IIT JEE (Advanced) 2016 Paper 2: PHYSICS€¦ · IIT JEE (Advanced) 2016 Paper 2: PHYSICS SECTION 1...

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IIT JEE (Advanced) 2016 Paper 2: PHYSICS

SECTION 1 (Maximum Marks: 18)

• This section contains SIX questions.• Each question has FOUR options (A), (B), (C) and (D). ONLY ONE

of these four options is correct.• For each question, darken the bubble corresponding to the correct option

in the ORS.• For each question, marks will be awarded in

one of the following categories:

Full Marks :+3 If only the bubble corresponding to the correctoption is darkened.

Zero Marks :0 If none of the bubbles is darkened.Negative Marks:-1 In all other cases.

Question 1. The electrostatic energy of Z protons uniformly distributedthroughout a spherical nucleus of radius R is given by

E =3

5

Z(Z − 1)e2

4πε0R.

The measured masses of the neutron, 11H, 15

7N, 158O are 1.008665 u, 1.007825 u,

15.000109 u and 15.003065 u, respectively. Given that the radii of both the157N and 15

8O nuclei are same, 1 u = 931.5 MeV/c2 (c is the speed of light)and e2/(4πε0) = 1.44 MeV-fm. Assuming that the difference between thebinding energies of 15

7N and 158O is purely due to the electrostatic energy,

the radius of either of the nuclei is [ 1 fm = 10−15 m ] (2016)

(A) 2.85 fm (B) 3.03 fm (C) 3.42 fm (D) 3.80 fm

Solution. The binding energies of 157N and 15

8O are given by

BEN = ∆mNc2 = (8mn + 7mp −MN )c2

= (8× 1.008665 + 7× 1.007825− 15.000109)× 931.5

= 115.49 MeV,

BEO = ∆mOc2 = (7mn + 8mp −MO)c2

= (7× 1.008665 + 8× 1.007825− 15.003065)× 931.5

= 111.95 MeV.

The difference in binding energies of 157N and 15

8O is

∆BE = BEN − BEO = 115.49− 111.95 = 3.54 MeV. (1)


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The electrostatic energies of 157N and 15

8O are given by

EN =3

5

Z(Z − 1)e2

4πε0R=

3

5

7(7− 1)1.44

R=

36.288

RMeV (2)

text−fm

EO =3

5

8(8− 1)1.44

R=

48.384

RMeV-fm.

The difference in electrostatic energies of 157N and 15

8O is

∆E = EO − EN = (12.096/R) MeV-fm. (3)

Since ∆E = ∆BE, equations (1) and (3) give R = 12.096/3.54 = 3.42 fm.

Question 2. An accident in a nuclear laboratory resulted in depositionof a certain amount of radioactive material of half life 18 days inside thelaboratory. Tests revealed that the radiation was 64 times more than thepermissible level required for safe operation of the laboratory. What is theminimum number of days after which the laboratory can be considered safefor use? (2016)

(A) 64 (B) 90 (C) 108 (D) 120

Solution. The activity of a radioactive sample at a time t is given by A =A0e

−λt, where A0 is the initial activity and λ is the decay rate. The decayrate is related to the half life of the sample by λ = 0.693/t1/2 = (ln 2)/t1/2.Let A1 be the activity measured in the laboratory test at time t1. Let thelaboratory becomes safe at time t2 i.e., activity at time t2 is A2 = A1/64.Apply the formula for activity to get

A1 = A0e−λt1 , A2 = A0e

−λt2 .

Divide A1 by A2 to get

A1/A2 = 64 = eλ(t2−t1). (1)

Take logarithm on both sides of equation (1) and substitute λ = (ln 2)/t1/2 =(ln 2)/18 to get t2 − t1 = 6(18) = 108 days. Thus, the laboratory can beconsidered safe after 108 days of the test.

Question 3. A gas is enclosed in a cylinder with a movable frictionlesspiston. Its initial thermodynamic state at pressure pi = 105 Pa and vol-ume Vi = 10−3 m3 changes to a final state pf = (1/32)× 105 Pa andVf = 8× 10−3 m3 in an adiabatic quasi-static process, such that p3V 5 =constant. Consider another thermodynamic process that brings the systemfrom the same initial state to the same final state in two steps: an isobaricexpansion at pi followed by an isochoric (isovolumetric) process at volumeVf . The amount of heat supplied to the system in the two-step process isapproximately (2016)

(A) 112 J (B) 294 J (C) 588 J (D) 813 J


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V

p

Vi

pf

pi

Vf

i

f

1

2

Solution. The equation, p3V 5 = constant, ofgiven adiabatic quasi-static process can be writ-ten as pV 5/3 = constant. Compare this withequation for adiabatic process, pV γ = constant,to get ratio of the specific heats γ = 5/3. Letf be the degree of freedom of the molecules ofthe gas. The internal energy of one mole gas attemperature T is given by U = f

2RT . The specific heats at constant vol-ume Cv, specific heat at constant pressure Cp, and the ratio of the specificheats are given by

Cv =dU

dT=f

2R, Cp = Cv +R =

f + 2

2R, γ =

CpCv

=f + 2

f.

