Jee Advanced 2

8/12/2019 Jee Advanced 2

1/36

Sri Chaitanya IIT Academy., India

2013

JEE-AD

VANCED

Create PDF files without this message by purchasing novaPDF printer ( http://www.novapdf.com )
http://www.novapdf.com/http://www.novapdf.com/http://www.novapdf.com/http://www.novapdf.com/


2/36

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

Sri Chaitanya IIT Academy., India#Corporate Office : Plot No 304, Kasetty Heights, Ayyappa Society Madhapur, Hyderabad

E-Mail - [email protected], Web - www.srichaitanya.net 2 / 36

IMPORTANT INSTRUCTIONS : CODE :1A. General:* This booklet is your Question 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 elec-

tronic gadgets are NOT allowed inside the examinatioin 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 par ts should only be separated at the end of the examination when in-structed by the invigilator. The upper sheet is a 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 suff icient pressure sothat the impression is created on the bottom duplicate sheet.

* DO NOT TAMPER WITH/ MUTILATE THEORS OR THE BOOKLET.* On breaking the seals of the booklet check that it contains 44 pages and all the 60 questions and

corresponding answer choices are legible. Read carefully the instruction printed at the beginning of each section.

* B. Filling the right part of the ORSThe ORS also has a CODEprinted on its left and right parts.

* Check that the CODEprinted on the ORS (on both sheets) is the same as that on this booklet and put yoursignature affirming that you have verified this.

* IF THE CODES DO NOT MATCH, ASK FOR A CHANGE OF THE BOOKLET.Write your Name, Rol No. and the name of centre and sign with pen in the boxes provided on the rightpart of ORS. Do not write any of this anywhere else. Darken the appropriate bubble UNDER each digit

of your Roll No. in such a way that the impression is created on the bottom sheet.C. Question Paper Format

The question paper consisits of three parts ( Physics, Chemistry and Mathematics). Each part consists of three sections.Section 1 contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which 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.

D. Marking SchemeFor each question in section 1, you wil be awarded 3 marks if you darken all the bubble(s) correspondingto only the correct answer(s) and zero mark if no bubbles are darkened. In all other cases, minus one (-1) mark will be awarded.

For each question Section 2 and 3, you will be awarded 3 marks if you darken the bubble correspond-ing to only the correct answer and zero mark if no bubbles are darkened. In all other cases, minus one(-1) mark will be awarded.

mailto:[email protected],http://www.srichaitanya.net/http://www.novapdf.com/http://www.novapdf.com/http://www.novapdf.com/http://www.novapdf.com/http://www.srichaitanya.net/mailto:[email protected],


3/36




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 90 0. The wavelength is exactly known and the error in

is constant for all values of . As increases from 0 0

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 isdd d d tan

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

Again d tan dd 2sin

Absolute error in d is

2

d cosdd

2sin

d and are constant (given). Clearly dd decreases with increase in .

02. Two non-conducting spheres of radii R 1 and R 2 and carrying uniform volume charge densi-

ties and respectively, are placed such that they partially overlap, as showin in the

figure. At all points in the overlapping region.



4/36




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 r

3

due to 2nd sphere, 2 20

E r 3

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 required for increasing the temperature from 400-500K

C) there is no change in the rate of heat absorption in the range 400-500K D) the rate of heat absirption increases in the range 200-300K

Ans : B,C,D



5/36




Sol. Rate of heat absorptiondQ

R dt

dTR mC

dt

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 correct

From 400K to 500K, C and hence R is constant.From 200K to 300K, C and hence R increases.

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

Bohr radius. Its orbital angular momentum is3h2

. 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)

916R

C)9

5R D)

43R

Ans : A,C

Sol. Radius of nth orbit2

n 0

nr a

Z

2 2

0 0

n n4.5a a 4.5 1

Z Z

nh 3h n 3 22 1

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

From Ridbergs equation,2

2 21 2

1 1 1RZ

n n



6/36




n = 3 to 1 transition :- 11

1 1 94R 1

9 32R


1 1 1 94R

4 9 5R


1 14R 1 3R

4

05. Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m is projected from the midpoint of the line joining their centres, perpendicular to the line. 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 isGM

2L

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

bodies is 2GML

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


7/36




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 u 0.Whenthe speed of the particle is 0.5 u 0, 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 u 0B) 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 is4 m

t3 k

.

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

time is5 m

t 3 k

Ans : A,D

Sol. O to A : 0v u cos wt

00

u 2 t 2 t Tu cos t

2 T 3 T 6

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

time.First time of crossing mean position

1

T 1 m 2 mt 2t 2

3 3 K 3 K

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

2

T T 7T 7 m 7 mt 2

3 4 12 12 K 6 K

second time mean position crossing time

3

T T 5 5 m 5 mt T 2

3 2 6 6 K 3 K



8/36




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, centered on the axis.

