iit physics mock paper

7/27/2019 iit physics mock paper

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Physics

IIT JEE 2012

Paper I

Single choice (One option is correct)

Q.1 A body of mass 3.14 Kg is suspended from one end of a wire of length 10 m. The

radius of the wire is changing uniformly from 9.8 x 104 m at one end to 510-4 at

the other end. Find the change in length of the wire in mm Young's modulus of

the material of the wire is 2 x 1011 NP m2. (Mass of the wire is negligible)

(a) 1 mm (b) 3 mm

(c) 2 mm (d) 4 mm

Q.2 Two block of mass m and M are connected by means of a metal wire passing over

a frictionless fixed pulley. The area of cross-section of the wire is 6.5 x 10-9

m2

and its breaking stress is 2 x 109 Nm2. If m=1 Kg Find the maximum value of M

in Kg for which the wire will not break. (g=10 m/s2)

(a) 1.50 Kg (b) 1.86 Kg

(c) 1.76 Kg (d) 1.96 Kg

Q.3 A metal rod of cross area 2.0 cm2 is being heated at one end. At some instant the

temperature gradient is 5.0 0C/cm at cross section A is 20 C/cm at cross section B.

The heat capacity of the part AB of the rod is 0.5 J/0C Thermal conductivity of the

material of the rod = 200 W/m0C. Neglect any loss of heat to the atmosphere.

Calculate the rate at which the temperature is increasing in the part AB of the rod.

(a) 24 0C/sec (b) 40 0C/sec

(c) 16 0C/sec (d) 60 0C/sec

Q.4 If 49 division of the vernier scale coincides with 50 divisions on the main scale of

vernier caliper. The least count of the instrument is, if graduation on the main

scale is 3 mm.

(a)1

49mm (b)

1

50mm


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(c)3

49mm (d)

2

49mm

Q.5 A vertical pencil of rays come from bottom of a tank filled with a liquid. When it

is accelerated with an acceleration of 7.5 m/s2. The ray is seen to be totally

reflected by liquid surface. What is minimum possible refractive index of liquid ?

(a) slightly greater than 4/3 (b) slightly greater than 5/3

(c) slightly grater than 1.5 (d) slightly greater than 1.75

Q.6 Copper and aluminum wire have equal resistances and masses which of the two

wires is longer and how many times ? densities of copper and aluminum are 8.93

x 103 Kg m-3 and 2.7 x 103 Kg m-3. Their resistivities are 0.17 W m and 0.028W m respectively.(a) Aluminum wire is 14.5 times longer than copper.

(b) Copper wire is 7.3 times longer than aluminum.

(c) Aluminum wire is 4.5 times longer than copper.

(d) Copper wire is 2.3 times longer than aluminum.

Q.7 The force exerted by a photon of intensity 1.4 KW/m2

. If it falls on perfectabsorber of radius 2 m is

(a) 2.35 x 10-4 N (b) 10.8 N

(c) 8.352 x 104 N (d) 8.8 x 104 N

Q.8 Find the period of small oscillation of a simple pendulum of length 'l' if its point

of suspension O moves relative to the earth with a constant acceleration ' ar

' given

l=30 cm and ar

=2

gand the angle between the vectors g

r

and ar

is = 1200. The

period of oscillation of the pendulum

(a) 0.08 S (b) 1.2 S (c) 0.95 S (d) 1.05 S

Q.9 The ratio of the energy required to raise a satellite upto a height h above the earth

to that required to put it into orbit there is


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(a) h : R (b) 2h : R (c) R: h (d) R : 2h

Q.10 In the potentiometer circuit shown, the length of potentiometer wire AB is 400 cm

long and resistance per unit length is 0.2 W /cm. If a battery of emf E and internalresistance 'r' is connected across XY as shown, then for

(a) E=1.5 V , r = 16 W the balance point would exist on the potentiometerwire

(b) E=1 V, r = 25W the balance point is at 250 cm from A(c) We cannot compare emf of two cells which are more than 1.6 V using this

potentiometer.

