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VMC/Solutions 2 Paper 2 Code 0 JEE Advanced 2013
JEE Advanced 2013
Paper 2
Answer
Key
Code 0
PHYSICS CHEMISTRY MATHS 1 A,B,C,D 21 C,D 41 B,C,D 2 A,C 22 A,B,D 42 A,B 3 D 23 A,C,D 43 C,D 4 C,D 24 A,B 44 A,B,C 5 A,D 25 B 45 A,D 6 A,B 26 C 46 B,D
7 B,D 27 B 47 B,D 8 A,D 28 B,D 48 A,C 9 B 29 A 49 A
10 A 30 D 50 C 11 B 31 B 51 B 12 A 32 A 52 D 13 B 33 C 53 B 14 B 34 B 54 C 15 C 35 A 55 A 16 A 36 A 56 D 17 A 37 A 57 C 18 C 38 D 58 A 19 D 39 D 59 B
20 C 40 A 60 A
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Part 1 : 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 whichONE or MORE are correct
1. The figure below shows the variation of specific heat capacity (C) of a solid as a function of temperature (T). Thetemperature is increased continuously from 0 to 500 K at a constant rate. Ignoring any volume change, thefollowing statement(s) is (are) correct to a reasonable 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 400 500 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)
Q mCdT m = = area under C T graphdQ dT
mC dt dt
=
if C increases rate of heat absorption increases. ( asdT dt
is constant)
2. The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a 0, where a 0 is the Bohr radius. Its orbital
angular momentum is32
h
. It is given that h is Planck constant and R is Rydberg constant. The possible
wavelength(s), when the atom de-excites, is /are
(A) 9
32 R (B)
916 R
(C) 9
5 R (D)
43 R
Ans.(A,C) 22 22 1
1 1 1 RZ
n n
=
13
32h
L n = =
2 21
0 0 03 9
4 5 24 5
na a . a a . Z
z z .= = = =
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22 22
1 1 1
3 R( Z )
n
=
29
25
n , R
= =
29
1 32n , R = =
3. Using the expression 2 =d sin , one calculates the values of d by measuring the corresponding angles
in the range 0 to 90 o . The wavelength is exactly known and the error in is constant for all values
of . As increases from 0o , (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) 2d sin =
( is exactly known)2
d sin
=
2d cos ec
= .....(1)
2d
( cosec cot )
=
2d cos ec cot ( )
= ...(2)
Equation 2/1 ( ) is constantd cot ( )d
=
both andd
d d
are decreasing with .
4. Two non-conducting spheres of radii R 1 and R 2 carrying uniform volume charge densities and+ ,respectively, are placed such that they partially overlap as shown in the figure. At all points in theoverlapping region,
(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) 0 03 3
O' P OP E
=
r rr
=03
O' O( O' O OP O' P )
+ =r r r r
Q
So, E is constant in magnitude as well as direction.
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5. A Steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinder is placedcoaxilly inside an infinite solenoid of radius 2 R. The solenoid has n turns per unit length and carries a steadycurrent 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 is non-zero(B) In the region R < r < 2 R, the magnetic field is along the common axis.(C) In the region R < r < 2 R, the magnetic field is tangential to the circle of radius r, centered on the axis.(D) In the region r > 2 R, the magnetic field is non-zero.
Ans. (A,D)
6. Two vehicles, each moving with speed u on the same horizontal straight road, are approaching each other. Wind blows the road with velocity w. One of these vehicles blows a whistle of frequency f 1. An observer in the other vehicle hearts the frequency of the whistle to be f 2. 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, f 2 > f 1.
(B) If the wind blows from the source to the observer, f 2 > f 1 (C) If the wind blows from observer to the source, f 2 < f 1 (D) If the wind blows from the source to the observer, f 2 < f 1
Ans.(A,B) If the wind blows fromsource to observer
2 1 2 1( v w ) ( u ) v w u
f f f f ( v w ) ( u ) v w u
+ + += = >+ + +
If the wind blows from observer to source
2 1 1 2 1( v w ) ( u ) v w u f f f f f ( v w ) ( u ) v w u += = > +
7. Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m is projected from themidpoint of the line joining centres, perpendicular to the line. The gravitational constant is G. The correctstatement(s) is (are)
(A) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 4GM L
(B) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 2GM 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.
