Triumph- JEE Advanced Physics Paper 2

1. PHYSICS
[1] JEE Advanced – 2013/ Paper - 2
Part I
Section–I : One or More Options Correct Type
TheSection contains 8 multiplechoicequestions.Eachquestion has fourchoices (A),(B),(C) and(D)out of
which ONE or MORE are correct.
1. Asteadycurrent Iflowsalonganinfinitelylonghollowcylindricalconductorofradius R.Thiscylinderis
placedcoaxiallyinsideaninfinitesolenoidofradius2R.Thesolenoidhasnturnsperunitlengthandcarries
asteadycurrent I.ConsiderapointP atadistancerfromthecommonaxis.Thecorrect statement(s)is are
(A) In the region 0 < r < R, the magnetic field is non-zero.
(B) In the region R < r < 2R, themagnetic field is alongthe common axis.
(C) In the region R< r < 2R, themagnetic field is tangential to the circle ofradius r, centeredon the axis.
(D) In the region r > 2R, the magnetic field is non-zero.
Ans. [A, C, D]
For cylinder, B r RC  0, ,
  


0
2
I
r
r Rto axis
For solenoid, B ni r RS   0 0 2along the axis, ;
 0 2, r R
Accordingly, options A, C and D are correct.
2. Twovehicles,eachmovingwithspeeduonthesamehoizontal straightroad,areapproachingeachother.
Wind blows along the road with velocity w. One of these vehicles blows a whistle of frequency f1
. An
observerintheothervehiclehears thefrequencyofthewhistletobef2
.Thespeedofsoundinstill airis V.
The correct statement(s) is are
(A) If the wind blows from the observer to the source, f2
> f1
.
(B) If the wind blows from the source to the observer, f2
> f1
.
(C) If the wind blows from observer to the source, f2
> f1
.
(D) If the wind blows from the source to the observer, f2
> f1
.
Ans. [A, B] When wind blows from observer to source,
f
V v u
V v u
f fw
w
2 1 1
 
 
F
HG I
KJ 
When wind blows from source to observer,
f
V v u
V v u
f fw
w
2 1 1
 
 
F
HG I
KJ  .

2. PHYSICS
[2] JEE Advanced – 2013/ Paper - 2
3. Two bodies, eachof mass M, are kept fixedwith a separation 2L. A particleof mass m is projected from
themidpointofthelinejoiningtheircentres,perpendiculartotheline.ThegravitationalconstantisG.The
correct statement(s)is (are)
(A) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is
4
GM
L
.
(B) The miminum velocityofthe mass mto escapethe gravitational field of the two bodies is 2
GM
L
.
(C) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is
2GM
L
.
(D) The energyof the mass m remains constant.
Ans. [B, D] For m to escape, its total energy must become zero or positive
× m
v
o MM
2L
i.e.
1
2
2
02
mv m
GM
L
 
L
NM O
QP
or v
GM
L
 2
Also during the motion, its total energy remains constant.
4. Aparticleofmass mis attachedtooneendofamass-less springofforceconstant k,lyingonafrictionless
horizontal plane. The other end of the spring is fixed. The particle starts moving horizontallyfrom its
equilibrium position at time t = 0 with an initial velocity u0
. When the speed of the particle is 0.5 u0
, it
collideselasticallywitharigidwall.Afterthiscollision
(A) the speedoftheparticlewhen itreturns toits equilibriumpositionis u0
.
(B)thetimeat whichtheparticlepasses throughtheequilibriumpositionforthefirsttimeis t
m
k
  .
(C)thetimeat whichthemaximumcompressionofthespringoccurs is t
m
k

4
3

.
(D)thetimeatwhichtheparticlepassesthroughtheequilibriumpositionforthesecondtimeis t
m
k

5
3

.

3. PHYSICS
[3] JEE Advanced – 2013/ Paper - 2
Ans. [A, D] Since the particle collides elastically so its speed first after collision is 0.5 u0
.
So option (A) is correct.
v t u t
u
t
T
b g   0
0
2 6
cos  .
So the particle passes through the equilibrium position first time at
t
T T m
k

F
HG I
KJ  2
6 3
2
3

so option (B) is wrong.
Maximum compression of the spring occurs at
t
T T T m
k
   
4 3
7
12
7
6

so option (C) is wrong.
Particle passes through the equilibrium position for the second time at
t
T T T m
k
   
