sri chaitanya iit academy., india...email: #corporate office: plot no: 304, kasetty heights, ayyappa...

Sri Chaitanya IIT Academy., India

#Corporate Office: Plot No: 304, Kasetty Heights, Ayyappa Society, Madhapur – Hyd

Email: [email protected], Web: www.srichaitanya.net

2014 JEE-ADVANCED PAPER-2_Q’PAPER WITH SOLUTIONS

INSTRUCTIONS

A. General

1. This booklet is your Question Paper. Do not break the seal of this booklet before being instructed to do so by the

invigilators.

2. The question paper CODE is printed on the left hand top corner of this sheet and on the back cover page of this booklet.

3. Blank spaces and black pages are provided in the question paper for your rough work. No additional sheets will be

provided for rough work.

4. Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers and electronic gadget of

any kind are NOT allowed inside the examination hall.

5. Write your Name and Roll number in the space provided on the back cover of this booklet.

6. Answers to the questions and personal details are to be filled on an Optical Response Sheet, which is provided

separately. The ORS is a doublet of two sheets-upper and lower, having identical layout. The upper sheet is a machine-

gradable Objective Response Sheet (ORS) which will be collected by the invigilator at the end of the examination. The

upper sheet is designed in such a way that darkening the bubble with a ball point pen will leave an identical impression at

the corresponding place on the lower sheet. You will be allowed to take away the lower sheet at the end of the

examination. (see Figure-1 on the back cover page for the correct way of darkening the bubbles for valid answers).

7. Use a black ball point pen only to darken the bubbles on the upper original sheet. Apply sufficient pressure so that

the impression is created on the lower sheet. See Figure-1 on the back cover page for appropriate way of darkening the

bubbles for valid answers.

8. DO NOT TAMPER WITH /MUTILATE THE ORS OR THIS BOOKLET.

9. On breaking the seal of the booklet check that it contains 28 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 ORS

10. The ORS also has a CODE printed on its left and right parts.

11. Verify that the CODE printed on the ORS (on both the left and right parts) is the same as that on this booklet and put

your signature in the Box designated as R4.

12. IF THE CODES DO NOT MATCH, ASK FOR A CHANGE OF THE BOOKET / ORS AS APPLICABLE.

13. Write your Name, Roll No. and the name of centre and sign with pen in the boxes provided on the upper sheet of ORS.

Do not write any of this anywhere else. Darken the appropriate bubble UNDER each digit of your Roll No. in such

way that the impression is created on the bottom sheet. (see example in Figure 2 on the back cover)

C. Question Paper Format

The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists of two sections.

14. Section 1contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which

ONE is correct.

15. Section 2 contains 3 paragraphs each describing theory, experiment and data etc. Six questions relate to three paragraphs with two questions on each paragraph. Each question pertaining to a particular passage should have only one correct answer among the four given choices (A), (B), (C) and (D).

16. Section 3 contains 4 multiple choice questions. Each question has two lists (List-1: P,Q,R and S; List-2: 1,2,3 and

4). The options for the correct match are provided as (A), (B), (C) and (D) out of which ONLY ONE is correct.




PART I : PHYSICS SECTION – 1

(Only One Option Correct Type)

This section contains 10 multiple choice questions. Each question has four choices (A),(B),(C),and

(D) out of which ONLY ONE option is correct

1. If Cu is the wavelength of K X ray line of copper (atomic number 29) and Mo is

the wavelength of the K X ray line of molybdenum (atomic number 42), then the

ratio /Cu Mo is close to

A) 1.99

B) 2.14

C) 0.50

D) 0.48

Key: B

Sol: 2

1(z 1)

2 2

Cu MO

Mo Cu

Z 1 41 2.14Z 1 28

2. A metal surface is illuminated by light of two different wavelengths 248 nm and 310

nm . The maximum speeds of the photoelectrons corresponding to these wavelengths

are 1u and 2 ,u respectively. If the ratio 1 2: 2 :1u u and 1240hc eV ,nm the work

function of the metal is nearly

A) 3.7eV

B) 3.2eV

C) 2.8eV

D) 2.5eV




Key: A

Sol: 21 0

1

1 cmu ..................(1)2

22 0

2

1 Cmu ..................(2)2

2 0

0 0

0

1240(1) 2 248 (4 )4 51240(2) 1

310

0 3.7eV

3. Parallel rays of light of intensity 2912I Wm are incident on a spherical black body

kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant 8 2 45.7 10 Wm K and assume that the energy exchange with the surroundings is

only through radiation. The final steady state temperature of the black body is close to

A) 330 K

B) 660 K

C) 990 K

D) 1550 K

Key: A

Sol:

RR

2 4 40I( R ) Ae(T T )

2 2 4 4

0I( R ) (4 R )(1)(T T )

4 40

IT T4

8

8 8912 10 81 10 121 104 5.67

1/4 2T (121) 10 330K




4. During an experiment with a metre bridge, the galvanometer shows a null point when

the jockey is pressed at 40.0 cm using a standard resistance of 90 , as shown in the

figure. The least count of the scale used in the metre bridge is 1 mm. The unknown

resistance is

R 90

40.0cm

A) 60 0.15

B) 135 0.56

C) 60 0.25

D) 135 0.23

Key: C

Sol: R 40R 90 6090 100 60

C) Again, (90)R nR n90 n n(100 )100

dR d d0R 100

1 1 1 1 1dR (R)(d ) 60 0.1 0.25100 40 60 4

R 60 0.25




5. A wire, which passes through the hole in a small bead, is bent in the form of quarter of

a circle. The wire is fixed vertically on ground as shown in the figure. The bead is

released from near the top of the wire and it slides along the wire without friction. As

the bead moves from A to B, the force it applies on the wire is

090

A

B

A) always radially outwards

B) always radially inwards

C) radially outwards initially and radially inwards later

D) radially inwards initially and radially outwards later

Key: D

Sol:

00 mg

mgcos

2mvrP

At P, let 2

0mvmg cos

r

2

0v g cos ...................(1)r

And 20

1mgr(1 cos ) mv2

2

0v 2g(1 cos ).....................(2)r

02(1) (2) cos3

At this angle 0 , force between wire and bead is zero

2

0mvmg cos

r force by the wire on the bead will be outwards and by the bead on the

wire will be radially inwards.

