NARAYANA IIT/PMT ACADEMY  IIT JEE COACHING ... ADVANCE2013 (PAPER...NARAYANA IIT/PMT ACADEMY JEE...
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NARAYANA IIT/PMT ACADEMY JEE ADVANCE : 2013
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This booklet contains 40 printed pages XYZ Test Booklet Code
PAPER 1 : PHYSIC, CHEMISTY & MATHEMATICS
Please Read the Instructions carefully You are allotted 5 minutes specifically for
this purpose
Important Instructions:
INSTRUCTIONS
A General :
This booklet is your Questions Paper Do not break the seals of this booklet before being instructed to do
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The question paper CODE is printed on the right hand top corner of this sheet and on the back page
(Page No 44) of this booklet
Blank spaces and blank pages are provided in the question paper for your rough work No additional
sheets will be provided for rough work
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Write your name and roll number in the space provided on the back cover of this booklet
Answers to the questions and personal details are to be filled on a two – part carbon –less paper, which
is provided separately These parts should only be separated at the end of the examination when
instructed by the invigilator The upper sheet is machine –gradable objective Response Sheet (ORS)
which will be retained by the invigilator You will be allowed to take away the bottom sheet at the end of
the examination
Using a black ball point pen darken the bubbles on the upper original sheet Apply sufficient pressure so
that the impression is created on the bottom duplicate sheet
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PART I : PHYSICS
SECTION 1 : (One or more options correct Type)
The section contains 8 multiple choice questions Each question has four choices (A), (B), (C) and (D)
out of which ONE and MORE are correct
1 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 reasonable
approximation
(A) the rate at which heat is absorbed in the range 0100 K varies linearly with temperature T
(B) heat absorbed in increasing the temperature from 0100 K is less than the heat required for
increasing the temperature from 400500 K
(C) there is no change in the rate of heat absorption in the range 400500 K
(D) the rate of heat absorption increases in the range 200300 K
C
100 200 300 400 500
T(K) Ans A, B, C, D
Sol: dQ CdT and C varies linearly with T
(A) is correct
Area under the graph in the range O100 k is less than area under 400500 K range
(B) is correct
In the range 400 – 500 K, C is constant
(C) is correct
In the range 200 – 300 K, C is increasing
(D) is correct
2 The radius of the orbit of an electron in a hydrogenlike atom is 45 a0, where a0 is the Bohr
radius Its orbital angular momentum is 3h
2 It is given that h is Planck constant and R is Rydberg
constant The possible wavelength(s), when the atom deexcites, is (are)
(A) 9
32R (B)
9
16R
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(C) 9
5R (D)
4
3R
Ans A, C
Sol: 2
oa n4.5
Zao
nh 3h
2 2
n 3, z 2
Now, 2
2 2
1 2
1 1 1Rz
x x
Possible wavelengths are
9 9 1
, and5R 32R 3R
3 Using the expression 2d sin = , one calculates the values of d by measuring the corresponding
angles in the range 0 to 90o 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
Sol: d2sin
d cosec cot2
Now d
cotd
As increases, fractional error decreases
4 Two nonconducting spheres of radii R1 and R2 and carrying uniform volume charge densities +
and – , respectively, are placed such that they partially overlap, as shown in the figure At all
points in the overlapping 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

R1 R2
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Ans C, D
Sol: In the overlapping region
1 2
o o
r rE
3 3
1rr
2rr
lr
o
lE
3
C and D are correct
5 A steady current I flows along an infinitely long hollow cylindrical conductor of radius R This
cylinder is placed coaxialy inside an infinite solenoid of radius 2 R 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 is nonzero
(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 of radius r, centered on
the axis
(D) In the region r > 2R, the magnetic field is nonzero
Ans A, C, D
Sol: hollow solenoidconductor
B B B
B due to hollow conductor is zero inside the conductor and nonzero outside (coaxial) while B
due to solenoid is zero outside the solenoid and nonzero inside the solenoid (axial)
R 2R
A, C & D
6 Two vehicles, each moving with speed u on the same horizontal straight road, are approaching
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 source to the observer, f2<f1
Ans A, B
o
2 1
s
v w vf f
v w v
A & B are correct
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7 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
bodies is GM
4L
(B) The minimum initial velocity of the mass m to escape the gravitational field of the two
bodies is GM
2L
(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
Sol: 2
e
1 2GMmmv 0
2 L
e
GMv 2
L
M Mve
8 A particle of mass m is attached to one end of a massless spring of force constant k, lying on a
frictionless horizontal 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 u0 When the speed
of the particle is 05 u0, it collides elastically with a rigid wall After this collision,
(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
mt .
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
mt .