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

Ans : A,D

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

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

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

08. Two vehicles, each moving with speed u on the same horizontal straight road, are approach-ing each other. Wind blows along the road with velocity w. One of these vehicles blows a

whistle of frequency f 1. An observer in the other vehicle hears the frequency of the whistleto be f 2. The speed of sound in still air is V. The correct statement(s) is (are)

A) If the wind blows from the observer to the source, f 2 > f 1

B) If the wind blows from the source to the observer, f 2 > f 1

C) If the wind blows from observer to the source, f 2 < f 1

D) If the wind blows from the course to the observer, f 2 < f 1

Ans : A,B

Sol. f 2 > f 1 for any direction of motion of wind as there will be relative velocity of approach in both the cases.



9/36




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 an

angular velocity . This can be considered as equivalent to a loop carrying a steady

currentQ2

. 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)BR 4

B)BR 2

C) BR D) 2BR

Ans : B

Sol. 2B 0 BR

E 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

2C)

2BQR 2

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 QVR R I R 2 R 2

R,Q are constantQR

d dv2

, 2

dv d 2QR



10/36




Substituting (2) in (1) 2

EQ m d QR

BR 1 d Q

2 r R [ It is given that, r L 2m

]

2rBR d 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 m

1 and m

2 only if (m

1 +

m 2) < M. Also two light nuclei of mass m 3 and m 4 can undergo complete fusion andform a heavy nucleus of mass M 1 only if (m 3 + m 4) > M

1. The masses of some neutralatoms are given in the table below.

11 H 1.007825 u

21 H 2.014102u

31 H 3.016050u

42 He 4.002603 u

63 Li 6.015123 u

73 Li 7.016004 u

7030 Zn 69.925325 u

8234 Se 81.916709 u

152

64Gd 151.919803 u 206

82Pb 205.974455 u 209

83Bi 208.980388 u 210

84Po 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 and8234 Se 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



11/36




B : 210 1 2090 1 83P H Bi

This is also not possible as sum of masses 11 H and209

83 Bi is more than that of210

84 0P .

C : 2 4 61 2 3H Li

Sum of the masses of 21 H and4

2 is more than that of6

3 Li . Hence this reaction is pos-

sible.70 82 152

30 34 64Zn Se Gd

This is not possible.

12. The kinetic energy (in keV) of the alpha particle, when the nucleus 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.

2Pb a

Pb

K MPK

2m K M

Pb Pb PbPb Pb

K M M1 1 K K K

K M M M

Pb

Pb

M 205.974455K E 5.419467 5.31616 MeV 5316.16KeV

M 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 instantaneous

velocity.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).



12/36




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 21

mgR 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 mv N mg cos 60R

1 1001 N 1 10 7.5N

2 40

Paragraph for Questions 15 to 16 A 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.



13/36




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 produced5

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

225

0

150 8I R 100 30

P 6 10

16. In the method using the transformers assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1:10.If the power to the con-sumers has to be supplied at 200V, the ratio of the number of turns in the primary to that 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 v 1 is given by41

13

v 10v 4 10 v

4 10 1

This is input voltage to the step down transformed . Its output voltage v 2 is given as 200v.

The ratio number of turns in step-down transformed44 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) [ML 2T 1]Q) Coefficient of viscosity 2) [ML 1T 1]R) Plank constant 3) [MLT -3K -1]

S) Thermal conductivity 4) [ML2

T 2

K 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



14/36




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

index 2 , which is surrounded by a medium of refractive index 3 , as shown in the figure.A ray of light e enters the rectangular block at normal incidence. Depending upon the

relationships between 1 2, and 3 it takes one of the four possible paths ef, eg, eh or ei.

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

P) e f 1) 1 22 Q) e g 2) 2 1 and 2 3 R) e h 3) 1 2

S) e i 4) 2 1 22 and 2 3 P Q R S P Q R 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 given below the lists.LIST - I LIST - II

P) Alpha decay 1) 15 158 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



15/36




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 P 0V0 ln2

Q) G H 2) 36 P 0V0

R) F H 3) 24 P 0V0

S) F G 4) 31 P 0 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



16/36




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 correct21. The correct statement(s) about O 3 is (are)

A) O O bond lengths are equalB) Thermal decomposition of O 3 is endothermicC) O 3 is diamagnetic in nature D) O 3 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) , pAns : A,B

Sol. 9 0 8 14 0 4 0Be Be n 9 0 8 2

4 0 4 1Be P Be D

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

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

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

Ans : C,D

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

But 2 3Al O and dolamite cannot be reduced

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

3 2CaCO s CaO s CO gFor this equilibrium, the correct statement(s) is (are)

A) H is dependent on TB) K is independent of the initial amount of 3CaCO

C) K is dependent on the pressure of 2CO at given T

D) H is independent of the catalyst, if anyAns : A,B,D



17/36




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 10Ans : B

Sol. 2 2sp 4K Ag CrO

212 241.1 10 0.1 CrO

24CrO solubility = 1.1 10 1026. In the following reaction, the product(s) formed is (are)

OH

CH 3

CHCl 3

OH -?

