Triumph JEE Advanced Physics- Paper 1

1. PHYSICS
[1] IIT JEE ADVANCED – 2013 / Paper-1
JEE ADVANCED -2013
Paper - I
Part I: Physics
Section – I (Single Correct Answer Type)
This section contains 10 multiplechoice questions. Eachquestion has 4 choices(A), (B), (C) and(D), out of
which ONLYONE is correct.
1. ThediameterofacylinderismeasuredusingaVerniercalliperswithnozeroerror.Itisfoundthatthezero
of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50
divisions equivalent to2.45 cm. The 24th
divisionof the Vernier scale exactlycoincides with one of the
mainscaledivisions.Thediameterofthecylinderis
(A) 5.112 cm (B) 5.124 cm (C) 5.136 cm (D) 5.148 cm
Ans. [B] Value of one
Vernier division
cm = 0.049 cmV 
2 45
50
.
Value of main
scale division
cm, .S  005
 Least count, V C S V   0001. cm
Hence, diameter   510 24 0 001 5124. . .b g cm
2. Arayoflighttravellinginthedirection
1
2
3 i jd i isincidentonaplanemirror.Afterreflection,ittravels
alongthedirection
1
2
3 i jd i.Theangleofincidenceis
(A) 30o
(B) 45o
(C) 60o
(D) 75o
Ans. [A] Angle  between the incident and reflected ray will be
cos 
F
HG I
KJ F
HG I
KJ
F
HG
I
KJ F
HG
I
KJ
 
1
2
1
2
3
2
3
2
1 1
1
2b gb g
   600
Hence, angle made by incident ray with normal,   300

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3. IntheYoung'sdoubleslitexperimentusingamonochromaticlightofwavelength  ,thepathdifference(in
terms ofan integern)correspondingtoanypointhavinghalfthepeakintensityis
(A) 2 1
2
n b g
(B) 2 1
4
n b g
(C) 2 1
8
n b g
(D) 2 1
16
n b g 
Ans. [B] Intensity at a point, I Im
F
HG I
KJcos2
2

So, for I
Im

F
HG I
KJ
2 2
1
2
2
, cos

or    2 1
2
2
n Lb g 

or L n 2 1
4
b g
4. Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their
partial pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their
densitiesis
(A) 1 : 4 (B) 1 : 2 (C) 6 : 9 (D) 8 : 9
Ans. [D] From pM RT0   , p
M


0
(as T = constant)
p
p
M
M
1
2
1
2
0
0
2
1
8
9
  


5. Two rectangular blocks, having identical dimensions, can be arranged either in configuration Ior in
configuration IIas shown in the figure.One of the blocks has thermal conductivity k andthe other 2k .
The temperature difference between the ends along the x-axis is the same in both the configurations. It
takes 9 s to transport a certainamount of heat from the hot end to thecold end in theconfiguration I. The
timetotransport thesame amountof heat intheconfiguraitonIIis
(A) 2.0 s (B) 3.0 (C) 4.5 s (D) 6.0 s
Ans. [A] R
L
kA
L
kA
L
kA
I   
2
3
2
1 2 3
3R
kA
L
kA
L
kA
L
R
L
kAII
II    
Q Ht
T
R
t t R   

so
t
t
R
R
tI
II
1
2
2
9
2
2    s

3. PHYSICS
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6. A pulse of light of duration 100 ns is absorbedcompletelybya small object initiallyat rest.Power of the
pulse is 30 mW and thespeed of light is 3 × 108
m s–1
. The final momentum of the object is
(A) 03 10 17 1
.   
kgms (B) 10 10 17 1
.   
kgms (C) 30 10 17 1
.   
kgms (D) 9 0 10 17 1
.   
kgms
Ans. [B] The total energy absorbed,
E Pt        
30 10 100 10 3 103 9 9
J
Hence momentum delivered i.e. the final momentum of object,
p
E
c
 


 

 3 10
3 10
10 10
9
8
17 1
. kg m s
7. Aparticleofmassmisprojectedfromthegroundwithaninitialspeed u0 atanangle  withthehorizontal.
Atthehighestpointofitstrajectory,itmakesacompletelyinelasticcollisionwithanotheridenticalparticle,
which was thrownverticallyupward from theground with the same initial speed u0 . The angle that the
compositesystemmakeswiththehorizontalimmediatelyafterthecollisionis
(A)

4
(B)


4
 (C)


4
 (D)

2
Ans. [A] At the highest point,

v u i1  cos 

v u gh j u g
u
g
j u j2
2 2
2 2
1
2
2 2
2
   
F
HG
I
KJ
L
N
MM
O
Q
PP
e j sin  cos 

Applying conservation of linear momentum,

 
v
v v

1 2
2
so tan    
v
v
2
1
0
1 45
8. The work done on a particle of mass m by a force, K
x
x y
i
y
x y
j
2 2 3 2 2 2 3 2



