PAPER 2 Questions & Solutions

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22-05-2016JEE (Advanced) 2016

®

DATE: 03-10-2021

JEE (ADVANCED) 2021

PAPER 2

Questions & Solutions

Time : 2:30 pm – 05:30 pm | Duration : 3 Hrs. | Total Marks : 180

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mailto:[email protected]

http://www.resonance.ac.in/reso/results/jee-main-2014.aspx





PAPER-2 : INSTRUCTIONS TO CANDIDATES

Question Paper-2 has three (03) parts: Physics, Chemistry and Mathematics.

Each part has a total Nineteen (19) questions divided into four (04) sections (Section-1, Section-2, Section-3 & Section-4)

Total number of questions in Question Paper-2 are Fifty Seven (57) and Maximum Marks are One Hundred and Eighty (180).

Type of Questions and Marking Schemes SECTION-1

• This section contains SIX (06) questions. • Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four

option(s) is (are) correct answer(s). • For each question, choose the option(s) corresponding to (all) the correct answer(s). • Answer to each question will be evaluated according to the following marking scheme: Full Marks : +4 If only (all) the correct option(s) is(are) chosen; Partial Marks : +3 If all the four options are correct but ONLY three options are chosen; Partial Marks : +2 If three or more options are correct but ONLY two options are chosen, both of

which are correct; Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a correct

option; Zero Marks : 0 If unanswered; Negative Marks : −2 In all other cases. • For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct

answers, then choosing ONLY (A), (B) and (D) will get +4 marks; choosing ONLY (A) and (B) will get +2 marks; choosing ONLY (A) and (D) will get +2marks; choosing ONLY (B) and (D) will get +2 marks; choosing ONLY (A) will get +1 mark; choosing ONLY (B) will get +1 mark; choosing ONLY (D) will get +1 mark; choosing no option(s) (i.e. the question is unanswered) will get 0 marks and choosing any other option(s) will get −2 marks.

SECTION 2 • This section contains THREE (03) question stems. • There are TWO (02) questions corresponding to each question stem. • The answer to each question is a NUMERICAL VALUE. • For each question, enter the correct numerical value corresponding to the answer in the designated place using the mouse and the on-screen virtual numeric keypad. • If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal

places. • Answer to each question will be evaluated according to the following marking scheme: Full Marks : +2 If ONLY the correct numerical value is entered at the designated place; Zero Marks : 0 In all other cases

SECTION 3 • This section contains TWO (02) paragraphs. Based on each paragraph, there are TWO (02) questions. • Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct

answer. • For each question, choose the option corresponding to the correct answer. • Answer to each question will be evaluated according to the following marking scheme:

Full Marks : +3 If ONLY the correct option is chosen; Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);

Negative Marks : −1 In all other cases.







SECTION 4

• This section contains THREE (03) questions.

• The answer to each question is a NON-NEGATIVE INTEGER.

• For each question, enter the correct integer corresponding to the answer using the mouse and the on-

screen virtual numeric keypad in the place designated to enter the answer.

• Answer to each question will be evaluated according to the following marking scheme:

Full Marks : +4 If ONLY the correct integer is entered;

Zero Marks : 0 In all other cases.



| JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-2 | PHYSICS

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PAGE # 1

7340010333

PART-I : PHYSICS

SECTION 1

• This section contains SIX (06) questions.

• Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four

option(s) is (are) correct answer(s).

• For each question, choose the option(s) corresponding to (all) the correct answer(s).


Full Marks : +4 If only (all) the correct option(s) is(are) chosen;

Partial Marks : +3 If all the four options are correct but ONLY three options are chosen;

Partial Marks : +2 If three or more options are correct but ONLY two options are chosen, both of

which are correct;

Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a correct

option;

Zero Marks : 0 If unanswered;

Negative Marks : −2 In all other cases. • For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct

answers, then choosing ONLY (A), (B) and (D) will get +4 marks; choosing ONLY (A) and (B) will get +2 marks; choosing ONLY (A) and (D) will get +2marks; choosing ONLY (B) and (D) will get +2 marks; choosing ONLY (A) will get +1 mark; choosing ONLY (B) will get +1 mark; choosing ONLY (D) will get +1 mark; choosing no option(s) (i.e. the question is unanswered) will get 0 marks and choosing any other option(s) will get −2 marks.

