IIT JEE Advanced2 2013

2013 JEE-ADVANCED_P2_Code::1 QPaper with & Key &
Sol Solutions
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Sol Solutions
PART-I_PHYSICS
Section-1
(One or more options correct type)
This section contains 8 Multiple Choice questions. Each Question has Four choices
(A), (B), (C) and (D). Out of Which One or MORE are correct
01. Using the expression 2dsin   , one calculates thevalues of d by measuring the corre-
sponding angles  in therange 0 to 900
. Thewavelength  is exactly known and the error in
 is constant for all values of . As  increases from 00
A) the absolute error in d remains constant.
B) the absolute error in d increases
C) the fractional error in d remains constant
D) the fractional error in d decreases.
Ans : D
Sol.   2dsin d sin      constant K
     d cos d sin dd 0     
    d d tan dd  
Fractional error in d is
dd d
d tan



d constant (given). tan increases with increase in   the above fraction decreases.
Again    d tan dd
2sin

  

Absolute error in d is  
 
2
d cos
dd
2sin
  


d and  are constant (given). Clearly dd decreases with increase in .
02. Two non-conducting spheres of radii R1
and R2
and carrying uniform volume charge densi-
ties  and  respectively, are placed such that they partially overlap, as showin in the
figure. At all points in the overlapping region.

Sol Solutions
A) the electrostatic field is zero
B) the electrostatic potential is constant
C) the electrostatic field is constant in magnitude
D) the electrostatic field has same direction.
Ans : C,D
Sol. At P, due to 1st sphere, 1 1
0
E r
3



due to 2nd sphere, 2 2
0
E r
3



 P 1 2 1 2
0 0
E E E r r x
3 3
 
    
 
PE is independent of 1r or 2r but depends upon constant vector x .
E = constant  V is variable.
03. The figure below shows the variation of specific heat capacity(C) of a solid as a function of
temperature (T).The temperature is increased continuously from 0 to 500 K at a constant
rate. Ignoring any volume change, the following statement(s) is (are) correct to a reason-
able approximation.
A) the rate at which heat is absorbed in the range 0-100 K varies linearly with temperature
T
B) heat absorbed in increasing the temperature from 0-100 K is less than the heat required
for increasing the temperature from 400-500K
C) there is no change in the rate of heat absorption in the range 400-500K
D) the rate of heat absirption increases in the range 200-300K
Ans : B,C,D

Sol Solutions
Sol. Rate of heat absorption
dQ
R
dt

dT
R mC
dt

dT
dt
is constant R C 
From T = 0 to T = 100K. The graph is not a straight line passing through origin.
c and hence R does not vary linearly with T
dQ m CdT 
Q = m[area under C - T graph]
Q  area.
From graph, the area from0 to 100 K is less than the area from400 to 500 K. Hence option
B is correct
From 400K to 500K, C and hence R is constant.
From 200K to 300K, C and hence R increases.
04. The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 0 , where 0 is the
Bohr radius. Its orbital angular momentum is
3h
2
. It is given that h is Planck constant and R
is Rydbergconstant.The possible wavelength(s), when the atom de-excites, is(are)
A)
9
32R
B)
9
16R
C)
9
5R
D)
4
3R
Ans:A,C
Sol. Radius of nth orbit
2
n 0
n
r a
Z

 
2 2
0 0
n n
4.5a a 4.5 1
Z Z
   
 
nh 3h
n 3 2
2 1
   
 
Substituting (2) in (1), Z = 2
From Ridberg’s equation,
2
2 2
1 2
1 1 1
RZ
n n
 
    

Sol Solutions
n = 3 to 1 transition :- 1
1
1 1 9
4R 1
9 32R
 
      
2
1 1 1 9
4R
4 9 5R
 
      
3
1 1
4R 1 3R
4
 
      
05. Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m is
projected from the midpoint of the line joining their centres, perpendicular to the line. The
gravitational constant is G. The correct statement(s) is (are)
A) The minimum initial velocity of the mass m to escape the gravitational field of the two
bodoies is
GM
4
L
B) The minimum initial velocity of the mass m to escape the gravitational field of the two
bodies is
GM
2
L
C) The minimum initial velocity of the mass m to escape the gravitational field of the two
bodies is
2GM
L
D) The energy of the mass m remains constant
A ns : B,D or B
Sol. W KE 
2GMm 1 GM
2 mv v 2
L 2 2
   
If t he gravitational PE of system of M,M and m is considered as P.E of m itself, (which is
reasonable only when m <<M)
Total energy of m can be taken as constant.

