IIT JEE Advanced2 2013

2013 JEE-ADVANCED_P2_Code::1 QPaper with & Key &Sol Solutions

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C. Question Paper FormatThe question paper consisits of three parts ( Physics, Chemistry and Mathematics). Each part consists ofthree sections.Section 1 contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out ofwhich ONE or MORE are correctSection 2 contains 4 paragraphs each describing theory, experiment, data etc. Eight questions relate tofour paragraph with two question on each paragraph. Each question of a paragraph has ONLY ONEcorrect answer among the four choices (A), (B), (C) and (D).Section 3 Contains 4 multiple choice questions. Each question has matching lists. The codes for the listshave choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

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PART-I_PHYSICSSection-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 correct01. Using the expression 2dsin , one calculates thevalues of d by measuring the corre-

sponding angles in the range 0 to 900. The wavelength is exactly known and the error in

is constant for all values of . As increases from 00

A) the absolute error in d remains constant.B) the absolute error in d increasesC) the fractional error in d remains constantD) the fractional error in d decreases.

Ans : D

Sol. 2d sin d sin constant K

d cos d sin dd 0

d d tan dd

Fractional error in d is dd dd tan

d constant (given). tan increases with increase in the above fraction decreases.

Again d tan dd2sin

Absolute error in d is 2

d cosdd

2sin

d and are constant (given). Clearly dd decreases with increase in .02. Two non-conducting spheres of radii R1 and R2 and carrying uniform volume charge densi-

ties and respectively, are placed such that they partially overlap, as showin in thefigure. At all points in the overlapping region.









A) the electrostatic field is zeroB) the electrostatic potential is constantC) the electrostatic field is constant in magnitudeD) the electrostatic field has same direction.

Ans : C,D

Sol. At P, due to 1st sphere, 1 10

E r3

due to 2nd sphere, 2 20

E r3

P 1 2 1 20 0

E E E r r x3 3

PE is independent of 1r or 2r but depends upon constant vector x .

E = constant V is variable.03. The figure below shows the variation of specific heat capacity(C) of a solid as a function of

temperature (T).The temperature is increased continuously from 0 to 500 K at a constantrate. Ignoring any volume change, the following statement(s) is (are) correct to a reason-able approximation.

A) the rate at which heat is absorbed in the range 0-100 K varies linearly with temperatureTB) heat absorbed in increasing the temperature from 0-100 K is less than the heat requiredfor increasing the temperature from 400-500KC) there is no change in the rate of heat absorption in the range 400-500KD) the rate of heat absirption increases in the range 200-300K

Ans : B,C,D









Sol. Rate of heat absorption dQRdt

dTR mCdt

dTdt is constant R C

From T = 0 to T = 100K. The graph is not a straight line passing through origin.c and hence R does not vary linearly with T

dQ m CdT Q = m[area under C - T graph]Q area.From graph, the area from 0 to 100 K is less than the area from 400 to 500 K. Hence optionB is correctFrom 400K to 500K, C and hence R is constant.From 200K to 300K, C and hence R increases.

04. The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 0 , where 0 is the

Bohr radius. Its orbital angular momentum is 3h2 . 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 C) 9

5R D) 4

3R

Ans : A,C

Sol. Radius of nth orbit 2

n 0n

r aZ

2 2

0 0n n

4.5a a 4.5 1Z Z

nh 3h n 3 22 1

Substituting (2) in (1), Z = 2

From Ridberg’s equation, 2

2 21 2

1 1 1RZn n









n = 3 to 1 transition :- 11

1 1 94R 19 32R


1 1 1 94R4 9 5R


1 14R 1 3R4

05. Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m isprojected from the midpoint of the line joining their centres, perpendicular to the line. Thegravitational constant is G. The correct statement(s) is (are)

A) The minimum initial velocity of the mass m to escape the gravitational field of the two

bodoies is GM4

L

B) The minimum initial velocity of the mass m to escape the gravitational field of the two

bodies is GM2

L

C) The minimum initial velocity of the mass m to escape the gravitational field of the two

bodies is 2GM

L

D) The energy of the mass m remains constant

A ns : B,D or B

Sol. W KE

2GMm 1 GM2 mv v 2L 2 2

If t he gravitational PE of system of M,M and m is considered as P.E of m itself, (which isreasonable only when m <<M)Total energy of m can be taken as constant.









06. A particle of mass m is attached to one end of a mass-less spring of force constant k, lyingon a frictionless horizontal plane. The other end of the spring is fixed. The particle startsmoving horizontally from its equilibrium position at time t = 0 with an initial velocity u0.Whenthe speed of the particle is 0.5 u0, it collides elastically with a rigid wall. After this colli-sion,A) the speed of the particle when it returns to its equilibrium position is u0

B) the time at which the particle passes through the equilibrium position for the first time is

mtk

C) the time at which the maximum compression of the spring occurs is 4 mt3 k

.