3 k
Ans A, D
Sol: using 2 2v A x , we get 3 u
x2
(x : distance between mean position & the wall)
Now using x = A sin t
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2 m
t3 k
time for option (B) = 2 m 2 m 4 m
3 k 3 k 3 k
time for option (C) 4 m m 7 m
3 k 2 k 6 k
time for option (D) = 7 m m 5 m
6 k 2 k 3 k
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)
A 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 instantaneous velocity 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, g = 10 ms–2
)
R P
R
O
Q
y
x
30o
9 The speed of the block when it reaches the point Q is
(A) 5 ms–1
(B) 10 ms–1
(C) 110 3ms (D) 120ms
Ans B
Sol: From workkinetic energy theorem
o 2
f
1mgR sin30 W mv
2
v = 10 m/s
10 The magnitude of the normal reaction that acts on the block at the point Q is
(A) 75 N (B) 86 N
(C) 115 N (D) 225 N
Ans A
Sol: At Q:
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2
o mvN mgcos60
R
N 7.5N
Paragraph for questions 11 and 12
A thermal power plant produces electric power of 600 kW at 4000 V, 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 stepup and stepdown
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 stepup transformer 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 consumers 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 currents and voltages mentioned are rms values
11 If the direct transmission method with a cable of resistance 04 km1
is used, the power
dissipation (in%) during transmission is
(A) 20 (B) 30 (C) 40 (D) 50
Ans (B)
Sol 3P 600 10 watt
rms
V 4000 volt
Total resistance R = 04 20 = 80
Power lost 2
L rmsP I R
2
rms
PR
V
23600 10
84000
= 180000 watt
= 180 kW
So % loss = 180
100 30600
= 30%
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 stepup transformer is 1 : 10 If the power to the consumers has to
be supplied at 200 V, the ratio of the number of turns in the primary to that in the secondary in
the stepdown transformer is
(A) 200 : 1 (B) 150 : 1 (C) 100 : 1 (D) 50 : 1
Ans (A)
Sol Step – up
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NP NS
V =?S
4000V
S
S
V 10V 40,000volt
4000 1
For Step – down
S out
P input
N V 200 1
N V 40,000 200
So NP : NS = 200 : 1
Paragraph for Questions 13 and 14 A point charge Q is moving in a circular orbit of radius R in the xy plane with an angular velocity
This can be considered as equivalent to a loop carrying a steady current 2
A uniform magnetic field
along the positive zaxis 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 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
13 The magnitude of the induced electric field in the orbit at any instant of time during the time
interval of the magnetic field change is
(A) BR
4 (B)
BR
2 (C) BR (D) 2BR
Ans(B)
Sol Induced emf = d
dt
2dBe R
dt
2e B R
2E.2 R B. R
BR
E2
14 The change in the magnetic dipole moment associated with the orbit, at the end of the time
interval of the magnetic field change, is
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(A) 2BQR (B) 2BQR
2 (C)
2BQR
2 (D) 2BQR
Ans (C or B)
Sol 2Be R
t
2BE.2 R R
t
2B R
E2 R. t
L t
m L
2QBR
m2
Sign may be taken as +ve/ve also depending upon direction of angular velocity
Paragraph for questions 15 and 16
The mass of a nucleus A
ZX is less than the sum of the masses of (AZ) number of neutrons and Z number
of protons in the nucleus The energy equivalent to the corresponding mass difference is known as the
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 masses m3 and m4 can undergo complete fusion
and form a heavy nucleus of mass M ' only if (m3 + m4) > M ' The masses of some neutral atoms are
given in the table below :
1
1H 1007825 u 2
1H 2014102 u 3
1H 3016050 u 4
2He 4002603 u
6
3Li 6015123 u 7
3Li 7016004 u 70
30Zn 69925325 u 82
34Se 81916709 u
152
64Gd 151919803 u 206
82Pb 205974455 u 209
83Bi 208980388 u 210
84Po 209982876 u
15 The correct statement is
(A) The nucleus 6
3Li can emit an alpha particle
(B) The nucleus 210
84Po can emit a proton
(C) Deuteron and alpha particle can undergo complete fusion
(D) The nuclei 70
30Zn and
82
34Se can undergo complete fusion
Ans (C)
Sol from data given only option C is correct
3 4
m m M '
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16 The kinetic energy (in keV) of the alpha particle, when the nucleus 210
84Po at rest undergoes alpha
decay, is
(A) 5319 (B) 5422 (C) 5707 (D) 5818
Ans (A)
Sol From data given
210 206 4
84 82 2Po Pb He
m=00058184 u
E = 0005818 932
=5422376 MeV
KE of particle E 206
210
= 5319 MeV = 5319 KeV
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 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 paths 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
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S F G 4 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)
Sol = 5/3
FH is steeper therefore it is adiabatic process FG is isothermal
FH F F H H
P V P V
0 0 H H
(32P )V P V H 0
V 8V
FG PFVF = PGVG
(32P0)V0 = P0VG VG = 32 V0
GE Isobaric Process
Work = P V = P0 (32V0 – V0) = 31 P0V0
GH Work = P0 (32V0 – 8V0) = 24 P0V0
FH Work 0 0 0 01 1 2 2
0 0
(32P )V P 8VPV P V36P V
1 5/ 3 1
FG G
F F 0 0
F
VWork P V ln 32P V ln 32 160ln 2
V
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 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
98 90U Th ....
