JEE ADVANCE : 2013

NARAYANA IIT/PMT ACADEMY JEE ADVANCE : 2013

(PAPER – II) CODE: 0

1

NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS - INDIA

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 so by the invigilators 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 Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers and electronic gadgets are NOT allowed inside the examination hall 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 DO NOT TAMPER WITH/MUTILATE THE ORS OR THE BOOKLET On breaking the seals of the booklet check that it contains 44 pages and all the 60 questions and corresponding answers choices are legible Read carefully the instruction printed at the beginning of each section and TOTAL MARKS 180 B, Filling the right part of the ORS The ORS also has a CODE printed on its left and right parts Check that the CODE printed on the ORS (on both sheets) is the same as that on this booklet and put your signature affirming that you have verified this IF THE CODES DO NOT MATCH, ASK FOR A CHANGE OF THE BOOKLET

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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 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

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 O-100 k is less than area under 400-500 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 hydrogen-like 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 de-excites, is (are)

(A) 9

32R (B)

9

16R



3


(C) 9

5R (D)

4

3R

Ans A, C

Sol: 2

oa n4.5

Zao

nh 3h

2 2

n 3, z 2

Now, 22 21 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 non-conducting 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



4


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 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 of radius r, centered on

the axis (D) In the region r > 2R, the magnetic field is non-zero Ans A, C, D

Sol: hollow solenoidconductor

B B B

B due to hollow conductor is zero inside the conductor and non-zero outside (co-axial) while B due to solenoid is zero outside the solenoid and non-zero 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

o2 1

s

v w vf f

v w v

A & B are correct



5


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: 2e

1 2GMmmv 0

2 L

e

GMv 2

L

M Mve

8 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 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 work-kinetic energy theorem

o 2f

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:



7


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 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 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 km-1 is used, the power

dissipation (in%) during transmission is (A) 20 (B) 30 (C) 40 (D) 50 Ans (B) Sol 3P 600 10 watt rmsV 4000 volt

Total resistance R = 04 20 = 80 Power lost 2

L rmsP I R

2

rms

PR

V

23600 108

4000

= 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 step-up 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 step-down transformer is

(A) 200 : 1 (B) 150 : 1 (C) 100 : 1 (D) 50 : 1 Ans (A) Sol Step – up



8


NP NS

V =?S

4000V

SS

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 x-y 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 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 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



9


(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

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 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

1 H 1007825 u 2

1 H 2014102 u 3

1 H 3016050 u 4

2 He 4002603 u 6

3 Li 6015123 u 7

3 Li 7016004 u 70

30 Zn 69925325 u 82

34Se 81916709 u 152

64 Gd 151919803 u 206

82 Pb 205974455 u 209

83 Bi 208980388 u 210

84 Po 209982876 u

15 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 from data given only option C is correct 3 4m m M '



10


16 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 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 HP V P V

0 0 H H(32P )V P V H 0V 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 20 0

(32P )V P 8VPV P V36P V

1 5/ 3 1

FG GF 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



12


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 3and , 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 22

Q e g 2 2 1 2 3and

R e h 3 1 2

S e i 4 2 1 2 2 32 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



13


o2C 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 [ML-1T-1] R Planck constant 3 [MLT-3K-1] 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 = ML2T-2K-1

Plank’s constant = energy time = ML2T-2 T = ML2T-1

Co-efficient 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



14


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 carbon-based 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



15


(B) 9 1 8 24 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

OHBr

PBr

Br

OHBr

Q

OH

BrBr

Br

R

OH

BrSO3H

Br

Br

S (A) P (B) Q (C) R (D) S Ans (B)

SO3H

OH

Br

Br

OHBr

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)



16


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 10-12 at 298 K The solubility (in mol/L) of Ag2CrO4 in a 01 M

AgNO3 solution is (A) 11 10-11 (B) 11 10-10 (C) 11 10-12 (D) 11 10-9 Ans (B)

Sol 2

2 4 4Ag CrO 2Ag CrO

s ' 0.1 2s ' s '

Ksp = (01 + 2s )2 s 11 10-12 = 10-2 s s = 11 10-10 28 In the following reaction, the product(s) formed is(are)

CH3

OH

3CHCl

OH?

P Q R S

CH3

CHO

OHCHO

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



17


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) Hg2+

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)



18


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

+ VAlCl3

(anhydrous) 1. Zn-Hg/HCl

2. H3PO4

W

O

O

O

and

VO

W

(A)

CH2OH

CH2OH

and

V W

(B)



19


O

O

O

and

V W

(C)

HOH2C

CH2OH

and

V

CH2OH

W

(D)

Ans (A)

HOOC

COOH

H2/NiO

O

O

(V)

Fumaric acid (Q)

Sol.