(d) All the above

5 Questions multiple correct answers one or more than correct.

Q.11 Which of the following quantities are always zero in a simple harmonic motion

Given Fr

force ar

acceleration rr

displacement and vr

velocity.

(a) Fr

x ar

(b) Vr

x rr

(c) ar

x rr

(d) Fr

x rr

Q.12 The figure shows a ray incident at an angle i=3

p. The ray is refracted and angle of

refraction is r. The value of r changes between 0 to

2

pwhen the ratio 1

2

H

H

= K

changed.

If the plot drawn shown the variation of r i versus 1

2

H

H= K. (r = angle of

refraction )


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(a) the value of 1K is2

3

(b) the value of q1 =6

p

(c) the value of q2 =3

p

(d) the value of K2 is 1

Q.13 In a Young's double slit experiment green light is incident on the two slits. The

interference pattern is observed on a screen which of the following changes would

cause the observed fringes to be more closely spaced ?

fringes

(a) Reducing the separation between the slits

(b) using blue light instead of green light

(c) used red light instead of green light

(d) moving the light source further away from the slits

Q.14 Sound passes through a gas with a velocity of 1260 ms-1 at 00C. The ratio of

specific heat from this gas is 1.4

(a) ratio of the pressure to the density of the gas at 00C is 11.4 x 105

(b) Molecular weight of the gas is 2 g

(c) density of the gas cannot be determined from the given data

(d) All the above


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Q.15 The torque t on a body about a given point is found to be equal to Ar x Lr whereAr

is constant vector andLr

is angular momentum about that point from this is

follows that

(a)dL

dt

r

is perpendicular to Lr

at all constant of time.

(b) The component ofLr

in the direction ofAr

does not change with time.

(c) The magnitude ofLr


(d) Lr


SECTION III

7 Integer type question single digit answers.

Q.16 A closed organ pipe and an open pipe of same length are set into vibration

simultaneously. The beat frequency is 4. If the length of each of them is doubled,

then what is the beat frequency ?

Ans. 2

Q.17 A particle is executing SHM. If the displacement at any instant is given by

x=3cos2t+4sin2t. What is the time period of the particle ? (Answer in Sec)

Ans. 3

Q.18 In a photoelectric effect experiment the maximum energy of the emitted electrons

is 1 ev for incoming radiation of frequency V0 and 3 eV for incoming radiation of

frequency 3 V0/2. What is the maximum energy of the electrons emitted for

incoming radiation of frequency 9 V0/4 ? (Answer in eV)

Ans. 6

Q.19 A battery is connected between points P and Q such that a current I flows from P

to Q. From point A to point C, the current flows through two circular areas ABC

and ADC as shown in the figure. The arc ADC makes an angle 900 at the centre.

The entire circle is made up of the same wire. What is the magnetic induction at

the centre of the circle of radius r.


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Ans. 0

Q.20 Three resistors 1 W , 2 W and 3 W are connected to from a triangle across 3 W

resistor a 3V battery is connected. The current through 3 W resistor is

Ans. 1

Q.21 A door 1.6 m wide requires a force of 1N to be applied at the free end to open or

close it. The force that required at a point 0.4 m distance from the hings for

opening or closing the door is (Answer in N)

Ans. 4

Q.22 y=3sin2 4

t x -

represents an equation of a progressive wave, where t is in

second and x is in meter. The distance travelled by the wave in 2.5 seconds.