2

0mvmg cos

r , force by the bead on the wire will be radially outwards.




6. A planet of radius 110

R (radius of Earth) has the same mass density as Earth.

Scientists dig a well of depth 5R on it and lower a wire of the same length and of

linear mass density 3 110 kgm into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth

66 10 m and the acceleration due to gravity on Earth is 210ms ) A) 96 N B) 108 N C) 120 N D) 150 N Key: B

Sol:

Ry

dy

C

gp

planet

On the surface, 4g GR3

p p p 1p e

e e e

g R R 1g g 10 1msg R R 10

All a depth ‘y’ from the surface of planet,

1p

yg g 1R

Gravitational force on element of length ‘dy’ is 1p

ydF g (dm) (dy) 1 gR

R /5R/5 2

p p0 0

y yF dF g 1 dy g yR 2R

2

pR 1 R

g5 2R 5

3 6

2p p

R 1 R 9 10 6 10 5.4g 1 g 0.9 10 108N5 10 5 10 5 10 5




7. Charges , 2Q Q and 4Q are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii / 2R , R and 2R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance R from the centre of spheres 1, 2 and 3 are 1 2,E E and 3E respectively, then

R

2RQ

Sphere 1 Sphere 2 Sphere 3

2Q

R4Q

PR

2R

P P

A) 1 2 3E E E B) 3 1 2E E E C) 2 1 3E E E D) 3 2 1E E E Key: C Sol:

1 2 12 2

0 0

1 Q 1 2QE , E 2E4 R 4 R

3 130

1 4Q 1E R E4 (2R) 2

2 1 3E E E

8. A glass – capillary tube is of the shape of a truncated cone with an apex angle so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height ,h where the radius of its cross section is .b If the surface tension of water is ,S its density is , and its contact angle with glass is , the value of h will be (g is the acceleration due to gravity)

h

A) 2 cosS

b g

B) 2 cosSb g

C) 2 cos / 2Sb g

D) 2 cos / 2Sb g




Key: D

Sol:

R

hb

/2

/2

/2

bcos

2 R

bR

cos2

2Sh gR

2ShR g

cos

2S 2hg b

9. A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is

BLOCK

LIQUID

S A) 1.21

B) 1.30

C) 1.36

D) 1.42




Key: C

Sol:

2 2g

22rel

2h 4hD 1

D1

22.72 4 10 2.721 4 2 1.36

11.54 2

10. A tennis ball is dropped on a horizontal smooth surface. It bounces back to its

original position after hitting the surface. The force on the ball during the collision is

proportional to the length of compression of the ball. Which one of the following

sketches describes the variation of its kinetic energy K with time t most

appropriately? The figures are only illustrative and not to the scale

A) K

t

B) K

t

C) K

t

D) K

t




Key: B

Sol: Before it collides, 2 2 21 1KE mv m(gt) KE t2 2

Contact time is very small and in that time, KE drops to zero very quickly and increases and after

that it rises with decreasing KE under gravity. This is best described by graph (B)

SECTION – 2 Comprehension Type (Only One Option Correct)

This section contains 3 paragraphs, each describing theory, experiments, data etc. Six questions relate to the three paragraphs with two questions on each paragraph. Each question has ONLY ONE correct answer among the four given options (A), (B), (C), and (D).

Paragraph For Questions 11 & 12

In the figure a container is shown to have a movable (without friction) piston on top.

The container and the piston are all made of perfectly insulating material allowing no

heat transfer between outside and inside the container. The container is divided into

two compartments by a rigid partition made of a thermally conducting material that

allows slow transfer of heat. The lower compartment of the container is filled with 2

moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2

moles of an ideal diatomic gas at 400 K . The heat capacities per mole of an ideal

monatomic gas are 3 5, ,2 2V PC R C R and those for an ideal diatomic gas are

5 7,2 2V PC R C R .




11. Consider the partition to be rigidly fixed so that it does not move. When equilibrium

is achieved, the final temperature of the gases will be

A) 550 K

B) 525 K

C) 513 K

D) 490 K

Key: D

Sol: 1 21 v 2 pn C R(700 T) n C (T 400) T 490K

12. Now consider the partition to be free to move without friction so that the pressure of

gases in both compartments is the same. Then total work done by the gases till the

time they achieve equilibrium will be

A) 250R

B) 200R

C) 100 R

D) 100 R

Key: D

Sol:

1 1 2 2

1 2 1 2 initial final

( U W ) ( U W ) 0( W W ) ( U U ) U U

Calculation for Tf

1 2

1 2 1 2

1 P f 2 P f f

gases 1 v 2 v 1 V 2 V f

gas

6300n C (700 T ) n C (T 400) T K12

( W) n C (700) n C (400) (n C n C )T

so, W 100R




Paragraph For Questions 13&14

A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin

tube of uniform cross section is connected to the nozzle. The other end of the tube is

in a small liquid container. As the piston pushes air through the nozzle, the liquid from

the container rises into the nozzle and is sprayed out. For the spray gun shown, the

radii of the piston and the nozzle ar 20mm and 1mm respectively. The upper end of the

container is open to the atmosphere.