D) the time at which the particle passes through the equilibrium position for the second

time is 5 mt3 k

Ans : A,D

Sol. O to A : 0v u cos wt

00

u 2 t 2 t Tu cos t2 T 3 T 6

From conservation of energy, it will have a speed 0u while crossing mean position every

time.First time of crossing mean position

1T 1 m 2 mt 2t 23 3 K 3 K

Compression is max. when the particle reaches left extreme position. That time

2T T 7T 7 m 7 mt 23 4 12 12 K 6 K

second time mean position crossing time

3T T 5 5 m 5 mt T 23 2 6 6 K 3 K









07. A steady current I flows along an infinitely long hollow cylindrical conductor of radius R.This cylinder is placed coaxially inside an infinite solenoid of radius 2R.The solenoid has nturns per unit length and carries a steady current I. Consider a point P at a distance r from thecommon axis. The correct statement(s) is (are)

A) In the region 0 < r < R, the magnetic field in non-zero

B) In the region R < r < 2R, the magnetic field is along the common axis.

C) In the region R < r < 2R, the magnetic field is tangential to the circle radius r, centeredon the axis.

D) In the region r > 2R, the magnetic field is non-zero

Ans : A,D

Sol. At P : 0 < r < R, B due to current in pipe is zero but due to solenoid, 0B nI 0

At Q : R < r < 2R, B due to solenoid is parallel to the common axis and that due to I in pipeis tangential to the circle of radius r i.e. the two fields are perpendicualr to each other. Theirresultant will make some angle with common axis.

At S : r > 2R, B = 0 due to solenoid but it is non zero due to current in pipe.

08. Two vehicles, each moving with speed u on the same horizontal straight road, are approach-ing each other. Wind blows along the road with velocity w. One of these vehicles blows awhistle of frequency f1. An observer in the other vehicle hears the frequency of the whistleto be f2. The speed of sound in still air is V. The correct statement(s) is (are)

A) If the wind blows from the observer to the source, f2 > f1

B) If the wind blows from the source to the observer, f2 > f1

C) If the wind blows from observer to the source, f2 < f1

D) If the wind blows from the course to the observer, f2 < f1

Ans : A,B

Sol. f2 > f1 for any direction of motion of wind as there will be relative velocity of approach inboth the cases.









Section-2(Paragraph Type)

This section contains 4 paragraphs each describing theory, experiment, data etc. Eightquestions relate to four paragraphs with two questions on each paragraph. Each questionof a paragraph has only one correct answer among the four choices A,B,C and D.

Paragraph for Questions 9 to 10A point charge Q is moving in a circular orbit of radius R in the x-y plane with anangular velocity . This can be considered as equivalent to a loop carrying a steady

current Q2

. 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 raduisof the orbit remains constant.The application of the magnetic field induces an emf inthe orbit. The induced emf is defined as the work done by an induced electric field inmoving a unit positive charge around a closed loop.It is known that, for an orbitingcharge, the magnetic dipole moment is proportional to the angular momentum witha proportionality constant .

09. The magnitude of the induced electric field in the orbit at any instant of time during the timeinterval of the magnetic field change is

A) BR4

B) BR2

C) BR D) 2BR

Ans : B

Sol. 2B 0 BRE 2 R R E1 2

10. The change in the magnetic dipole moment associated with the orbit, at the end of the timeinterval of the magnetic field change, is

A) 2BQR B) 2BQR

2 C)

2BQR2

D) 2BQR

Ans : B or CSol. Induced electric filed will do work and it leads to change in KE of the particle

Fdt mdv EQ m dv 1 dt 1

Dipole moment 2 2 QV QVRR I R2 R 2

R,Q are constant QRd dv2

, 2dv d 2QR









Substituting (2) in (1) 2EQ m dQR

BR 1 dQ2 r R

[ It is given that, rL 2m ]

2rBRd

2

. d may be +ve or –ve depending upon sense of revolution (clockwise or

anticlockwise) of the particle.

Paragraph for Questions 11 to 12

The mass of a nucleus AZ X is less than the sum of the masses of(A-Z) number of neu-

trons and Z number of protons in the nucleus. The energy equivalent to the corre-sponding mass difference is known as ther binding energy of the nucleus. A heavynucleus of mass M can break into two light nuclei of masses m1 and m2 only if (m1 +m2) < M. Also two light nuclei of mass m3 and m4 can undergo complete fusion andform a heavy nucleus of mass M1 only if (m3 + m4) > M1. The masses of some neutralatoms are given in the table below.