Ans. 5

PhysicsSolution Paper I

Sol: 1

Ans. (a)

/

/

F Ay

dy dx

2

b ar b x

- = -

dy = change in length of dx

tanr b xq= -


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2( tan )A b xp q= -

20 0 ( tan )

L LF dx

dy y b xp q=

-

V

310 1L m mm-D = =

Sol: 2

Ans. (b)

Mg T ma

T mg ma

- =

- =

2

( )

Mmg

T M m= +

2

( )

T Mmg

A A M m=

+

92 2 10

(1 )

mg

mA

M

= +

M=1.86kg

Sol 3 Ans. (a)

A B

dq dQ dT C

dt dt dt - =

1

x A B

dT dT dT kA

dt C dx dt =

= -

= 240C/S

Sol 4 Ans. (c)

least count of cornier scale =

11 1 xn

+ -

lest count of min scale

13

49mm

3/49mm

Sol 5


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27.5 /a m s- =

7.5 3tan

10 4q = =

0sin 1sin 90H q =

31

5H =

5

3H=

Sol 6 Ans. (c) .l l

Ra l

r=

2l

vr=

Further

m

v d=

2l d

R mr=

Resistance and man being the samer Cu l2 d2 = r Al l2 dAl2

2

Al Cu Cu

Cu Al Al

l d

l d

r

r=

2 6 3

2 6 3

0.17 10 8.93 1020.08

0.028 10 2.7 10

Al

Cu

l

l

-

-

= =

20.08 4.48 4.5Al

Cu

l

l = = =

4.5Al Cul l=

Sol 7 Ans. (a) Force exerted by a photon beam on the surface of a perfect absorber

F = PA

Intensity of the photoelectron beam

I = 1.4 kw/m2 = 1400 W/m2


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Pressure due to photon beam

2814003 10

IP NMC

-= =

Surface area of a spherical surface24A rp=

2 24 (2) 16A mp p= =

Force exerted by the photon beam on the surface

8

140016

3 10F P A p= =

42.35 10 N-=

Sol 8 ans. (c)

with respect to the point O, the bob has two acceleration ar

and g at an

angel

of 600 as shown in big the net acceleration of f the bob is

2 2( ) 2 cos 60g a b ag= + +

22 2 1

4 2

gg g= + +

74

g=

22

2

lT

gp=

2 0.32 0.955

7 9.8p

= =

Sol. 9Ans. (b)

work done W to tae satellite to height h = Change in potential energy =

Energy required say E1

1

1 GMm GMmE

R h R

- - - +

2

.( ) ( )

GMm mgR hh

R h R R h R=

+ +

1

mgh

h

R

+


10/35

Energy required to put it into orbit 10 say E2 is given by

2

2

1

2oE mv= (because energy of a body in orbit in

21

2omv

Now2

2( )

omv GMm

R h R h=

+ +

Or2

2

0

1

gR gRV

hR h

R

= + +

2

1

2 1

ghE m

h

R

= =+

2 1

mgRh

R

=

+

1

2

1

2 1

mgh

h

E R

mghE

h

R

+

=

+

2h

R=

E1:E2 = 2h:R

Sol. 10 Ans. (d)

Current in the primary circuit

20.02

20 (400 0.2)I A= =

+

Potential drop across AB

V= I R=0.02x400x0.2=1.6VHence (A) in current, because the value of E in the secondary

circuit is 1.5V for E=1.0V the balance point is let us say at distance c cm

from A Then

1 1

2 2

E l

E l=

1.6 400

1 x=

400250

1.6x cm= =

Hence (B) is correct since potentiometer or compare upto 1.6V , we


11/35

Cannot compare the emf of two cells which are more than 1.6 V . hence

Choice

(c) is also correct choice is (D)

MCQSol 11

Ans. (A,B,C,D)

All the variables in the choices such as F aur

etc, are always either parallel

or anti parallel to each other hence their crass product is zero.