R Fission 3 185 184
83 82Bi Pb ....
S Proton emission 4 239 140
94 57Pu La ....
Codes :
P Q R S
(A) 4 2 1 3
(B) 1 3 2 4
(C) 2 1 4 3
(D) 4 3 2 1
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Ans (C)
Sol
1 15 15 0
8 7 1O N X x is particle
2 238 234 4
92 90 2U Th x is particle
3 185 184 1
83 82 1Bi Pb X Xis Proton
4 239 140 99
94 57 47Pu La X Fission
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 3
and , 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 :
List I List II
P e f 1 1 2
2
Q e g 2 2 1 2 3
and
R e h 3 1 2
S e i 4 2 1 2 2 3
2 and
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)
Sol
S  1 e – i case of TIR
which is possible only when o
ci 45 is i
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o2
C 1 2 C
1
sin i if 2 , i is 45
Q – 3 e g is 1 2
[ ]rays undergo undeviated for any angle of incidence
R – 4 e h is 1 2
and not undergoing (TIR) light ray bends towards base of
prism, (prism block)
P – 2 e f 1 2
and 2 3
(Snell’s law)
20 Match List I with List II and select the correct answer using the codes given below the lists :
List I List II
P Boltzmann constant 1 [ML2T
1]
Q Coefficient of viscosity 2 [ML1
T1
]
R Planck constant 3 [MLT3
K1
]
S Thermal conductivity 4 ML2T
2K
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)
Sol
Boltzmann constant = 2 2energy ML T
Kelvin K = ML
2T
2K
1
Plank’s constant = energy time = ML2T
2 T = ML
2T
1
Coefficient of viscosity = 2
1 1
1 2 1
F MLTML T
L LT L T
Thermal conductivity = Energy Length
Area Kelvin TimeMLT
3K
1
P 4
Q 2
R 1
S 3
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PART II : 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 carbonbased reduction method is NOT used for the extraction of
(A) tin from SnO2 (B) iron from Fe2O3
(C) aluminium from Al2O3 (D) magnesium from MgCO3CaCO3
Ans (C), (D)
Sol Al is extracted by electrolytic reduction
MgCO3, CaCO3 are extracted by electrolytic reduction
22 The thermal dissociation equilibrium of CaCO3 (s) is studied under different conditions
CaCO3 (s) CaO (s) + CO2 (g)
For this equilibrium, the correct statement(s) is(are)
(A) H is dependent of T
(B) K is independent of the initial amount of CaCO3
(C) K is dependent on the pressure of CO2 at a given T
(D) H is independent of the catalyst, if any
Ans (A), (B), (D)
Sol Conceptual
23 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), (B), (C), (D)
Sol (A) – bond length are equal due to resonance
O
O+
O
No unpaired electrons so diamagnetic
24 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 (A) 9 8 1
4 4 0Be Be n
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(B) 9 1 8 2
4 1 4 1Be p Be D
25 The major product(s) of the following reaction is(are)
SO3H
OH
2aqueous Br 3.0 equivalents?
SO3H
Br
Br
OH
Br
P
Br
Br
OH
Br
Q
OH
BrBr
Br
R
OH
BrSO3H
Br
Br
S (A) P (B) Q (C) R (D) S
Ans (B)
SO3H
OH
Br
Br
OH
Br
2aqueous Br 3.0 equivalentsSol.
26 After completion of the reactions (I and II), the organic compound(s) in the reaction mixtures
is(are)
Reaction I: CH3 CH3
O2Br 1.0mol
aqueous NaOH
1.0 mol
Reaction II: CH3 CH3
O
1.0 mol
2
3
Br 1.0mol
CH COOH
CH3 CH2Br
O
CH3 CBr 3
O
Br 3C CBr 3
O
CH2Br CH2Br
O
CH3 ONa
O
CH3 ONa
O
P Q R S T U (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)
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Sol Reaction I: 1/3 mole of 3 3CH C CH
O
react with 1 mole of Br2 to give Bromofarm and sodium
acetate and 2/3 mole of acetone itself
Reaction II: Mono bromination at carbon in acidic medium
27 The Ksp of Ag2CrO4 is 11 1012
at 298 K The solubility (in mol/L) of Ag2CrO4 in a 01 M
AgNO3 solution is
(A) 11 1011
(B) 11 1010
(C) 11 1012
(D) 11 109
Ans (B)
Sol 2
2 4 4Ag CrO 2Ag CrO
s ' 0.1 2s ' s '
Ksp = (01 + 2s )2 s
11 1012
= 102
s
s = 11 1010
28 In the following reaction, the product(s) formed is(are)
CH3
OH
3CHCl
OH?