11H 1.007825 u 2

1 H 2.014102u 31 H 3.016050u 4

2 He 4.002603 u

63 Li 6.015123 u 7

3 Li 7.016004 u 7030 Zn 69.925325 u 82

34Se 81.916709 u

15264 Gd 151.919803 u 206

82 Pb 205.974455 u 20983 Bi 208.980388 u 210

84 Po 209.982876 u

11. 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 7030 Zn and 82

34Se can undergo complete fusion

Ans : C

Sol. A : 6 4 23 2 1Li H

Here sum of masses of products is more than the mass of 63 Li . Hence this is not possible









B : 210 1 2090 1 83P H Bi

This is also not possible as sum of masses 11H and 209

83 Bi is more than that of 21084 0P .

C : 2 4 61 2 3H Li

Sum of the masses of 21H and 4

2 is more than that of 63 Li . Hence this reaction is pos-

sible.70 82 152

30 34 64Zn Se Gd

This is not possible.

12. The kinetic energy (in keV) of the alpha particle, when the nucleus 21084 Po at rest undergoes

alpha decay, isA) 5319 B) 5422 C) 5707 D) 5818

Ans : A

Sol. 210 4 20684 0 2 82P Pb

mass deffect 1 2M M m m 209.982876 4.002603 205.974455 0.00582amu

aviable energy E 0.00582 931.5 5.419467 MeV

Momentum is same in magnitude for and Pb.2

Pb a

Pb

K MPK

2m K M

Pb PbPb

Pb Pb

K M M1 1 K K KK M M M

Pb

Pb

M 205.974455K E 5.419467 5.31616MeV 5316.16KeVM M 209.977058

Paragraph for Questions 13 to 14A small block of mass 1kg is released from rest at the top of a rough track.The trackis a circular arc of radius 40 m. The block slides along the track without topplingand 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 thefigure below, is 150J. (Take the acceleration due to gravity, g = 10 ms-2).









13. The speed of the block when it reaches the point Q is

A) 5 ms-1 B) 10 ms-1 C) 110 3ms D) 20 ms-1

Ans: B

Sol. 0 21mgR sin 30 150 mv2

2 11 11 10 40 150 1 v v 10 ms2 2

14. The magnitude of the normal reaction that acts on the block at the point Q isA) 7.5 N B) 8.6 N C) 11.5 N D) 22.5 N

Ans : A

Sol.2

0 mvN mg cos60

R

1 1001N 1 10 7.5N

2 40

Paragraph for Questions 15 to 16A thermal power plant produces electric power of 600kW at 4000V, which is to betransported to a place 20 km away from the power plant for consumers usage. It canbe transported either directly with a cable of large current carrying capacity or byusing a combination of step-up and step-down transformers at the two ends. Thedrawback of the direct transmission is the large energy dissipation. In the methodusing transformers, the dissipation is much smaller. In this method, a step-up trans-former is used at the plant side so that the current is reduced to a smaller value. ATthe consumers’ end a step-down transformer is used to supply power to the consum-ers at the specified lower voltage. It is reasonable to assume that the power cable ispurely resistive and the transformers are ideal with a power factor unity. All thecurrent and voltages mentioned are rms values.









15. If the direct transmission method with a cable of resistance 10.4 km is used, the power

dissipation (in %) during transmission isA) 20 B) 30 C) 40 D) 50

Ans : B

Sol. Current produced 5

03

0

P 6 10I 150A

V 4 10

Resistance of cable R 0.4 20 8 .Percentage of power disspation when compared with produced power

22

50

150 8I R 100 30P 6 10

16. In the method using the transformers assume that the ratio of the number of turns in theprimary to that in the secondary in the step-up transformer is 1:10.If the power to the con-sumers has to be supplied at 200V, the ratio of the number of turns in the primary to that inthe secondary in the step- down transformer isA) 200 : 1 B) 150 : 1 C) 100 : 1 D) 50 : 1

Ans : A

Sol. At the step-up transformed, output voltage v1 is given by 4113

v 10 v 4 10 v4 10 1

This is input voltage to the step down transformed . Its output voltage v2 is given as 200v.

The ratio number of turns in step-down transformed 44 10

200200

.

Section-3(Matching List Type)

This section contains 4 multiple choice questions. Each question has matching lists. Thecodes for lists have choices A,B,C and D out of which ONLY ONE is correct.17. Match List I with List II and select the correct answer using the codes given below the List:

LIST - I LIST - IIP) Boltzmann constant 1) [ML2T–1]Q) Coefficient of viscosity 2) [ML–1T–1]R) Plank constant 3) [MLT-3K-1]S) Thermal conductivity 4) [ML2T–2K–1] P Q R S P Q R SA) 3 1 2 4 B) 3 2 1 4C) 4 2 1 3 D) 4 1 2 3

Ans : CSol. Conceptual









18. A right angled prism of refractive index 1 is placed in a rectangular block of refractive

index 2 , which is surrounded by a medium of refractive index 3 , as shown in the figure.A ray of light ‘e’ enters the rectangular block at normal incidence. Depending upon therelationships between 1 2, and 3 it takes one of the four possible paths ‘ef’, ‘eg’, ‘eh’ or‘ei’.