Sol Solutions
06. A particle of mass m is attached to one end of a mass-less spring of force constant k, lying
on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts
movinghorizontallyfrom itsequilibrium positionat timet =0 withan initialvelocity u0
.When
the speed of the particle is 0.5 u0
, it collides elastically with a rigid wall. After this colli-
sion,
A) the speed of the particle when it returns to its equilibrium position is u0
B) the time at which the particle passes through the equilibrium position for the first time is
m
t
k
 
C) the time at which the maximum compression of the spring occurs is
4 m
t
3 k

 .
D) the time at which the particle passes through the equilibrium position for the second
time is
5 m
t
3 k


Ans:A,D
Sol. O toA : 0v u cos wt
0
0
u 2 t 2 t T
u cos t
2 T 3 T 6
  
    
From conservation of energy, it will have a speed 0u while crossing mean position every
time.
First time of crossing mean position
1
T 1 m 2 m
t 2t 2
3 3 K 3 K

     
Compression is max. when the particle reaches left extreme position. That time
2
T T 7T 7 m 7 m
t 2
3 4 12 12 K 6 K

      
second time mean position crossing time
3
T T 5 5 m 5 m
t T 2
3 2 6 6 K 3 K

      

Sol Solutions
07. A steady current I flows along an infinitely long hollow cylindrical conductor of radius R.
This cylinder is placed coaxially inside an infinite solenoid of radius 2R.The solenoid has n
turns per unit length and carries a steady current I. Consider a point P at a distance r from the
common axis. The correct statement(s) is (are)
A) In the region 0 < r < R, the magnetic field in non-zero
B) In the region R < r < 2R, the magnetic field is along the common axis.
C) In the region R < r < 2R, the magnetic field is tangential to the circle radius r, centered
on the axis.
D) In the region r > 2R, the magnetic field is non-zero
Ans:A,D
Sol. At P : 0 < r < R, B due to current in pipe is zero but due to solenoid, 0B nI 0  
At Q : R < r < 2R, B due to solenoid is parallel to the common axis and that due to I in pipe
is tangential to the circle of radius r i.e. the two fields are perpendicualr to each other. Their
resultant will make some angle with common axis.
At S : r > 2R, B = 0 due to solenoid but it is non zero due to current in pipe.
08. Two vehicles, each moving with speed u on the same horizontal straight road, are approach-
ing each other. Wind blows along the road with velocity w. One of these vehicles blows a
whistle of frequency f1
.An observer in the other vehicle hears the frequency of the whistle
to be f2
. The speed of sound in still air is 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 course to the observer, f2
< f1
Ans: A,B
Sol. f2
> f1
for any direction of motion of wind as there will be relative velocity of approach in
both the cases.

Sol Solutions
Section-2
(Paragraph Type)
This section contains 4 paragraphs each describing theory, experiment, 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 to 10
A point charge Q is moving in a circular orbit of radius R in the x-y plane with an
angular velocity  . This can be considered as equivalent to a loop carrying a steady
current
Q
2


. A uniform magnetic field along the positive z-axis is now switched on,
which increases at a constant rate from 0 to B in one second. Assume that the raduis
of the orbit remains constant.The application of the magnetic field induces an emf in
the orbit. The induced emf is defined as the work done by an induced electric field in
moving a unit positive charge around a closed loop.It is known that, for an orbiting
charge, the magnetic dipole moment is proportional to the angular momentum with
a proportionality constant  .
09. The magnitude of the induced electric field in the orbit at any instant of time duringthe time
interval of the magnetic field change is
A)
BR
4
B)
BR
2
C) BR D) 2BR
Ans : B
Sol.   2B 0 BR
E 2 R R E
1 2
 
      
10. The change in the magnetic dipole moment associated with the orbit, at the end of the time
interval of the magnetic field change, is
A) 2
BQR B)
2
BQR
2
 C)
2
BQR
2
 D) 2
BQR
Ans : B or C
Sol. Induced electric filed will do work and it leads to change in KE of the particle
   Fdt mdv EQ m dv 1     dt 1
Dipole moment
2 2 QV QVR
R I R
2 R 2
      

R,Q are constant
QR
d dv
2
 
      ,    
2
dv d 2
QR
   

Sol Solutions
Substituting (2) in (1)    
2
EQ m d
QR
 
BR 1 d
Q
2 r R
   
       [ It is given that, r
L 2m
 
  ]
2
rBR
d
2

  . d may be +ve or –ve depending upon sense of revolution (clockwise or
anticlockwise) of the particle.
The mass of a nucleus A
Z X is less than the sum of the masses of(A-Z) number of neu-
trons and Z number of protons in the nucleus. The energy equivalent to the corre-
sponding mass difference is known as ther binding energy of the nucleus. A heavy
nucleus of mass M can break into two light nuclei of masses m1
and m2
only if (m1
+
m2
) < M. Also two light nuclei of mass m3
and m4
can undergo complete fusion and
form a heavy nucleus of mass M1
only if (m3
+ m4
) > M1
. The masses of some neutral
atoms are given in the table below.
1
1H 1.007825 u 2
1 H 2.014102u 3
1 H 3.016050u 4
2 He 4.002603 u
6
3 Li 6.015123u 7
3 Li 7.016004u 70
30 Zn 69.925325u 82
34Se 81.916709u
152
64 Gd 151.919803 u 206
82 Pb 205.974455 u 209
83 Bi 208.980388 u 210
84 Po 209.982876u
11. The correct statement is
A) The nucleus 6
3 Li can emit an alpha particle
B) The nucleus 210
84 Po can emit a proton
C) Deuteron and alpha particle can undergo complete fusion
D) The nuclei 70
30 Zn and 82
34Se can undergo complete fusion
Ans : C
Sol. A : 6 4 2
3 2 1Li H  
Here sum of masses of products is more than the mass of 6
3 Li . Hence this is not possible