P Q R S
CH3
CHO
OH
CHO
CHCl 2
O
CH3 CHCl 2
OH
CH3 CH3
CHO
OH
(A) P (major) (B) Q (minor) (C) R (minor) (D) S (major)
Ans (B), (D)
Sol It is Reimer Tiemann Reaction
So
Q
CHCl 2
O
CH3
is minor and
S
CH3
CHO
OH
is major
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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 29 and 30
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 H2S in a dilute mineral acid medium However, it gave a precipitate (R)
with H2S in an ammoniacal medium The precipitate R gave a coloured solution (S), when treated with
H2O2 in an aqueous NaOH medium
29 The precipitate P contains
(A) Pb2+
(B) 2
2Hg (C) Ag+ (D) Hg
2+
Ans (A)
Sol Salt will be of PbCl2 as it is soluble in hot water
30 The coloured solution S contains
(A) Fe2(SO4)3 (B) CuSO4 (C) ZnSO4 (D) Na2CrO4
Ans (D)
Sol Cr3+
+ H2S/NH4OH, NH4Cl Cr(OH)3
Cr(OH)3 + H2O2 + NaOH Na2CrO4
(yellow colour)
Paragraph for Question 31 and 32
P and Q are isomers of dicarboxylic acid C4H4O4 Both decolorize Br2/ H2O On heating, P forms the
cyclic anhydride
Upon treatment with dilute alkaline KMnO4, P as well as Q could produce one or more than one from S,
T and U
31 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
31 (B)
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Sol ‘P’ and ‘Q’ are
C C
COOH
HH
HOOC
C C
H
COOHH
HOOC,
(Maleic acid) (Fumaric acid)
Baeyer’s reagent, ie dilute alkaline KMnO4 will give syn addition product of alkene
C C
COOH
HH
HOOC
KMnO4/OH
dilute
COOH
COOH
H OH
H OH
(meso)(cis)
C C
H
COOHH
HOOC
(trans)
KMnO4/OH
dilute
COOH
COOH
H OH
OH H
COOH
COOH
OH H
H OH
(Racemic mixture)
32 In the following reaction sequences V and W are, respectively
2 /H NiQ V
+ VAlCl
3 (anhydrous) 1. ZnHg/HCl
2. H3PO
4
W
O
O
O
and
VO
W
(A)
CH2OH
CH2OH
and
V W
(B)
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O
O
O
and
V W
(C)
HOH2C
CH2OH
and
V
CH2OH
W
(D)
Ans (A)
HOOC
COOH
H2/Ni
O
O
O
(V)
Fumaric acid
(Q)
Sol.
+ O
O
O
AlCl3
anhydrousCH2
CH2C
O
OH
O
1. ZnHg/HCl
2. H3PO
4
O
Paragraph for Question 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
M
Volume
Pressure
N
K L
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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)
Sol
M
Volume
Pressure
N
K L
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)
Sol According to following graph the proven L M isochoric and N K also isochoric
Paragraph for Question 35 and 36
The reactions of Cl2 gas with colddilute and hotconcentrated 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 phosphorous to give a compound S On hydrolysis, S
gives an oxoacid of phosphorus, T
35 P and Q, respectively, are the sodium salts of
(A) hypochlorus and chloric acids (B) hypochlorous and chlorous acids
(C) chloric and perchloric acids (D) chloric and hypochlorous acids
Ans (A)
Sol Cl2 + cold NaOH(aq) NaCl + NaOCl, (oxiacid HOCl) hypochlorous acid
Cl2 + (hot) NaOH NaCl + NaClO3, (oxiacid HClO3) chloric acid
P – HOCl
Q – HClO3
Cl2 + SO2 + Coke SO2Cl2 4P PCl5 2H O H3PO4
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36 R, S and T, respectively, are
(A) SO2Cl2, PCl5 and H3PO4 (B) SO2Cl2, PCl3 and H3PO3
(C) SOCl2, PCl3 and H3PO2 (D) SOCl2, PCl5 and H3PO4
Ans (A)