Substitute γ = 5/3 to get f = 3 (given gas is monatomic). The increasein internal energy of n moles of the gas when it goes from the initial state(pi, Vi, Ti) to the final state (pf , Vf , Tf ) is given by

∆U = Uf − Ui = f2 (nRTf − nRTi) = f

2 (pfVf − piVi)= 3

2

(132 × 105 × 8× 10−3 − 105 × 10−3

)= −112.5 J. (1)

Negative sign of ∆U indicates that the internal energy of final state is lessthan that of the initial state. The work done by the gas in the isochoricprocess is zero and in the isobaric process is

W = pi(Vf − Vi) = 105(8× 10−3 − 10−3) = 700 J. (2)

The change in internal energy ∆U is same in the two processes because itis a state function. Apply first law of thermodynamics, ∆Q = ∆U +W , toget heat supplied to the gas as

∆Q = ∆U + ∆W = −112.5 + 700 = 587.5 J.

Question 4. The end Q and R of two thin wires, PQ and RS, are soldered(joined) together. Initially each of the wires has a length of 1 m at 10 C.Now the end P is maintained at 10 C, while the end S is heated and main-tained at 400 C. The system is thermally insulated from its surroundings.If the thermal conductivity of wire PQ is twice that of the wire RS and thecoefficient of linear thermal expansion of PQ is 1.2× 10−5 K−1, the changein length of the wire PQ is (2016)

(A) 0.78 mm (B) 0.90 mm (C) 1.56 mm (D) 2.34 mm

P

10C

2κ Q,R

T

κ S

400C

∆x = 1m ∆x = 1m

Solution. Let steady state temperature of thejunction (Q,R) be T . In steady state, the rateof heat flow from the end S to R is equal to therate of heat flow from the end Q to P becausethe system is insulated from the surrounding.


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Thus,

dQSRdt

=dQQP

dt, i.e.,

κA(400− T )

∆x=

2κA(T − 10)

∆x,

which gives T = 140 C.

0m 1m 2mx10C

140C

400C

T (x)Let P be the origin and x-axis points to-

wards Q. In steady state, the temperature onthe wire varies linearly with the distance x i.e.,T (x) = mx+c, where m is the slope and c is theintercept of T axis. In wire PQ, the tempera-ture T = 10 C at x = 0 m and the temperatureT = 140 C at x = 1 m. Substitute these values in T (x) = mx + c to getc = 10 C and m = 130 C/m. Thus, steady state temperature on the wirePQ is given by

T (x) = 130x+ 10.

P Q

xdx

Now, consider a small element dx of the wirePQ located at a distance x from the end P. Ini-tial temperature of this element was 10 C andfinal temperature (in steady state) is T (x) = 130x+ 10. The coefficient oflinear thermal expansion of PQ is 1.2× 10−5 K−1. The increase in lengthof this element due to thermal expansion is given by

dl = α∆Tdx = α(130x+ 10− 10)dx = 130αxdx

Integrate from x = 0 m to x = 1 m to get increase in length as

l =

∫ 1

0

130αxdx = 130α

[x2

2

]10

= 130(1.2× 10−5)1

2= 0.78 mm.

Detailed analysis of heat transfer reveals that temperature obeys the

differential equation 1c2∂T∂t = d2T

dx2 , where c is a constant. In steady state,

temperature is independent of time i.e., ∂T∂t = 0 which gives d2T

dx2 = 0.Integrate it twice to get T (x) = mx+ c.

Question 5. A small object is placed 50 cm to the left of a thin convexlens of focal length 30 cm. A convex spherical mirror of radius of curvature100 cm is placed to the right of the lens at a distance of 50 cm. The mirroris tilted such that the axis of the mirror is at an angle θ = 30 to the axisof the lens, as shown in the figure.

x(−50, 0)

f = 30 cm

(0, 0)

(50 + 50√

3,−50)

θ

R=

100cm

50 cm


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If the origin of the coordinate system is taken to be at the centre of the lens,the coordinates (in cm) of the point (x, y) at which the image is formedare (2016)

(A) (0, 0) (B) (50− 25√

3, 25) (C) (25, 25√

3) (D) (125/3, 25/√

3)

Solution. The object distance for the lens of focal length f = 30 cm isu = −50 cm. Apply lens formula, 1

v −1u = 1

f , to get the image distancev = 75 cm. Thus, image I1 by the lens is formed 25 cm behind the mirror.

xO

50 cm 25 cm50 cm

I1I2

Let us consider the case when mirror is not tilted. The image I1 will actas a virtual object with u = 25 cm for a convex mirror of focal lengthfm = R/2 = 50 cm. Apply the mirror formula, 1

v + 1u = 1

fm, to get the

image distance v = −50 cm. Thus, image I2 by the mirror is formed 50 cmleft of the mirror (see figure).

x(−50, 0) (0, 0)

50cm

I′2

I2 θθ

50 cm

x

y

What happens to the image if mirror is tilted by an angle θ? Consider aray along the principal axis of the lens. If mirror is not tilted then thisray starts from the object, refracts through the lens without deviation,incident normally on the mirror and finally comes to the image I2 afterreflection from the mirror. If the mirror is tilted by an angle θ then thisray is incident on the mirror at an angle of incidence θ and reflected by themirror with an angle of reflection θ and finally forms a new image at I ′2 (seefigure). Thus, the axis on which image lies makes an angle 2θ = 60 withthe principal axis of the lens. Also, the image distance will remain sameupto the first order of approximation (image distance will remain exactlysame in case of plane mirror). Thus, (x, y) coordinates of the image are

x = 50− 50 cos 60 = 25 cm, y = 50 sin 60 = 25√

3 cm.