Match the paths in List I with conditions of refractive indices in List II and select the cor-rect answer using the codes given below the lists:LIST -I LIST - II

P) e f 1) 1 22

Q) e g 2) 2 1 and 2 3

R) e h 3) 1 2

S) e i 4) 2 1 22 and 2 3 P Q R S P Q R SA) 2 3 1 4 B) 1 2 4 3C) 4 1 2 3 D) 2 3 4 1

Ans : DSol. Conceptual19. 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 givenbelow the lists.LIST - I LIST - IIP) Alpha decay 1) 15 15

8 7O N ......

Q) decay 2) 238 23492 90U Th ..........

R) Fission 3) 185 18483 82Bi Pb ...........

S) Proton emission 4) 239 14094 57Pu La ..........

P Q R S P Q R SA) 4 2 1 3 B) 1 3 2 4C) 2 1 4 3 D) 4 3 2 1

Ans : CSol. Conceptual









20. One mole of a monatomic ideal gas is taken along two cyclic processes E F G E and E F H E as shown in the PV diagram.The processes involved are purelyisochoric. isobaric, isothermal or adiabatic.

Match the path in List I with the magnitudes of the work done in List II and select thecorrect answer using the codes given below the lists.

LIST - I LIST - II

P) G E 1) 160 P0V0 ln2

Q) G H 2) 36 P0V0

R) F H 3) 24 P0V0

S) F G 4) 31 P0 V0

P Q R S P Q R S

A) 4 3 2 1 B) 4 3 1 2

C) 3 1 2 4 D) 1 3 2 4

Ans : A

Sol. E H : 5 /3

r r 0 H1 1 2 2 H 0

0 0

32P VP V P V V 8VP V

G E 0 0 0 0 0W P 32V V 31P V

G H 0 0 0 0 0W P 32V 8V 24P V









PART-II_CHEMISTRYSection-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 correct21. The correct statement(s) about O3 is (are)

A) O – O bond lengths are equalB) Thermal decomposition of O3 is endothermicC) O3 is diamagnetic in nature D) O3 has a bent structure

Ans : A,C,D

Sol. 3O exihibits resonance ; its formation is endothermic but decomposition is exothermic ; it

is diamagnetic with ben strucuture22. In the nuclear transmutation

9 84 4Be X Be Y (X,Y) is (are)

A) , n B) (p,D) C) (n,D) D) , p

Ans : A,B

Sol. 9 0 8 14 0 4 0Be Be n

9 0 8 24 0 4 1Be P Be D

23. The carbon – based reduction method is NOT used for the extraction of

A) tin from 2SnO B) iron from 2 3Fe O

C) aluminium from 2 3Al O D) magnesium from 3 3MgCO .CaCO

Ans : C,D

Sol. 2SnO C 8n CO 2 3Fe O C Fe CO

But 2 3Al O and dolamite cannot be reduced

24. The thermal dissociation equilibrium of 3CaCO s is studied under different conditions.

3 2CaCO s CaO s CO g

For this equilibrium, the correct statement(s) is (are)A) H is dependent on T

B) K is independent of the initial amount of 3CaCO

C) K is dependent on the pressure of 2CO at given TD) H is independent of the catalyst, if any

Ans : A,B,D









Sol. (1) BH depends on T but not an catalyst(2) K is independenc of conc.

25. The spK of 2 4Ag CrO is 121.1 10 at 298K. The solubility (in mol/L) of 2 4Ag CrO in 0.1M

3AgNO solution is

A) 111.1 10 B) 101.1 10 C) 121.1 10 D) 91.1 10

Ans : B

Sol. 2 2sp 4K Ag CrO

212 241.1 10 0.1 CrO

24CrO solubility = 1.1 × 10–10

26. In the following reaction, the product(s) formed is (are)

OH

CH3

CHCl3

OH- ?

(P)

OH

CH3

OHC CHO

(Q)

O

CHCl2H3C

(R)

OH

CHCl2H3C

(S)

OH

CH3

CHO

A) P (major) B) Q (minor) C) R (minor) D) S (major)Ans : DSol. Reimer teimann reaction

OH

CH3

CHCl3OH-

OH

CH3

CHO









27. The major product(s) of the following reaction is (are)

OH

SO3H

aqueous Br2 (3.0 equivalents)?