Sol Solutions
B : 210 1 209
0 1 83P H Bi 
This is also not possible as sum of masses 1
1H and 209
83 Bi is more than that of 210
84 0P .
C : 2 4 6
1 2 3H Li  
Sum of the masses of 2
1 H and 4
2  is more than that of 6
3 Li . Hence this reaction is pos-
sible.
70 82 152
30 34 64Zn Se Gd 
This is not possible.
12. The kinetic energy (in keV) of the alpha particle, when the nucleus 210
84 Po at rest undergoes
alpha decay, is
A) 5319 B) 5422 C) 5707 D) 5818
Ans:A
Sol. 210 4 206
84 0 2 82P Pb  
mass deffect    1 2M M m m 209.982876 4.002603 205.974455 0.00582amu       
aviable energy E 0.00582 931.5 5.419467MeV   
Momentum is same in magnitude for  and Pb.
2
Pb a
Pb
K MP
K
2m K M
  
 Pb Pb
Pb
Pb Pb
K M M
1 1 K K K
K M M M

 
 
 
      
Pb
Pb
M 205.974455
K E 5.419467 5.31616MeV 5316.16KeV
M M 209.977058


 
       
A small block of mass 1kg is released from rest at the top of a rough track.The track
is a circular arc of radius 40 m. The block slides along the track without toppling
and a frictional force acts on it in the direction opposite to the instantaneous
velocity.The work done in overcoming the friction up to the point Q, as shown in the
figure below, is 150J. (Take the acceleration due to gravity, g = 10 ms-2
).

Sol Solutions
13. The speed of the block when it reaches the point Q is
A) 5 ms-1
B) 10 ms-1
C) 1
10 3ms D) 20 ms-1
Ans: B
Sol.
0 21
mgR sin30 150 mv
2
 
  2 11 1
1 10 40 150 1 v v 10ms
2 2

      
14. The magnitude of the normal reaction that acts on the block at the point Q is
A) 7.5 N B) 8.6 N C) 11.5 N D) 22.5 N
Ans :A
Sol.
2
0 mv
N mgcos60
R
 
  1 1001
N 1 10 7.5N
2 40
    
A thermal power plant produces electric power of 600kW at 4000V, which is to be
transported to a place 20 km away from the power plant for consumers usage. It can
be transported either directly with a cable of large current carrying capacity or by
using a combination of step-up and step-down transformers at the two ends. The
drawback of the direct transmission is the large energy dissipation. In the method
using transformers, the dissipation is much smaller. In this method, a step-up trans-
former is used at the plant side so that the current is reduced to a smaller value. AT
the consumers’ end a step-down transformer is used to supply power to the consum-
ers at the specified lower voltage. It is reasonable to assume that the power cable is
purely resistive and the transformers are ideal with a power factor unity. All the
current and voltages mentioned are rms values.

Sol Solutions
15. If the direct transmission method with a cable of resistance 1
0.4 km
 is used, the power
dissipation (in %) during transmission is
A) 20 B) 30 C) 40 D) 50
Ans : B
Sol. Current produced
5
0
3
0
P 6 10
I 150A
V 4 10

  

Resistance of cable R 0.4 20 8   .
Percentage of power disspation when compared with produced power
 22
5
0
150 8I R
100 30
P 6 10

   

16. In the method using the transformers assume that the ratio of the number of turns in the
primary to that in the secondary in the step-up transformer is 1:10.If the power to the con-
sumers has to be supplied at 200V, the ratio of the number of turns in the primary to that in
the secondary in the step- down transformer is
A) 200 : 1 B) 150 : 1 C) 100 : 1 D) 50 : 1
Ans:A
Sol. At the step-up transformed, output voltage v1
is given by
41
13
v 10
v 4 10 v
4 10 1
   

This is input voltage to the step down transformed . Its output voltage v2
is given as 200v.
The ratio number of turns in step-down transformed
4
4 10
200
200

  .
Section-3
(Matching List Type)
This section contains 4 multiple choice questions. Each question has matching lists. The
codes for lists have choices A,B,C and D out of which ONLY ONE is correct.
17. Match List I with List IIand select the correct answer using the codes given below the List:
LIST- I LIST-II
P) Boltzmann constant 1) [ML2
T–1
]
Q) Coefficient of viscosity 2) [ML–1
T–1
]
R) Plank constant 3)[MLT-3
K-1
]
S) Thermal conductivity 4) [ML2
T–2
K–1
]
P Q R S 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
Sol. Conceptual

Sol Solutions
18. A right angled prism of refractive index 1 is placed in a rectangular block of refractive
index 2 , which is surrounded by a medium of refractive index 3 , as shown in the figure.
A ray of light ‘e’ enters the rectangular block at normal incidence. Depending upon the
relationships between 1 2,  and 3 it takes one of the four possible paths ‘ef’, ‘eg’, ‘eh’or
‘ei’.
Match the paths in List I with conditions of refractive indices in List II and select the cor-
rect answer using the codes given below the 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  
P Q R S 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
Sol. Conceptual
19. Match List I of the nuclear processes with List II containing parent nucleus and one of the
end products of each process and then select the correct answer using the codes given
below the lists.
LIST- I LIST-II
P)Alpha decay 1) 15 15
8 7O N ...... 
Q) 
 decay 2) 238 234
92 90U Th .......... 
R) Fission 3) 185 184
83 82Bi Pb ........... 
S) Proton emission 4) 239 140
94 57Pu La .......... 
P Q R S 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
Sol. Conceptual