Ans. (B,D) In order to escape to infinity T. E = 0 2
21 GMm GMmu u 22 L L
= =
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8. A particle of mass m is attached to one end of a mass-less spring of force constant k, lying on a frictionlesshorizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium
position at time t = 0 with an initial velocity u 0. when the speed of the particle is 0.5 u 0, it collides elastically witha rigid wall. After this collision,(A) the speed of the particle when it returns to its equilibrium position is u 0
(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 is43
= mt k
(D) the time at which the particle passes through the equilibrium position for the second time is53
= mt k
Ans. (A,D)
x Asin t=
v (A ) cos t=
00
uu cos t cos t 1 / 2 t / 3
2= = =
K mt / 3 t
m 3 K = =
(time taken to reach wall from mean position)
m 2 mtime of passing equilibrium for first time 2
3 K 3 K = =
2 m 2 m 7 mtime of maximum compression
3 K 4 K 6 K = + =
2 m m 5 mtime of passing equilibrium for second time
3 K K 3 K = + =
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Section 2: (Paragraph Type)This section contains 4 paragraphs each describing theory, experiment, data etc. Eight questions related 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 Question 9-10A small block of mass 1 kg 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 instantaneousvelocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J . (Take the
acceleration due to gravity, 210 )=g m s .
9. The speed of the block when it reaches the point Q is.
(A) 15 ms (B) 110 ms (C) 110 3 ms (D) 120 ms
Ans. (B)
against f Loss of P.E.=gain of K.E + W 2
2
R 1mg mv 150
2 240 1
1 10 1 v 150 v 10 m / s2 2
= +
= + =
10. 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)
20
2
At Q
mv N mg cos60
R
1 1 10 N 1 10
2 405
N 5 7.5 N2
=
=
= + =
Paragraph for Question 11-12A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 20 km awayfrom 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 two ends. The drawback of the directtransmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In thismethod, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumersend, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to
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assume that the power cable is purely resistive and the transformers are ideal with a power factor unity. All the currentsand voltages mentioned are rms values.
11. If the direct transmission method with a cable of resistance 10 4 km . is used, the power dissipation (in %)
during transmission is.
(A) 20 (B) 30 (C) 40 (D) 50Ans.(B)
P Vi (production)= 3
2 2 2
600 10 I 4000 I 150 amp
Power dissipation I R 150 (0.4 20) 150 8 180 kW
180 100Percentage 30%
600
= =
= = = == =
12. 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 consumers has to be supplied at 200 V, the ratioof 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)Atstepup
p ps
s s
s
p p
s s
V N 1V 40000 V
V N 10
At Step down V 200 V
N V 40000200:1
N V 200
= = =
=
= = =
Paragraph for Question 13-14
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 current2Q
. 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 radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf isdefined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It isknown that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a
proportionally constant .
13. The magnitude of the induced electric field in the orbit at any instant of time during the time interval of themagnetic field change is.
(A) 4
BR (B)
2 BR
(C) BR (D) 2 BR
Ans.(B) 2d
E. Rdt
=
22d
E. R Rdt
=
22 E. R R1
0 =
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2 R
E =
14. The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of themagnetic field change is.
(A) 2 BQR (B) 2
2 BQR (C)
2
2 BQR
(D) 2 BQR
Ans.(B)
22
M L
.( t )
.(QER1)
R.(Q. .R)
Q R.
=
=
=
2
==
Since angular momentum increases in downward direction (-ve z)
Hence, M =2Q R
.
2
Paragraph for Question 15-16
The mass of a nucleus A Z X is less than the sum of the masses of (A Z) number of neutrons and Z number of
protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the bindingenergy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m 1 and m 2 only if (m 1 + m 2) < M. Also two light nuclei of masses m 3 and m 4 can undergo complete fusion and form a heavynucleus of mass M only if (m
3+ m
4) > M. The masses of some neutral atoms are given in the table below:
15. The correct statement is.
(A) The nucleus 63 Li can emit an alpha particle
(B) The nucleus 21084 Po can emit a proton
(C) Deuteron and alpha particle can undergo complete fusion
(D) The nuclei 70 8230 34and Zn Se can undergo complete fusion
Ans.(C)6 4 23 2 1Li He H + is not feasible as4.002603+2.014102=6.016705u i.e. total mass of product is more than 6.015123 u ie the mass of reactant.Similarly reaction mention in option (B) and (D) are not feasible.On this basis, the only feasible reaction is that deuteron and alpha particle can undergo complete fusion.