2 3
5
6
5
3

Hence, option (D) is correct.
5. The figure below shows the variation of specific heat capacity(C) of a solid as afunction of temprature
(T). The temperature is increased continuouslyfrom 0 to 500 K at a constant rate. Ignoringanyvolume
change,thefollowingstatement(s)is (are)correct toa reasonable approximation.
(A) the rate at which heat is absorbed in the range 0-100 K varies linearlywith temperature T.
(B) heat absorbedin increasingthetemperaturefrom 0-100Kisless thanthe heatrequiredforincreasing
the temperature from 400-500 K.
(C) there is no change in the rate of heat absorption in the range 400-500 K.
(D) the rate of heat absorption increases in the range 200-300 K.
Ans. [A, B, C, D] From observing the figure, options B, C and D are correct. Also C is varying approximately linearly with
temperature in range 0-100 K.

dQ
dt
mC
dT
dt
 and since
dT
dt
= constant
dQ
dt
C and is varying linearly with temperature T.

4. PHYSICS
[4] JEE Advanced – 2013/ Paper - 2
6. The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a0
, where a0
is the Bohr radius. Its
orbital angular momentum is
3
2
h

. It is given that h is Planck constant and R is Rydberg constant. The
possible wavelength(s),whentheatom de-excites,is(are)
(A)
9
32R
(B)
9
16R
(C)
9
5R
(D)
4
3R
Ans. [A, C] L
nh

2
suggests electron is in third orbit.
Also using r
n
an 
2
0
2
 z = 2.
Hence possible transitions are
(i) 3  1,
1
4 1
1
9
 
L
NM O
QPR   
9
32R
(ii) 3  2,
1
4
1
4
1
9
 
L
NM O
QPR   
9
5R
(iii) 2  1,
1
4 1
1
4
 
L
NM O
QPR   
3
R
Hence [A, C] are correct.
7. Usingtheexpression2dsin=,onecalculatesthevalues ofdbymeasuringthecorrespondingangles 
in the range 0to 90°. The wavelength is exactlyknown and the errorin is constant forall values of .
As increases from 0°,
(A) the absolute error in dremains constant. (B) the absolute error in d increases.
(C)thefractional errorin dremains constant. (D) the fractional error in d decreases.
Ans. [D] Given d 

2sin so
d d
d
b g


 
2
cosec cot
Absolute error in d, d d db g 

  
2
cosec cot
Fractional error in d,
d d
d
b g  cot d 
Both decrease with increase in  .

5. PHYSICS
[5] JEE Advanced – 2013/ Paper - 2
8. Two non-conducting spheres of radii R1
and R2
and carrying uniform volume charge densities  and
 , respectively, are placed such that theypartiallyoverlap, as shown in the figure. At all points in the
overlappingregion,
(A) the electrostaticfieldis zero. (B) the electrostaticpotential is constant.
(C)theelectrostaticfieldis constantinmagnitude. (D)theelectrostaticfield has same direction.
Ans. [C, D] The field at any point say P is given by
r1
r2
d
P
  
E
r r
 




1
0
2
03 3
b g







  
r r d1 2
0 03 3
i.e. field is same at all points.
Hence, its magnitude and direction both remains same. As electric field is non-zero so potential is not constant.

6. PHYSICS
[6] JEE Advanced – 2013/ Paper - 2
Section–II : Paragraph Type
This section contains 4 paragraphs each describing theory, excperiment, data etc. Eight questions relate to
four paragraphs with two questions on each paragraph. Each question of a paragraph has only one correct
answer among the four choices. (A), (B), (C) and (D).
Paragraph for Questions 9 and 10
A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place
20kmawayfrom thepowerplant forconsumers usage.It canbetransportedeitherdirectlywithacableoflarge
currentcarryingcapacityorbyusingacombinationofstep-upandstep-downtransformers at thetwoends.The
drawbackofthedirecttransmissionisthelargeenergydissipation.Inthemethodusingtransformers,thedissipation
is much smaller.Inthis method, astep-uptransformeris usedat the plant side sothat thecurrent is reducedtoa
smaller value. At the consumers end, astep-down transformer is used tosupplypower to theconsumers at the
specified lower voltage.It is reasonabletoassumethat the powercable is purelyresistiveand the transformers
are ideal with apower factor unity.All the currents andvoltages mentioned are rms values.
9. If the direct transmission method with a cable of resistance 0 4 1
.  km
is used, the power dissipation
(in%)duringtransmissionis
(A) 20 (B) 30 (C) 40 (D) 50
Ans. [B]
10. Inthemethod usingthetransformers,assumethat theratioofthe numberofturns intheprimarytothat in
the secondaryin the step-up transformer is 1 : 10. If the power to the consumers has to be supplied at
200 V ,theratioofthenumberofturnsintheprimarytothatinthesecondaryinthestep-downtransformer
is
(A) 200 : 1 (B) 150 : 1 (C) 100 : 1 (D) 50 : 1
Ans. [A] For step up transformer at plant,
V
V
N
N
P
S
P
S
  V VS  40 000,
At consumer side
40 000
200
,