1. One end of a horizontal uniform beam of weight W and length L is hinged on a vertical wall at point O

and its other end is supported by a light inextensible rope. The other end of the rope is fixed at point Q,

at a height L above the hinge at point O. A block of weight 𝛼W is attached at the point P of the beam, as

shown in the figure (not to scale). The rope can sustain a maximum tension of ( 22 )W. Which of the

following statement(s) is(are) correct?

(A) The vertical component of reaction force at O does not depend on 𝛼

(B)The horizontal component of reaction force at O is equal to W for 𝛼 = 0.5

(C) The tension in the rope is 2W for 𝛼 = 0.5

(D) The rope breaks if 𝛼 > 1.5



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PAGE # 2

7340010333

Ans. (ABD)

Sol.

Ny 45°

W W

Nx

T

W

2W45sinT o

W2

W

2T

2

2

1WT

(D) To break the string T > W22

2

2

1W > W22

2

3

(C) = 0.5 = 1/2 ,

2

12

2

1WT = W2

(B) ox 45cosTN = W2 ×

2

1 Nx = W

(A) Nv + Tsin45° = W+W Nv + 2

W+ W = W + W Nv =

2

W It does not depend on .

2. A source, approaching with speed u towards the open end of a stationary pipe of length 𝐿, is emitting a

sound of frequency fs. The farther end of the pipe is closed. The speed of sound in air is v and f0 is the fundamental frequency of the pipe. For which of the following combination(s) of u and fs, will the sound reaching the pipe lead to a resonance? (A) u = 0.8 v and fs = f0 (B) u = 0.8 v and fs = 2 f0 (C) u = 0.8 v and fs = 0.5 f0 (D) u = 0.5 v and fs = 1.5 f0

Ans. (AD)

Sol.

u

fs

Source

The appeared frequency of the sound will be

uv

0vf'f s , and for resonance, it should match with any

of the natural frequency of the closed organ pipe. fundamental frequency of the closed organ pipe is f0, then its natural frequencies will be f0, 3f0, 5f0…….(2n – 1)f0 for resonance

0s f)1n2(uv

vf

0s f)1n2(v

u1f

(A, B, C) u = 0.8 v fs = (1 – 0.8)(2n – 1)f0



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PAGE # 3

7340010333

fs = 0f5

1n2

n = 1, fs = 5

f0

n = 2, fs = 5

f3 0

n = 3, fs = 00 f

5

f5

n = 4, fs = 5

f7 0

n = 5, fs = 5

f9 0

n = 6, fs = 5

f11 0

So option (A) is correct, (B) and (C) are in correct (D) u = 0.5 v fs = (1 – 0.5)(2n – 1)f0

fs = 0f2

1n2

n = 1 fs = 2

f0

n = 2 fs = 00 f5.1

2

f3

n = 3 fs = 2

f5 0

3. For a prism of prism angle = 60°, the refractive indices of the left half and the right half are,

respectively, n1 and n2 (n2 ≥ n1) as shown in the figure. The angle of incidence i is chosen such that the

incident light rays will have minimum deviation if n1 = n2 = n = 1.5. For the case of unequal refractive

indices, n1 = n and n2 = n + ∆ n (where ∆n << n1), the angle of emergence e = i + ∆e. Which of the

following statement(s) is(are) correct?