16. the kinetic energy (in keV) of the alpha particle, when the nucleus 21084 Po at rest undergoes alpha decay, is
(A) 5319 (B) 5422 (C) 5707 (D) 5818
Ans.(A)210 206 484 82 2Po Pb He +
m 209.982876 (205.974455 4.002603) 0.005818 = + = Total energy released = 0.005818 x 932 MeV=E
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Po
Po
Po Po
Po Po
k k E (i)
p p (from conservation of momentum)
2m k 2m k
m k m k (ii)
+ ==
= =
From (i) and (ii)
po
po
m 206K E xE 5319 KeVm m 210
= = =+
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Section 3: (Matching list Type)This section contains 4 multiple choice questions . Each question has matching lists.
The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
17. One mole of a monatomic ideal gas is taken along two cyclic processes
and E F G E E F H E as shown in the PV diagram. The processes involved are purelyisochoric, isobaric, isothermal of adiabatic.
List -I List -II(P) G E (1) 160 P 0V0 ln2(Q) G H (2) 36 P 0V0 (R) F H (3) 24 P 0V0 (S) F G (4) 31 P 0V0
Code:
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) 0 0GE GW P (V V )= For 032F F G G GFG, P V P V V V = =
0 0 0 0 032 31GE W P (V V ) P V = =
0 0 0 0 08 32 24GH W P ( V V ) P V = = for F to H
r PV const =
0 032 0 H ( P )(V ) P (V ) =
5 35 3
00 0
32 2 2 8 8 /
H H H
V V V V
V V
= = = = =
2 2FH F H F F H H f f
W nR(T T ) ( P V P V )= =
0 0 0 03
32 82
( P V P V )= 0 036 P V =
21
FH V
W PV ln V
=
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00 0
0
3232
V PV ln
V
=
0 0160 2P V ln( )=
4 3 2 1P Q R S
18. Match List-I of the nuclear processes with List-II containing parent nucleus and one of the end products of each process and then the correct answer using the codes given below the lists:
List-I List -II(P) Alpha decay (1) 15 15
9 7 +O N ..... (Q) decay+ (2) 238 23492 90 +U Th ..... (R) Fission (3) 185 184
83 82 + Bi Pb ..... (S) Proton emission (4) 239 140
94 57 +Pu La .....
Code: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)238 234 492 90 2P : Alpha Decay: u Th +
15 15 08 7 1
239 140 99
94 94 37185 184 183 82 1
Q : Decay : O N e
R : Fission : Pu La X
S :Pr oton Emission : Bi Pb p
P 2, Q 1, R 4, S 3
++ +
+
+
19. 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 correct answer using the
codes given below the lists :
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Part 2 : CHEMISTRY
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
21. 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 3 Al O (D) magnesium from 3 3 MgCO .CaCO
Ans.(C,D) Strong metals like Na, Mg, Al are extracted by electrolytic reduction method but not by carbon reduction.
22. The thermal dissociation equilibrium of 3 ( )CaCO s is studied under different conditions.
3 2( ) ( ) ( )CaCO 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 a given T
(D) H is independent of the catalyst, if anyAns.(A,B,D) (A) H is dependent on T
(B) K is independent of the initial amount of CaCO 3
(D) A catalyst does not changes H
23. The correct statement(s) about 3O is (are)
(A) O O bond lengths are equal(B) Thermal decomposition of 3O is endothermic.
(C) 3O is diamagnetic in nature.
(D) 3O has a bent structure.
Ans.(A,C,D) (A) In ozone, bond lengths are equal because of resonance.
(B) Thermal decomposition of O 3 is exothermic.
(C) Ozone is diamagnetic in nature.
(D) Structure of ozone is bent.