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 reaction 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 (C2H5)3N + CH3COOH
X Y
1 Conductivity decreases and then increases
Q KI(01M) + AgNO3(001M)
X Y
2 Conductivity decreases and then doest not
change much
R CH3COOH + KOH
X Y
3 Conductivity increases and then does not
change much
S NaOH + HI
X Y
4 Conductivity does not change much and then
increases
Code :
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) Factual
38 The standard reduction potential data at 250C is given below
E0 (Fe
3+, Fe
2+) = +077 V
E0 (Fe
2+, Fe) = 044 V
E0 (Cu
2+, Cu) = +034 V;
E0 (Cu
2+, Cu) = +052 V
E0[O2(g) + 4H
+ + 4e 2H2O] = +123 V
E0[O2(g) + 2H2O + 4e 4OH ] = +040V
E0 (Cr
3+, Cr) = 074 V;
E0 (Cr
2+, Cr) = 091 V;
Match E0 of the redox pair in List I with the values given in List II and select the correct answer
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using the code given below the lists:
List I List II
P E0 ( Fe
3+, Fe) 1 018 V
Q E0 (4H2O 4H
+ + 4OH ) 2 04 V
R E0 (Cu
2+ + Cu 2Cu
+) 3 004 V
S E0 (Cr
3+, Cr
2+) 4 083 V
Code :
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 (P) E0 (Fe
3+, Fe)
Fe3+
+ e Fe2+
0
1E 0.77V
Fe2+
+ 2e Fe 0
2E 0.44V
Fe3+
+ 3e Fe 0
3E
3 1 2G G G 0
33E 1 0.77 (2 0.44)
0
3
0.11E 0.0366 0.04
3V
(Q) E0 (4H2O 2H
+ + 4OH )
2H2O O2 + 4H+ + 4e E
0 = 123 V
O2 + 2H2O + 4e
4OH E0 = +04 V
4H2O 4H+ + 4OH E
0 = 083 V
(R) E0 (Cu
2+ + Cu 2Cu
+)
Cu2+
+ 2e Cu E0 = +034 V
2(Cu Cu+ + e) E
0 = 052 V
Cu2+
+ Cu 2Cu+ E
0 = 034 – 052 = 018 V
(S) E0(Cr
3+, Cr
2+)
Cr3+
+ 3e Cr 0
1E = 074 V
Cr2+
+ 2e Cr 0
2E = 091V
Cr3+
+ e Cr2+
0
3E
3 1 2G G G
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1 0
3E = 3 (074) ( 2 091) 0
3E = 222 – 182 0
3E = 04V
39 The unbalanced chemical reactions given in List I show missing reagent or condition (?) which
are provided 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 PbO2 + H2SO4 ? PbSO4 + O2 + other product 1 NO
Q Na2S2O3 + H2O ? NaHSO4 + other product 2 I2
R N2H4 ? N2 + other product 3 warm
S XeF2 ? Xe + other product 4 Cl2
Code :
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 (i) PbO2 + H2SO4 warm PbSO4 + H2O2 H2O + 2
1O
2
(ii) Na2S2O3 + H2O 2Cl 2NaHSO4 + NaCl
(iii) N2H4 2I N2 + 2HI
(iv) XeF2 + NO Xe + NOF
40 The unbalanced chemical reactions given in List I show missing reagent or condition (?) which
are provided 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. Cl
1 (i) Hg(OAc)2; (ii) NaBH4
Q. ONa OEt
2 NaOEt
R.
OH
3 Et – Br
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S.
OH
4 (i) BH3; (ii) H2O2/NaOH
Code :
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)
P. ClNaOEt
EliminationSol.
Q. ONaEtBr
SubstitutionOEt
R.
(i) Hg(OAc)2
(ii) NaBH4
Follows markownikov's product
OH
S.
(i) BH3
(ii) H2O2/NaOH
Follows antimarkownikov's product
OH
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PART – III : MATHEMATICS
SECTION – 1 (Only or more options correct Type)
This section contains 8 multiple choice questions Each question has four choices (A) (B), (C) and (D)
out of the 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 with
i j
ijp
Then 2P 0 , when n =
(A) 57 (B) 55 (C) 58 (D) 56
Ans (B,C,D)
Sol
2 2 2 2
2 2
2 2
2 2
1 1 ... ... 1 1 ... ...
1 ... 1 ...
1 ... 1 ...
... ... ... ... ... ...
1 ... 1 ...
..... .....
... ...