(A) The value of ∆e (in radians) is greater than that of ∆n

(B) ∆e is proportional to Δn

(C) Δe lies between 2.0 and 3.0 milliradians, if ∆n = 2.8 × 10−3

(D) Δe lies between 1.0 and 1.6 milliradians, if ∆n = 2.8 × 10−3

Ans. (BC)



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PAGE # 4

7340010333

Sol.

ie

A=60°

rm=30° rm=30°

n1 n2

n1 = n2 = n = 2

3 for minimum deviation

302

60

2

Arm

and i = e =

2

Asin

2

Asin

n

min

4

7)cos(,

4

3)sin(

2

60sin

sin

2

3

If n1 = n = 2

3and n2 = n + n

e = +

(1) sin = n sin30°

(n + n) sin30° = (1) sin( + )

Solving 2

1 (n) = sin( + ) – sin

sin)sin(

2

n

cosd

)(sind

cos2

n

4

7

2

n

n

2

64.2

2

7

134.1n

n , so option (A) is incorrect

n = (1.34) n option (B) is correct

2

7108.2

2

7n 3

33

108.077

106.5

= (2.64 × 0.8) × 10–3 = 2.11 × 10–3 rad

option (C) is correct



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PAGE # 5

7340010333

4. A physical quantity S

is defined as

0

BES

, where E

is electric field, B

is magnetic field and 𝜇0 is the

permeability of free space. The dimensions of S

are the same as the dimensions of which of the

following quantity(ies) ?

(A) CurrenteargCh

Energy

(2)

TimeLength

Force

(3)

volume

Energy (4)

Area

Pow er

Ans. (BD)

Sol. 0

BEs

0

BE

is pointing vector and its dimension is same as that of intensity (w/m2)

Dimension = 2

22

TL

TML MT

–3

Alternate Solutions

Dimension of Electric field (E) , [E] = MA–1

LT–3

Dimensions of Magnetic field (B); [B] = MA–1

T–2

Dimensions of magnetic permeability (0) ; [0] = MA–2

T–2

L

[S] = LTMA

)TMA)(LTMA(EB22

2131

0

= MT–3

5. A heavy nucleus N, at rest, undergoes fission N → P + Q, where P and Q are two lighter nuclei. Let 𝛿 = MN – MP – MQ, where MP, MQ and MN are the masses of P, Q and N, respectively. EP and EQ are

the kinetic energies of P and Q, respectively. The speeds of P and Q are vP and vQ, respectively. If c is

the speed of light, which of the following statement(s) is(are) correct?

(A) EP + EQ = c2 (B)

2

QP

PP c

MM

ME (C)

P

Q

Q

P

M

M

v

v

(D) The magnitude of momentum for P as well as Q is 2c , where 𝜇 = )MM(

MM

QP

QP

Ans. (ACD)

Sol.

N P P

P Q

(Ptotal) = 0 (Ptotal)f = 0

So momentum of one nucleus is P in forward direction then momentum of the other nucleus will be P in

backwards direction

Energy released = (m)C2 = C2 = KEP + KEQ



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PAGE # 6

7340010333

P

2

Pm2

PKE ,

Q

2

Qm2

PKE

QPQP

m

1:

m

1KE:KE = mQ : mP

2

QP

QP C

mm

mKE

2

QP

PQ C

mm

mKE

KEP + KEQ = C2

2

Q

2

P

2

Cm2

P

m2

P

QP

QP

mm

mm2CP

6. Two concentric circular loops, one of radius R and the other of radius 2R, lie in the xy-plane with the

origin as their common center, as shown in the figure. The smaller loop carries current 1 in the anti-

clockwise direction and the larger loop carries current 2 in the clockwise direction, with 2 > 21. )y,x(B

denotes the magnetic field at a point (x, y) in the xy-plane. Which of the following statement(s) is(are) correct?

(A) )y,x(B

is perpendicular to the xy-plane at any point in the plane

(B) | )y,x(B

| depends on x and y only through the radial distance 22 yxr

(C) | )y,x(B

| is non-zero at all points for r < R

(D) )y,x(B

points normally outward from the xy-plane for all the points between the two loops

Ans. (AB)

Sol.

x

y

i1,R

i2, 2R



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PAGE # 7

7340010333

(A) Magnetic field at the plane of the ring is perpendicular to the plane. However they bend as they

move forward.