13. If the piston is pushed at a speed of 15mms the air comes out of the nozzle with a

speed of

A) 10.1ms

B) 11ms

C) 12ms

D) 18ms Key: C

Sol: 1 1 2 2 2A V A V V 2m / s

14. If the density of air is a and that of the liquid ,l then for give piston speed the rate

(volume per unit time) at which the liquid is sprayed will be proportional to

A) a

l

B) a l

C) l

a

D) l




Key: A

Sol:

P

Consider a fluid element at point P due to flow in horizontal pipe pressure energy of

liquid in vertical pipe at ‘P’ reduces = 20 a 0

1P V2

But net energy of fluid element remains same

22 1 1 a0 0 a 0 0

1 1P P V V V V2 2

Paragraph For Questions 15 & 16

The figure shows a circular loop of radius a with two long parallel wires (numbered 1

and 2) all in the plane of the paper. The distance of each wire from the centre of the

loop is d . The loop and the wires are carrying the same current I . The current in the

loop is in the counterclockwise direction if seen from above.

a

P R

Q S

Wire1 Wire2

d d

15. When d a but wires are not touching the loop, it is found that the net magnetic field

on the axis of the loop is zero at a height h above the loop. In that case

A) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h a

B) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h a

C) current in wire 1 and wire 2 is the direction PQ and SR, respectively and 1.2h a

D) current in wire 1 and wire 2 is the direction PQ and RS, respectively and 1.2h a




Key: C

Sol:

a

P R

Q S

Wire1 Wire2

d d

Bwire

Bclosed loop

Bwire

h

a a

2 2h a

2

0 0wire closed loop 1/2 2 2 3/22 2

2 i ia2B cos B cos2 2(a h )2 d h

2 2

2 2 1/2 2 2 3/2 2 2

2 1cos a 2 acos(a h ) [a h ] [a h ]

2 2

2 2 2 2 2 2

2 h a aha h (a h ) 2 a h

2 2 2 2 44h (a h ) a

22h x

On solving we get

2ax (2 11 1)

2

2

2 a2h 2 11 1 h 1.2a2

16. Consider ,d a and the loop is rotated about its diameter parallel to the wires by 030

from the position shown in the figure. If the currents in the wires are in the opposite

directions, the torque on the loop at its new position will be (assume that the net field

due to the wires is constant over the loop)

A) 2 2

oI ad

B) 2 2

2oI a

d

C) 2 23 oI a

d

D) 2 23

2oI ad




Key: B

Sol:

2 00net

2 i| | M B (i a ) sin 302 d

2 2

0i a 1d 2

SECTION-3

Matching List Type (Only One Option Correct)

This section contains four questions, each having two matching lists. Choices for the correct

combination of elements from list-I and List-II are given as options (A), (B),(C) and (D), out of

which ONE is correct.

17. Four charges 1 2, 3,Q Q Q and 4Q of same magnitude are fixed along the x axis at x =–2a,

-a, +a and +2a, respectively. A positive charge q is placed on the positive y axis at a

distance b >0. Four options of the signs of these charges are given in List I. The

direction of the forces on the charge q is given in List II. Match list I with List II and

select the correct answer using the code given below the lists.

Q1 Q3Q2 Q4

q(0,b)

(-2a,0) (-a,0) (+a,0) (+2a,0) List I List II

P. 1 2 3 4, , ,Q Q Q Q all positive 1. x

Q. 1 2,Q Q positive; 3 4,Q Q negative 2. x

R. 1, 4Q Q positive; 2, 3Q Q negative 3. y S. 1 3,Q Q positive; 2 4,Q Q negative 4. y Code : A) P-3, Q-1, R-4 S-2 B) P-4, Q-2, R-3, S-1 C) P-3, Q-1, R-2, S-4 D) P-4, Q-2, R-1, S-3




Key: A

Sol: Q1 Q3Q2 Q4

q(0,b)

(-2a,0) (-a,0) (+a,0) (+2a,0)

1 2 3 4 y

1 2 3 4 x

1 4 2 3 y

1 3 2 4 x

ˆ ˆP) Q Q Q Q ve; E 0i E jˆ ˆQ) Q Q ve; Q Q ve; E E i 0 j

ˆ ˆR) Q Q ve; Q Q ve; E 0i E jˆ ˆS) Q Q ve; Q Q ve; E E i 0j

P 3;Q 1; R 4;S 2

18. Four combinations of two thin lines are given in List I. The radius of curvature of all curved surfaces is r and the refractive index of all the lenses is 1.5. Match lens combinations in list I with their focal length in List II and select the correct answer using the code given below the lists.

List I List II

P. 1. 2r

Q. 2. 2r

R. 3. r

S. 4. r Code: A) P-1, Q-2, R-3,S-4 B) P-2, Q-4, R-3, S-1

C)P-4, Q-1, R-2, S-3

D) P-2, Q-1, R-3, S-4




Key: B

Sol: for convex lens 1 1 1 1(1.5 1) f rf r r r

For plano-convex it is f1 = 2r

For concave lens 1 1 1 1(1.5 1) f rf r r r

For plao-concave it is 11f 2r

P. eqeq

1 1 1 2 rff r r r 2

Q. eqeq

1 1 1 1 f rf 2r 2r r

R. eqeq

1 1 1 1 f rf 2r 2r r

S. eqeq

1 1 1 f 2rf r 2r

P 2;Q 4; R 3;S 1

19. A block of mass 1 1m kg another mass 2 2 ,m kg are placed together (see figure) on an inclined plane with angle of inclination . Various values of are given in List I. The coefficient of friction between the block 1m and the plane is always zero. The coefficient of static and dynamic friction between the block 2m and the plane are equal to 0.3. In List II expressions for the friction on block 2m are given. Match the correct expression of the friction in List II with the angles given in List I, and choose the correct option. The acceleration due to gravity is denoted by g.