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Aliter: The image I1 acts as a virtual object for the tilted mirror. Theobject distance (along the principal axis of the mirror) is u = 25 cos 30 =25√

3/2 and height of the object is h = 25 sin 30 = 25/2. The mirrorformula gives the image distance v = −50

√3/(4−

√3). The magnification

by the mirror is m = −v/u = 4/(4 −√

3). Thus, the image height isVI′2 = mh = 50/(4 −

√3). In right angled triangle PVI′2, angle θ = 30

and distance PI′2 = 100/(4−√

3) cm.

x252

25√

32

50√

34−√

3

V

504−√

3

(0, 0)

θ

θ

50 cm

I′2(x, y)

P

1004−√

3

The x and y coordinates of the image I′2 are given by

x = 50− 100

4−√

3cos 60 = 50

(1− 1

4−√

3

)= 28 cm,

y =100

4−√

3sin 60 =

50√

3

4−√

3= 38 cm.

C1

2 3 4

0 5 10

C2

2 3 4

0 5 10

Question 6. There are two Vernier calipersboth of which have 1 cm divided into 10 equaldivisions on the main scale. The Vernier scaleof one of the calipers (C1) has 10 equal divi-sions that correspond to 9 main scale divisions.The Vernier scale of the other caliper (C2) has10 equal divisions that correspond to 11 mainscale divisions. The readings of the two calipersare shown in the figure. The measured values(in cm) by calipers C1 and C2, respectively are (2016)

(A) 2.85 and 2.82 (B) 2.87 and 2.83(C) 2.87 and 2.86 (D) 2.87 and 2.87

Solution. In both calipers C1 and C2, 1 cm is divided into 10 equal di-visions on the main scale. Thus, 1 division on the main scale is equalto xm1 = xm2 = 1 cm/10 = 0.1 cm. In calipers C1, 10 equal divisionson the Vernier scale are equal to 9 main scale divisions. Thus, 1 divi-sion on the Vernier scale of C1 is equal to xv1 = 9xm1/10 = 0.09 cm. Incalipers C2, 10 equal divisions on the Vernier scale are equal to 11 main


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scale divisions. Thus, 1 division on the Vernier scale of C2 is equal toxv2 = 11xm2/10 = 0.11 cm.

C1

MSR1 7xm1

7xv1

x1

X1

2 3 4

0 5 10

C2

MSR2 8xm2

7xv2

x2

X2

2 3 4

0 5 10

Let main scale reading be MSR and vth division of the Vernier scale coin-cides with mth division of the main scale (m is counted beyond MSR). Thevalue measured by this calipers is

X = MSR + x = MSR +mxm − vxv. (1)

In calipers C1, MSR1 = 2.8 cm, m1 = 7 and v1 = 7 and in calipers C2,MSR2 = 2.8 cm, m2 = 8 and v2 = 7. Substitute these values in equation (1)to get

X1 = MSR1 +m1xm1 − v1xv1 = 2.8 + 7(0.1)− 7(0.09) = 2.87 cm.

X2 = MSR2 +m2xm2 − v2xv2 = 2.8 + 8(0.1)− 7(0.11) = 2.83 cm.

The Vernier calipers are generally of type C1 having m = v and least countLC = xm − xv. For these calipers, equation (1) gives X = MSR + v × LC.


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• This section contains EIGHT questions.• Each question has FOUR options (A), (B), (C) and (D). ONLY OR

MORE THAN ONE of these four option(s) is(are) correct.• For each question, darken the bubble(s) corresponding to the correct op-

tion(s) in the ORS.• For each question, marks will be awarded in


Full Marks :+4 If only the bubble(s) corresponding to all the cor-rect option(s) is(are) darkened.

Partial Marks :+1 If darkening a bubble corresponding to eachcorrect option, provided NO incorrect optionis darkened.

Zero Marks :0 If none of the bubbles is darkened.Negative Marks:-2 In all other cases.

• For example, if (a), (C) and (D) are all the correct options for a question,darkening all these three will result in +4 marks; darkening only (A) and(D) will result in +2 marks; and darkening (A) and (B) will result in −2marks, as a wrong option is also darkened.

O

z

a

2a

ω4m

m

l

l

Question 7. Two thin circular discs of massm and 4m, having radii of a and 2a, respec-tively, are rigidly fixed by a massless, rigid rodof length l =

√24a through their centers. This

assembly is laid on a firm and flat surface, andset rolling without slipping on the surface sothat the angular speed about the axis of the rod is ω. The angular mo-mentum of the entire assembly about the point ‘O’ is ~L (see the figure).Which of the following statement(s) is(are) true? (2016)

(A) The center of mass of the assembly rotates about the z-axis with anangular speed ω/5.

(B) The magnitude of angular momentum of center of mass of the assemblyabout the point O is 81ma2ω.

(C) The magnitude of angular momentum of the assembly about its centerof mass is 17ma2ω/2.

(D) The magnitude of the z-component of ~L is 55ma2ω.