A) P B) Q C) R D) SAns : B

Sol.

OH

SO3H

+ 3Br2

OH

Br

Br Br

28. After completion of the reactions (I and II), the organic compound(s) in the reaction mix-tures is(are)

Reaction I : H3C CH3

O

(1.0 mol)

Br2 (1.0 mol)aqueous/NaOH

Reaction II : H3C CH3

O

(1.0 mol)

Br2 (1.0 mol)CH3COOH

(P) H3C CH2Br

O(Q)

H3C CBr3

O(R)

Br3C CBr3

O(S)

BrH2C CH2Br

O

(T) H3C ONa

O(U) CHBr3

A) Reaction I : P and Reaction II : PB) Reaction I : U, acetone and Reaction II : Q, acetoneC) Reaction I : T,U, acetone and Reaction II : PD) Reaction I : R, acetone and Reaction II : S, acetone









Ans : CSol. Acetone + Br + NaOh 3 3CH COONa CHBr

3CH COOH3 3 2 3 2CH C CH Br CH C CH Br

O O



Paragraph for Question 29 and 30A fixed mass ‘m’ of a gas is subjected transformation of states from K to L to M to Nand back to K as shown in the figure

K L

MN

Pressure

Volume

29. The succeeding operations that enable this transformation of states areA) Heating, cooling, heating, colling B) Cooling, heating, cooling, heatingC) Heating, cooling, cooling, heating D) Cooling, heating, hetaing, cooling

Ans : CSol. K L – expantion which needs heating

L M – Isochloric which needs heating process

M N – Isobaric which needs cooling process

N K – Isochoric which needs heating process30. The pair of isochoric processes among the transformation of state is

A) K to L and L to M B) L to M and N to KC) L to M and M to N D) M to N and N to K

Ans : BSol. Conceptual









Paragraph for Question 31 and 32The reactions of Cl2 gas with cold-dilute and hot-concentrated NaOH in water givesodium 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 withwhite phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phos-phorus T.

31. P and Q, respectively, are the sodium salts ofA) hypochlorus and chloric acids B) hypochlorus and chlorus acidsC) chloric and perchloric acids D) chloric and hypochlorus acids

Ans : A

Sol.

2 2NaOH Cl NaCl NaClO H O

Salt of HOCl

cold

Salt of HOCl3

hot2 3 2NaOH Cl NaCl NaClO H O

32. R,S and T respectively, are

A) 2 2 5SO Cl , PCl and 3 4H PO B) 2 2 3SO Cl , PCl and 3 3H PO

C) 2 3SOCl ,PCl and 3 2H PO D) 2 5SOCl ,PCl and 3 4H PO

Ans : ASol. 2 2 2 2SO Cl SO Cl

2 2 5SO Cl P PCl

5 2 3 4PCl H O H PO HCl

Paragraph for Question 33 and 34An aqueous solution of a mixture of two inorganic salts, when treated with diluteHCl, gave a precipitate (P) and a filtrate (Q). The precipitate P was found to dissolvein hot water. The filtrate (Q) remained unchanged, when treated with H2S in a dilutemineralacid medium. However, it gave a precipitate (R) with H2S in an ammoniacalmedium. The precipitate R gave a coloured solution (S), when treated with 2 2H O inan aqueous NaOH medium.









33. The precipitate P contains

A) 2Pb B) 22Hg C) Ag D) 2Hg

Ans : A

Sol. 2PbCl is soluble is hot water

34. The coloured solution S contains

A) 2 4 3Fe SO B) 4CuSO C) 4ZnSO D) 2 4Na CrO

Ans : D

Sol. 2 4Na CrO is coloured solution.

Paragraph for Question 35 and 36

P and Q are isomers of dicarboxylic acid 4 4 4C H O . Both decolorize 2 2Br / H O . On heat-ing P forms the cyclic anhydride.

Upon treatment with dilute alkaline 4KMnO , P as well as Q could produce one ormore than one from S,T and U.