[Useful information : tan 5.5 0.1; tan 11.5 0.2; tan 16.5 0.3; ]




1m2m

List I List II

P. 5 1. 2 sinm g

Q. 10 2. 1 2 sinm m g

R. 15 3. 2 cosm g

S. 20 4. 1 2 cosm m g

Code :

A) P-1, Q-1, R-1, S-3

B) P-2, Q-2, R-2, S-3

C) P-2, Q-2, R-2, S-4

D) P-2, Q-2, R-3, S-3

Key: D

Sol: L 1 2f (m m )gsin

22 1 2

1 2

mm g cos (m m )gsin tan 0.2(m m )

Where is angle of repose = 11.50

0 0

2

15 & 20 ; blocks are in motionHence friction is m g cos

0 0

1 2

5 & 10 ; blocks are at rest.Hence friction (m m )g sin

P 2;Q 2;R 3;S 3

20. A person in a lift is holding a water jar, which has a small hole at the lower end of its

side. When the lift is at rest, the water jet coming out of the hole hits the floor of the

lift at a distance d of 1.2 m from the person. In the following, state of the lift’s

motion is given in List I and the distance where the water jet hits the floor of the lift is

given in List II. Match the statements from List I with those in List II and select the

correct answer using the code given below the lists.




List I List-II

P. Lift is accelerating vertically up. 1. 1.2d m

Q. Lift is accelerating vertically down with an 2. 1.2d m

acceleration less than the gravitational acceleration

R. Lift is moving vertically up with constant speed 3. 1.2d m

S. Lift is falling freely 4. No water leaks out of the jar

Code:

A) P-2, Q-3, R-2, S-4

B) P-2, Q-3, R-1, S-4

C) P-1, Q-1, R-1, S-4

D) P-2, Q-3, R-1, S-1

Key: C

Sol: Of the lift is accelerations upward are downward with acceleration less than

acceleration due to gravity d = 1.2

Lift is moving vertically up with constant speed

d = 1.2

When lift is falling freely, water does not leak.

Hence ‘C’ is the correct option

P 1;Q 1;R 1;S 4




PART II: CHEMISTRY SECTION – 1

(Only One Option Correct Type) This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE option is correct. 21. The product formed in the reaction of 2SOCl with white phosphorous is A) 3PCl

B) 2 2SO Cl

C) 2SCl D) 3POCl Key: A

Sol: 4 2 3 2 2 28 4 4 2 P SOCl PCl SO S Cl

22. The major product in the following reaction is

[Figure]

Cl

3CH

O1 dry ether, 03

2 aq. acid.CH MgBr, C

.

A) 3H C

3CH

O

B) 2H C

3CH

OH

3CH

C) O

2CH

D) O

3CH

3CH




Key: D

Sol:

ClO

3CH MgBr

ClO-

o

23. Hydrogen peroxide in its reaction with 4KIO and 2NH OH respectively, is acting as a

A) reducing agent, oxidising agent

B) reducing agent, reducing agent

C) oxidising agent, oxidising agent

D) oxidising agent, reducing agent

Key: A

Sol: 2 2 2 2 22 4 NH OH H O N H O

24. Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is

CompleteHydrolysis

6 other productXeF P

2OH / H O

Q

2slow disproportionation in OH / H O

products A) 0 B) 1 C) 2 D) 3 Key: C

Sol: 3 2 2 4 6 24 6 .8 XeO NaOH H O Xe O Na XeO H O

Gaseous products are Xe and 2O




25. The acidic hydrolysis of ether (X) shown below is fastest when

[Figure]

acid OH ROH

X

OR

A) one phenyl group is replaced by a methyl group.

B) one phenyl group is replaced by a para-methoxyphenyl group.

C) two phenyl groups are replaced by two para-methoxyphenyl groups.

D) no structural change is made to X.

Key: D

Sol: Additional stability of carbocation is attained by 4 methoxy phenyl substitution than

simple phenyl substitution.

26. Isomers of hexane, based on their branching, can be divided into three distinct classes

as shown in the figure.

[Figure]

andI andII

III

The correct order of their boiling point is

A) I > II > III

B) III > II > I

C) II > III > I

D) III > I > II




Key: B

Sol: Boiling points of isomeric alkanes are dependent on surface area [branching decreases

surfaces area increases ] with increasing branching bp decreases

So order is III>II>I

27. For the identification of -naphthol using dye test, it is necessary to use

A) dichloromethane solution of -naphthol.

B) acidic solution of -naphthol.

C) neutral solution of -naphthol.

D) alkaline solution of -naphthol.

Key: D

Sol: Alkaline solution of -naphthol increases the electron density at position and

thereby it enhances the rate of reaction .