Sol Solutions
20. One mole of a monatomic ideal gas is taken along two cyclic processes E F G E  
and E F H E   as shown in the PV diagram.The processes involved are purely
isochoric. isobaric, isothermal or adiabatic.
Match the path in List I with the magnitudes of the work done in List II and select the
correct answer using the codes given below the lists.
LIST- I LIST-II
P) G E 1) 160 P0
V0
ln2
Q) G H 2) 36 P0
V0
R) F H 3) 24 P0
V0
S) F G 4) 31 P0
V0
P Q R S 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
Sol. E H :
5/3
r r 0 H
1 1 2 2 H 0
0 0
32P V
P V P V V 8V
P V
 
      
 G E 0 0 0 0 0W P 32V V 31P V   
 G H 0 0 0 0 0W P 32V 8V 24P V   

Sol Solutions
PART-II_CHEMISTRY
Section-1
21. The correct statement(s) about O3
is (are)
A) O – O bond lengths are equal
B) Thermal decomposition of O3
is endothermic
C) O3
is diamagnetic in nature D) O3
has a bent structure
Ans:A,C,D
Sol. 3O exihibits resonance ; its formation is endothermic but decomposition is exothermic ; it
is diamagnetic with ben strucuture
22. In the nuclear transmutation
9 8
4 4Be X Be Y   (X,Y) is (are)
A)  ,n B) (p,D) C) (n,D) D)  ,p
Ans: A,B
Sol. 9 0 8 1
4 0 4 0Be Be n   
9 0 8 2
4 0 4 1Be P Be D  
23. The carbon – based reduction method is NOT used for the extraction of
A) tin from 2SnO B) iron from 2 3Fe O
C) aluminium from 2 3Al O D) magnesium from 3 3MgCO .CaCO
Ans : C,D
Sol. 2SnO C 8n CO   2 3Fe O C Fe CO  
But 2 3Al O and dolamite cannot be reduced
24. The thermal dissociation equilibrium of  3CaCO s is studied under different conditions.
     3 2CaCO s CaO s CO g
For this equilibrium, the correct statement(s) is (are)
A) H is dependent on T
B) K is independent of the initial amount of 3CaCO
C) K is dependent on the pressure of 2CO at given T
D) H is independent of the catalyst, if any
Ans: A,B,D

Sol Solutions
Sol. (1) BH depends on T but not an catalyst
(2) K is independenc of conc.
25. The spK of 2 4Ag CrO is 12
1.1 10
 at 298K. The solubility (in mol/L) of 2 4Ag CrO in 0.1M
3AgNO solution is
A) 11
1.1 10
 B) 10
1.1 10
 C) 12
1.1 10
 D) 9
1.1 10

Ans : B
Sol.  
2 2
sp 4K Ag CrO 
   
   212 2
41.1 10 0.1 CrO 
 
 2
4CrO
 solubility = 1.1 × 10–10
26. In the following reaction, the product(s) formed is (are)
OH
CH3
CHCl3
OH-
?
(P)
OH
CH3
OHC CHO
(Q)
O
CHCl2
H3C
(R)
OH
CHCl2
H3C
(S)
OH
CH3
CHO
A) P (major) B) Q (minor) C) R (minor) D) S (major)
Ans : D
Sol. Reimer teimann reaction
OH
CH3
CHCl3
OH-
OH
CH3
CHO

Sol Solutions
27. The major product(s) of the following reaction is (are)
OH
SO3H
aqueous Br2 (3.0 equivalents)
?
A) P B) Q C) R D) S
Ans : B
Sol.
OH
SO3H
+ 3Br2
OH
Br
Br Br
28. After completion of the reactions (I and II), the organic compound(s) in the reaction mix-
tures is(are)
Reaction I : H3C CH3
O
(1.0 mol)
Br2 (1.0 mol)
aqueous/NaOH
Reaction II : H3C CH3
O
(1.0 mol)
Br2 (1.0 mol)
CH3COOH
(P)
H3C CH2Br
O
(Q)
H3C CBr3
O
(R)
Br3C CBr3
O
(S)
BrH2C CH2Br
O
(T)
H3C ONa
O
(U) CHBr3
A) Reaction I : P and Reaction II : P
B) Reaction I : U, acetone and Reaction II : Q, acetone
C) Reaction I : T,U, acetone and Reaction II : P
D) Reaction I : R, acetone and Reaction II : S, acetone