Solution. Let x, y, z be the axes as shown in the figure. Without any lossof generality, we can analyse the problem, when points on the disc which arein contact with the ground, lie on the x-axis. As discs are rolling withoutslipping, the points on the discs which are in contact with the ground, areat rest i.e., ~vP = ~0 and ~vR = ~0. The angular velocity of the discs about anaxis joining their centres Q and S is ~ω. The velocities of the points Q and


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S are

~vQ = ~vP + ~ω × ~PQ

= (−ω cos θ ı− ω sin θ k)× (−a sin θ ı+ a cos θ k) = ωa j, (1)

~vS = ~vR + ~ω × ~RS

= (−ω cos θ ı− ω sin θ k)× (−2a sin θ ı+ 2a cos θ k) = 2ωa j. (2)

O

z

Q

S

a

P

2a

R

ω

4m

m

~ω

ω cos θωsin

θ

x

y

~Labt,cm

~Lof,cm

~Ω

Ωcosθ

Ω sin θ

C

4l/5

l

l

~vcm

~vQ

~vS

θθ

θ

θ

The points Q and S are located on a rigid rod connecting two discs. Thediscs rotate (without slipping) with an angular velocity ~ω about the axis

QS. The axis QS itself rotates with an angular velocity ~Ω. Note that thedirection of ~Ω is perpendicular to axis QS. If it is not so then ~Ω will havea non-zero component along QS. This component will make the discs torotate faster or slower, which will violate the no-slip condition we haveused in equations (1) and (2). Mathematically, ~Ω · ~QS = 0 i.e., (Ωx ı +

Ωy + Ωz k) · (l cos θı+ l sin θk) = 0 which gives

Ωxl cos θ + Ωzl sin θ = 0. (3)

The velocities of Q and S are related by ~vS = ~vQ + ~Ω × ~QS which givesΩy = 0 and

−Ωxl sin θ + Ωzl cos θ = ωa. (4)

Solve equations (3) and (4) to get

Ωx = −(ωa/l) sin θ = −ω/(5√

24), (5)

Ωz = (ωa/l) cos θ = ω/5, (6)

where we have used l =√

24a, cos θ =√

24/5 and sin θ = 1/5. Thus, the

angular velocity ~Ω = (ω/√

24)(− sin θı+ cos θk).The distance of the centre of mass of the system (C) from the origin

O and its position vector are given by

lcm =ml + 4m(2l)

m+ 4m=

9

5l, ~rcm =

9

5l(cos θı+ sin θk) (7)


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The velocity and linear momentum of the centre of mass C are

~vcm =m~vQ + 4m~vSm+ 4m

=9

5ωa , ~pcm = (m+ 4m)~vcm = 9mωa . (8)

The angular momentum of the centre of mass about the origin is given by

~Lof cm = ~rcm × ~pcm =81√

24

5mωa2(− sin θı+ cos θk). (9)

The angular momentum of the system about the centre of mass (it is theangular momentum of the system due to rotation of the discs about theaxis QS) is given by

~Labout cm = Icm~ω =

(1

2ma2 +

1

2(4m)(2a)2

)~ω

=17

2mωa2(− cos θı− sin θk). (10)

Total angular momentum of the system is given by

~L = ~Lof cm + ~Labout cm =

(−247

√6

25ı+

3803

50k

)mωa2. (11)

−V+

Electrons

LightQuestion 8. Light of wavelength λph falls ona cathode plate inside a vacuum tube as shownin the figure. The work function of the cath-ode surface is φ and the anode is a wire mesh ofconducting material kept at a distance d fromthe cathode. A potential difference V is main-tained between the electrodes. If the minimumde Broglie wavelength of the electron passingthrough the anode is λe, which of the following statement(s) is(are) true?

(2016)

(A) λe decreases with increase in φ and λph.

(B) λe is approximately halved, if d is doubled.

(C) For large potential difference (V φ/e), λe is approximately halvedif V is made four times.

(D) λe increases at the same rate as λph for λph < hc/φ.

Solution. In photo-electric effect, the maximum kinetic energy of thephoto-electron ejected at the cathode, is given by

Kmax,c = hc/λph − φ, (1)

where λph is the wavelength of the ejected photo-electron and φ is the workfunction of the material. The ejected photo-electrons are accelerated from


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the cathode to the anode by a potential V . Thus, the maximum kineticenergy of the photo-electron at the anode is

Kmax,a = Kmax,c + eV = hc/λph − φ+ eV. (2)

The linear momentum of the electron of mass m and kinetic energy K isgiven by p =

√2mK. Thus, the minimum de-Broglie wavelength of the

electron at the anode is

λe =h

p=

h√2mKmax,a

=h√

2m(hc/λph − φ+ eV ). (3)

The wavelength λe increases with increase in φ and λph as denominator ofequation (3) decreases with increase in φ and λph. Also, λe is independentof d. The order of magnitudes of hc/λph and φ is almost same. Thus,

if V φ/e then equation (3) reduces to λe ≈ h/√

2meV . Hence, λe isapproximately halved if V is made four times.

λph

λeThe equation (3) is not linear in λph and λe.

Thus, the rates of increase of λe and λph will bedifferent. This can be verified by differentiatingequation (3). Show that the slope dλe/dλph isdifferent at different values of λph.

Question 9. In an experiment to determine the acceleration due to grav-ity g, the formula used for the time period of a periodic motion is T =

2π√

7(R−r)5g . The values of R and r are measured to be (60± 1) mm and

(10± 1) mm, respectively. In five successive measurements, the time pe-riod is found to be 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The least countof the watch used for the measurement of time period is 0.01 s. Which ofthe following statement(s) is(are) true? (2016)

(A) The error in the measurement of r is 10%.

(B) The error in the measurement of T is 3.57%.

(C) The error in the measurement of T is 2%.

(D) The error in the determined value of g is 11%.

Solution. The relative percentage error in the measurement of r is givenby

er =∆r

r× 100 =

1

10× 100 = 10%.