28. Assuming 2 2s p mixing is NOT operative, the paramagnetic species among the

following is

A) 2Be

B) 2B

C) 2C

D) 2N Key: C

Sol: If there is no s-p mixing M.O electro configuration of 2C is

2 * 2 2 2 1 11 1 2 2 2z x ys s s sp p p

So 2C is paramagnetic




29. For the elementary reaction M N , the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of the reaction with respect to M is

A) 4 B) 3 C) 2 D) 1 Key: B

Sol: nr kM 30. For the process

2 2H O l H O g at 0100T C and 1atmosphere pressure, the correct choice is

A) 0systemS and 0surroundingsS

B) 0systemS and 0surroundingsS

C) 0systemS and 0surroundingsS

D) 0systemS and 0surroundingsS

Key: B

Sol: qST

SECTION – 2

Comprehension Type (Only One Option Correct) This section contains 3 paragraphs, each describing theory, experiments, data etc.Six questions relate to the three paragraphs with two questions on each paragraph. Each question has only ONE correct answer among the four given options (A), (B), (C)and(D)

Paragraph For Questions 31 and 32 Schemes 1 and 2 describe sequential transformation of alkynes M and N. Consider

only the major products formed in each step for both the schemes.

HO

H1 (excess)22.CH (1 equivalent)3 2

3 (1 equivalent)34 Lindlar's catalyst2

Scheme-1.NaNH

CH I.CH I.H .

X

M




H

21 2equivalent.NaNH

OHBr2.

33 mild.H O.

24.H ,Pd / C

35.CrO

Y Scheme-2

N

31. The product X is

A) 3H CO

H H

B)

3H CO

H

H

C) 3 2CH CH O

H H

D) 3 2CH CH O

H

H

Key: A

Sol: Since acetalyde carbanion is more nucleophilic in nature it firstly react with ethyl

iodide and give the product accordingly

32. The correct statement with respect to product Y is

A) It gives a positive Tollens test and is a functional isomer of X.

B) It gives a positive Tollens test and is a geometrical isomer of X.

C) It gives a positive iodoform test and is a functional isomer of X.

D) It gives a positive iodoform test and is a geometrical isomer of X.




Key: C

Sol: As 2 /H Pd C reduces triple bond to single bond in absence of poison catalyst like

Barium sulphate and 3CrO oxidises secondary alcohol to ketone finally we get the

functional isomer of alcohol which gives iodoform test

Paragraph For Questions 33 and 34

An aqueous solution of metal ion M1 reacts separately with reagents Q and R in

excess to give tetrahedral and square planar complexes, respectively. An aqueous

solution of another metal ion M2 always forms tetrahedral complexes with these

reagents.

Aqueous solution of M2 on reaction with reagent S gives white precipitate which

dissolves in excess of S. The reactions are summarized in the scheme given below:

SCHEME:

Tetrahedral Q Rexcess excessM1 Squareplanar

Tetrahedral Q Rexcess excessM2 Tetrahedral

S, stoichiometric amount

White precipitate Sexcess precipitatedissolves

33. M1, Q and R, respectively are

A) 2Zn , KCN and HCl B) 2 ,Ni HCl and KCN C) 2 ,Cd KCN and HCl

D) 2 ,Co HCl and KCN

Key: B




34. Reagent S is

A) 4 6K Fe CN

B) 2 4Na HPO

C) 2 4K CrO

D) KOH Key:-D

Sol: 33 & 34

2Ni with Cl gives tetrahedral 24NiCl and with CN gives square planar

24

Ni CN

2Zn gives tetrahedral complexes with both Cl and CN

2Zn gives white ppt with KOH but dissolve in excess


X and Y are two volatile liquids with molar weights of -110g mol and 140g mol

respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are

simultaneously placed at the ends of a tube of length 24L cm , as shown in the

figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature

of 300K. Vapours of X and Y react to form a product which is first observed at a

distance d cm from the plug soaked in X. Take X and Y to have equal molecular

diameters and assume ideal behaviour for the inert gas and the two vapours.

Initial formation ofthe product

Cotton woolsoaked in Y

Cotton woolsoaked in X d

24L cm




35. The value of d in cm (shown in the figure), as estimated from Graham’s law is

A) 8

B) 12

C) 16

D) 20

Key: C

Sol: 4024 10

d

d

36. The experimental value of d is found to be smaller than the estimate obtained using

Graham’s law. This is due to

A) larger mean free path for X as compared to that of Y.

B) larger mean free path for Y as compared to that of X.

C) increased collision frequency of Y with the inert gas as compared to that of X

D) increased collision frequency of X with the inert gas as compared to that of Y with

the inert gas.

Key: D

Sol: Collision frequency increases then d decreases

SECTION -3

Matching List Type (Only One Options Correct)

This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List-I and List-II are given as options (A), (B), (C) and (D), out of which ONE is correct 37. Different possible thermal decomposition pathways for peroxyesters are shown below.

Match each pathway from List -1 with an appropriate structure from List-II and select the

correct answer using the code given below the lists.




OO

R'R

O

Peroxyester

P

2-CO R R' O

Q

2-CO R R' O carbonyl compoundR X

R2RCO R' O

2-CO carbonyl compoundR X '

S2RCO R' O

2-CO R R' O

List-I List-II

P. Pathway P 1. 6 5 2C H CH O O3CH

O

Q. Pathway Q 2. 6 5C H O O

3CH

O

R. Pathway R 3.

6 5 2C H CH O O 3CHO

3CH2 6 5CH C H

S. Pathway S 4.

6 5C H O O 3CHO

3CH6 5C H

Codes:

P Q R S

A) 1 3 4 2

B) 2 4 3 1

C) 4 1 2 3

D) 3 2 1 4




Key: A

Sol: P

6 5 2 3 2 3 21C H CH C O O CH Ph CH CH O CO

O

Q

6 5 2 3 2 3 23CH CH C O O C CH Ph CH O C CH CO

O CH3

CH2Ph

CH3

CH2Ph

2Ph CH

O

R

6 5 2 3 34CH CH C O O C CH Ph C O O C CH

O CH3

Ph

CH3

Ph

Ph

O

OH

+

S

6 5 3 6 5 32C H C O O CH C H C O CH O

O

Ph

O2CO

38. Match the four starting materials (P, Q, R, S) given in List-1 with the corresponding reaction

schemes (I, II, II, IV) provided in List-II and select the correct answer using the code given

below the lists.