(B) By symmetry, we can say that B

will be same at all the points having the same radial distance. so B

(x,y) will depend only the radial distance r = 22 yx

(C) (Bnet)centre = 1201020 i2iR4R2

i

)R2(2

i

Since i2 > 2i1 so Bnet at the centre will be non zero in

direction. But at some other point, Bnet may be zero.

From the graph , it is clear that Bnet = 0 at for r (O, R). So option (C) is incorrect

=

i1,R

i2, 2R

= –

ve

ve

B

r=0 r=R r=2R r

(D) For the graph , it is clear that B = –Ve indirection for r(R to 2R), So option D is also incorrect.



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7340010333

SECTION 2

• This section contains THREE (03) question stems.

• There are TWO (02) questions corresponding to each question stem.

• The answer to each question is a NUMERICAL VALUE.

• For each question, enter the correct numerical value corresponding to the answer in the designated place

using the mouse and the on-screen virtual numeric keypad.

• If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal

places.


Full Marks : +2 If ONLY the correct numerical value is entered at the designated place;

Zero Marks : 0 In all other cases

Question Stem for Question Nos. 7 and 8

Question Stem

A soft plastic bottle, filled with water of density 1 gm/cc, carries an in be with some airverted glass test-tu

(ideal gas) trapped as shown in the figure. The test-tube has a mass of 5 gm made of a thick, and it is

glass of density 2.5 gm/cc. Initially the bottle is sealed at atmospheric pressure 𝑝0 0= 1 5 Pa so that the

volume of the trapped air is v0 side at constant= 3.3 cc. When the bottle is squeezed from out

temperature, the pressure inside rises and the volume of the trapped air reduces. It und that the testis fo

tube begins to sink at pressure p0 + e, the volume ofp without changing its orientation. At this pressur

the trapped air is v0 − v.

Let v = X cc and p = Y × 103 Pa.

7. The value of X is .

Ans.

8.

Ans.

0.3

The value of Y is .

10.00



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PAGE # 9

7340010333

Sol. As soon as, the upwards and downward force on the tube are balanced, the tube will just start sinking.

(w) (Vtube + Vair) g = (mtube) g

(2cm3 + Vair) g = (5 gm) g

mg

B = (w) (Vtube + Vair) g

Vair = 3 cm, while the initial volume of the air was 3.3 cm3 so the decrease in the volume

V = 3.3 – 3 = 0.3 cm3

Vtube = 3

tube

tube

cm/gm5.2

gm5m

Vtube = 2 cm3

Since the temperature of the tapped air remains constant, so PV = constant

(P+P) (V – V) = PV

P = 10 × 103 = Y × 103

Y = 10


Question Stem

A pendulum consists of a bob of mass m = 0.1 kg and a massless inextensible string of length

L = 1.0 m. It is suspended from a fixed point at height H = 0.9 m above a frictionless horizontal floor.

Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A

horizontal impulse P = 0.2kg-m/s is imparted to the bob at some instant. After the bob slides for some

distance, the string becomes taut and the bob lifts off the floor. The magnitude of the angular

momentum of the pendulum about the point of suspension just before the bob lifts off is J kg-m2/s. The

kinetic energy of the pendulum just after the lift- off is K Joules.

9. The value of J is .

Ans. 0.18

10. The value of K is .

Ans. 0.16



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PAGE # 10

7340010333

Sol.

1m

J V = 2m/s

0.9m

Angular momentum about top point = mv × 0.9

= 0.2 × 0.9

0.18 kg m2/s

When string get taut a Impulse act by string on particle.