The mean value of time period is given by

T =

∑51 Ti5

=0.52 + 0.56 + 0.57 + 0.54 + 0.59

5= 0.556 s ≈ 0.56 s,


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where we have truncated at 2 significant figures. The absolute error andthe relative percentage error in the measurement of T are given by

∆T =

∑51 |Ti − T |

5=

0.04 + 0 + 0.01 + 0.02 + 0.03

5= 0.02.

eT =∆T

T× 100 =

0.02

0.56× 100 = 3.57%.

Given equation for time period can be written as g = 28π2

5R−rT 2 . Differen-

tiate it to get the relative error in g as

∆g

g=

∆(R− r)R− r

+ 2∆T

T=

∆R+ ∆r

R− r+ 2

∆T

T

=1 + 1

60− 10+ 2

0.02

0.56= 0.11 = 11%.

It is interesting to note that the time period T = 2π√

7(R−r)5g is same as

the time period of a small spherical ball of mass m and radius r when itrolls without slipping on a rough concave surface of large radius R. Thereaders are encouraged to derive it.

Question 10. Consider two identical galvanometers and two identical re-sistors with resistance R. If the internal resistance of the galvanometersRc < R/2, which of the following statement(s) about any one of the gal-vanometers is(are) true? (2016)

(A) The maximum voltage range is obtained when all the components areconnected in series.

(B) The maximum voltage range is obtained when the two resistors andone galvanometer are connected in series, and the second galvanometeris connected in parallel to the first galvanometer.

(C) The maximum current range is obtained when all the components areconnected in parallel.

(D) The maximum current range is obtained when the two galvanometersare connected in series and the combination is connected in parallelwith both the resistors.

R R

ig

Rc Rc

A B

Solution. Let ig be the current that gives fulldeflection of the galvanometer. When all com-ponents are connected in series (see figure), ef-fective resistance of the circuit is Re = 2R + 2Rc and maximum currentallowed in the circuit is ig. Thus, the voltage between A and B is

VAB = igRe = 2ig(R+Rc).


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R R ig Rc

A

ig Rc

B

Consider the case when two resistors andone galvanometer are connected in series andthe second galvanometer is connected in paral-lel (see figure). The maximum current througheach galvanometer is ig and the maximum current through the resistors is2ig. Apply Kirchhoff’s law to get the voltage between A and B as

V ′AB = 2igR+ 2igR+ igRc = 2ig(R+Rc) + 2ig(R−Rc/2)

= VAB + 2ig(R−Rc/2) > VAB (∵ Rc < R/2).

IAB

A

i R

i R

ig Rc

ig Rc

B

Consider the case when all four componentsare connected in parallel (see figure). Let i andig be the currents through the resistors and thegalvanometers. By Kirchhoff’s law, iR = igRc,which gives i = igRc/R. The current betweenA and B is

IAB = 2i+ 2ig = 2ig(1 +Rc/R).

I′AB

A

i R

i R

ig Rc Rc B

Consider the case when the two galvanome-ters are connected in series and the combinationis connected in parallel with both the resistors(see figure). By Kirchhoff’s law, iR = ig(2Rc),which gives i = 2igRc/R. The current betweenA and B is

I ′AB = ig + 2i = ig + 2ig(2Rc/R) = 2ig(1 +Rc/R)− (1− 2Rc/R)ig

= IAB − (1− 2Rc/R)ig < IAB (∵ Rc < R/2).

A

40µF 25kΩ

5VKey

50kΩ 20µF

V+

−

Question 11. In the circuit shown in the fig-ure, the key is pressed at time t = 0. Which ofthe following statement(s) is(are) true? (2016)

(A) The voltmeter displays −5 V as soon as the key is pressed, and displays+5 V after a long time.

(B) The voltmeter will display 0 V at time t = ln 2 seconds.

(C) The current in the ammeter becomes 1/e of the initial value after1 second.

(D) The current in the ammeter becomes zero after a long time.


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A

25kΩ

5V

Q

50kΩP

V+

−

Solution. The impedance (effective resistance)of the capacitors is zero immediately after thekey is pressed (t → 0+). Thus, capacitors willact as short circuit elements (see figure). Thus,potential at the node Q is VQ = 5 V and poten-tial at the node P is VP = 0 V. The voltmeterwill display V = VP − VQ = 0− 5 = −5 V.

A

25kΩ

5V

Q

50kΩP

V+

−

The impedance of the capacitors is infinitein steady state (t → ∞) and they act as opencircuit elements (see figure). We assume volt-meter to be ideal with infinite resistance. Thus,no current flows through the circuit and henceammeter will read zero. Also, potential at thenode Q is VQ = 0 V and potential at the nodeP is VP = 5 V. The voltmeter will displayV = VP − VQ = 5− 0 = 5 V.

A

i1

40µF

Q

25kΩ

5V

i2

50kΩ

P

20µF

The capacitor C1 = 40 µF is connectedacross a V = 5 V battery by a series resis-tance R1 = 25 kΩ. Similarly, the capacitorC2 = 20 µF is connected across the same bat-tery by a series resistance R2 = 50 kΩ. We canremove voltmeter from the circuit for currentcalculation as it has infinite resistance. Thiscircuit is used to charge both the capacitors. The currents through thecapacitors at time t (charging of the capacitor) are given by

i1 = (V/R1)e−t

R1C1 = 200e−t µA, (1)

i2 = (V/R2)e−t

R2C2 = 125e−t µA. (2)

From equations (1) and (2), the current through the ammeter at timet = 0 s is i0 = i1 + i2 = 325 µA. The current through the ammeter at timet = 1 s is i = i1 + i2 = 200e−1 + 125e−1 = 325/e = i0/e.

