in 1ms− is

A) 5 B) 13 C) 1 D) 19 2. A screw guage with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the

thickness of thin sheet. What is the thickness of the sheet, if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line? (zero error of screw guage is zero)

A) 0.50 mm B) 0.70 mm C) 0.75 mm D) 0.8 mm 3. A constant power ‘P’ is applied to a particle of mass ‘m’. The distance travelled by the particle when

its velocity increases from 1V to 2V is (neglect friction)

A) ( )2 22 1

3PV V

m− B) ( )2 13

mV V

P− C) ( )3 3

2 13

mV V

P− D) ( )2 2

2 13

mV V

P−

4. A ball of mass ‘m’ collides with the ground at an angle ' 'α with the vertical. If the collision lasts for time ‘t’, the average force exerted by the ground on the ball is (e = coefficient of restitution between the ball and the ground)

A) cosemu

t

α B)

( )1 cose mu

t

α+ C)

( )2 1 cose mu

t

α+ D)

emu

t

5. The acceleration ‘a’ of the plank ‘P’ required to keep the centre ‘C’ of a cylinder in a fixed position during the motion is (no slipping takes place between cylinder and plank)

A) 2 sing θ B) sing θ C) sin2

g θ D) 2 sing θ

6. A particle ‘P’ lies on the axis of a ring of mass ‘M’ and radius ‘a’, at a distance ‘a’ from its centre ‘C’. A small particle starts from ‘P’ and reaches ‘C’ under gravitational attraction only. Its speed at ‘C’ will be

A) 2GM

a B)

2 11

2

GM

a −

C) ( )22 1

GM

a− D) Zero

7. A particle of mass 55 10 kg−× is placed at the lowest point of a smooth parabola having the equation 2 40x y= (x,y in cm) If it is displaced slightly and it moves such that it is constrained to move along

the parabola, the angular frequency of oscillation will be approximately A) 11S − B) 17S − C) 15S − D) 19S −

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8. A block of mass 2 kg and specific gravity 5

2 is attached with a spring of force constant 1100K Nm−=

and is half dipped in the water. If extension in the spring is 1 cm. The force exerted by the bottom of

the tank on block is ( )210g ms−=

A) 19 N B) 20 N C) 16 N D) 17 N 9. A soap bubble of radius ‘r’ is placed on another bubble of radius 2r. The radius of the surface common

to the both the bubbles is

A) 2

3

r B) 3r C) 2r D) r

10. Speed of sound wave is ‘V’. If a reflector moves towards a stationary wave emitting waves frequency ‘f’ with speed ‘U’, the frequency of reflected wave will be

A) V U

fV U

+ −

B) V U

fV U

− +

C) V U

fV

+

D) V U

fV

−

11. Two radioactive materials 1x and 2x have decay constants 10λ and λ respectively. Initially they have

same number of nuclei. The ratio of the number of nuclei 1x to that of 2x will be 1

e after a time

A) 1

10λ B)

1

11λ C)

11

10λ D)

1

9λ

12. A ring consisting of two parts ADB and ACB of same conductivity ‘K’ carries an amount of heat ‘H’. The ADB part is now replaced with another metal keeping the temperatures 1T and 2T constant. The

heat carried increases to ‘2H’. What should be the conductivity of the new part ADB part?

(given 3ACB

ADB= )

A) 7

3K B) 2 K C)

5

2

K D) 3 K

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Page No.3

13. A thin magnetic iron rod of length 30 cm is suspended in a uniform magnetic field. Its time period of

oscillation is 4 S. It is broken into three equal parts parallel to its breadth. The time period in seconds of oscillation of one part when suspended in the same magnetic field is

A) 1

3 B)

4

3 C) 12 D)

4

3

14. Two charged spheres ‘A’ and ‘B’ charged with charges of + 10 C and + 20 C respectively and separated by a distance of 80 cm. The electric field at a point on the line joining the centres of the two spheres will be zero at a distance from sphere ‘A’ is

A) 20 cm B) 33 cm C) 55 cm D) 60 cm 15. In the Young’s double slit experiment apparatus shown in figure, the ratio of maximum to minimum

intensity on the screen is 9. The wavelength of light used is ' 'λ , then the value of ‘Y’ is

A) D

d

λ B)

2

D

d

λ C)

3

D

d

λ D)

4

D

d

λ

16. A circuit as shown in figure, the value of currents ( )1 2i i+ is

A) Zero B) 2A− C) 2A D) 4 A 17. When a object is kept at distances x and y from lens, a real and virtual images are formed respectively

having same magnification. The focal length of lens is

A) 2

x y− B)

2

x y+ C) xy D)

2x

y

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Page No.4

18. The angular frequency of damped oscillator is given by 2

24

k r

m mω = − where k is the spring constant,

‘m’ is the mass of the oscillator and ‘r’ is the damping constant. If the ratio 2r

mk is 8% then the

undamped (ω ) oscillator is approximately as follows: A) decreases by 1% B) increases by 8% C) decreases by 8% D) increases by 1% 19. Two long concentric cylindrical conductors of radii a and b (b < a) are maintained at a potential

difference ‘V’ from static charge and carry equal and opposite currents l. An electron with a particular velocity ‘U’ parallel to the axis will travels undeviated in the evacuated region between the conductors. The ‘U’ is equal to