Clearly P2 will be O if n is a multiple of 3, then only, the elements of P
2 will be reduced to
21 0
2If P 0 n is not multiple of 3
42 The function y 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 y 2  x   x 2   x 2  2  x 
Case (i) x 2
y 2x x 2  x 2 2x 
y 3x 2  x 2 
y 2x 4 … (1)
Case (ii) 2 x 0
y x 2  3x 2 
So for 2
x3
y 2x 4 … (2)
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And for 2
x ,03
y 4x … (3)
Case (iii) 0 x
y 2x x 2  x 2 2x 
y 3x 2  x 2 
For 0 x 2
y 4x … (4)
And for, x 2
y 2x 4 … (5)
So graph is
(0,8/3) y=4x
y = 2x+4(0,8)
y= 2x4y= 2x+4
y= 4x
(2, 0) (2/3,0) (2,0)x
(0, 0)
From graph
x 2,0 , are point of minimum,
& 2
x3
is point of maxima
43 Let 3 i
w2
and nP : n 1,2,3,... Further 1
1H z C : Rez
2 and
2
1H 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 O z
(A) 2
(B) 6
(C) 2
3 (D)
5
6
Ans (C, D)
Sol
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Clearly from the figure, 1 2
2 5z Oz or or
3 6
44 If x x 13 4 , then x =
(A) 3
3
2log 2
2log 2 1 (B)
2
2
2 log 3 (C)
4
1
1 log 3 (D) 2
2
2log 3
2log 3 1
Ans (A, B, C)
Sol x x 13 4
Taking log both sides with base 3
x 1
3 3log 3 log 4
3 3x log 3 x 1 log 4
3x x 1 log 4
3 3x x log 4 log 4
3 3log 4 x log 4 1
3
3
log 4x
log 4 1
3
3
2log 2x
2log 2 1
x x 13 4
Taking log both sides with base 4
x x 1
4 4log 3 log 4
4x log 3 x 1
4 41 x x log 3 x 1 log 3 1
4 2
1 2x or x
1 log 3 2 log 3
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45 Two lines 1
y zL : x 5,
3 2 and
2
y zL : x ,
1 2 are coplanar Then can take
value(s)
(A) 1 (B) 2 (C) 3 (D) 4
Ans (A, D)
Sol 1
x 5 y zL :
0 3 2
2
x y zL :
0 1 2
5 0 0
0 3 2 0
0 1 2
5, 3 2 2 0
25;6 5 2 0
2 5 4 0
1 4 0
1,4,5
46 In a triangle PQR, P is the largest angle and 1
cos P3
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
Ans (B, D)
Sol 1
Cos P3
2 2 24n 2 4n 4 4n 61
3 2 4n 2 4n 4
n 4 or n 1
Hence, the possible length of side of PQR will be 18, 20 or 22
47 For a R (the set of all real numbers), a 1, a a a
a 1n
1 2 .... n 1Lt
60n 1 na 1 na 2 ... na n then a =
(A) 5 (B) 7 (C) 15
2 (D)
17
2
Ans (B, D)
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Sol
na
r 1
a 1 nn
r 1
r1
Ltn 1 na r
an
a
r 1
a 1 nn
r 1
r
n nLt
rn 1 n an
an
a 1
r 1
a 1 nn
r 1
1 r
n n nLt
1 rn 1 an n
an
r 1
a 1 nn
r 1
1 r
1 n nLt
1 r1 a1n nn
1
a
20
1
0
x dx1
2a 3a 119 060
a x dx
17
a 7,a2
48 Circles(s) touching x – axis at a distance 3 from the origin and having an intercept of length
2 7 on yaxis is (are)
(A) 2 2x y 6x 8y 9 0 (B)
2 2x y 6x 7y 9 0
(C) 2 2x y 6x 8y 9 0 (D)
2 2x y 6x 7y 9 0
Ans (A, C)
Sol Let circle be
2 2x y 2gx 2fy c 0
2 2g c,2 f c 2 7
g 3,c 9,f 4
So circle will be
2 2x y 6x 8y 9 0
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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 49 to 50
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 f x 2f ' x f x f x x 0,1
49 Which of the following is true for 0 < x < 1 ?
(A) 0 f x (B)1 1
f x2 2
(C)1
f x 14
(D) f x 0
Ans (D)
Sol x x xe f " x 2f ' x e e f x 1
x x x xe f " x f ' x e e f ' x e f x 1
x xd de f ' x e f x 1
dx dx
xd de f x 1
dx dx
Let xe f x g x
g" x 1
Since g 0 g 1 0 and g" x positive
Possible graph for g(x) is
O1
Since g x 0for x 0,1
Therefore f x is negativefor x 0,1
Which satisfies f x 0
50 If the function xe f x assumes its minimum in the interval {0, 1] at
1x
4, which of the
following is true?