24. In the nuclear transmutation9 84 4 Be X Be Y + + ( ) X ,Y is(are)
(A) ( ) ,n (B) ( ) p, D (C) ( )n, D (D) ( ) , p
Ans.(A,B) 9 0 8 1
4 0 4 0Be Be n( (n)+ +)
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9 1 8 24 1 4 1Be H Be H
(p (D)
+ +)
25. The major product(s) of the following reaction is(are)
(A) P (B) Q (C) R (D) S
Ans.(B)
26. After completion of the reactions (I and II), the organic compounds(s) in the reaction mixtures is(are)
(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
Ans.(C) Reaction 1 : 2Br (1mol )3 3 NaOHCH C CH
1/3rd of acetone reacts with 1 mole of Br 2, then products are [Bromoform reaction]
3 3
O||
CH C O Na , CHBr + and remaining acetone
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(cis) 4P dil. KMnO Meso (S)+
(Trans) 4Q dil. KMnO Racemic (T U)+ +
32. In the following reaction sequences V and W are, respectively
(A)
(B)
(C)
(D)
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Ans.(A)
Paragraph for questions 33 and 34
A fixed mass m of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the
figure
33. The succeeding operations that enable this transformation of states are
(A) Heating, cooling, heating, cooling
(B) Cooling, heating, cooling, heating
(C) Heating, cooling, cooling, heating
(D) Cooling, heating, heating, cooling
Ans.(C) Using ideal gas equation
PV = nRT
From K to L, heating occurs
From L to M, cooling occurs
From M to N, cooling occursFrom N to K, heating occurs
34. The pair of isochoric processes among the transformation of states 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) Isochoric processes are those in which volume remain constant. So isochoric processes are L to M and N to K.
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Paragraph for Questions 35 and 36
The reactions of Cl 2 gas with cold-dilute and hot-concentrated NaOH in water give sodium salts of two (different) oxoacids
of chlorine, P and Q, respectively. The Cl 2 gas reacts with SO 2 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 phosphorus, T .
35. P and Q, respectively, are the sodium slats of (A) hypochlorus and chloric acids
(B) hypochlorus and chlorus acids
(C) chloric and perchloric acids
(D) chloric and hypochlorus acids
Ans.(A) 2 3 NaOH Cl NaOCl NaClO+ +
Salt of Salt of Chloric acidhypochlorous acid
36. R, S and T , respectively, are
(A) SO2Cl 2, PCl 5 and H 3PO 3 (B) SO2Cl 2, PCl 3 and H 3PO 3
(C) SOCl 2, PCl 3 and H 3PO 2
(D) SOCl 2, PCl 5 and H 3PO 4
Ans.(A) 2 2 2 2SO Cl SO Cl+
2 2 5 25SO Cl 2P 2PCl 5O+ +
5 2 3 4PCl H O H PO 5HCl+ +
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Section 3: (Matching list Type)This section contains 4 multiple choice questions . Each question has matching lists.
The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
37. An aqueous solution of X is added slowly to an aqueous solution of Y as 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 thecode given below the lists:
List I List II
P. 2 5 3 3(C H ) N CH COOH
X Y
+1. Conductivity decreases and then increases
Q. 3KI(0.1M) AgNO (0.01M)
X Y
+2. Conductivity decreases and then does not change much
R. 3(CH COOH) KOH
X Y
+3. Conductivity increases and then does not changes much
S. NaOH HI
X Y
+4. Conductivity does not change much and then increases
Codes
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) When Et 3 N solution is added to CH 3COOH, ' ' of CH 3COOH will increase, so though H + ions ions will be
consumed by Et 3 N but overall no. of ions in the solution will increase and hence conductivity increases initially.KI (aq)
(aq) 3(aq) (aq) 3(aq)Ag NO AgI(s) K NO+ + ++ + +
Number of ions remain the same so, conductivity does not change much. After all Ag + ions have been precipitated
out, then the conductivity will increase.
3CH COOH(aq)(aq) (aq) 2 (aq) 3 (aq)K OH H O( ) K CH COO+ + + + +l
Number of ions remains the same, but number of OH ions decreases and hence conductivity decreases.
NaOH2 (aq) (aq)H I H O( ) Na I
+ + + + +l
Number of ions remains the same, but number of H+
decreases and hence conductivity decreases.
