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Amity Institute for Competitive Examinations : Phones: 24336143/44, 25573111/2/3/4, 95120-2431839/42- 1 -

AMITYInstitute for Competitive Examinations

Paper - II Code-2Time: 3 Hours Maximum Marks: 198Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

INSTRUCTIONSA. General:

1. This booklet is your Question Paper. Do not break the seals of this booklet before being instructed to doso by the invigilators.

2. The question paper CODE is printed on the right hand top corner of this sheet and also on the back page(page no. 28) of this booklet.

3. Blank spaces and blank pages are provided in this booklet for your rough work. No additional sheets willbe provided for rough work.

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5. Answers to the questions and personal details are to be filled on a two-part carbon-less paper, which isprovided separately. You should not separate these parts. The invigilator will separate them at the end ofexamination. The upper sheet is a machine-gradable Objective Response Sheet (ORS) which will betaken back by the invigilator. You will be allowed to take away the bottom sheet at the end of the examination.

6. Using a black ball point pen, darken the bubbles on the upper original sheet. Apply sufficientpressure so that the impression is created on the bottom sheet.

7. DO NOT TAMPER WITH/MUTILATE THE ORS OR THIS BOOKLET.8. On breaking the seals of the booklet check that it contains 28 pages and all the 60 questions and

corresponding answer choices are legible. Read carefully the instructions printed at the beginning ofeach section.

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11. Write your Name, Registration No. and the name of examination centre and sign with pen in the boxesprovided on the right part of the ORS. Do not write any of this information anywhere else. Darken theappropriate bubble UNDER each digit of your Registration Number in such a way that the impression iscreated on the bottom sheet. Also darken the paper CODE given on the right side of ORS (R4).

C. Question paper formatThe question paper consists of 3 parts (Physics, Chemistry and Mathematics). Each part consists of threesections.12. In Section I contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D)

out of which ONLY ONE is correct.13. In Section II contains 3 Paragraphs each describing theory, experiment, data etc. There are 6 multiple

choice questions relating to three paragraphs with 2 questions on each paragraph. Each question of aparticular paragraph has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

14. In Section III contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and(D) out of which ONLY ONE or MORE are correct.

D. Marking Scheme15. For each question in Section I and Section II, you will be awarded 3 marks if you darken the bubble

corresponding to the correct answer ONLY and zero (0) marks if no bubbles are darkened. In all othercases, minus one (–1) mark will be awarded in these sections.

16. For each question in Section III, you will be awarded 4 marks if you darken ALL the bubble(s)corresponding to the correct answer(s) ONLY. In all other cases zero (0) marks will be awarded. Nonegative marks will be awarded for incorrect answer(s) in this section.


Physics, Chemistry & Mathematics (Paper-II) Code-2 IIT-JEE-2012


PART I : PHYSICS

SECTION- I (Total Marks : 24)Single Correct Answer Type

This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) outof which ONLY ONE is correct

1. An infinitely long hollow conducting cylinder with inner radius R/2 and outer R carries a uniform current

density along its length. The magnitude of the magnetic field, | |B

as a function of the radial distance r

from the axis is best represented by

(A)| |B

R/2 R r

(B)| |B

R/2 R r

(C)| |B

R/2 R r

(D)| |B

R/2 R r

Sol: [C] Factual

2. A thin uniform cylindrical shell, closed at both ends, is partially filled with water. It is floating vertically in

water in half-submerged state. If c is the relative density of the material of the shell with respect to

water, then the correct statement is that the shell is

(A) more than half-filled if c is less than 0.5 (B) more than half-filled if c is more than 1.0

(C) half-filled if c is more than 0.5 (D) less than half-filled if c is less than 0.5

Sol: [D] Since, cylinder is half submerged

it must be filled less than half for any value of C.