Sol 12 Ans. (B,C,D)

0r i- = if 12

1HH =

2 1k

If k = H1/H2 is large, then refracted ray will pan close to the normal and

3r i

p-

23

Qp

=

If the refracted ray graces the surface

6

r ip

-

In this case k1

0

1 sin 90 sin3

H Hp

=

1

2

3

2

Hk

H= =

Sol 13 Ans. (b)

fringe width/ D

B

d

l=

lblue < l green

Sol 14 Ans. (B,c)


12/35

we havep

cg

r=

1.41260

p

r=

0 50 11.34 10p

at cr

=

Future p RTM

r= is in perfect has equation in term of density

p RT

mr=

5

8.3 273

11.34 10

RTm

p

r

= =

Density of the has connect be determined from the given data

Solution 15 Ans. (A,B,C,)

d LA L

dtt = =

d Ldt

in perpendicular to both LA an L

Because A andL ar in the same plane here L cosq is the

component ofL L parallel to A i.e 11L

11d L

dt= 0, Consequently, L11 is a constant further as

11d L

dtis perpendicular

to L

magnitude ofL does not change with time.

Section III integers type question

Sol 16 .Ans (2)

It is assumed that beats are obtained when the tubes are

sounded in their fundamental mode

V for closed tube4

V

l=


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V for opened tube2

V

l=

Beats 42 4 4V V Vl l l

= - == =

If the length are doubled

Beats obtained4 8

V V

l l= -

1

8 4 2

V V

l l= =

2 beats are obtained

Sol 17 Ans. (3)

3 2 4sin 2x cas t t= +

As of the form 2sinx coswt wtb= +

Let cosra q= and sinrb q=

=4 = 3

cos sin sin cosx r wt r wtq q\ = +

2 2r a b= +

sin( )x r wt q= +

5sin( )x wt q= +

W=2; w is the some period2

3.142

T sp

= =

Nearest integer (3) Ans.

Sol 18

Ans. (6) Einsteins photoelectric equation is

0hv w Te= +

Where hv =incident energy

W0 = work function and Te isKinetic energy

0 0hv w ev= +

00

33

2

vh w ev= +

002 4

2

hvev hv ev= =

If the energy of incident beam is


14/35

09 9 , 9 3 64

hvev Te ev= = - =

Sol 19 Ans. (0)

Let l1l2 be the length of the two parts ABC and ADC of the conductor

andr be resistance per unit length of the conductorThe resistance of the part ABC

Will be R1 = r l1The resistance of the part ADC

Wiil be R2 = r l2Potential difference across

1 1 2 2AC I R I R= =

1 1 2 2 1 1 2 21I l I l I I lr r= = =

OR

Magnetic field at 0 due to a current I. flowing in ABC is given by

01 1 1 11 2

1

4 4

HHoI IB

r

q

p p p= = (l1= Q1r)

Magnetic field at 0 due to current I2 flowing in ADC is given by

2 22 22 24 4

oH I lHoIBr rq

p p= =

B2 =0 1 1

2

1

4

H I

rp(using I )

The resultant Magnetic field Induction at 0 is

2 1 0B B B= - =

2 1B B= (as B2=B1)


15/35

Sol 20.

Ans .(1)

=

The resistances 1W and2W are in series of arm ABC are in parallel

to resistance 3p in arm AC

Their effective resistance is

(1 2) 3 3

(1 2) 3 2

+ = W

+ +

Current in the circuit3

2(3/ 2)

vI A= =

W

Current in arm AC

2 3, 1

3 3

AI A

W= =

W + W

Q.21 1.6 1 1.6L NM-

= =

Ans. (4)

1.6

40.4

L

F Nd

-

= = =

Sol 22

Ans. (5) The given equation of a progressive wave is

3sin2 4

t xy p

= -

3sin24 8

t xp

= -

(i)

3

B

A

1W

3W 2W

3W

A C

3v

3v

A C

32W


16/35

The standard equation of a progressive wave is

0 sin 2 (11)t x

y y

T l

= = -

Comparing eq (i) & (ii) we get

T=45

8ml= Velocity of wave

18 24

V msT

l -= = =

Distance travelled by wave in time t is

S vt=(t is 2.5 sec.)