N
N
P
S

N
N
P
S

200
1
Also current flowing during transmission
i
P
V
T
T
 (assuming first transformer is 100% efficient)



600 10
40 000
15
3
,
A
Hence P i RLoss T    2 2
15 0 4 20 1800b g b g. W
 % loss 

 
1800
600 10
100% 30%3 .

7. PHYSICS
[7] JEE Advanced – 2013/ Paper - 2
Paragraph for Questions 11 and 12
A small block of mass 1 kg is released from rest at the topof a rough track. Thetrack is a circular arc of radius
40 m.Theblock slides alongthetrack without topplingandafrictional forceacts onit in thedirectionopposite
totheinstantaneousvelocity.Theworkdoneinovercomingthefrictionuptothepoint Q,asshowninthefigure
below, is 150 J. (Take the acceleration due to gravity, g = 10 ms–2
).
11. The speed of the block when it reaches the point Qis
(A) 5 1
ms
(B) 10 1
ms
(C) 10 3 1
ms
(D) 20 1
ms
Ans. [B]
12. The magnitudeof the normal reactionthat acts on theblock at the point Q is
(A) 7.5 N (B) 8.6 N (C) 11.5 N (D) 22.5 N
Ans. [A] For the motion till point Q, applying work energy theorem
1
2
30 1502
mV W W mg R JQ mg f   sinb g
 VQ  10m / s.
For the circular motion,
N mg sin30
mv
R
2
  
 N mg
mv
R
  sin N30 7 5
2
. .

8. PHYSICS
[8] JEE Advanced – 2013/ Paper - 2
Paragraph for Questions 13 and 14
The mass of a nucleus Z
A
X is less than the sum of the masses of (A – Z) number of neutrons and Z number of
protons in the nucleus. The energyequivalent to the corresponding mass difference is known as the binding
energy of nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m1 and m2 only if
m m M1 2 b g . Also two light nuclei of masses m3 and m4 can undergo complete fusion and form a heavy
nucleus of mass M onlyif m m M3 4  b g . Themasses of someneutral atoms aregiven in thetable below.
1
2
1
2
1
3
2
4
3
6
3
7
30
70
34
82
64
152
82
206
83
209
84
210
1007825 2 014102 3016050 4 002603
6 015123 7 016004 69 925325 81916709
151919803 205974455 208 980388 209 982876
H u H u H u He u
Li u Li u Zn u Se u
Gd u Pb u Bi u Po u
. . . .
. . . .
. . . .
.
1 932 2
u MeV c /c h
13. The correct statement is
(A) The nucleus 3
6
Li can emitanalphaparticle
(B) The nucleus 84
210
Po canemit a proton
(C) Deuteronandalphaparticle canundergocompletefusion.
(D) The nuclei 30
70
Zn and 34
82
Se canundergo complete fusion
Ans. [C]
14. Thekineticenergy(in keV )ofthe alphaparticle,whenthenucleus 84
210
Po atrest undergoesalphadecay,
is
(A) 5319 (B)5422 (C) 5707 (D) 5818
Ans. [A] For  decay of 84
214
Po , energy released
Q m m m e c  84
214
82
206
2
4 2
Po Pb He j e j e j =5422KeV.
Now energy taken
away by a particle
KeVK
A
A
Q 
F
HG I
KJ 
4
5319
For second part, only deutron and alpha particle undergo fission is feasible.