List-I List-II

P. H H 1. Scheme I

4 22 3

( ) , , ,

7 6 2 3?i KMnO HO heat ii H H Oiii SOCl iv NH C H N O

Q.

OH

OH 2. Scheme II

3 2 4

3 2 4

( ) / .. ,

6 6 2 2?i Sn HCl ii CH COCl iii conc H SOiv HNO v dil H SO heat vi HO C H N O

R.

NO2

3. Scheme III

3 2 4

2 3 2 2 4

,873 fuming HNO , ,. ,

6 5 3?i red hot iron K ii H SO heatiii H S NH iv NaNO H SO v hydrolysis C H NO

S.

NO2

CH3 4. Scheme IV

2 4

3 2 4 2 4

. ,60. , . .

6 5 4?i conc H SO Cii conc HNO conc H SO iii dil H SO heat C H NO

Codes:

P Q R S

A) 1 4 2 3

B) 3 1 4 2

C) 3 4 2 1

D) 4 1 3 2




Key: C

Sol: 3P

3/873

redHotFe K

min/ /3 2 4

Fu gHNO H SO

2NO

2NO

/3 2NH H S2NO

2NH

2 /NaNO HCl Hydrolysis

2NO

OH

6 5 3C H NO

4Q

OH

oy

2 4060

ConH SOC

OH

3SOH

2 43

ConH SOHNO

2NO

3SOH

OH

2 4

dilH SO

OH

OH

2NO

6 5 4C H NO

OH OH

R

2NO

(2) /Sn HCl2NH

3 CH C Cl

O NH

O

3CH

2 4 /ConH SO

NH

O

3CH

3SOH

3HNO

NH

O

3CH

3SOH

2 4dilH SO2NO

3SOH

2NH2NO

OH

2NH2NO

6 6 2 2C H N O

2

S 1

2NO

3CH

/ /4

KMnO OH Heat

2NO

2CO H

2SOCl

2NO

COCl

3NH

2NO

2CONH




39. Match each coordination compound in List-I with an appropriate pair of characteristics from

List-II and select the correct answer using the code given below the lists.

2 2 2 2en=H NCH CH NH ;atomicnumbers:Ti=22; 24;Co=27;Pt=78Cr

List-I List-II

P. 3 24Cr NH Cl Cl 1. Paramagnetic and exhibits ionisation

isomerism

Q. 2 35 2Ti H O Cl NO 2. Diamagnetic and exhibits cis-trans

isomerism

R. 3 3Pt en NH Cl NO 3. Paramagnetic and exhibits cis-trans

isomerism

S. 3 3 34 2Co NH NO NO 4. Diamagnetic and exhibits ionisation

isomerism

Codes:

P Q R S

A) 4 2 3 1

B) 3 1 4 2

C) 2 1 3 4

D) 1 3 4 2

Key: B

Sol: 3 24 Cr NH Cl Cl exhibit paramagnetic and geometrical isomerism.

2 35 2 Ti H O Cl NO exhibit paramagnetism and ionisation isomerism

3 3 Pt en NH Cl NO exhibit diamagnetism and ionsation isomers

3 3 34 2 Co NH NO NO exhibit diamagnetism and as-trans isomers




40. Match the orbital overlap figures shown in List-I with the description given in List-II and

select the correct answer using the code given below the lists.

List-I List-II

P. 1. p d antibonding

Q. 2. d d bonding

R. 3. p d bonding

S. 4. d d antibonding

Codes:

P Q R S

A) 2 1 3 4

B) 4 3 1 2

C) 2 3 1 4

D) 4 1 3 2

Key: C




List –I List-II

Sol: P. d-d (possible ovelap)

Q. p-d (possible-side wise where so p-d bond )

R. p-d (Negative side wise overlap)

S. d-d (Negative overlap) antibonds d-d anti bond




PART –III_MATHEMATICS SECTION-1

(Only One Option Correct Type) This section contains 10 multiple choice questions. Each question has four choices A,B,C and D out of which ONLY ONE is correct. 41. Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in

envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be done is

A) 264 B) 265 C) 53 D) 67 Key: C

Sol: Req. no. of ways = 41

1 1 1 34 2 1 2 532 3 4 4

C

42. In a triangle the sum of two sides is ‘x’ and the product of the same two sides is ‘y’. If 2 2 , x c y where ‘c’ is the third side of the triangle, then the ratio of the in-radius to

the circum-radius of the triangle is

A) 3

2 y

x x c

B) 3

2 y

c x c

C) 3

4 y

x x c

D) 3

4 y

c x c

Key: B

Sol: 2 2 2 1 22 2 2 cos4 3

a b c ab s s c ab c c

1 sin 321 22 3

ab Cr ycR sR c x cc x




43. The common tangents to the circle 2 2 2 x y and the parabola 2 8y x touch the

circle at the points P, Q and the parabola at the points R, S. Then the area of the

quadrilateral PQRS is

A) 3

B) 6

C) 9

D) 15

Key: D

Sol: 2 2 22 tan 8 2 Let y mx be a gent to y x if also touches x ym

1 m

2, 4 , 1, 1 R P

Req. area = (1 + 4) (1 + 2) = 15

44. Three boys and two girls stand in a queue. The probability, that the number of boys

ahead of every girl is at least one more than the number of girls ahead of her, is

A) 12

B) 13

C) 23

D) 34

Key: A

Sol: Required probability = 5 3! 2!5!