2 2.5 5S m= =

Physics

IIT JEE 2012

Paper II

SECTION I8 Question single option is correct.

Q.1 Two identical trucks each of mass M (excluding sacks of rice) move in national

highways with speeds V1 and V2 towards each other. When they meet each other a

sack of rice of mass m is thrown from one truck to the other and an identical sack

of rice is thrown from the second to the first. Calculate their velocities V1 and V2

after the exchange of sacks given m=50 Kg, m=200 Kg,1

Vr

=50 m/s and 2Vr

=200

m/s. Friction of the road may be neglected.

(a) 0, -150 m/s (b) 150 m/s, -50 m/s

(c) 100 m/s, 150 m/s (d) 100 m/s, 50 m/s

Q.2 The diagram shown an iron wire AB mounted in a brass frame and attached to the

frame at both ends A and B. The length of AB at 00C is 300 cm and the diameter

of the wire is 0.6 mm. What extra tension will be set up in the stretched wire when

the temperature of the system is raised to 400C in power of 10

6dyne ?

Given brass = 8 x 10-6 / 0C

iron = 12 x 10-6 / 0C

iron = 21 x 1011 dyne/ cm2


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(a) 0.95 (b) 0.165 (c) 0.142 (d) 0.125

Q.3 A particle having charge to mass ratioq

m

=k is projected in a magnetic field B

with a speed

V= 2 kBd as shown in figure. Where d is the width of the region in which

magnetic field is present. For this situation mark the incorrect statement.

(a) Particle remains in the magnetic field for a time of6kB

p

(b) It travels a distance of3

dpin the magnetic field.

(c) The angle of deviation from its initial direction is 300

(d) The displacement of particle in magnetic field is 2d sin 15

Q.4 Two plane mirror are placed as shown in Fig. A point object O is approaching the

intersection point A of mirrors with a speed of 100 cm/s. The velocity of image ofthe object formed by M2 with respect to velocity of image of object formed by M1

is

(a) -128 i + 96 j cm/s

(b) -28 i + 48 j cm/s

(c) 128 i + 48 j cm/s

(d) 100 i + 48 j cm/s

Q.5 For a post office box, the graph of galvanometer deflection versus R (resistance

pulled out of the resistance arm) for the ratio of resistances in ratio arms equal to

100:1 is given as shown. A careless student pulls out two non corrective values of


18/35

R (320 ohm and 326 ohm) as shown in the fig. Find the value of unknown

resistance.

(a) 3.2 Ohm (b) 3.24 Ohm(c) 3.206 Ohm (d) 3.9 Ohm

Deflection (in division)

Q.6 Find the voltage (in KV) applied to an X-Ray tube with Nickel (z=28) as a target

material. if wavelength difference between the K2 line cut off wavelength ofcontinuous X-Ray spectrum is equal to 84 PM (here ionization energy off H-atom

13.6 ev & hc=12420 ev A)

(a) 15 KV (b) 75 KV

(c) 1 KV (d) 500 KV

Q.7 A nut is screwed onto a bolt with 12 turns per cm and diameter more than 1 cm.

The bolt is lying in a horizontal position. The nut spins at 216 rpm.Time taken by

the nut to cover 3 cm along the bolt.

(a) 10 S (b) 12 S

(c) 14 S (d) 16 S

Q.8 Special bacteria of radius 1 x 10-6 m and density 1.1 x 103 Kg m-3 is allowed to

settle uniformly in an aqueous suspension. Density of water is 10-3 kg m-3,

viscosity of water 1 x 10-3 pa-s and depth of settlement 1 x 10-2 m. The time for

settlement is

(a) 10 hrs 25 min (b) 12 hrs 45 min

(c) 11 hrs 10 min (d) settlement will not occur without

centrifugal

SECTION II

3 paragraph 2 question each

Paragraph I

An energy of 68.0 ev is required to excite a hypothetical hydrogen like atom from

its second Bohr orbit to third. The nuclear charge is Ze.