(S)

COOH

H OH

COOH

H OH (T)

COOH

H OH

COOH

HO H (U)

COOH

HO H

COOH

H OH

35. Compounds formed from P and Q are, respectively

A) Optically active S and optically active pair (T,U)

B) Optically inactive S and optically inactive pair (T,U)

C) Optically active pair (T,U) and optically active S

D) Optically inactive pair (T,U) and optically inactive S

Ans : B

Sol. P – malie acid (cis compound)

Q – fumeric acid (trans compound)

4KMnOOHP meso compound S

4KMnOOHQ resmic mixture (T & U)









36. In the following reaction sequences V and W are, respectively2 /H NiQ V

+ V AlCl3 (anhydrous) 1.Zn-Hg/HCl2.H3PO4

W

A) O

O

OV

and OW

B) CH2OH

CH2OH

V and

W

C) O

O

OV

and W

D)

HO H 2C

CH 2O HV and

CH2OHW

Ans : A

Sol. Fumeric acid 2H / NiButane dioic acid

O

O

O

O

O

O

+ 3AlClacylation

2 2CH CH COOH

An Hg / HCl2 2CH CH CH COOH

3 4H PO

O










This section contains 4 multiple choice questions. Each question has matching lists. Thecodes for lists have choices A,B,C and D out of which ONLY ONE is correct.37. Match the chemical conversions in List I with the appropriate reagents in List II and select

the correct answer using the code given below the lists : List I List II

P) Cl 1) 42i Hg OAc ; ii NaBH

Q) ONa OEt 2) NaOEt

R) OH

3) Et–Br

S) OH

4) 3 2 2i BH ; ii H O / NaOH

Codes : P Q R S P Q R SA) 2 3 1 4 B) 3 2 1 4C) 2 3 4 1 D) 3 2 4 1

Ans : A

Sol. ClC2H5ONa

;

ONa + C2H5Br

OC2H5

Oxymerurationdemeruration OH

; Hydroboration

oxidationOH









38. 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 thecode given below the lists : List I List II

P) ?2 2 4 4 2PbO H SO PbSO O other product 1) NO

Q) ?2 2 3 2 4Na S O H O NaHSO + other product 2) 2I

R) ?2 4 2N H N other product 3) Warm

S) ?2XeF Xe other product 4) 2Cl


Ans : D

Sol. warm2 2 4 4 2 2PbO H SO PbSO O H O

2Cl2 2 3 2 4Na S O H O NaHSO S HCl

2 4 2 2N H I N HI

2XeF NO Xe NOF

39. The standard reduction potential data at 250C is given below.

0 3 2E Fe ,Fe 0.77V ;

0 2E Fe , Fe 0.44V

0 2E Cu ,Cu 0.34V ;

0E Cu ,Cu 0.52V

02 2E O g 4H 4e 2H O 1.23V

02 2E O g 2H O 4e 4OH 0.40V

0 3E Cr ,Cr 0.74V

0 2E Cr ,Cr 0.91V

Match 0E of the redox pair in List I with the values given in List II and select the correctanswer using the code given below the lists :









List I List II

P) 0 3E Fe , Fe 1) –0.18V

Q) 02E 4H O 4H 4OH 2) –0.4V

R) 0 2E Cu Cu 2Cu 3) –0.04V

S) 0 3 2E Cr ,Cr 4) –0.83V


Ans : DSol. Conceptual40. 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 andselect the correct answer using the code given below the lists : List I List II

P) 2 5 33C H N CH COOH

X Y 1) Conductivity decreases and then increases

Q) X Y 3KI 0.1M AgNO 0.01M

2) Conductivity decreases and then does not change much

R) X Y

3CH COOH KOH 3) Conductivity increases and then does not chnage much

S) X YNaOH HI

4) Conductivity does not change much and then increases


Ans : ASol. Conceptual









PART-III_MATHEMATICSSection-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 3 iw2

and nP w : n 1, 2,3,... . Further 11H z C : Re z2

and

21H 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) 23

D) 56

Ans: C,D

Sol :From diagram 1 22 5z z ,3 6

42 If x x 13 4 then x=

A) 3

3

2log 22 log 2 1 B)

2

22 log 3 C)

2

11 log 3 D) 2

2

2 log 32 log 3 1

Ans:- A,B,C

Sol: 3

3 2

2log 2 2x log3 (x 1)log 4 x2log 2 1 2 log 3

43. Let be a complex cube root of unity with 1 and ijP p be a n n

matrix with i jijp . Then 2p 0 , when n=

A) 57 B) 55 C) 58 D) 56









Ans:- B,C,D

Sol: 4 6 8 2n 22P 0 w w w .............. w 0

4 2n2n

2

w (1 w ) 0 w 1 n 55,58,561 w

44. The function f x 2 x x 2 x 2 2 x has a local minimum or alocal maximum at x=

A) -2 B) 2

3

C) 2 D) 23

Ans:- A,B

Sol: f(x) = 0 for 2x 2, ,0,2

3

f(x) has extremum at 2x 2,

3

45. For a R (the set of all real numbers), a 1,

a a a

a 1n

1 2 ... n 1lim60n 1 na 1 na 2 ... na n

Then a=

A) 5 B) 7 C) 152

D) 172

Ans:- B,D

Sol: -

1a

01n

0

x dx1 2 1 17Lt a 7 or

(a 1) (2a 1) 60 2(a x)dx

46. Circle(s) touching x-axis at a distance 3 from the origin and havingan 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: 2 2 2x 2gx c (x 3) x 6x 9