(A)1 3
f ' x f x , x4 4
(B)1
f ' x f x ,0 x4
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(C)1
f ' x f x ,0 x4
(D) 3
f ' x f x , x 14
Ans (C)
Sol For 1
x 0, g ' x 04
x xe f ' x e f x 0
f ' x f x for 1
x 0,4
Paragraph for Question 51 and 52
Let PQ be a focal chord of the parallel 2 4 .y ax The tangents of the parabola at P and Q meet at a point
lying on the line 2 , 0.y x a a
51 Length of chord PQ is
(A) 7a (B) 5a (C) 2a (D) 3a
Ans (B)
Sol
y=2x+a P(at
2, 2at)
S(0, 0)
(a, a)
Q
21
PQ a tt
………(i)
Eq of tangent at P
2ty x at
As it passes through (–a, –a)
– at = – a + at2
t2 + t – 1 = 0
5 1
2t (for p, t > 0)
2
5 1 2 5 1
2 5 1 5 1PQ a
2
2,
a aQ
t t
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5 1 5 1
2 2a = 5a
52 If chord PQ subtends an angle at the vertex of 2 4 ,y ax then tan
(A) 2
73
(B) 2
73
(C) 2
53
(D) 2
53
Ans (D)
Sol m1 = slope of 2
5 1OPt
m2 = slope of 2 5 1OQ t
1 2
1 2
tan1
m m
m m
25
3
Where is Acute angle between line OP and OQ but as POQ obtuse
2
tan 53
Paragraph for Question 53 and 54
Let 1 2 3 ,S S S S where
1 2
1 3: 4 , : Im 0
1 3
z iS z z S z
i
3 : Re 0 .S z z
53 Area of S =
(A) 10
3 (B)
20
3 (C)
16
3 (D)
32
3
Ans (B)
Sol
z=4
A
B
600
(1, 3)
3 0x y
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S1 represent the inside part of circle z = 4
By putting z = x + iy, we get 2 3 0S x y
S3 represent the right part of yaxis
S is the shaded region in above diagram
Now 2 2 22 20
4 3 4 3 4 3
r r rA
54 min 1 3z S
i z
(A) 2 3
2 (B)
2 3
2 (C)
3 3
2 (D)
3 3
2
Ans (C)
Sol Min 1 – 3i – z = minimum distance of point 1, 3A z S from region S
= AB distance in diagram
3 3 3 3
2 2
Paragraph for Question 55 and 56
A box B1 contains 1 white ball, 3 red balls and 2 block balls Another box B2 contains 2 white balls, 3 red
balls and 4 black balls A third box B3 contains 3 white balls, 4 red balls and 5 black balls
55 If 1 balls is drawn from each of the boxed B1, B2 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 (A)
Sol Required probability
1 2 3 3 3 4 2 4 5
6 9 12 6 9 12 6 9 12
82
648
56 If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is
white and the other balls is red, the probability that these 2 balls are drawn from box B2 is
(A) 116
181 (B)
126
181 (C)
65
181 (D)
55
181
Ans (D)
Sol Required probability
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2 3
1 1
9
2
1 3 2 3 3 4
1 1 1 1 1 1
6 9 12
2 2 2
.1
3
. . .1
3
C C
C
C C C C C C
C C C
55
181
SECTION – 3: (Matching list Type)
This section contains 4 multiple choice question Each question has matching lists The codes for the lists
have choice (A), (B), (C) and (D) out of which ONLY ONE is correct
57
ListI ListII
P Volume of parallelepiped determined by vector
,a b and c is 2 Then the volume of the
parallelepiped determined by vectors
2 , 3a b b c and c a is
1 100
Q Volume of parallelepiped by vectors ,a b and c is
5 Then the volume of the parallelepiped determined
by vectors 3 , 3a b b c and 2 c a is
2 30
R Area of a triangle with adjacent sides determined
by vectors a and b is 20 Then the area of the
triangle with adjacent sides determined by vectors
2 3a b and a b is
3 24
S Area of a parallelogram with adjacent sides
determined by vectors a and b is 30 Then the area
of the parallelogram with adjacent sides determined
by vectors a b and a is
4 60
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 (C)
Sol (P) volume of parallelepiped = 2
Volume 2 . 3a b b c c a
6 . . .a b b c a c b c c a
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6 .a b a b c c
2
6 a b c
= 24
(Q) 5a b c
Volume 3 . 2a b b c c a
6 .a b b c b a c a
= 60
(R) Area of triangle 1
2a b
40 a b
Required area1
2 32
a b a b
1
2 2 3 32
a a a b b a b b
= 100
(S) Area of pallelogram a b
30 a b
Required area a b a
a a b a
0 a b
= 30
58 Consider the lines 1 2
1 3 4 3 3: , :
2 1 1 1 1 2
x y z x y zL L and the planes
1 2: 7 2 3, : 3 5 6 4.P x y z P x y z Let ax by cz d be the equation of the plane passing