The potentials at points P and Q at a time t are given by

VP = V − i2R2 = 5(1− e−t), (3)

VQ = 0 + i2R1 = 5e−t. (4)

From equations (3) and (4), the voltmeter at time t = ln 2 will displayV = Vp − VQ = 5− 10e− ln 2 = 0 V.

Question 12. A block with mass M is connected by a massless springwith stiffness constant k to a rigid wall and moves without friction on ahorizontal surface. The block oscillates with small amplitude A about anequilibrium position x0. Consider two cases: (1) when the block is at x0;and (2) when the block is at x = x0 +A. In both the cases, a particle withmass m (< M) is softly placed on the block after which they stick to each


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other. Which of the following statement(s) is(are) true about the motionafter the mass m is placed on the mass M? (2016)

(A) The amplitude of oscillation in the first case changes by a factor of√M

m+M , whereas in the second case it remains unchanged.

(B) The final time period of oscillation in both the cases is same.(C) The total energy decreases in both the cases.(D) The instantaneous speed at x0 of the combined masses decreases in

both the cases.

Solution. The angular frequency of the spring mass system depends onthe spring constant and the attached mass. Initial angular frequency of thesystem (ω), final angular frequency of the system in case 1 (ω′1), and finalangular frequency of the system in case 2 (ω′2) are given by

ω =√k/M, ω′1 =

√k/(M +m), ω′2 =

√k/(M +m). (1)

The final time period of oscillation (T = 2π/ω) in both the cases is samebecause ω′1 = ω′2. For SHM in horizontal plane, mean position x0 is equalto the natural length of the spring. Thus, the mean position x0 remainssame in both the cases.

M

x0

kv

before

after

M

m

x0

kv′1

Let the initial amplitude of SHM be A, finalamplitude in case 1 be A′1, and final amplitudein case 2 be A′2. In case 1, the speed of themass M at the mean position just before massm is softly placed over it is v = ωA. Let thespeed just after mass m is placed over it be v′1.The linear momentum of the system along thedirection of motion is conserved because there isno external force on it in the direction of motion.Apply conservation of linear momentum, Mv = (M +m)v′1, to get

v′1 = Mv/(M +m) = MωA/(M +m). (2)

From equation (2), the instantaneous speed at the mean position is de-creased i.e., v′1 < v. The speed at the mean position v′1 is related to theamplitude A′1 by v′1 = A′1ω

′1. Substitute for v′1 and ω′1 to get

A′1 =v′1ω′1

=MA

M +m

√k/M√

k/(M +m)= A

√M

M +m. (3)

Let T be the initial total energy, T ′1 be the final total energy in case 1, andT ′2 be the final total energy in case 2. The final energy in case 1 is given by

T ′1 =1

2kA′1

2=

1

2kA2 M

M +m= T

M

M +m. (using equation (3)).


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M

x0 A

k V = 0

before

after

M

m

x0 A

kV ′ = 0

In case 2, the mass m is placed over theblock when it is at extreme position x = x0 +A. Thus, velocity of the block just before mis placed over it is zero. The conservation oflinear momentum gives that the velocity of thecombined masses just after m is placed over theblock is also zero. Hence the position x = x0 +A remains as the extreme position for SHM ofcombined masses. Thus, the amplitude of theSHM remains unchanged in case 2 i.e., A′2 = A. The total energy of the

system also remains unchanged in case 2 i.e., T ′2 = 12kA

′22

= 12kA

2 = T .The instantaneous speed at the mean position in case 2 is given by

v′2 = ω′2A′2 =

√k

M +mA =

√M

M +mωA =

M

M +mv.

yx

O

z Scr

een

S1 S2

d

D

Question 13. While conducting the Young’sdouble slit experiment, a student replaced thetwo slits with a large opaque plate in the x-yplane containing two small holes that act as twocoherent point sources (S1, S2) emitting lightof wavelength 600 nm. The student mistak-enly placed the screen parallel to the x-z plane(for z > 0) at a distance D = 3 m from the mid-point of S1S2, asshown schematically in the figure. The distance between the sources isd = 0.6003 mm. The origin O is at the intersection of the screen and theline joining S1S2. Which of the following is(are) true of the intensity pat-tern on the screen? (2016)

(A) Straight bright and dark bands parallel to the x-axis.(B) The region very close to the point O will be dark.(C) Hyperbolic bright and dark bands with foci symmetrically placed about

O in the x-direction.(D) Semi circular bright and dark bands centered at point O.

Solution. The waves from the sources S1 and S2 interfere at the origin O.The path difference between these waves is

∆ = S1O− S2O = d = 0.6003 mm = 2001(λ/2),

where λ = 600 nm. The path difference at O is odd integral multiple ofλ/2. Thus there will be dark fringe at O.

xO

z

P (x, z)

It can be easily seen that the path differenceis same for all points on the screen which lie on acircle centered at O. It can be proved by taking apoint P (x, z) on the screen. The path differenceat P is given by

∆=S1P− S2P=√x2 + z2 + (D + d

2 )2 −√x2 + z2 + (D − d

2 )2. (1)


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Simplify equation (1) to get (you need to square it twice)

x2 + z2 =D2d2

∆2+

∆2

4−D2 − d2

4= r2, (2)

which is an equation of a circle with centre O and radius r. Note that ∆is constant at a fringe. Since the screen covers upper half of the x-z plane,the student will see semi-circular bright and dark bands centred at point O.