Q.9 Kinetic energy of the electron in the first Bohr orbit is

(a) 490 ev (b) 520 ev


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(c) 620 ev (d) 310 ev

Q.10 Wavelength of the electron magnetic radiation required to eject the electron from

the first orbit to infinity is(a) 22.2

0

A (b) 280

A

(c) 25.30

A (d) 33.10

A

Paragraph II

An alternating current flows through a coil of inductance 0.015 henry and ohmic

resistance 1 W . Another ohmic resistance of 3 W is connected in series with the

coil. The rms value of current and voltage are 2 ampere and 10 volt respectively.

Q.11 The frequency of the alternating current

(a) 60 Hz (b) 50 Hz

(c) 32 Hz (d) 82 Hz

Q.12 Effective voltage across the coil is

(a) 11.2 V (b) 8.6 V

(c) 6.3 V (d) 5.2 V

Paragraph III

Two blocks of masses 3 m and 4 m are connected by a light in

extensible string passing over a smooth pulley. The mass 4 m is

held with the string just taut and mass 3 m just stationary on a

horizontal table below the pulley. The system is released and 70

cm from above the table, the 3m mass picks up another mass 7

m (g= 9.8 m/s2)

Q.13 Velocity of the 4 m mass when 3 m mass picks up 7 m is

(a) 1.9 m/s (b) 1.7 m/s(c) 1.4 m/s (d) 1.2 m/s

Q.14 The time that elapses from the start of the motion to the instant. When the combined

mass strikes the table is

(a)7 13

2s

+

(b)7 13

6s

+


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(c)7 13

6s

+

(d)6 12

4s

+

SECTION III

6 question one or more options correct

Q.15 A vessel of mass M filled with water has a hole at its bottom and is fastened on a

cart. The dimensions of the vessel are as shown in fig. The cart is pulled with a

force F so that maximum amount of water remains in the vessel.

(a) The amount of water retained is bcAP)/2L

(b) Inclination of the water surface with the horizontal is such that

acceleration a=gb

c

(c) Force 'F' acting on the vessel to retain maximum water in the vessel is

bcAPM g

L

+

(d) All the above

Q.16 In an endoergic nuclear reaction an incoming particle collides with stationary

nucleus.

(a) Kinetic energy of incoming particle is greater the Q-value of reaction in

ground frame.

(b) Kinetic energy of incoming particle is equal to the Q-value of reaction in

center of mass frame.

(c) Linear momentum of particle nucleus system is conserved.

(d) Energy is released.

Q.17 A ring of mass M and radius R sliding with a velocity V0 suddenly enters into

rough surface where coefficient of friction is H as shown

(a) The linear distance moved by the centre of mass before ring starts rolling

is2

03

8

V

Hg


21/35

(b) The net work alone by friction of ring is3

8

-MV0

2

(c) The loss in kinetic energy of ring is2

0

4MV

(d) The gain in rotational kinetic energy is2

0

8

MV

Q.18 The total translational kinetic energy of all molecules of 5 liters of nitrogen

exerting a pressure P is 3000 J. Then

(a) The total kinetic energy of 10 liters of N2 at pressure of 2P is 3000 J.

(b) The total kinetic energy of 10 liters of He pressure of 2P is 3000 J.

(c) The total kinetic energy of 10 liters of O2 at pressure of 2P is 20000J.

(d) The total kinetic energy of 10 liters of Ne at pressure of 2P is 12000 J.

Q.19 A region has uniform electric field and magnetic field along the positive x-

direction. An electron is fired from the origin at an angle (


22/35

SOLUTION Physics

Sol (1) Let the first truck move to the right and the second to the left let us take

the rightward direction as positive. In the horizontal direction, friction

being absent, no external force is acting on the system. Hence momentum

is conserved.