22 f 9 2 7 f 4

47. Two lines 1y 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,DSol: lines are x - 5 = 0, 2y (3 )z 0

x , (2 )y z 0

since lines are coplanar

0 3 20 1 2 0

5 0 0

2( 5)( 5 4) 0 1,4

48. In a triangle PQR, P is the largest angle and 1cos P3

. Further the

incircle of the triangle touches the sides PQ, QR and RP at N,L andM respectively, such that the lengths of PN, QL and RM are con-secutive even integers. Then possible lengths of the sides of thetriangle is (are)A) 16 B) 18 C) 24 D) 22

Ans:- B,D

P

RQ

N M

L

2K

2K+2

2K+42K+2

2K+4

2K









Sol:2 2 21 (4k 4) (4k 4) (4k 6)cos P

3 2(4k 2)(4k 4)

2 2 2 2 2

2(2k 1) (2k 2) (2k 3) 4k 4 k 14

8(2k 1)(2k 2) 4(2k 1)(k 1) 2k 3k 1

2 23k 3 2k 3k 1 2k 3k 4 0

(k-4)(k+1)=0k = 4sides: 18, 22But B D options are correct



Let 1 2 3S S S S , where

1 2z 1 3iS z C : z 4 ,S z C : Im 0

1 3i

and

3S z C : Re z 0

49. Area of S=

A) 10

3

B) 20

3

C) 16

3

D) 32

3

50. z Smin 1 3i z

A) 2 32 B) 2 3

2 C) 3 3

2 D) 3 3

2

Sol: Q.No.: 49 & 502 2 2

1S x y 4 ,

2 m(x 1) i(y 3)S I 0

1 3i

3(x 1) y 3 0 3x y 0

3S x 0









P(1,-3)

x

y

3

3x y 0

49 Ans:- B

Area of 2 21 1S 4 4 204 2 3 3

50. Ans:- C

min 1 3i z perpendicular distance from (1, -3)

the line 3x y 0

3 3 3 32 2

Paragraph for Questions 51 and 52A box 1B contains 1 white ball, 3 red balls and 2 black balls. An-other box 2B contains 2 white balls, 3 red balls and 4 black balls.A third box 3B contains 3 white balls, 4 red balls and 5 blackballs.

51. If 1 ball is drawn from each of the boxes 1 2B , B and 3B , the probabil-ity that all 3 drawn balls are of the same colour is

A) 82648 B)

90648 C)

558648 D)

566648

52. If 2 balls are drawn (without replacement) from a randomly se-lected box and one of the balls is white and the other ball is red,the probability that these 2 balls are drawn from box 2B is

A) 116181 B)

126181 C)

65181 D)

55181









Sol: Q.No.: 51 & 52B1 B2 B3

1 2 33R 3R 4R

2B6

4B9

5B12

51. Ans:- AProb = www (or) RRR (or) BBB

= 1 2 3 3 3 4 2 4 5 826 9 12 6 9 12 6 9 12 648

52. Ans: D

92

6 9 122 2 2

1 2 3553 CPr ob 1 1 3 1 2 3 1 3 4 181

3 C 3 C 3 C

Paragraph for Questions 53 and 54Let f : 0,1 R (the set of all real numbers) be a function. Supposethe function f is twice differentiable, f 0 f 1 0 and satisfies

xf " x 2f ' x f x e , x 0,1 .53. Which of the following is true for 0<x<1 ?

A) 0 f x B) 1 1f x2 2

C) 1 f x 14

D) f x 0

54. If the function xe f x assumes its minimum in the interval 0,1 at1x4

, which of the following is true ?

A) 1 3f ' x f x , x4 4

B) 1f ' x f x ,0 x4

C) 1f ' x f x ,0 x4

D) 3f ' x f x , x 14









Sol: Q.No.: 53 & 5453 Ans: D

11 1 nf (x) 2f (x) f (x) e

x 11 1e (f (x) 2f (x) f (x) 1

x 11e g (x) 1

11g (x) 0

g(x) 0 x (0,1)

xe f (x) 0 x (0,1)

f (x) 0 x (0,1)

54. Ans: Cxg(x) e f (x)

1 x 1g (x) e (f (x) f (x)) 0 at x = 1/4

From graph

10 x4

g(x) is decreasing1g (x) 0

x 1 1e (f (x) f (x) 0 f (x) f (x) 0

1f (x) f (x)

Paragraph for Questions 55 and 56Let PQ be a focal chord of the parabola 2y 4ax . The tangents tothe parabola at P and Q meet at a point lying on the liney 2x a,a 0