0 L 2L 3L 4Lx

R

L

v0

Question 14. A rigid wire loop of square shapehaving side of length L and resistance R is mov-ing along the x-axis with a constant velocity v0in the plane of the paper. At t = 0, the rightedge of the loop enters a region of length 3Lwhere there is a uniform magnetic field B0 intothe plane of the paper, as shown in the figure. For sufficiently large v0, theloop eventually crosses the region. Let x be the location of the right edgeof the loop. Let v(x), i(x) and F (x) represent the velocity of the loop,current in the loop, and force on the loop, respectively as a function ofx. Counter-clockwise current is taken as positive. Which of the followingschematic plot(s) is(are) correct? [Ignore gravity.] (2016)

(A)

0 L 2L 3L 4Lx

v0

v(x) (B)

0 L 2L3L 4L x

i(x)

(C)

0 L 2L 3L 4Lx

i(x) (D)

0 L 2L 3L 4Lx

Fx)

Solution. Let the right edge of the loop be at x (< L) i.e., the right edgeof the loop is inside the magnetic field B0 and the left edge is outside B0.Let the velocity of the loop at this instant be v. By Faraday’s law, theinduced emf in the loop is e = B0Lv and induced current in the loop is

i = e/R = B0Lv/R. (1)

By Lenz’s law, the induced current should oppose the increase in magneticflux through the loop. Thus, the magnetic field due to the induced currentshould be coming out of the paper and hence current in the loop should becounterclockwise (see figure). The magnetic force acting on the right edge

is ~F = i~L × ~B0 = −iLB0 ı (towards left). Let m be the mass of the loop.Apply Newton’s second law to get

F = mdv/dt = mvdv/dx = −iLB0 = −(B20L

2/R)v, (2)


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where we have used equation (1) to replace i. Integrate equation (2) fromx = 0 to x (< L) to get

v(x) = v0 −B2

0L2

mRx. (3)

Substitute v(x) from equation (3) in equations (1) and (2) to get

i(x) =B0L

R

(v0 −

B20L

2

mRx

), (4)

F (x) = −B20L

2

R

(v0 −

B20L

2

mRx

). (5)

Thus, the velocity v(x) and the current i(x) decrease linearly with x butthe force F (x) increases linearly with x.

0 L 2L 3L 4Lx

iF R iF

L

v

Consider the case when loop is completely inside the magnetic filed i.e.,L < x < 3L. The magnetic flux through the loop is constant. Thus, theinduced emf e = 0, induced current i = 0, and magnetic force F = 0 (asi = 0). The velocity of the loop remains constant (as F = 0) at its valueat x = L i.e., v = v0 −B2

0L3/(mR).

Now, consider the case when the right edge of the loop is at x (> 3L).Let the loop velocity at this instant be v. The induced emf in the loopis e = B0Lv and induced current is i = B0Lv/R. By Lenz’s law, theinduced current should oppose the decrease in magnetic flux. Thus, themagnetic field due to the induced current should be into the paper andhence induced current is clockwise. The magnetic force acting on the leftedge is ~F = i~L × ~B0 = −(B2

0L2/R)v ı. Apply Newton’s second law and

integrate from x = 3L to x (< 4L) to get

v(x) = v0 −B2

0L3

mR− B2

0L2

mR(x− 3L). (6)

Substitute v(x) from equation (6) into the expressions for the induced cur-rent and the force to get

i(x) =B0L

R

(v0 −

B20L

3

mR− B2

0L2

mR(x− 3L)

), (7)

F (x) = −B20L

2

R

(v0 −

B20L

3

mR− B2

0L2

mR(x− 3L)

). (8)


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Thus, the velocity v(x) and the current i(x) decrease linearly with x butthe force F (x) increases linearly with x. The readers are encouraged toplot v(x), i(x), and F (x). What is the minimum value of v0 for the loopto come out of the magnetic field?


• This section contains TWO paragraphs.• Based on each paragraph, there are TWO questions.• Each question has FOUR options (A), (B), (C) and (D). ONLY ONE

of these four options is correct.• For each question, darken the bubble corresponding to the correct option

in the ORS.• For each question, marks will be awarded in


Full Marks :+3 If only the bubble corresponding to all the cor-rect option is darkened.

Zero Marks:0 In all other cases.

Paragraph for Question 15-16

A frame of reference that is accelerated with respect to an inertialframe of reference is called a non-inertial frame of reference. A coordinatesystem fixed on a circular disc rotating about a fixed axis with a constantangular velocity ω is an example of a non-inertial frame of reference. Therelationship between the force ~Frot experienced by a particle of mass mmoving on the rotating disc and the force ~Fin experienced by the particlein an inertial frame of reference is

~Frot = ~Fin + 2m(~vrot × ~ω) +m(~ω × ~r)× ~ω,

where ~vrot is the velocity of the particle in the rotating frame of referenceand ~r is the position vector of the particle with respect to the centre of thedisc.

m

R

R/2

ωNow consider a smooth slot along a di-ameter of a disc of radius R rotating counter-clockwise with a constant angular speed ω aboutits vertical axis through its center. We assign acoordinate system with the origin at the centerof the disc, the x-axis along the slot, the y-axisperpendicular to the slot and the z-axis alongthe rotation axis (~ω = ωk). A small block of mass m is gently placed inthe slot at ~r = (R/2) ı at t = 0 and is constrained to move only along theslot. (2016)