The momentum carried away by a sack thrown from the first is mv, the

momentum brought in by the second sack to the first is m(-v2)

\ by the low of conservation of momentum of truck + Sack of rice)

1 1 2( ) ( )M m v mv m v+ - + - = 1

1( )M m v+ , where1

1v is the new velocity of track 1

1

1 2 1( )Mv mv m m v\ - = +

1

1 1 2

M mv v v

M m M m\ = -

+ +

200 5050 200 0

250 250= - =

Similarly considering conservation of momentum of the second (truck + sacks of

rice) ,

we get for the second truck

1

2 2 1

M m

v v vM m M m

= - - - + +

200 50200 50

250 250

= - -

= - 150m/s

Sol. (2)

Ans. (a)

we have 0(1 2 )L L q= +

For brass 0 .

B B

B BL L Lq a q- =

For iron

0

iron iron

iron ironL L L

qa q- =

Effective increase in length of the iron wire.

0( )iron B

iron Bl L L

qa a qD = -

Exta tension Y strain x A2

iron

lY

Lqpp

D=

0 0

B ironL L =


23/35

2 B Biron iron

Y LL

q

q

app a q

= -

11 1 2 6 621 10 3.14 (0.3 10 ) (12.10 8 10 ) 4q- - - - 4 695 10 .945 10dynes dynes= @


24/35

Sol 3

Ans. (d)Radius of the circular

path

2mv

dqB

p = =

v

Now sin dp q =

So, 30q = Hence the particle

Remains in the magnetic field for

1 2

12t

kB

p=

6t

kB

p\ =

Distance travelled 2

6 3

dS r d

p pq= = =

Displacement of the particle in magnetic field = 2 sin2

rq

04 sin15d

Sol (4) the components of various

Ans. (A) velocities are shown in the figure

2ImVq is given by the vector sum of components of velocity of image

1

sin 2q\ =

Vm = 100cm/s

100cos 37o

100sin 37o


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with respect to M2 along the normal and perpendicular to the normal$2

2 0 0 0[100sin 37 100sin 37 . 37 ]imV i i cas j= + $

2 0 0 0 [ 100cos 37 100sin 37 cos37 ]i j+ - + $(28 96 ) /i j cm s= + 2

2 11 1 im immV m V V = = - =

$( 128 96 ) /i j cm s= - +

Another method

For the reflection at mirror M2 the incident ray axis reflected along OR

(On normal )0

74 2 )L OR q = -

Then $2 cos 2 sin 2imV V i V jq q= - +

Sol 5 Ans. (b)

5 33.75

6x

x x= =

-

323.75R = W \ unknown resistance

530

N

74o


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323.75

100x =

3.24= W

Sol 6

Ans. (a)2

2

13.6(2 1)E

n

-=

2 113.6(2 1) 14

hcev

l

= - -

2

12420

3(13.6)274

I A=

= 1.67A

I =12420hc

Av v

=

124201.67

v-

= 0.84

V= 15kvoH

Sol 7

Ans. (a)

pitch of the screw =1

12cm

Number of revelation mode to cover 3 cm =3

1

12

= 36

Spin frequency of the nut = 216 pm

=21660

Hz

angular speed 2w vp=

2162

60p=

= 7.2p rad/s.9 9

w tt w

= = =


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2 2 36

7.2

n

w

p p

p

=

= 10 s


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Sol 8

Ans. (b)

speed of bacteria v =2

1 22 ( )

9

r g r r

h

-

6 2 8

3

2 (1 10 ) 9.8 (1.1 1)10

9 1 10

-

-

= -

12 3

3

2 1 98 10.1 10 10

9 10

-

-

=

Time for settlement =2

7

tan 1 10

2.18 10

dis ce

speed

-

-

=

4459 10 s= 44.59 10

3600

=

12 hrs 45 min

Section II

Sol 9

Ans. (a)Total energy in the I bohr orbit is

2 2

1 13.6E Z ev=

13.6 36ev= -

489.6ev= -

\ kE in the I bohr orbit= - (Total energy)