55. Length of chord PQ isA) 7a B) 5a C) 2a D) 3a

56. If chord PQ subtends an angle at the vertex of 2y 4ax , then tan

A) 2 73 B)

2 73

C) 2 53 D)

2 53









Sol:55. Ans:- B

1a t at

,2

1t 1t

,21 1t t 4 5

t t

21PQ a t 5at

56. Ans:- D

1 22 2m , m 2t1t

t

,

2 2t2 1 2ttan t 5

1 4 3 t 3


This section contains 4 multiple choice questions. Each question has matching lists. Thecodes for lists have choices A,B,C and D out of which ONLY ONE is correct.57. A line L : y mx 3 meets y-axis at E(0,3) and the arc of the parabola

2y 16x,0 y 6 at the point 0 0F x , y . The tangent to the parabola at 0 0F x , y

intersects the y-axis at 1G 0, y . The slope m of the line L is chosen such that thearea of the triangle EFG has a local maximum.Match List I with List II and select the correct answer using the code given belowthe lists :List I List II

P) m = 1) 12

Q) Maximum area of EFG is 2) 4

R) 0y 3) 2

S) 1y 4) 1

P Q R S P Q R SA) 4 1 2 3 B) 3 4 1 2C) 1 3 2 4 D) 1 3 4 2

Ans:- A









Sol: 2 2 31 (3 4t)4t 2(3t 4t )2

2d 10 6t 12t 0 tdt 2

1 2t2

8t 3 4 3m [slope] 114t 44

m 1 P 4 , 1 1Q (1) 1 Q 12 2

, 0y 4 R 2 , 1y 2 S 3

A is the correct option

58. Match List I with List II and select the correct answer using the code given belowthe lists :List I List II

P)

21 14

2 1 1

cos tan y ysin tan y1 yy cot sin y tan sin y

1)

1 52 3

Q) If cos x cos y cos z 0 sin x sin y sin z then possible 2) 2

value of x ycos

2

is

R) If cos x cos 2x sin x sin 2x sec x cos x sin 2x sec x cos x4 4

cos 2x then possible value of secx is 3) 12

S) If 1 2 1cot sin 1 x sin tan x 6 , x 0 , then possible value 4) 1

of x is

P Q R S P Q R SA) 4 3 1 2 B) 4 3 2 1C) 3 4 2 1 D) 3 4 1 2

Ans:- B









Sol: cosx + cosy + cosz = sinx + siny + sinz

x x, y 120 x, z 240 x

0x y 1cos cos602 2

, S) 1 2 1cot sin 1 x sin tan x 6

2 2

x x 61 x 1 6x

, 2 2 2 1 51 6x 6 6x 12x 5 x2 3

B is correct answer

59. Consider the lines 1x 1 y z 3L : ,

2 1 1

2x 4 y 3 z 3L :

1 1 2

and the

planes 1 2P : 7x y 2z 3, P :3x 5y 6z 4 . Let ax by cz d be the equa-tion of the plane passing through the point of intersection of lines

1L and 2L , and perpendicular to planes 1P and 2P

Match List-I with List-II and select the correct answer using thecode given below the lists :List I List IIP) a= 1) 13Q) b= 2) -3R) c= 3) 1S) d= 4) -2 P Q R S

A) 3 2 4 1

B) 1 3 4 2

C) 3 2 1 4

D) 2 4 1 3

Ans:- A









Sol: (a, b, c) = (3, 5, -6) x (7, 1, 2) = (1, -3, -2)

A is the correct option60. Match List-I with List-II and select the correct answer using the code given below

the lists :List - I List - II

P) Volume of parallelepiped determined by vectors a, b 1) 100

and c is 2. Then the volume of the parallelepiped

determined by vectors 2 a b ,3 b c and c a is

Q) Volume of parallelepiped determined by vectors a, b 2) 30

and c is 5. Then the volume of the parallelepiped determined by

vectors 3 a b , b c and 2 c a is

R) Area of a triangle with adjacent sides determined by vectors 3) 24

a and b is 20. Then the area of the triangle with adjacent

sides determined by vectors 2a 3b and a b is

S) Area of a parallelogram with adjacent sides determined by 4) 60

vectors a and b is 30. Then the area of the parallelogram with

adjacent sides determined by vectors a b and a is

P Q R S P Q R SA) 4 2 3 1 B) 2 3 1 4C) 3 4 1 2 D) 1 4 3 2

Ans:- C

Sol:2

V 6 a b b c c a 6 a b c 6 4 24

P 3C is correct answerThis section contains 10 Multiple Choice questions. Each Ques-tion has Four choices (A), (B), (C) and (D). Out of Which Only One is correct






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IIT JEE Advanced2 2013