9. PHYSICS
[9] JEE Advanced – 2013/ Paper - 2
Paragraph for Questions 15 and 16
ApointchargeQismovinginacircularorbitofradiusRinthex-yplanewithanangularvelocity .Thiscanbe
considered as equivalenttoa loopcarryingasteadycurrent.
Q
2
.Auniform magneticfield alongthepositive
z-axis is nowswitchedon, whichincreasesat a constant ratefrom 0toBin one second.Assumethat the radius
oftheorbitremainsconstant.Theapplicationofthemagneticfieldinduces anemfintheorbit.Theinducedemf
is definedas the workdonebyan induced electricfield in movingaunit positive charge arounda closed loop.
It is knownthat, for an orbitingcharge, the magnetic dipolemoment is proportional tothe angular momentum
withaproportionalityconstant  .
15. Themagnitudeoftheinducedelectricfieldintheoribitatanyinstantoftimeduringthetimeintervalofthe
magneticfieldchangeis
(A)
BR
4
(B)
BR
2
(C) BR (D) 2BR
Ans. [B]
16. The changeinthemagneticdipolemoment associatedwiththeorbit,at theendofthetimeinterval ofthe
magneticfieldchange,is
(A)  BQR2
(B) 
BQR2
2
(C) 
BQR2
2
(D)  BQR2
Ans. [B] The time varying increasing magnetic field in + z direction produces induced electric field as shown. The magnitude
of induced electric field is E
R dB
dt
RB
 
2 2
.
Z
E
The angular impulse imported by induced electric field is equal to change in the angular momentum of moving charge
particle.
L QER dt Q
R dB
dt
Rdt
QR B
  
F
HG I
KJ 

zz 2 2
2
As   L so   

  L
QR B2
2

10. PHYSICS
[10] JEE Advanced – 2013/ Paper - 2
Section – III
Matching List Type
This section contains 4 questions. Eachquestion has fourstatements (A,B,C andD) given in ColumnI and
five statements (p, q, r, s and t) in Column II. Anygiven statement in Column I can have correct mathcing
with ONE or MORE statement(s) given in Column II. For example, if for a given question, statement B
matches with thestatements given in qand r, then for that particular question, against statement B, darken the
bubbles correspondingto q and r in the ORS.
17. MatchListIofthenuclearprocesses withList IIcontainingparent nucleus andoneoftheendproducts of
each process andthen select the correct answer usingthe codes givenbelow the lists:
List I List II
(P) Alpha decay (1) 8
15
7
15
O N ...
(Q) 
decay (q) 92
238
90
234
U Th ...
(R) Fission (r) 83
185
82
184
Bi Pb ...
(S) Protonemission (s) 94
239
57
140
Pu La ...
P Q R S
(A) 4 2 1 3
(B) 1 3 2 4
(C) 2 1 4 3
(D) 4 3 2 1
Ans. [C]

11. PHYSICS
[11] JEE Advanced – 2013/ Paper - 2
18. One mole of a monatomic ideal gas is taken along two cyclic processes E F G E   and
E F H E   as shownin the pV diagram.The processes involved arepurelyisochoric, isobaric,
isothermaloradiabatic
MatchthepathsinListIwiththemagnitudesoftheworkdoneinlistIIand selectthecorrectanswerusing
thecodes givenbelowthelists:
ColumnI Column II
(P) G E (1) 160 p0
V0
ln2
(Q) G H (q) 36 p0
V0
(R) F H (r) 24 p0
V0
(S) F G (s) 31 p0
V0
Codes
P Q R S
(A) 4 3 2 1
(B) 4 3 1 2
(C) 3 1 2 4
(D) 1 3 2 4
Ans. [A]

12. PHYSICS
[12] JEE Advanced – 2013/ Paper - 2
19. Math List Iwith List IIandselect the correct answer usingthe codes given below the lists :
List I List II
P. Boltzmannconstant 1. ML T2 1
Q. Coefficientofviscosity 2. ML T 1 1
R. Planckconstant 3. MLT K 3 1
S. Thermalconductivity 4. ML T K2 2 1 
Codes : P Q R S
(A) 3 1 2 4
(B) 3 2 1 4
(C) 4 2 1 3
(D) 4 1 2 3
Ans. [C]
20. Aright angledprismofrefractiveindex 1
is placedinarectangularblockofrefractiveindux 2
,whichis
surroundedbyamediumofrefractiveindex3
,asshowninthefigure.Arayoflight eenterstherectangular
blockatnormalincidence.Dependingupontherelationshipsbetween1
,2
and3
,ittakesoneofthefour
possible paths ef , eg , eh or ei .
Matchthepaths inList Iwithconditions ofrefractiveindices inList IIandselect thecorrectanswerusing
the codes givenbelowthe lists :
List I List II
P. e f 1.  1 22
Q. e g 2.  2 1 and  2 3
R. e h 3.  1 2
S. e i 4.   2 1 22  and  2 3
Codes : P Q R S
(A) 2 3 1 4
(B) 1 2 4 3
(C) 4 1 2 3
(D) 2 3 4 1
Ans. [D]