38. The standard reduction potential data at 25 oC is given below.0 3 2E (Fe , Fe ) 0.77 V;
+ + = +
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Codes
P Q R S
(a) 4 2 3 1
(b) 3 2 1 4
(c) 1 4 2 3
(d) 3 4 2 1Ans.(D) [P] Its disproportionation reaction, so just heating will do.
[Q] I2 does reaction as 2 22 3 4 6S O S O So Cl 2 is required for 22 3 4S O HSO
[R] 2 4 2 2 N H 2I N 4HI+ + [S] 2XeF NO Xe+
40. 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:
Codes
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)
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Part 3 : MATHS
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
41. Let be a complex cube root of unity with 1 and ijP p = be a n n matrix withi j
ij p += .
Then 2 0P , when n =(A) 57 (B) 55 (C) 58 (D) 56
Ans.
42. The function ( ) 2 2 2 2 f x x x x x= + + + has a local minimum or a local maximum at x =
(A) 2 (B) 2
3
(C) 2 (D)
2
3
Ans.
43. Let3
2
iw
+= and { }1 2 3nP w : n , , , . . .= = . Further 1 12 H z C : Re z
= >
and
21
2 H z C : Re z
= <
, where C is the set of all complex numbers. If 1 1 2 2 z P H , z P H and O
represents the origin, then 1 2 z Oz =
(A)2
(B)
6
(C)
2
3
(D)
5
6
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Ans.
44. If 13 4 x x= , then x =
(A)3
3
2 2
2 2 1
log
log (B) 2
2
2 3log (C) 4
1
1 3log (D)2
2
2 3
2 3 1
log
log
Ans.
45. Two lines L 1 : 53 2
y z x ,
= =
and L 2 :
1 2
y z x ,
= =
are coplanar. Then can take value(s) :
(A) 1 (B) 2 (C) 3 (D) 4
Ans.
46. In a triangle PQR, P is the largest angle and1
3cos P = . 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 consecutive even integers.
Then possible length(s) of the side(s) of the triangle is(are) :
(A) 16 (B) 18 (C) 24 (D) 22
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VMC/Solutions 26 Paper 2 Code 0 JEE Advanced 2013
Ans.
47. For a R (the set of all real numbers), 1a ,
( )( ) ( ) ( ) ( )1
1 2 1
601 1 2
a a a
an
. . . nlim
n na na . . . na n
+ + +=
+ + + + + + +
Then a =
(A) 5 (B) 7 (C) 15
2
(D)
17
2
Ans.
48. 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 6 8 9 0 x y x y+ + + = (B) 2 2 6 7 9 0 x y x y+ + + =
(C) 2 2 6 8 9 0 x y x y+ + = (D) 2 2 6 7 9 0 x y x y+ + = Ans.
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VMC/Solutions 27 Paper 2 Code 0 JEE Advanced 2013
Section 2: (Paragraph Type)This section contains 4 paragraphs each describing theory, experiment, data etc. Eight questions related 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 49 - 50
Let f : [ ]0 1 , R (the set of all real numbers) be a function. Suppose the function f is twice differentiable,
( ) ( )0 1 0 f f = = and satisfies ( ) ( ) ( ) [ ]2 0 1 x f x f x f x e , x , + .
49. Which of the following is true for 0 < x < 1 ?
(A) ( )0 f x< < (B) ( )1 1
2 2 f x < < (C) ( )
11
4 f x < < (D) ( ) 0 f x < <
Ans.
50. If the function ( ) xe f x assumes its minimum in the interval [0, 1] at1
4 x = , which of the following is true ?
(A) ( ) ( )1 3
4 4 f x f x , x < < < (B) ( ) ( )1
0 4 f x f x , x > < <
(C) ( ) ( )1
04
f x f x , x < < < (D) ( ) ( )3
14
f x f x , x < < <
Ans.
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VMC/Solutions 28 Paper 2 Code 0 JEE Advanced 2013
Paragraph for Questions 51 - 52
Let PQ be a focal chord of the parabola 2 4 y ax= . The tangents to the parabola at P and Q meet at a point lying on the line2 0 y x a, a= + > .
51. Length of chord PQ is :
(A) 7a (B) 5a (C) 2a (D) 3a
Ans.