Sol Solutions
Ans : C
Sol. Acetone + Br + NaOh 3 3CH COONa CHBr 
3CH COOH
3 3 2 3 2CH C CH Br CH C CH Br     
O O
Section-2
(Paragraph Type)
Paragraph for Question 29 and 30
A fixed mass ‘m’ of a gas is subjected transformation of states from K to L to M to N
and back to K as shown in the figure
K L
MN
Pressure
Volume
29. The succeeding operations that enable this transformation of states are
A) Heating, cooling, heating, colling B) Cooling, heating, cooling, heating
C) Heating, cooling, cooling, heating D) Cooling, heating, hetaing, cooling
Ans : C
Sol. K L – expantion which needs heating
L M – Isochloric which needs heating process
M N – Isobaric which needs cooling process
N K – Isochoric which needs heating process
30. The pair of isochoric processes among the transformation of state is
A) K to L and L to M B) L to M and N to K
C) L to M and M to N D) M to N and N to K
Ans : B
Sol. Conceptual

Sol Solutions
The reactions of Cl2
gas with cold-dilute and hot-concentrated NaOH in water give
sodium salts of two (different) oxoacids of chlorine, P and Q, respectively. The Cl2
gas reacts with SO2
gas, in presence of charcoal, to give a product R. R reacts with
white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phos-
phorus T.
31. P and Q, respectively, are the sodium salts of
A) hypochlorus and chloric acids B) hypochlorus and chlorus acids
C) chloric and perchloric acids D) chloric and hypochlorus acids
Ans:A
Sol.
2 2NaOH Cl NaCl NaClO H O   
Salt of HOCl
cold
Salt of HOCl3
hot
2 3 2NaOH Cl NaCl NaClO H O   
32. R,S and Trespectively, are
A) 2 2 5SO Cl ,PCl and 3 4H PO B) 2 2 3SO Cl ,PCl and 3 3H PO
C) 2 3SOCl ,PCl and 3 2H PO D) 2 5SOCl ,PCl and 3 4H PO
Ans:A
Sol. 2 2 2 2SO Cl SO Cl 
2 2 5SO Cl P PCl 
5 2 3 4PCl H O H PO HCl  
An aqueous solution of a mixture of two inorganic salts, when treated with dilute
HCl, gave a precipitate (P) and a filtrate (Q). The precipitate P was found to dissolve
in hot water. The filtrate (Q) remained unchanged, when treated with H2
S in a dilute
mineralacid medium. However, it gave a precipitate (R) with H2
S in an ammoniacal
medium. The precipitate R gave a coloured solution (S), when treated with 2 2H O in
an aqueous NaOH medium.

Sol Solutions
33. The precipitate P contains
A) 2
Pb  B) 2
2Hg 
C) Ag
D) 2
Hg 
Ans:A
Sol. 2PbCl is soluble is hot water
34. The coloured solution S contains
A)  2 4 3
Fe SO B) 4CuSO C) 4ZnSO D) 2 4Na CrO
Ans : D
Sol. 2 4Na CrO is coloured solution.
P and Q are isomers of dicarboxylic acid 4 4 4C H O . Both decolorize 2 2Br / H O . On heat-
ing P forms the cyclic anhydride.
Upon treatment with dilute alkaline 4KMnO , P as well as Q could produce one or
more than one from S,T and U.
(S)
COOH
H OH
COOH
H OH (T)
COOH
H OH
COOH
HO H (U)
COOH
HO H
COOH
H OH
35. Compounds formed from P and Q are, respectively
A) Optically active S and optically active pair (T,U)
B) Optically inactive S and optically inactive pair (T,U)
C) Optically active pair (T,U) and optically active S
D) Optically inactive pair (T,U) and optically inactive S
Ans : B
Sol. P – malie acid (cis compound)
Q – fumeric acid (trans compound)
4KMnO
OH
P  meso compound S
4KMnO
OH
Q  resmic mixture (T & U)

Sol Solutions
36. In the following reaction sequences V and W are, respectively
2 /H Ni
Q V

+ V AlCl3 (anhydrous) 1.Zn-Hg/HCl
2.H3PO4
W
A)
O
O
OV
and
O
W
B) CH2OH
CH2OH
V
and
W
C)
O
O
OV
and
W
D)
HO H2C
CH2O HV
and
CH2OH
W
Ans:A
Sol. Fumeric acid 2H / Ni
Butane dioic acid

O
O
O
O
O
O
+ 3AlCl
acylation

2 2CH CH COOH 
An Hg/HCl

2 2CH CH CH COOH  
3 4H PO

O

Sol Solutions
Section-3
37. Match the chemical conversions in List I with the appropriate reagents in List II and select
the correct answer using the code given below the lists :
List I List II
P) Cl 1)       42
i Hg OAc ; ii NaBH
Q) ONa OEt
2) NaOEt
R)
OH
3) Et–Br
S)
OH
4)    3 2 2i BH ; ii H O / NaOH
Codes :
P Q R S P Q R S
A) 2 3 1 4 B) 3 2 1 4
C) 2 3 4 1 D) 3 2 4 1
Ans:A
Sol. Cl
C2H5ONa
;
ONa + C2H5Br
OC2H5
Oxymeruration
demeruration OH
;
Hydroboration
oxidation
OH