489.6ev=

= 490 ev

Sol 10

Ans. (c)2

1hv E E = -

0 ( 490 )ev= - -

490ev=

19490 1.6 10c

h joulel

-\ =


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19490 1.6 10 )

hcl

-\ =

34 8

19

6.6 10 (3 10 )

490 (1.6 10 )

-

- =

= 25.28 Ao


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Sol 11

Ans. (c)10rmsV volt =

2rmsI amp=

105

I 2

VrmsimpedanceZ

rms= = = W

But 2 2( ) 7( )Z R r wL= +

2 25 (3 1) ( )wL= + +

Squaring we get2 2

25 4 ( )wL= + 3 3

200 /0.15

w rad seL

= = =

20032

2 2

wf hz

p p\ = = =

Sol 12

Ans. (c) Effective voltage across the coil

LV = effective current reactance

2 22 ( )r wL= +

2 1 9= +

2 10=

6.3V=

Sol 13

Ans. (c)

When 4m mass is let fall

the equations of motion are

4 mg - T = 4ma / (i)

T 3mg = 3ma ..(ii)

From (1) & (2)

4mg 3mg = 7m a

7

ga =

When mass 4 m reaches 0.7m above the table let the system be

moving with a velocity V


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Then

2 0.70 2

7

V g

= +

2 12 9.810

V\ =

= 1.96

1.96 1.4ms= =

Sol 14

Ans. (c)

because of the change in the mass of the system. The acceleration of thesystem has also changed. If new acceleration is 'a m/s2 in the opposite

direction

to the acceleration then

10mg 4mg = 14 ma

a = 4.2 m/s2

let the commanded weight rise further up-to a height hm before coming

momentarily to rest

then 0=v2 2ah2

0.7 0.7 0.0582 42 12

h m= = =

The maximum height reached is

H+h=70 +5.8 = 57.8 cm

Calculation of time period

If t1 is the time take before 3 m man picks up 7 m mass

then 1.4 =1

9.80

7t= +


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1

7 1.41

9.8t s

\ = =

after picking up 7m mass the motion of combined mass is2

2 2

11.4 0.7 4.2

2t t- = -

2

2 2

10.7 0.7 4.2

2t t- = -

2

2 23 1 0t t- - =

2

1 1 12 1 13

6 6t

+ +\ = =

1 2

7 13

6t t S

+

+ =

Section III

Sol 15

Ans. (A,B) As the last to draw by a force F the water in the vessel takes up a start

position rising upward at the back wall of the vessel.

To prevent water from from flowing out of the hole H the acceleration of

the

vessel should have such a value that it occupies a face area DBH and

width

of the vessel given byA

L

as shown in figure below

Area DBH =1/2 bc

If p in the density of water, then mass of water retained2

bcAe

L=

Ifq is the inclination of water surface with the horizontal.

b


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tanma a

mg gq = =

Or tan ba g gc

q = =

tanb

cq

\ =

The required force f=mass x acceleration

92

bcAP bM

L c

= +

Whose M is the mass of the vessel

So 16Ans. (A,B,C) In Grand Frame

Kinetic energyg

x

x

mK ( Q) 1

m

= - + Kx > (-Q)

in center of mass frame

kinetic energy : (kx) = (-q )

In nuclear reactions linear momentum is conserved

So 17

Ans. (A,D)

Let V is rolling speed and w in angular rolling speed

oV V Hgt = - andHg

W tr

=

0

2

VV =

o

Vt

2Hg= Now

2

200 2 .

2

VV Hg s

= -

2

03

8

VS

Hg

2

2 20100 0 0

1 1( )

2 2 2

Vk m V m mV

D = - =


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Sol 18Ans. (C,D)

3 3

2 2TK nRT PV = =

Sol 19

Ans. [B,C]

It will go an helical path of decreasing pitch

Sol 20

Ans. [ B,C,D ]


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iit physics mock paper

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