45. The quadratic equation 0p x with real coefficients has purely imaginary roots.

Then the equation 0p p x has

A) Only purely imaginary roots B) All real roots C) two real and two purely imaginary roots D) neither real nor purely imaginary roots Key: D Sol: Let p(x) = 2 4 21 2 2 x p p x x x

102

iIf p p x then x

Hence 0p p x has neither real nor purely imaginary roots




46. For 0, x , the equation sin 2sin 2 sin 3 3 x x x has

A) infinitely many solutions

B) three solutions

C) one solution

D) no solution

Key: D

Sol: Apply boundary conditions

47. The following integral

2

17

4

2cos

ec x dx is equal to

A) log 1 2 16

02

u ue e du

B) log 1 2 17

0

u ue e du

C) log 1 2 17

0

u ue e du

D) log 1 2 16

02

u ue e du

Key: A

Sol: Let cos2

u ue eecx

cos cot2

u ue eecx x dx du

2

u ududx

e e

021717

log 2 14

22cos

u uu u

duec x dx e ee e

log 2 1

16

0

2

u ue e du




48. Coefficient of 11x in the expansion of 4 7 122 3 41 1 1 x x x is

A) 1051

B) 1106

C) 1113

D) 1120

Key: C

Sol: 4 7 124 7 122 3 4 4 2 7 3 12 4

0 0 0

1 1 1

s r ts r t

s r t

x x x C x C x C x

2 3 4 11 Hence s r t

Req. value is 4 7 12 4 7 12 4 7 12 4 7 122 1 1 0 1 2 4 1 0 1 3 0. . . . . . . . 1113 C C C C C C C C C C C C

49. Let : 0, 2 f be a function which is continuous on [0, 2] and is differentiable on

(0, 2) with f(0) = 1. Let 2

0

x

F x f t dt

0, 2 . ' ' 0, 2 , 2 for x If F x f x for all x then F equals

A) 2 1e

B) 4 1e

C) 1e

D) 4e

Key: B

Sol: ' .2 ' F x f x x f x

2'2 log 0 0 1

f xx f x x c c f

f x

4 ' ' 1 0 0 f x e and F x f x F x f x C F x f x F

42 2 1 1 F f e




50. The function y f x is the solution of the differential equation

4

2 2

21 1

dy xy x xdx x x

in (-1, 1) satisfying f(0) = 0. Then

32

32

f x dx is

A) 33 2

B) 33 4

C) 36 4

D) 36 2

Key: B

Sol: 2 21. 1 . 1, 1 x dx

xI F e x x

5

2 4 21 2 , 0 0 05

xy x x x dx x c but c t

5 2

2 25 1 1

x xf xx x

3 3 322 2 2

1 2 1

2003

2

1 1 12 2 sin 1 sin2 21

x xf x dx dx x x xx

3 323 8 6 3 4




SECTION – 2

Comprehension Type (Only One Option Correct) This section contains 3 paragraphs, each describing theory, experiments, data etc .Six questions relate to the three paragraphs with two questions on each paragraph. Each question has only ONE correct answer among the four given options (A), (B), (C)and(D)


Box 1 contains three cards bearing numbers 1, 2, 3; box 2 contains five cards bearing

numbers 1, 2, 3, 4, 5; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6,

7. A card is drawn from each of the boxes. Let ix be the number on the card drawn

from the thi box, i = 1, 2, 3

51. The probability that 1 2 3 x x x is odd, is

A) 29105

B) 53105

C) 57105

D) 12

Key: B

Sol: Required probability 2 3 4 1 2 4 1 3 3 2 2 3 533 5 7 3 5 7 3 5 7 3 5 7 105

52. The probability that 1 2 3, ,x x x are in an arithmetic progression, is

A) 9105

B) 10105

C) 11105

D) 7105




Key: C

Sol: 2 1 32 x x x

Combinations are 1,1,1 2,2,2 3,3,3 1,2,3 3,2,1 2,3,4 3,4,5 3,5,7 2,4,6 1,3,5 1,4,7

Required probability 11105


Let a, r, s, t be non-zero real numbers. Let 2 2 2, 2 , , , 2 , 2P at at Q R ar ar and S as as

be distinct points on the parabola 2 4y ax . Suppose that PQ is the focal chord and lines QR and PK are parallel, where ‘K’ is the point (2a, 0).

53. The value of ‘r’ is

A) 1

t

B) 2 1t

t

C) 1t

D) 2 1t

t

Key: D

Sol: Given that 1 1 t and 2

121

2 2 2 1 12 2

at tt r t r t

t at a t t t

54. If st = 1, then the tangent at ‘P’ and the normal at S to the parabola meet at a point whose ordinate is

A) 22

3

12tt

B) 22

3

12a tt

C) 22

3

1a tt

D) 22

3

2a tt




Key: B

Sol: Equation of tangent at P is 2 xyt x at y att

&

Equation of normal at S is 3 32 2 xy sx as as at sax as ast

But 23 2

2 21 2 x a a ast at a att t t t

222

2

112 2 2 3

a tat ay a att t


Given that for each 0, 1a 1

1

0lim 1

haa

hh

t t dt

exists. Let this limit be g(a). In addition, it is given that the function g(a) is differentiable on (0, 1).