Sol Solutions
38. The unbalanced chemical reactions given in List I show missing reagent or condition (?)
which are provided in List II. Match List Iwith ListII and select the correct answer using the
code given below the lists :
List I List II
P) ?
2 2 4 4 2PbO H SO PbSO O    other product 1) NO
Q) ?
2 2 3 2 4Na S O H O NaHSO  + other product 2) 2I
R) ?
2 4 2N H N  other product 3) Warm
S) ?
2XeF Xe  other product 4) 2Cl
Codes :
P Q R S P Q R S
A) 4 2 3 1 B) 3 2 1 4
C) 1 4 2 3 D) 3 4 2 1
Ans : D
Sol. warm
2 2 4 4 2 2PbO H SO PbSO O H O   
2Cl
2 2 3 2 4Na S O H O NaHSO S HCl   
2 4 2 2N H I N HI  
2XeF NO Xe NOF  
39. The standard reduction potential data at 250
C is given below.
 0 3 2
E Fe ,Fe 0.77V 
  ;
 0 2
E Fe ,Fe 0.44V
 
 0 2
E Cu ,Cu 0.34V
  ;
 0
E Cu ,Cu 0.52V
 
 0
2 2E O g 4H 4e 2H O 1.23V 
      
 0
2 2E O g 2H O 4e 4OH 0.40V 
      
 0 3
E Cr ,Cr 0.74V
 
 0 2
E Cr ,Cr 0.91V
 
Match 0
E of the redox pair in List I with the values given in List II and select the correct
answer using the code given below the lists :

Sol Solutions
List I List II
P)  0 3
E Fe ,Fe
1) –0.18V
Q)  0
2E 4H O 4H 4OH 
 2) –0.4V
R)  0 2
E Cu Cu 2Cu 
  3) –0.04V
S)  0 3 2
E Cr ,Cr 
4) –0.83V
Codes :
P Q R S P Q R S
A) 4 1 2 3 B) 2 3 4 1
C) 1 2 3 4 D) 3 4 1 2
Ans : D
Sol. Conceptual
40. An aqueous solution of X is added slowly to an aqueous solution ofYas shown in List I. The
variation in conductivity of these reactions is given in List II. Match List I with List II and
select the correct answer using the code given below the lists :
List I List II
P)
 2 5 33
C H N CH COOH
X Y
1) Conductivity decreases and then increases
Q)
X Y
   3KI 0.1M AgNO 0.01M
2) Conductivity decreasesand then does not change much
R)
X Y
3CH COOH KOH
3) Conductivity increases and then does not chnage much
S)
X Y
NaOH HI
4) Conductivity does not change much and then increases
Codes :
P Q R S P Q R S
A) 3 4 2 1 B) 4 3 2 1
C) 2 3 4 1 D) 1 4 3 2
Ans:A
Sol. Conceptual

Sol Solutions
PART-III_MATHEMATICS
Section-1
41. Let
3 i
w
2

 and  n
P w : n 1,2,3,...  . Further 1
1
H z C :Rez
2
 
   
 
and
2
1
H z C: Rez
2
 
   
 
, where C is the set of all complex numbers. if 1 1z P H  ,
2 2z P H  and O represents the origin, then 1 2z Oz 
A)
2

B)
6

C)
2
3

D)
5
6

Ans: C,D
Sol :From diagram 1 2
2 5
z z ,
3 6
 

42 If x x 1
3 4 
 then x=
A)
3
3
2log 2
2log 2 1 B)
2
2
2log 3 C)
2
1
1 log 3 D)
2
2
2log 3
2log 3 1
Ans:- A,B,C
Sol:
3
3 2
2log 2 2
x log3 (x 1)log 4 x
2log 2 1 2 log 3
    
 
43. Let  be a complex cube root of unity with 1  and ijP p    be a n n
matrix with i j
ijp 
  . Then 2
p 0 , when n=
A) 57 B) 55 C) 58 D) 56

Sol Solutions
Ans:- B,C,D
Sol: 4 6 8 2n 2
2P 0 w w w .............. w 0
      
4 2n
2n
2
w (1 w )
0 w 1 n 55,58,56
1 w

    

44. The function  f x 2 x x 2 x 2 2 x      has a local minimum or a
local maximum at x=
A) -2 B)
2
3

C) 2 D)
2
3
Ans:- A,B
Sol: f(x) = 0 for
2
x 2, ,0,2
3

 
f(x) has extremum at
2
x 2,
3

 
45. For a R (the set of all real numbers), a 1, 
 
       
a a a
a 1n
1 2 ... n 1
lim
60n 1 na 1 na 2 ... na n

  

         
Then a=
A) 5 B) 7 C)
15
2

D)
17
2

Ans:- B,D
Sol: -
1
a
0
1
n
0
x dx
1 2 1 17
Lt a 7 or
(a 1) (2a 1) 60 2
(a x)dx


    
 



46. Circle(s) touching x-axis at a distance 3 from the origin and having
an intercept of length 2 7 on y-axis is (are)
A) 2 2
x y 6x 8y 9 0     B) 2 2
x y 6x 7y 9 0    
C) 2 2
x y 6x 8y 9 0     D) 2 2
x y 6x 7y 9 0    
Ans:- A,C