(P)

OH

CH 3

OHC CHO

(Q)

O

CHCl 2H3C

(R)

OH

CHCl 2H3C

(S)

OH

CH 3

CH O

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

OH

CH 3

CHCl 3OH -

OH

CH 3

CHO



18/36




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

OH

SO 3H

aqueous Br 2 (3.0 equivalents) ?

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

Sol.

OH

SO 3H

+ 3Br 2

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 CH 3

O

(1.0 mol)

Br 2 (1.0 mol)

aqueous/NaOH

Reaction II : H3C CH 3

O

(1.0 mol)

Br 2 (1.0 mol)

CH 3COOH

(P)H3C CH 2Br

O

(Q)H3C CBr 3

O

(R)Br 3C CBr 3

O

(S)BrH2C CH 2Br

O

(T)H3C ONa

O(U) CHBr 3

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



19/36




Ans : C

Sol. Acetone + Br + NaOh 3 3CH COONa CHBr

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

O O

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 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

M N

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 process

30. The pair of isochoric processes among the transformation of state isA) 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 : BSol. Conceptual



20/36




Paragraph for Question 31 and 32The reactions of Cl 2 gas with cold-dilute and hot-concentrated NaOH in water givesodium salts of two (different) oxoacids of chlorine, P and Q, respectively. The Cl 2gas reacts with SO 2 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 of A) hypochlorus and chloric acids B) hypochlorus and chlorus acidsC) chloric and perchloric acids D) chloric and hypochlorus acids

Ans : A

Sol.

2 2 NaOH Cl NaCl NaClO H O

Salt of HOCl

cold

Salt of HOCl 3

hot2 3 2 NaOH 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 : A

Sol. 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 dissolve

in hot water. The filtrate (Q) remained unchanged, when treated with H 2S in a dilutemineralacid medium. However, it gave a precipitate (R) with H 2S in an ammoniacal

medium. The precipitate R gave a coloured solution (S), when treated with 2 2H O in

an aqueous NaOH medium.



21/36




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 4 Na CrO

Ans : D

Sol. 2 4 Na 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 or

more than one from S,T and U.

(S)

COOH

H OH

COOH

H OH (T)

COOH

H OH

COOH

HO H (U)

COOH

HO H

COOH

H OH

35. Compounds formed from P and Q are, respectively

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

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

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

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

Ans : B

Sol. P malie acid (cis compound)

Q fumeric acid (trans compound)

4KMnO

OHP meso compound S

4KMnO

OHQ resmic mixture (T & U)



22/36




36. In the following reaction sequences V and W are, respectively

2 / H NiQ V

+ V AlCl 3 (anhydrous)1.Zn-Hg/HCl

2.H 3PO 4W

A)O

O

OV

and

OWB) CH 2OH

CH 2OH

V and

W

C)O

O

OV

andW

D)

HO H 2 C

C H 2 O HV

andCH 2OH

W

Ans : A

Sol. Fumeric acid 2H / Ni Butane dioic acid

O

O

O

O

O

O+

3AlCl

acylation

2 2CH CH COOH

An Hg / HC l 2 2CH CH CH COOH

3 4H PO

O



23/36





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) EtBr

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 4

C) 2 3 4 1 D) 3 2 4 1Ans : A

Sol. ClC2H5ONa

;

ONa + C 2H5Br

OC 2H5

Oxymeruration

demeruration OH ;

Hydroboration

oxidationOH



24/36




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

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

S) ?2XeF Xe other product 4) 2ClCodes : P Q R S P Q R SA) 4 2 3 1 B) 3 2 1 4C) 1 4 2 3 D) 3 4 2 1

Ans : D

Sol.warm

2 2 4 4 2 2PbO H SO PbSO O H O 2Cl

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

2 4 2 2 N H I N HI

2XeF NO Xe NOF 39. The standard reduction potential data at 25 0C is given below.