L
N
MM
O
Q
PPc h c h/ /
 
(K being a con-
stantofappropriatedimensions),whentheparticleis takenfromthepoint a, 0b g tothepoint 0, ab galong
a circular path of radius a about theorigin in the x y plane is
(A)
2K
a

(B)
K
a

(C)
K
a

2
(D) 0
Ans. [D] Here
 
F K
xi yj
x y
Kr
r



L
N
MMM
O
Q
PPP

 
/2 2 3 2 3
e j
i.e., at all points

F is acting radially outwards from origin while particle is
moving tengentially. Thus
 
F v and work done is zero.

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9. One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another
horizontalthincopperwireoflengthLandradiusR.Whenthearrangementisstretchedbyappylingforces
at twoends, the ratio oftheelongationin thethin wire tothethat in the thickwire is
(A) 0.25 (B) 0.50 (C) 2.00 (D) 4.00
Ans. [C] Both rod are joined end to end, hence tension is same in both elongation, 


F
AY
, so






1
2
1
2
2
1
1
2
050   
A
A
.





2
1
2
10. The image ofan object, formed bya plano-convex lens at a distance of 8 m behind the lens, is real and is
one third thesize of the object. Thewavelength of light inside thelens is
2
3
times the wavelengthin free
space. The radius of the curved surface ofthe lens is
(A) 1 m (B) 2 m (C) 3 m (D) 6 m
Ans. [C] Here, v n  8
3
20
,


and m  
1
3
. So u  24 cm, applying lens equation
1
8
1
24
1
1
1 1
1 2


   
F
HG
I
KJb g b gf
n
R R
1
6
1
2
1
3
F
HG I
KJ F
HG I
KJ 
R
R m

5. PHYSICS
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Section–II Multiple Correct Answer Type
This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of
which ONE or MORE THAN ONE is / are correct.
11. Ahorizontalstretchedstring,fixedattwoends,isvibratinginitsfifthharmonicaccordingtotheequation,
y x t x t, . sin . cosb g b g c h c h  
0 01 62 8 628 1
m m s1 . Assuming   314. , the correct statement(s) is
(are)
(A) The number of nodes is 5.
(B) The lengthofthestringis0.25 m.
(C) Themaximumdisplacementofthemidpointofthestring,fromitsequilibriumpositionis0.01m.
(D) Thefundamental frequencyis100Hz.
Ans. [B, C] For fifth harmonic number of nodes are 6.
f
v
L
 
5
2
2
628

 L  
1
4
025. m
midpoint of the string is antinode. So maximum displacement is 0.01 m
f
f
0
5
20  Hz.
x=L/2
x=Lx=0
12. A solid sphere of radius R and densityis attachedto one end of a mass-less spring of force constant k.
The other endofthe springis connectedto anothersolidsphereof radius R anddensity3.Thecomplete
arrangementisplacedinaliquidofdensity2andisallowedtoreachequilibrium.Thecorrectstatement(s)
is (are)
(A) thenet elongation of thespringis
4
3
3
 R g
k
. (B) thenet elongation of thespringis
8
3
3
 R g
k
(C)thelight sphereis partiallysubmerged. (D)thelight sphereis completelysubmerged.
Ans. [A, D] The average density of complete system is equal to that of liquid so both the spheres will be completely
submerged. For equilibrium of system shown in figure,
F Vg Vgsp  2 3 b g b g
i.e. F Vgsp   and hence extension x
Vg
k
R g
k
 
  4
3
3

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13. AparticleofmassMandpositivechargeQ,movingwithaconstantvelocity

u i1
1
4  ms ,entersaregion
of uniform static magnetic field normal to the x-y plane. The region of the magnetic field extends from
x = 0 to x = L for all values of y. After passingthrough this region,the particle emerges onthe other side
after 10milliseconds with a velocity

u i j2
1
2 3   d i ms . The correct statement(s) is (are)
(A) Thedirectionofthemagneticfieldis -z direction.
(B) Thedirectionofthemagneticfieldis +z direction.
(C) Themagnitudeofthemagneticfield
50
3
M
Q
units.
(D) Themagnitudeofthemagneticfieldis
100
3
M
Q
units.
Ans. [A, C] From the given situation it is apparent that the magnetic field is in - z direction.
Particle deflects by 30° so duration is
T
12
.