through the point of intersection of lines L1 and L2, and perpendicular to planes P1 and P2
Match ListI with ListII and selected the correct answer using the code given below the lists:
ListI ListII
P a = 1 13
Q b = 2 3
R c = 3 1
S d = 4 2 Codes
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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 Point of intersection of the lines L1 and L2
1
1 3
2 1 1
x y zr
2
4 3 3
1 1 2
x y zr
x = 2r1 + 1 x = r2 + 4
y =  r1 y = r2 – 3
z = r1 – 3 z = 2r2 – 3
2r1 + 1 = r2 + 4
2r1 – r2 = 3 ………(i)
 r1 = r2  3 ……………(ii)
Solving (i) and (2) r1 = 2, r2 = 1
So point is (5,  2, 1)
Now the equation of the plane is
a(x – 5) + b(y + 2) + c(z + 1) = 0
Also this plane is perpendicular to P1 & P2
So 7a + b + 2c = 0
3a + 5b – 6c = 0
1 3 2
a b c
 1(x – 5) + 3(y + 2) + 2(z – 1) = 0
 x + 3y + 2z + 13 = 0
x – 3y – 2z = 13
by comparing
we get, a = 1, b =  3, c =  2, d = 13
59 Match List I with List II and select the correct answer using the code given below the lists
ListI ListII
P 1/22
1 1
4
2 1 1
cos tan sin tan1
cot sin tan sin
y y yy
y y y
1 1 5
2 313
Q If cos cos cos 0 sin sin sinx y z x y z then
possible value of cos2
x y is
2 2
R If
cos cos 2 sin sin 2 cos sin 2 sec4
x x x xsexx x x x
3 1
2
NARAYANA IIT/PMT ACADEMY JEE ADVANCE : 2013
(PAPER – II) CODE: 0
37
NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS  INDIA
cos cos 24
x x then possible value of sec x is
S If 1 2 1cot sin 1 sin tan 6 , 0,x x x then
possible value of x is
4 1
Codes
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 (P)
1/22
1 1
4
2 1 1
cot tan sin tan1
cot sin tan sin
y y yy
y y y
1/22
2 2
4
2 2
2
1
1 11
1
1
y
y yy
y y y
y y
1/22
2
2
4
2 22
2
1
11
1
1
y
yy
y yy
y y
=1
(Q) cos x + cosy =  cos z, sin x + sin y =  sin z
2 2 2cos cos 2cos cos cosx y x y z
2 2 2sin sin 2sin sin siny x y z
1 1 2cos 1x y
1
cos2
x y
2 12cos 1
2 2
x y
2 1cos
2 4
x y
1
cos2 2
x y
NARAYANA IIT/PMT ACADEMY JEE ADVANCE : 2013
(PAPER – II) CODE: 0
38
NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS  INDIA
(R) cos sin cos sin
cos 2 sin 2 sec cos sin2 2 2 2
x x x xx x x x x
2sin cos
cos 2 2 sin cos sincos
x xx x x x
x
cos sin 2 sin 2 14
x x x x
sec 24
(S) 1 2 1cot sin 1 sin tan 6 .x x
1 1
2 2
6cot cos sin sin
1 6 1
x x
x x
2 2
1 6
1 6 1x x
12x2 = 5
1 5
2 3x
60 A line : 3L y mx meets yaxis at E(0, 2) and the arc of the parabola y2 = 16x, 0 6y at the
point F(x0, y0) The tangent to the parabola at F(x0, y0) intersects the yaxis at G(0, y1) The slop 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:
ListI ListII
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
(A) 4 1 2 3
(B) 3 4 1 2
(C) 1 3 2 4
(D) 1 3 4 2
Ans (A)
Sol
NARAYANA IIT/PMT ACADEMY JEE ADVANCE : 2013
(PAPER – II) CODE: 0
39
NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS  INDIA
0, 0
F(x , y )0 0
(0, 3)
E
y =16x2
G(0, 3)
(0, y )1
Let Coordinate of F= (x0, y0) = (4t
2 8t)
Coordinate of E = (0, 3) ……… given
Equation of tangent at F ty = x + 4t2
Coordinate of G (0, 4t)
Area of 2
0 3 11
0 4 12
4 8 1
EFG t
t t
2 3112 16
2t t
Let
2 3112 16
2t t t
21' 24 48
2t t t
' 0t 1
2t
1
" 24 962
t t
1 1 1
" 24 96 122 2 2
1
2at t , t will be maximum
Since (4t2, 8t) lies on the lie y = mx + 3 28 4 3t mt
2
8 3
4
tm
t
At t = 1
2
We get 1m
Maximum area will be at t = 1/2
maximum area of triangle = 1/2
y0 = 8t = 4
y1 = 4t = 2
NARAYANA IIT/PMT ACADEMY JEE ADVANCE : 2013
(PAPER – II) CODE: 0
40
NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS  INDIA
Answers Key
PHYSICS CHEMISTRY MATHEMATICS
Q. No. Answer Q. No. Answer Q. No. Answer
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,B,C,D 43 C,D
4 C,D 24 A,B 44 A,B,C
5 A,C,D 25 B 45 A,D
6 A,B 26 C 46 B,D
7 BD 27 B 47 B,D
8 A,D 28 B,D 48 A,C
9 B 29 A 49 D
10 A 30 B 50 C
11 B 31 B 51 B
12 A 32 A 52 D
13 B 33 C 53 B
14 C or 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