0 3 2E Fe ,Fe 0.77V ;

0 2

E Fe , Fe 0.44V

0 2E Cu ,Cu 0.34V ;

0E Cu ,Cu 0.52V

0 2 2E O g 4H 4e 2H O 1.23V

0 2 2E O g 2H O 4e 4OH 0.40V

0 3

E Cr ,Cr 0.74V

0 2E Cr ,Cr 0.91V Match 0E of the redox pair in List I with the values given in List II and select the correct

answer using the code given below the lists :



25/36




List I List II

P) 0 3E Fe , Fe 1) 0.18V

Q) 0 2E 4H O 4H 4OH 2) 0.4V

R) 0 2E Cu Cu 2Cu 3) 0.04V

S) 0 3 2

E Cr ,Cr

4) 0.83VCodes : P Q R S P Q R SA) 4 1 2 3 B) 2 3 4 1C) 1 2 3 4 D) 3 4 1 2

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 and select 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 Y NaOH HI

4) Conductivity does not change much and then increases

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

Ans : ASol. Conceptual



26/36




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. Let3 i

w

2

and nP w : n 1, 2,3,... . Further 11

H z C : Re z2

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)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

2 log 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 2

x log 3 (x 1) log 4 x 2 log 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 jij p . Then 2 p 0 , when n=

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



27/36




Ans:- B,C,D

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

2n2

w (1 w )0 w 1 n 55,58,56

1 w

44. The function f x 2 x x 2 x 2 2 x has a local minimum or a

local maximum at x=

A) -2 B)2

3 C) 2 D)23

Ans:- A,B

Sol: f(x) = 0 for2

x 2, ,0, 23

f(x) has extremum at2

x 2,3

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

a a a

a 1n

1 2 ... n 1lim

60n 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 17

Lt a 7 or (a 1) (2a 1) 60 2

(a x)dx

46. Circle(s) touching x-axis at a distance 3 from the origin and 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



28/36




Sol: 2 2 2x 2gx c (x 3) x 6x 9

22 f 9 2 7 f 4

47. Two lines 1y z

L : 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) 4Ans:- A,D

Sol: lines are x - 5 = 0, 2y (3 )z 0

x , (2 )y z 0 since lines are coplanar

0 3 2

0 1 2 0

5 0 0

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

48. In a triangle PQR, P is the largest angle and1

cosP3

. 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

R Q

N M

L

2K

2K+2

2K+42K+2

2K+4

2K



29/36


30/36




P(1,-3)

x

y

3

3x y 0

49 Ans:- B

Area of 2 21 1

S 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 52

A box 1B contains 1 white ball, 3 red balls and 2 black balls. An-

other box 2B contains 2 white balls, 3 red balls and 4 black balls.

A third box 3B contains 3 white balls, 4 red balls and 5 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)90

648C)

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)65

181D)

55181



31/36




Sol: Q.No.: 51 & 52B 1 B 2 B 31 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 C

Prob 1 1 3 1 2 3 1 3 4 1813 C 3 C 3 C


Let 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


32/36




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: C

xg(x) e f (x)

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

10 x

4

g(x) is decreasing1g (x) 0

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


Let 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



33/36




Sol:55. Ans:- B

1a t a

t , 2

1t 1

t ,

21 1

t t 4 5t t

21

PQ a t 5at

56. Ans:- D

1 2

2 2m , m 2t

1tt

,

22t

2 1 2ttan t 51 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 , yintersects 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



34/36


35/36




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 2x x 6

1 x 1 6x

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

2 3

B is correct answer

59. Consider the l ines 1x 1 y z 3

L : ,2 1 1

2x 4 y 3 z 3

L :1 1 2

and the

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

1L and 2L , and perpendicular to planes 1P and 2PMatch List-I with List-II and select the correct answer using thecode given below the lists :

List I List II

P) a= 1) 13

Q) b= 2) -3

R) c= 3) 1

S) d= 4) -2 P Q R S

A) 3 2 4 1

B) 1 3 4 2

C) 3 2 1 4

D) 2 4 1 3

Ans:- A



36/36

2013 JEE-ADVANCED_P2_Code::1 QPaper with & Key &Sol SolutionsSol: (a, b, c) = (3, 5, -6) x (7, 1, 2) = (1, -3, -2)

A is the correct option

60. Match List-I with List-II and select the correct answer using the code given belowthe 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 isQ) 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 isR) 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 isS) Area of a parallelogram with adjacent sides determined by 4) 60

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

adjacent sides determined by vectors a b and a is P Q R S P Q R S

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

C) 3 4 1 2 D) 1 4 3 2Ans:- C

Sol:2

V 6 a b b c c a 6 a b c 6 4 24 P 3C is correct answer This section contains 10 Multiple Choice questions. Each Ques-tion has Four choices (A), (B), (C) and (D). Out of Which Only One is correct

Documents

Jee Advanced 2