1m

J

vcos 0.9m

90–

vsin

Due to impulse, component of V along string get change but component perpendicular to v remain same

Vcos0.9

V1 = 0 (By string constraint)

New velocity = Vcos = V × 0.9 So new momentum = 0.9 × Pinitial = 0.9 × 0.2 = 0.18

KE = 1.02

)18.0(

m2

P 22

= 0.162 Joule


Question Stem

n a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C F across a

200 V, 50 Hz supply. The power consumed by the lamp is 500 W while the voltage drop across it is 100

V. Assume that there is no inductive load in the circuit. Take rms values of the voltages. The magnitude

of the phase- angle (in degrees) between the current and the supply voltage is . Assume, 3 5

11. The value of C is ___________

Ans. 100

12. The value of is ___________

Ans. 60



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PAGE # 11

7340010333

Sol. For lamp

R

VP

2

20500

100100R

i = A520

100

R

V

2c

2 xR

Vi

2

2

C502

1)20(

2005

1600C100

1400

2

1200C100

12

4

22101200

C

1

62

1012

1C 310

12

1C

V = 2R

2c VV

22C 100V200 3100VC

3100

3100

V

V

R

XCtan

R

C

= 60°SECTION 3

• This section contains TWO (02) paragraphs. Based on each paragraph, there are TWO (02) questions.

• Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct

answer.

• For each question, choose the option corresponding to the correct answer.


Full Marks : +3 If ONLY the correct option is chosen;

Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);

Negative Marks : −1 In all other cases.

PARAGRAPH 13 TO 14

A special metal S conducts electricity without any resistance. A closed wire loop, made of S, does not

allow any change in flux through itself by inducing a suitable current to generate a compensating flux.

The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a

magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of

radius 𝑎, with its center at the origin. A magnetic dipole of moment m is brought along the axis of this

loop from infinity to a point at distance r (>> a) from the center of the loop with its north pole always

facing the loop, as shown in the figure below.

The magnitude of magnetic field of a dipole m, at a point on its axis at distance r,is 3

0

r

m

2

, where 𝜇 is

the permeability of free space. The magnitude of the force between two magnetic dipoles with moments,



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PAGE # 12

7340010333

m1 and m2, separated by a distance r on the common axis, with their north poles facing each other, is

4

21

r

mkm, where k is a constant of appropriate dimensions. The direction of this force is along the line

joining the two dipoles.

13. When the dipole 𝑚 is placed at a distance 𝑟 from the center of the loop (as shown in the figure), the

current induced in the loop will be proportional to

(A) 3r

m (B)

2

2

r

m (C)

2r

m (D)

r

m2

Ans. (A)

14. The work done in bringing the dipole from infinity to a distance 𝑟 from the center of the loop by the given

process is proportional to :

(A) 5r

m (B)

5

2

r

m (C)

6

2

r

m (D)

7

2

r

m

Ans. (C)

Sol.

N S r

m

Since it is a super conducting loop, So net flux (self flux + external flux) passing through it will remain constant

(total)i = (total)f

L(0) + (0)(pa2) = Li –

3

0

r

m

2, Here L = self inductance of the super conducting loop.

33

0

r

mi

r

m

L2i

(ii) The current carring superconducting loop will also behave like a magnet, whose magnetic dipole

moment m1 = NiA =(1) 2

3

0 ar

m

L2

31

r

mm

The repulsive force felt be the magnet will be

F = 7

2

4

3

4

21

r

m

r

)m(r

m)K(

r

mKm

Wext = drr

m–Fdr–7

2

Wext 6

2

r

m



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7340010333

PARAGRAPH 15 TO 16

A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle,

as shown in the figure below. On each side of the partition, there is one mole of an ideal gas, with specific

heat at constant volume, CV = 2R. Here, R is the gas constant. Initially, each side has a volume V0 and

temperature T0. The left side has an electric heater, which is turned on at very low power to transfer heat

Q to the gas on the left side. As a result the partition moves slowly towards the right reducing the right

side volume to 2

V0 . Consequently, the gas temperatures on the left and the right sides become TL and TR,

respectively. Ignore the changes in the temperatures of the cylinder, heater and the partition.