A) ( )0

4

ln /

V

i b a

πµ

B) ( )0

2

ln /

V

i a b

πµ

C) ( )0

2

ln /

V

i b a

πµ

D) ( )0

8

ln /

V

i a b

πµ

20. A light is incident on face AB of an equilateral glass prism ABC. After refraction at AB, the ray is incident on face BC at an angle slightly greater than critical angle so that it gets reflected from face BC and finally emerges out from face AC. Net deviation angle of the ray is 0112 anticlock wise. The angle of incidence ‘i’ has value

A) 022 B) 024 C) 026 D) 028 21. An equilateral triangular loop ADC of uniform specific resistivity having some resistance is pulled

with a constant velocity ‘V’ out of a uniform magnetic field directed into the paper. At time 0t = , side DC of the loop is at the edge of the magnetic field. The induced current (i) versus time (t) graph will be as

A) B)

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C) D) 22. The P-V graph represent the thermodynamic cycle of an engine, operating with an ideal monoatomic

gas. The amount of heat, extracted from the source in a single cycle is

A) 0 0PV B) 0 0

1

2PV C) 0 0

11

2PV D) 0 04PV

23. A negative charge is given to a non conducting loop and the loop is rotated in the plane of paper about its centre as shown in figure. The magnetic field produced by the ring effects a small magnet placed above the ring in the same plane

A) The magnet rotates clockwise as seen from below B) No effect on magnet is there C) The magnet rotates anticlockwise as seen from below D) The magnet does not rotate 24. In an electric heater is rated at 1000 W, then the time required to heat one litre of water from 020 C to

060 C is (heat capacity C = 1 0 14184J Kg C− − ) A) 5 min 36 sec B) 2 min 48 sec C) 1 min 24 sec D) 4 min 17 sec 25. A galvanometer of resistance 5Ω is connected to a battery of 3V along with a resistance of 2950Ω

in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 division, the resistance in series should be

A) 5550Ω B) 4427Ω C) 6050Ω D) 5050Ω

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Page No.6

26. Find the amplitude of the electric field in a parallel beam of light of intensity 28.85 /W m .

A) 1100Vm− B) 1100 1.5Vm− C) 1100

1.5Vm− D) 1150Vm−

27. A photon of wavelength 0

1000Ahas energy 12.3 eV. If light of wavelength 0

5000A having intensity I, falls on a metal surface, the saturation current is 0.40 Aµ and the stopping potential is 1.36 V. The work function of metal is

A) 2.47 eV B) 1.36 eV C) 1.10 eV D) 0.43 eV 28. A hydrogen atom in ground state absorbs 10.2 eV of energy. The orbital angular momentum of the

electron is increased by A) 341.05 10 secJ−× − B) 342.11 10 secJ−× −

C) 343.16 10 secJ−× − D) 344.22 10 secJ−× −

29. The wave length of Kα X-ray of tungsten is 21.3 pm. It takes 11.3 keV to knock out an electron from the L-shell of a tungsten atom. What should be the minimum accelerating voltage across an X-ray tube having tungsten target which allows production of Kα X-ray?

A) 69.6 KV B) 34.8 KV C) 13.0 KV D) 75 KV

30. In an ac circuit ( )100sin 100V t= volt and ( )0100sin 100 60I t mA= + . The power dissipated in the

circuit is A) 410 W B) 10W C) 2.5W D) 5W

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Page No.7

PHYSICS KEY :

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

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

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

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Page No.8

JEE GRAND TEST – 2 Solutions

1. 1^

3bV j ms−=

1^ ^ ^ ^

2cos30 2sin 30 3s bV V i j i j ms− − = + = +

^ ^ ^

3 3sV j i j− = +

^ ^

3 4sV i j= +

( )22 13 4 19sV ms−= + =

119S GV V ms−− =

2. Total reading = P.S.R. + H.S.R x L.C But L.C = Pitch of the screw

no.of divisionson thehead scale

= 0.5 mm + 25 x 0.01 mm = 0.5

0.0150

mmmm=

= 0.5 + 0.25 = 0.75 mm 3. .P F V=

.m dV

P Vdt

=

.m dV ds V

Pdt ds

= ×

.m dV V

P Vds

= ×

2P ds mV dV=

2Pds mV dV=∫ ∫

2

1

3

.3

V

V

VP S m

=

( )3 32 13

mS V V

P= −

4. Impulse = change in linear momentum Ft = ( )cos cosm eu Uα α+

F = ( )cos 1mU e

t

α +

5. linear acceleration of cylinder is zero sinmg θ = frictional force ( along the inclined plane & upward)