2 1
12
10 10 3M
qB
F
HG I
KJ   
or B
M
Q

50
3

units.
14. Two non-conducting solid spheres of radii R and2R, havinguniform volume charge densities 1
and 2
respectively,toucheachother.Thenetelectricfieldat adistance2Rfromthecentreofthesmallersphere,
alongthe line joiningthe centres of the spheres, is zero. The ratio


1
2
can be
(A) – 4 (B)
32
25
(C)
32
25
(D) 4
Ans. [B, D] The situation described can correspond to two situations as shown in figure. For
first,
for Ep1
0 , field of both must be equal and opposite i.e. both must have same polarity of
charge
i.e.
 
 




1
3
0
2
2
0
1
4
3
4 2 3 2
4

L
NM O
QP   
R
R
R
b g
Similarly in second case, for fields to cancel, polarity of charge must be opposite on
them.
 
 
 
 
1
3
2
2
3
2
4
3
4 2
4
3
2
4 5

L
NM O
QP 

L
NM O
QPR
R
R
R0 0b g
b g
b g



1
2
32
25
 
y
x
30
o
v1
v2
P1
2R R
P2
2R
5R

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15. Inthecircuitshowninthefigure,therearetwoparallelplatecapacitors eachofcapacitanceC.Theswitch
S1
is pressed first to fully charge the capacitor C1
and then released. The switch S2
is then pressed to
charge the capacitor C2
. After some time, S2
is released and then S3
is pressed. After some time,
2V0
C1 C2
V0
S1 S2 S3
(A) the charge on the upper plate of C1
is 2CV0
.
(B) the charge on the upper plate of C1
is CV0
.
(C) the charge on the upper plate of C2
is 0.
(D) the charge on the upper plate of C2
is – CV0
.
Ans. [B, D] On pressing S1
, C1
gets fully charged with potential 2V0
. Now on releasing S1
and pressing S2
, the charge of C1
gets redistributed to get same potential on C2
and C1
given by
V
C V
C C
V


1 0
1 2
0
2b g
Thus charge on upper plates of C1
and C2
= + CV0
and on lower is – CV0
. Now on releasing S2
and pressing S3
, C1
becomes isolated and charge on its plates gets unchanged while due to battery C2
now gets charged by potential
V0
in a manner that its upper plate now has – CV0
charge

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Section – III: Integer Answer Type
This section contains 5 questions. The answer to each of the questions is a single-digit integer,rangingfrom
0 to9. (both enclusive)
16. The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the
stoppingpotential versus frequencyplot forSilver to that ofSodium is ______
Ans. [1] hv eVo  or V
h
e
v
eo  

slope of graph for both metals is same
h
e
F
HG I
KJ
 Ratio of slopes is 1.
17. Afreshlypreparedsampleofaradioisotopeofhalf-life1386shas activity103
disintegrationspersecond.
Giventhatln2=0.693,thefractionoftheinitialnumberofnuclei(expressedinnearestintegerpercentage)
that will decayinthe first 80 s afterpreparation of the sample is______
Ans. [4] T1 2 1386/ 
 t s T 80 1 2/ , so the activity remains nearly constant for the given duration. Percentage of nuclei decayed
 

N
N
A t
A T
t
0
100
2
100 4   
F
HG I
KJ 
/
ln

18. A particle ofmass 0.2 kgis movinginone dimensionunder aforce that delivers a constant power 0.5 W
to the particle. If theinitial speed (in m s–1
) ofthe particle is zero, the speed (inm s–1
) after5 s is_______
Ans. [5] Pt K mv 
1
2
2
or 05 5
1
2
0 2 2
. .  b gv
or v = 5 m / s.
19. A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocityof 10 rad s–1

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about its own axis, which is vertical. Two uniform circularrings, each of mass 6.25 kgand radius 0.2 m,
are gentlyplacedsymmetricallyon the discin such a manner that theyare touchingeachother along the
axis of the disc and are horizontal. Assume that thefriction is large enough such that the rings are at rest
relative to the disc and the system rotates about the original axis. The new angularvelocity(in rad s–1
)of
the system is_________
Ans. [8] Applying angular momentum conservation
o
M
m
MR MR
mr
2
0
2
2
2 2
2 2
F
HG
I
KJ  
F
HG
I
KJ e j '
or  = 8 rad / s.
20. Abobofmass m,suspendedbyastringoflength l1 isgivenminimum velocityrequiredtocompleteafull
circleintheverticalplane.Atthehighestpoint,itcollideselasticallywithanotherbobofmassmsuspended
bya string of length l2 , which is initiallyat rest. Both the strings are mass-less and inextensible. If the
second bob, after collision acquires the minimum speed required tocomplete a full circle in the vertical
plane, the ratio
l
l
1
2
is________.
Ans. [5] At the highest point, speed of bob 1 is v g1 1  . For 1 D, elastic collision between objects of equal masses,
velocities get interchanged.
v gl gl2 1 25  (for complete circle),
or
l
l
1
2
5 .