52. If chord PQ subtends an angle at the vertex of 2 4 y ax= , then tan =
(A)2
73
(B)2
73
(C)
25
5(D)
25
3
Ans.
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VMC/Solutions 29 Paper 2 Code 0 JEE Advanced 2013
Paragraph for Questions 53 - 54Let 1 2 3S S S S = , where
{ }1 21 3
4 01 3
z iS z C : z , S z C : Im
i
+ = < = > and { }3 0S z C : Re z= > .
53. Area of S =
(A)10
3
(B)
20
3
(C)
16
3
(D)
32
3
Ans.
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54. 1 3 z S min i z =
(A)2 3
2
(B)
2 3
2
+(C)
3 3
2
(D)
3 3
2
+
Ans.
Paragraph for Questions 55 - 56
A box B 1 contains 1 white ball, 3 red balls and 2 black balls. Another box B 2 contains 2 white balls, 3 red balls and 4 black
balls. A third box B 3 contains 3 white balls, 4 red balls and 5 black balls.
55. If 1 ball is drawn from each of the boxes B 1, B2 and B 3, the probability that all 3 drawn balls are of the same
colour is :
(A)82
648(B)
90
648(C)
558
648(D)
566
648
Ans.
56. If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the
other ball is red, the probability that these 2 balls are drawn from box B 2 is :
(A)116
181(B)
126
181(C)
65
181(D)
55
181
Ans.
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VMC/Solutions 31 Paper 2 Code 0 JEE Advanced 2013
Section 3: (Matching list Type)This section contains 4 multiple choice questions . Each question has matching lists.
The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
57. 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 , bur ur
and cur
is 2. Then the
volume of the parallelepiped determined by vectors ( ) ( )2 3a b , b c ur ur ur ur
and
( )c aur ur
is
1. 100
(Q) Volume of parallelepiped determined by vectors a , bur ur and cur is 5. Then the
volume of the parallelepiped determined by vectors ( ) ( )3 3a b , b c+ +ur ur ur ur
and
( )2 c a+ur ur
is
2. 30
(R) Area of a triangle with adjacent sides determined by vectors aur
and bur
is 20. Then
the area of the triangle with adjacent sides determined by vectors ( )2 3a b+ur ur
and
( )a bur ur
is
3. 24
(S) Area of a parallelogram with adjacent sides determined by vectors aur
and bur
is 30.Then the area of the parallelogram with adjacent sides determined by vectors
( )a b+ur ur
and aur
is4. 60
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Codes :
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.
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58. Consider the line, L 1 : 21 3 4 3 3
2 1 1 1 1 2
x y z x y z , L
+ + += = = = =
and the planes P 1 : 7 2 3 x y z+ + = ,
P2 : 3 5 6 4 x y z+ = . Let ax by cz d + + = be the equation of the plane passing through the point of intersection of lines L 1 and L 2, and perpendicular to planes P 1 and P 2. 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
Codes :P Q R S 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.
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VMC/Solutions 35 Paper 2 Code 0 JEE Advanced 2013
59. Match List I with List II and select the correct answer using the code given below the lists :
List I List II
(P)( ) ( )
( ) ( )
1 221 1
42 1 1
1 cos tan y y sin tan y y
y cot sin y tan sin y
+ + +
takes value 1.1 5
2 3
(Q If 0cos x cos y cos z sin x sin y sin z+ + = = + + then possible value of 2
x ycos
is 2. 2
(R)If 2 2 2 2
4 4cos x cos x sin x sin x sec x cos sin x sec x cos x cos x
+ = + +
then
possible value of sec x is
3.1
2
(S) If ( )( )1 2 11 6 0cot sin x sin tan x , x = , then possible value of x is 4. 1
Codes :
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.
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60. A line L : 3 y mx= + meets y-axis at E(0, 3) and the arc of the parabola 2 16 0 6 y x, y= at the point
( )0 0F x , y . The tangent to the parabola at ( )0 0F x , y intersects the y-axis at ( )10G , y . The slope m of the lineL is chosen such that the area of the triangle EFG has a local maximum. Match List I with List II and select thecorrect 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] y0 = 3. 2[S] y1 = 4. 1
Codes :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.
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