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Page No.9

& a

RRf Iτ α α= = =

2

2sin mR a

RR mg θ× =

2 sina g θ=

6. KET PE∆ = ∆ ↓

21

2 2

GMm GMmmv

a a= + −

2 1

12

GM

aV −

=

7. E (energy) = 21

2mv mg y+

Equation of parabola 2 4ayx =

0dE

dt=

1720 20

g ga x sω −−= ⇒ = =

8. Upward force = downward force

N kx mg+ =

2 2 10100 10N − = ×+ × 19N N=

9. 1 2PP P −=

0 1 2

4 4 4T T T

r r r= −

0

1 1 1

2r r r= −

0 2rr =

10. at reflector1V U

fV

f + =

1reflected wave2V V V U

f fV U V U V

f + = = × − −

2

V Uf

V Uf + = −

11. 0teN N λ−= , 0N → given same

0

0

101 t

tN e

e N e

λ

λ

−

−=

1 9 tee λ− −= 1

9t

λ⇒ =

12. 1 2

4

3

KA TH KA T

l lHI ∆+ = = ∆→

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Page No.10

2 1

72

3

KA K A TH H H T

l lII

′ ∆= − = ∆ =→

7

3

KK =′

13. 2 4I

T SMB

π= = given

2

0 ,12

m LI I M m L→ = = ×

( )2

00

/ 3

3 12 27each part

Lm III I→ = =

3 3

L MM m= × =

0

/ 27 42 2

9 393

I I TT S

M MBBπ π= = = =

14. 0E =

01 2 2

&q

F F Fr

α=

( )22

10 20

80x x=

−

80

2x

x

− =

80

1 2x

− =

80

2 1x

= +

80

2 1x =

+

( )80 2 1 80 0.414 33x cm= − = × ≃

15.

2

1 2max

min 1 2

9I II

I I I

+= = −

1

2

4

1

I

I⇒ =

1 04I I⇒ =

2 0I I=

( )204 cos / 2I I φ=

2 2

3 3

yd DY

D d

π π λλ

= ⇒ =

16.

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R 2eff = Ω

12

62

Vi

R= = =

1 2 0i i+ =

17. 1 1 1

f v U= −

U U U

f V U= −

1

1U U

f V m+ = =

1

1x

m f= −

− and

11

y

m f= −

2

x yf

+∴ =

18. 1/22 2

21

4

k r k r

m m m mkω

= − = −

( )1/2

0.081 0.99

4

k k

m mω = − =

ω decreases by 1% . Hence ‘T’ increases by 1%.

19. 02

Er

λπ

=∈

0

ln ,2 ln

a VV E

ar br

b

λπ

= = ∈

0

2

iB

r

µπ

=

qUB qE= (Particle undeviated)

0

2

ln

E VU

aBi

b

π

µ= =

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Page No.12

20. deviation d = ( ) ( )180 2 Ci r i rθ− + − + −

( )1/ 2 2 180 2Crπ θ= − + −

026i⇒ =

21. AE b= AB AC DC a= = =

AH GH

AE EC=

AH

GH ECAE

= ×

b vt a

a a Vtb b

− = × = −

2 2a

FG GH a vtb

= = −

( ) 2a

e BV FG BV a tvb

= = −

2e BV at

i a VR R b

= = −

1 2i K K t= −

i t− graph → straight line with negative slope

22. 2 2 1 1

1 1

PV PVnRdTU

r r

−∆ = =− −

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0 0 0 00 0

2 40 3

51

3

CD CD CD

PV PVQ U PV

−∆ = ∆ + ∆ = + = −−

0 0 0 00 0 0 0

2 55 213

DA

PV PVQ PV PV

− −∆ = − =−

0 0reflected

11

2

PVQ∆ =

23. The rotating loop behaves like a magnet whose north pole will be present out of the plane. 24. Pt mc T= ∆

25. i for 30 divisions = 3

2955A

i for 20 divisions = 3 20 3

44272955 30 5

RR

× = ⇒ = Ω+

26. 20 0

1

2I E c= ∈

( )12 2 80

18.85 8.85 10 3 10

2E−= × × ×

4

20

10

1.5E =

0

100/

1.5E V m=

27. max

12.32.46 , 1.36

5E eV K eV= − =

2.46 1.36 1.10E W eVφ = − = − = 28. After absorbing 10.2 eV, electron goes to 2n = from 1n =

( ) ( )2 1 2 12 2

h hL n n

π π∆ = − = −

34

346.6 101.05 10 .sec

2 3.14J

−−×= = ×

×

29. K LK

hcE E

αλ

− =

21.3 0.0213K pm mα

λ = =

( ) ( )1242 1242

58.30.0213

E eV keVnmλ

= = =

K LE E E− = 11.3 58.3KE − =

69.6KE =

Acceleration potential = 69.6 kV 30. 3 0

0 0100 , 100 100 10 0.1 , 60V V i mA A φ−= = = × = =

0 0

1 1 1cos 100 0.1 2.5

2 2 2P V i Wφ= = × × × =