55. The value of 12

g is

A) B) 2

C) 2

D) 4

Key: A

Sol: 1

1

0

1 aag a t t dt

1

0

12 1

dtg

t t

56. The value of 1'2

g is

A) 2

B)

C) 2

D) 0




Key: D

Sol: 1

1

0

1 aag a t t dt

1' 02

g

SECTION-3

Matching List Type (Only one Option Correct)

This section contains four questions, each having two matching lists. Choices for the correct

combination of elements from List-II are given as options (A), (B), (C) and (D), out of which ONE

is correct.

57. List-I List-II

P. The number of polynomials f x with non-negative integer 1. 8

Coefficients of degree 2 , satisfying 0 0f and

1

01, f x dx is

Q. The number of points in the interval 13, 13 at which 2. 2

2 2sin cos f x x x attains its maximum value, is

R.

22

2

31 x

x dxe

equals 3. 4

S.

121212

0

1cos 2 log1

1cos 2 log1

xx dxx

xx dxx

equals 4. 0

P Q R S

(A) 3 2 4 1

(B) 2 3 4 1

(C) 3 2 1 4

(D) 2 3 1 4




Key: D Sol: P) Let 2 f x a x bx

1 2 3 6 0, 2 3,03 2

a b a b

No. of 2f x

Q) 22 sin4

f x x from the graph of ,f x no. of max is 4

R) 2 2 22 2

2

2 2 2

3 3 2 3 81 1

x x

x xI dx dx I x dx Ie e

S)

12

12

1cos 2 log 01

xx dxx

58. List-I List-II

P. Let 1 3cos 3cos , 1,1 ,2

y x x x x .Then 1. 1

22

2

1 1

d y x dy xx x

y x dx dx equals

Q. Let 1 2, ,... 2nA A A n be the vertices of a regular polygon 2. 2

of n sides with its centre at the origin. Let

ka be the position

vector of the point , 1,2,...,kA k n . If

1 11 11 1

. ,

n nk k k kk k

a a a a then the minimum value of n is

R. If the normal from the point ,1P h on the ellipse 2 2

16 3

x y 3. 8

is perpendicular to the line 8, x y then the value of h is

S. Number of positive solutions satisfying the equation 4. 9

1 1 12

1 1 2tan tan tan2 1 4 1

x x x is

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




Key: A

Sol: P) 1 1 2 2 2112 2

3cos 3cos 1 9 11 1

yy x y x yy x

2 21 2 1 12 1 11 2 18 y y xy yy

22 11 9 y x xy y

Q) 1 2 2 3 1 1 2 2 3 1...... . . ...... . n n n na a a a a a a a a a a a

2 21 sin 1 cos

n nn n

2tan 1 n

2tan 1 n

2 3,4 4

or

n

2 1 34 4 4 or

8n

R) Equation of normal 6 3 31

x y

h

Slope 6 / 23

h

h but 2 1 2 h

h

S) 2

1 112 1 4 1

1 112 1 4 1

x xx

x x

2 2

2 3 1 28 6

x

x x x

3 1 18 6

xx x

23 7 6 0 x x

3x only one solution




59. Let 1 2 3: , :[0, ) , : f f f and 4 : [0, ) f be defined by

10,0;

x

x if xf x

e if x

22 ;f x x

3

sin 0,0

x if xf x

x if x

And

2 1

42 1

0,

1 0.

f f x if xf x

f f x if x

List-I List-II

P. 4f is 1. Onto but not one-one

Q. 3f is 2. Neither continuous nor one-one

R. 2 1f of is 3. Differentiable but not one-one

S. 2f is 4. Continuous and one-one

P Q R S

(A) 3 1 4 2

(B) 1 3 4 2

(C) 3 1 2 4

(D) 1 3 2 4




Key: D

Sol :

1

, 0, 0

x

x xf x

e x

y = -xy = ex

2 f x

y

x

2y x

3f x

siny x y x

2

4 2

, 01, 0

x

x xf xe x

2y x 2 1 xy e

p) From the graph range of 4 [0, ) f , but not 1 – 1 q) from graph range is not R, but not 1 – 1 and continuous, differentiable at ‘0’ r) 2

2 1 , 0 f f x x 2 , 0xe x 0at x L. L = 0, R.L = 1 Disc at x = 0, Clearly from graph not 1 – 1 s) Clearly from graph one one and continuous




60. Let 2 2cos sin ; 1, 2,...,910 10

kk kZ i k .

List-I List-II

P. For each kZ there exists a jZ such that . 1k jZ Z 1. True

Q. There exists a 1,2,....,9k such that 1 . kZ Z Z has no 2. False

solution Z in the set of complex numbers.

R. 1 2 91 1 ... 110

z z z equals 3. 1

S. 9

1

21 cos10

k

k equals 4. 2

P Q R S A) 1 2 4 3

B) 2 1 3 4

C) 1 2 3 4

D) 2 1 4 3

Key: C Sol: 1, 1 2 3 9, , ,....z z z z are 10th roots of units

p) 10

2 10 2 1c s10 10

k k

k

k kz i cis zz

, . 1 k j k jfor each z a z such that z z

q) If 11

. kk

zz z z zz

19 1 2 9 10

. , .... ,1

kk

z z z z z z z z

r) 1 2 3 91 , 1 , 1 ,... 1 z z z z are the roots of 101 1 0 x

10 10 91 1 10 ..... 10 0 x x x x 1 2 91 1 ...... 1 z z z = product of roots for 9 910 .... 10 0 x x s) sum of 10th roots of units = 0

9

1

21 010

k

kcis

9

1

2 110

k

kcos

Ans = 2

sri chaitanya iit academy., india...email: #corporate office: plot no: 304, kasetty heights, ayyappa...

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