+ O

O

O

AlCl3

anhydrousCH2

CH2C

OOH

O

1. Zn-Hg/HCl

2. H3PO4

O


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



20


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


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 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



21


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 (Fe3+, Fe2+) = +077 V E0 (Fe2+, Fe) = 044 V E0 (Cu2+, Cu) = +034 V; E0 (Cu2+, Cu) = +052 V E0[O2(g) + 4H+ + 4e 2H2O] = +123 V

E0[O2(g) + 2H2O + 4e 4OH ] = +040V

E0 (Cr3+, Cr) = 074 V; E0 (Cr2+, Cr) = 091 V; Match E0 of the redox pair in List I with the values given in List II and select the correct answer



22


using the code given below the lists:

List I List II P E0 ( Fe3+, Fe) 1 018 V

Q E0 (4H2O 4H+ + 4OH ) 2 04 V

R E0 (Cu2+ + Cu 2Cu+) 3 004 V S E0 (Cr3+, Cr2+) 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 (Fe3+, Fe)

Fe3+ + e Fe2+ 01E 0.77V

Fe2+ + 2e Fe 02E 0.44V

Fe3+ + 3e Fe 0

3E

3 1 2G G G 033E 1 0.77 (2 0.44)

03

0.11E 0.0366 0.04

3V

(Q) E0 (4H2O 2H+ + 4OH ) 2H2O O2 + 4H+ + 4e E0 = 123 V O2 + 2H2O + 4e 4OH E0 = +04 V

4H2O 4H+ + 4OH E0 = 083 V (R) E0 (Cu2+ + Cu 2Cu+) Cu2+ + 2e Cu E0 = +034 V 2(Cu Cu+ + e) E0 = 052 V Cu2+ + Cu 2Cu+ E0 = 034 – 052 = 018 V (S) E0(Cr3+, Cr2+)

Cr3+ + 3e Cr 01E = 074 V

Cr2+ + 2e Cr 02E = 091V

Cr3+ + e Cr2+ 0

3E

3 1 2G G G



23


1 03E = 3 (074) ( 2 091)

03E = 222 – 182

03E = 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



24


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. ONaEt-Br

SubstitutionOEt

R.

(i) Hg(OAc)2

(ii) NaBH4

Follows markownikov's product

OH

S.

(i) BH3

(ii) H2O2/NaOH

Follows antimarkownikov's product

OH



25


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 jijp

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 P2 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)



26


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= -2x-4y= 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



27


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



28


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 2

4n 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)



29


Sol

na

r 1a 1 nn

r 1

r1

Ltn 1 na r

an

ar 1

a 1 nn

r 1

rn n

Ltrn 1 n an

an

a 1r 1

a 1 nn

r 1

1 rn n n

Lt1 rn 1 an n

an

r 1a 1 nn

r 1

1 r1 n n

Lt1 r1 a1n nn

1a

201

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 y-axis 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



30


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 1

x4

, which of the

following is true?

(A)1 3

f ' x f x , x4 4

(B)1

f ' x f x ,0 x4



31


(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


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(at2, 2at)

S(0, 0)(-a, -a)

Q

2

1PQ a t

t ………(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



32


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


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



33


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 y-axis 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 3

z Si 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


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



34


2 31 1

92

1 3 2 3 3 41 1 1 1 1 1

6 9 122 2 2

.13

. . .13

C CC

C C C C C CC 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

List-I List-II 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



35


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 List-I with List-II and selected 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



36


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

List-I List-II P 1/2

21 1

42 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



37


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)

1/221 1

42 1 1

cot tan sin tan1

cot sin tan sin

y y yy

y y y

1/22

2 24

2 2

2

1

1 11

1

1

y

y yy

y y yy y

1/222

24

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



38


(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 y-axis 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 y-axis 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: 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 (A) 4 1 2 3 (B) 3 4 1 2 (C) 1 3 2 4 (D) 1 3 4 2 Ans (A) Sol



39


0, 0

F(x , y )0 0

(0, 3)

E

y =16x2

G(0, 3)

(0, y )1

Let Co-ordinate of F= (x0, y0) = (4t2 8t) Co-ordinate of E = (0, 3) ……… given Equation of tangent at F ty = x + 4t2 Co-ordinate 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



40


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

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JEE ADVANCE : 2013

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