15. The value of

0

R

T

T is

(A) 2 (B) 3 (C) 2 (D) 3

Ans. (A)

16. The value of 0RT

Q is

(A) 1224 (B) 1224 (C) 125 (D) 125

Ans. (B)

Sol.

gas (1) gas (2)

Left Right

P0, T0, V0, n = 1 P0 , T0, V0, n = 1

TL , f0 P,

2

V3 fR

0 P,T,2

V



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PAGE # 14

7340010333

R2R2

fCV f = 4,

2

3

4

21

8

21

The gas of the right portion is undergone through an adiabatic process so

1

ff1

ii VTVT

12

3

0R

12

3

002

VTVT

0R T2T 2T

T

0

R

Work done by the gas (2) = – U = – nCV (Tf – Ti)

= – (1) (2R) 00 TT2

= 0RT222

So work done by the gas (1) = + 0RT222

PV = nRT V

TP

for gas (2) T 2 times, volume 2

1times

so 222/1

2P times Pf = 0P22

so pressure of the gas (1) will also be 22 times and volume of gas (1) becomes 2

3 times

PVT )2

322(T

= 23 times

TL = 0T23

U for the left gas = n CV (Tf – Ti)

= (1) (2R) 000 RT226TT23

Heat given to gas (1) = W + U

= 00 RT226RT222

= 122RT4 0

SECTION 4

• This section contains THREE (03) questions.

• The answer to each question is a NON-NEGATIVE INTEGER.

• For each question, enter the correct integer corresponding to the answer using the mouse and the on-

screen virtual numeric keypad in the place designated to enter the answer.


Full Marks : +4 If ONLY the correct integer is entered;

Zero Marks : 0 In all other cases.



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17. In order to measure the internal resistance r1 of a cell of emf E, 𝐸, a meter bridge of wire resistance

R0 = 50, a resistance R0/2, another cell of emf E/2 (internal resistance r) and a galvanometer G are used

in a circuit, as shown in the figure. If the null point is found at =72 cm, then the value of r1 = ___

Ans. 3

Sol.

72 cm

100 m

/236

G

r/2

R0/2=25r1

14

i = 0

O

Resistance of potential wire is R0 = 50

Resistance of 100 m wire = 50

So Resistance of 72 cm wire = 3672100

50

Current

36r

2/

2514

2/

1

r1 = 39 – 36 = 3

18. The distance between two stars of masses 3MS and 6MS is 9R. Here R is the mean distance between

the centers of the Earth and the Sun, and MS is the mass of the Sun. The two stars orbit around their

common center of mass in circular orbits with period nT, where T is the period of Earth’s revolution

around the Sun. The value of n is ___

Ans. 9



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Sol. GMs

R2T

2/3

for binary star

)MM(G

r2'T

21

2/3

Put M1 = 3Ms

M2 = 6Ms

r = 9R

)Ms6Ms3(G

R92'T

2/32/3

2/1

2/32/3

9GMs

R92

GMs

9R2 2/3

T' = 9T

value of n = 9

19. In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals P, Q, and

R are EP, EQ and ER, respectively, and they are related by EP = 2EQ = 2ER. In this experiment, the

same source of monochromatic light is used for metals P and P while a different source of

monochromatic light is used for the metal R. The work functions for metals P, Q and R are 4.0 eV,

4.5 eV and 5.5 eV, respectively. The energy of the incident photon used for metal R, in eV, is :

Ans. 6

Sol. ppp0 k4kE ……..(1)

qqq0 k5.4kE ……..(2)

rr' kE

from (1) and (2)

4 + kp = 4.5 + kq

kp = 2kr = 2kq

4 + kp = 4.5 + kq

4 + 2kq = 4.5 + kq

kq = 0.5 eV

kp = 2 × 0.5 = 1eV

kr =2

1

2

kp = 0.5

E = r + kr

= 5.5 + 0.5

= 6.0 eV



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