Sol Solutions
Sol: 2 2 2
x 2gx c (x 3) x 6x 9      
2
2 f 9 2 7 f 4    
47. Two lines 1
y z
L : x 5,
3 2
 
  
and 2
y z
L : x ,
1 2
  
  
are coplanar. Then
 can take value(s)
A) 1 B) 2 C) 3 D) 4
Ans:- A,D
Sol: lines are x - 5 = 0, 2y (3 )z 0   
x , (2 )y z 0     
since lines are coplanar
0 3 2
0 1 2 0
5 0 0
 
    

2
( 5)( 5 4) 0 1,4          
48. In a triangle PQR, P is the largest angle and
1
cos P
3
 . Further the
incircle of the triangle touches the sides PQ, QR and RP at N,L and
M respectively, such that the lengths of PN, QL and RM are con-
secutive even integers. Then possible lengths of the sides of the
triangle is (are)
A) 16 B) 18 C) 24 D) 22
Ans:- B,D
P
RQ
N M
L
2K
2K+2
2K+42K+2
2K+4
2K

Sol Solutions
Sol:
2 2 2
1 (4k 4) (4k 4) (4k 6)
cosP
3 2(4k 2)(4k 4)
    
 
 
2 2 2 2 2
2
(2k 1) (2k 2) (2k 3) 4k 4 k 1
4
8(2k 1)(2k 2) 4(2k 1)(k 1) 2k 3k 1
      
  
     
2 2
3k 3 2k 3k 1   
2
k 3k 4 0  
(k-4)(k+1)=0
k = 4
sides: 18, 22
But B D options are correct
Section-2
(Paragraph Type)
Let 1 2 3S S S S ,   where
 1 2
z 1 3i
S z C: z 4 ,S z C: Im 0
1 3i
    
       
   
and
 3S z C : Rez 0  
49. Area of S=
A)
10
3

B)
20
3

C)
16
3

D)
32
3

50. z S
min 1 3i z

  
A)
2 3
2

B)
2 3
2

C)
3 3
2

D)
3 3
2

Sol: Q.No.: 49 & 50
2 2 2
1S x y 4   ,
2 m
(x 1) i(y 3)
S I 0
1 3i
   
  
 
3(x 1) y 3 0 3x y 0       
3S x 0 

Sol Solutions
P(1,-3)
x
y
3

3x y 0 
49 Ans:- B
Area of
2 21 1
S 4 4 20
4 2 3 3
 
     
50. Ans:- C
min 1 3i z   perpendicular distance from (1, -3)
the line 3x y 0 
3 3 3 3
2 2
 
 
Paragraph for Questions 51 and 52
A box 1B contains 1 white ball, 3 red balls and 2 black balls. An-
other box 2B contains 2 white balls, 3 red balls and 4 black balls.
A third box 3B contains 3 white balls, 4 red balls and 5 black
balls.
51. If 1 ball is drawn from each of the boxes 1 2B ,B and 3B , the probabil-
ity that all 3 drawn balls are of the same colour is
A)
82
648
B)
90
648
C)
558
648
D)
566
648
52. If 2 balls are drawn (without replacement) from a randomly se-
lected box and one of the balls is white and the other ball is red,
the probability that these 2 balls are drawn from box 2B is
A)
116
181
B)
126
181
C)
65
181
D)
55
181

Sol Solutions
Sol: Q.No.: 51 & 52
B1
B2
B3
1 2 3
3R 3R 4R
2B
6
4B
9
5B
12
51. Ans:- A
Prob = www (or) RRR (or) BBB
=
1 2 3 3 3 4 2 4 5 82
6 9 12 6 9 12 6 9 12 648
         
52. Ans: D
9
2
6 9 12
2 2 2
1 2 3
553 C
Prob
1 1 3 1 2 3 1 3 4 181
3 C 3 C 3 C


 
  
    
Let  f : 0,1 R (the set of all real numbers) be a function. Suppose
the function f is twice differentiable,    f 0 f 1 0  and satisfies
       x
f " x 2f ' x f x e ,x 0,1    .
53. Which of the following is true for 0<x<1 ?
A)  0 f x   B)  
1 1
f x
2 2
   C)  
1
f x 1
4
   D)  f x 0  
54. If the function  x
e f x
assumes its minimum in the interval  0,1 at
1
x
4
 , which of the following is true ?
A)    
1 3
f ' x f x , x
4 4
   B)    
1
f ' x f x ,0 x
4
  
C)    
1
f ' x f x ,0 x
4
   D)    
3
f ' x f x , x 1
4
  

Sol Solutions
Sol: Q.No.: 53 & 54
53 Ans: D
11 1 n
f (x) 2f (x) f (x) e  
x 11 1
e (f (x) 2f (x) f (x) 1
   
x 11
e g (x) 1
  
11
g (x) 0 
g(x) 0 x (0,1)  
x
e f (x) 0 x (0,1)
  
f(x) 0 x (0,1)  
54. Ans: C
x
g(x) e f (x)

1 x 1
g (x) e (f (x) f (x)) 0
   at x = 1/4
From graph
1
0 x
4
 
g(x) is decreasing
1
g (x) 0
x 1 1
e (f (x) f (x) 0 f (x) f (x) 0
     
1
f (x) f (x) 
Let PQ be a focal chord of the parabola 2
y 4ax . The tangents to
the parabola at P and Q meet at a point lying on the line
y 2x a,a 0  
55. Length of chord PQ is
A) 7a B) 5a C) 2a D) 3a
56. If chord PQ subtends an angle  at the vertex of 2
y 4ax , then tan 
A)
2
7
3
B)
2
7
3