Question 15. The distance r of the block at time t is(A) R

4 (eωt + e−ωt) (B) R2 cosωt (C) R

4

(e2ωt + e−2ωt

)(D) R

2 cos 2ωt


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O x

z

y

mg

N2

N1

r

Solution. We will solve this problem in a ro-tating frame (x, y, z) attached to the disc. Letthe block is at a distance r from the centre of thedisc. The real forces (~Fin) acting on the blockare its weight mg, normal reaction N2 from thebottom surface of the groove, and normal reac-tion N1 from the side surface of the groove (seefigure). Since the groove is smooth, there is nofrictional force along the groove. Thus, the total force on the block is

~Fin = N1 + (N2 −mg) k. (1)

The force in the rotating frame is given by

~Frot = ~Fin + 2m(~vrot × ~ω) +m(~ω × ~r)× ~ω. (2)

In the rotating frame (x, y, z), position vector of the block is ~r = r ı, velocityof the block is ~vrot = (dr/dt) ı (because the block is constrained to movealong the groove), acceleration of the block is ~arot = d2r/dt2 ı, (becausethe block is constrained to move along the groove), and angular velocity

of the frame (disc) is ~ω = ω k. Newton’s second law in rotating frame is

written as ~Frot = m~arot. Substitute these parameters in equation (2) toget

md2r

dt2ı = (N1 + (N2 −mg) k) + 2m(dr/dt ı× ω k)

+m(ω k × r ı)× ω k

= mω2r ı+ (N1 − 2mω dr/dt) + (N2 −mg) k. (3)

Compare x, y and z components on two sides of equation (3) to get

d2r/dt2 = ω2r, (4)

N1 = 2mω dr/dt, (5)

N2 = mg. (6)

The equation (4) is a linear differential equation of second order. In higherclasses, you will learn various techniques to solve this type of equations.It is easy to show (by substitution) that r(t) = R

4 (eωt + e−ωt) is the onlysolution from the given options that satisfies equation (4). Also, it satisfiesthe initial condition i.e., r = R/2 at t = 0.

Question 16. The net reaction of the disc on the block is(A) 1

2mω2R(e2ωt − e−2ωt

)+mg k

(B) 12mω

2R (eωt − e−ωt) +mg k

(C) −mω2R cosωt −mg k(D) mω2R sinωt −mg k


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Solution. The net reaction on the block is

~N = N1 +N2k = −2mω dr/dt +mg k

= 2mω (Rω/4)(eωt − e−ωt) +mg k

= 12mω

2R(eωt − e−ωt) +mg k,

where we have used expressions of N1, N2, and r(t) from the previousquestion.

Paragraph for Question 17-18

A

HV

Consider an evacuated cylindrical chamberof height h having rigid conducting plates at theends and an insulating curved surface as shownin the figure. A number of spherical balls madeof light weight and soft material and coated witha conducting material are placed on the bottomplate. The balls have a radius r h. Now ahigh voltage source (HV) is connected across the conducting plates suchthat the bottom plate is at +V0 and the top plate at −V0. Due to theirconducting surface, the balls will get charged, will become equipotentialwith the plate and are repelled by it. The balls will eventually collide withthe top plate, where the coefficient of restitution can be taken to be zerodue to the soft nature of the material of the balls. The electric field in thechamber can be considered to be that of a parallel plate capacitor. Assumethat there are no collisions between the balls and the interaction betweenthem is negligible. [Ignore gravity.] (2016)

Question 17. Which one of the following statements is correct?

(A) The balls will stick to the top plate and remain there.

(B) The balls will bounce back to the bottom plate carrying the samecharge they went up with.

(C) The balls will bounce back to the bottom plate carrying the oppositecharge they went up with.

(D) The balls will execute simple harmonic motion between the two plates.

Solution. The distance between the two plates is h. The potential of thebottom plate is V0 and that of the top plates is −V0. The electric fieldbetween the plates is E = 2V0/h (directed upwards). The radius of eachball is r ( h). Let m be the mass and C be the capacitance of each ball.


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+V0

−V0

~E

+q

F

−qF

h

When the ball touches the bottom plate,it gets a positive charge q = CV0 (we assumethat the charge transfer is instantaneous). Thispositively charged ball experiences an upwardforce, F = qE = 2CV 2

0 /h, which acceleratesthe ball upwards. Since the force is constant,the ball cannot do SHM (for SHM, the forceshould be proportional to the displacement anddirected towards the centre).

When the ball hits the top plate, it transfers the positive charge tothe plate and gets negative charge q = −CV0. This negatively chargedball again experience a force F = qE (downward) and starts acceleratingdownwards. Thus, the ball keeps moving between the bottom and the topplates carrying a charge +q upwards and −q downwards.

Question 18. The average current in the steady state registered by theammeter in the circuit will be(A) zero (B) proportional to the potential V0

(C) proportional to V1/20 (D) proportional to V 2

0

Solution. The average current when a charge q moves from the bottomplate to the top plate in time T is i = q/T . The ball moves with a uniformacceleration a = F/m = 2CV 2

0 /(mh). The distance travelled in time T is

h =1

2aT 2 =

1

2

2CV 20

mhT 2, i.e., T =

h

V0

√m

C.

The average current carried by each ball is

i =q

T=C

h

√C

mV 20 .

IIT JEE (Advanced) 2016 Paper 2: PHYSICS€¦ · IIT JEE (Advanced) 2016 Paper 2: PHYSICS SECTION 1...

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