C)
2
5
3
D)
2
5
3


Sol Solutions
Sol:
55. Ans:- B
1
a t a
t
 
     ,
2
1
t 1
t
   ,
2
1 1
t t 4 5
t t
   
          
2
1
PQ a t 5a
t
 
    
56. Ans:- D
1 2
2 2
m , m 2t
1t
t
   
 ,
2
2t
2 1 2t
tan t 5
1 4 3 t 3
 
    
      
Section-3
57. A line L : y mx 3  meets y-axis at E(0,3) and the arc of the parabola
2
y 16x,0 y 6   at the point  0 0F x ,y . The tangent to the parabola at  0 0F x ,y
intersects the y-axis at  1G 0,y . The slope m of the line L is chosen such that the
area of the triangle EFG has a local maximum.
Match List I with List II and select the correct answer using the code given below
the lists :
List I List II
P) m = 1)
1
2
Q) Maximum area of EFG is 2) 4
R) 0y  3) 2
S) 1y  4) 1
P Q R S P Q R S
A) 4 1 2 3 B) 3 4 1 2
C) 1 3 2 4 D) 1 3 4 2
Ans:- A

Sol Solutions
Sol:
2 2 31
(3 4t)4t 2(3t 4t )
2
    
2d 1
0 6t 12t 0 t
dt 2

     
1 2
t
2
8t 3 4 3
m [slope] 1
14t 4
4

 
   

m 1 P 4    ,
1 1
Q (1) 1 Q 1
2 2
     , 0y 4 R 2   , 1y 2 S 3  
A is the correct option
58. Match List I with List II and select the correct answer using the code given below
the lists :
List I List II
P)
   
   
2
1 1
4
2 1 1
cos tan y ysin tan y1
y
y cot sin y tan sin y
 
 
  
  
   
1)
1 5
2 3
Q) If cos x cos y cosz 0 sin x sin y sin z      then possible 2) 2
value of
x y
cos
2
 
   is
R) If cos x cos2x sin xsin 2xsecx cos xsin 2xsec x cos x
4 4
    
          
cos2x then possible value of secx is 3)
1
2
S) If     1 2 1
cot sin 1 x sin tan x 6 , x 0 
   , then possible value 4) 1
of x is
P Q R S P Q R S
A) 4 3 1 2 B) 4 3 2 1
C) 3 4 2 1 D) 3 4 1 2
Ans:- B

Sol Solutions
Sol: cosx + cosy + cosz = sinx + siny + sinz
x x, y 120 x, z 240 x     
0x y 1
cos cos60
2 2

  , S)    1 2 1
cot sin 1 x sin tan x 6 
 
2 2
x x 6
1 x 1 6x
 
 
,
2 2 2 1 5
1 6x 6 6x 12x 5 x
2 3
       
B is correct answer
59. Consider the lines 1
x 1 y z 3
L : ,
2 1 1
 
 

2
x 4 y 3 z 3
L :
1 1 2
  
  and the
planes 1 2P :7x y 2z 3,P :3x 5y 6z 4      . Let ax by cz d   be the equa-
tion of the plane passing through the point of intersection of lines
1L and 2L , and perpendicular to planes 1P and 2P
Match List-I with List-II and select the correct answer using the
code given below the lists :
List I List II
P) a= 1) 13
Q) b= 2) -3
R) c= 3) 1
S) d= 4) -2
P Q R S
A) 3 2 4 1
B) 1 3 4 2
C) 3 2 1 4
D) 2 4 1 3
Ans:-A

Sol Solutions
Sol: (a, b, c) = (3, 5, -6) x (7, 1, 2) = (1, -3, -2)
A is the correct option
60. Match List-I with List-II and select the correct answer using the code given below
the lists :
List - I List - II
P) Volume of parallelepiped determined by vectors a,b 1) 100
and c is 2. Then the volume of the parallelepiped
determined by vectors    2 a b ,3 b c  and  c a is
Q) Volume of parallelepiped determined by vectors a,b 2) 30
and c is 5. Then the volume of the parallelepiped determined by
vectors    3 a b , b c  and  2 c a is
R) Area of a triangle with adjacent sides determined by vectors 3) 24
a and b is 20. Then the area of the triangle with adjacent
sides determined by vectors  2a 3b and  a b is
S) Area of a parallelogram with adjacent sides determined by 4) 60
vectors a and b is 30. Then the area of the parallelogram with
adjacent sides determined by vectors  a b and a is
P Q R S P Q R S
A) 4 2 3 1 B) 2 3 1 4
C) 3 4 1 2 D) 1 4 3 2
Ans:- C
Sol:
2
V 6 a b b c c a 6 a b c 6 4 24             
P 3
C is correct answerThis section contains 10 Multiple Choice questions. Each Ques-
tion has Four choices (A), (B), (C) and (D). Out of Which Only One is correct

IIT JEE Advanced2 2013

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