3. In the given circuit, a charge of +80 C is given to the upper

plate of the 4 F capacitor. Then in the steady state, the

charge on the upper plate of the 3 F capacitor is

(A) + 32 C (B) + 40 C

(C) + 48 C (D) + 80 C




Sol: [C] 1 2 80q q ...(i)

1

2

23

qq

+q2

–q2

+q2

–q2

Solving q2 = 48 C

4. Two moles of ideal helium gas are in a rubber balloon at 30°C. The balloon is fully expandable and can

be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly

changed to 35°C. The amount of heat required in raising the temperature is nearly

(take R = 8.31 J/mol.K)

(A) 62 J (B) 104 J (C) 124 J (D) 208 J

Sol: [D] Process is isobaric

Q = nCpT

52 8.31 52

= 208 J

5. Consider a disc rotating in the horizontal plane with a constant angular speed about its centre O. The

disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown

in the figure. When the disc is in the orientation as shown, two pebbles P and Q are simultaneously

projected at an angle towards R. The velocity of projection is in the y-z plane and is same for both

pabbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed

18

rotation, (ii) their range is less than half the disc radius, and (iii) remains constant throughout. Then

(A) P lands in the shaded region and Q in the unshaded region

(B) P lands in the unshaded region and Q in the shaded region

(C) Both P and Q land in the unshaded region

y

x

R

QO

P(D) Both P and Q land in the shaded region.

Sol: [C] Horizontal velocity of P is more than velocity of any point of disc lying between P and O. Also,

horizontal velocity of Q is less than velocity of any point of disc between QR.

6. A student is performing the experiment of Resonance Column. The diameter of the column tube is 4 cm.

The frequency of the tuning fork is 512 Hz. The air temperature is 38°C in which the speed of sound is

336 m/s. The zero of the meter scale coincides with the top end of the Resonance Column tube. When

the first resonance occurs, the reading of the water level in the column is

(A) 14.0 cm (B) 15.2 cm (C) 16.4 cm (D) 17.6 cm




Sol: [B] 4( )vf

L e

2 2

336 3365124[ 4] 10 4[ 0.3 4] 10L L

Solving L = 16.414 – 1.2 = 15.2 cm

7. Two identical discs of same radius R are rotating about their axes in opposite directions with the sameconstant angular speed . The discs are in the same horizontal plane. At time t = 0, the points P and Qare facing each other as shown in the figure. The relative speed between the two points P and Q is vr. Inone time period (T) of rotation of the discs, vr as a function of time is best represented by

R RQP

(A)

tTO

vr

(B)

tTO

vr

(C)

tTO

vr

(D)

tTO

vr

Sol: [A] Relation velocity will be maximum twice is one time period in following situation,

P Q

P Q

8. A loop carrying current I lies in the x-y plane as shown in the

figure. The unit vector k̂ is coming out of the plane of the

paper. The magnetic moment of the current loop is

(A) 2 ˆa Ik (B) 2 ˆ12

a Ik

I

a

y

xa

a(C) 2 ˆ12

a Ik

(D) 2 ˆ(2 1) a Ik




Sol: [B] Magnetic momentˆM Aik

22 ˆ

2a a i k

= 2 ˆ1

2a i k

SECTION- II (Total Marks - 15)Paragraph Type

This Section contains 6 multiple choice questions relating to three paragraph with two questions on eachparagraph. Each questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Question 9 to 10The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre ofmass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass.These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed)horizontally at its rim to a massless stick, as shown in the figure. When the disc-stick system is rotatedabout the origin on a horizontal frictionless plane with angular speed , the motion at any instant can betaken as a combination of (i) a rotation of the centre of mass of the disc about the z-axis, and (ii) arotation of the disc through an instantaneous vertical axis passing through its centre of mass (as is seenfrom the changed orientation of points P and Q). Both these motions have the same angular speed inthis case.

P QQ P y

z

xNow consider two similar systems as shown in the figure: Case (a) the disc with its face vertical andparallel to x-z plane; Case (b) the disc with its face making an angle of 45° with x-y plane and itshorizontal diameter parallel to x-axis. In both the cases, the disc is welded at point P, and the systems arerotated with constant angular speed about the z-axis.

P

Q

y

z

x

P

Q

y

z

x

45°

Case (a) Case (b)

9. Which of the following statements regarding the angular speed about the instantaneous axis (passing

through the centre of mass) is correct?

(A) It is 2 for both the cases (B) It is for case (a) and 2

for case (b)

(C) It is for case (a) and 2 for case (b) (D) It is for both the cases




Sol: [D]

P

Q

45°

Instantaneous axis Instantaneous axis

10. Which of the following statements about the instantaneous axis (passing through the centre of mass) is

correct?

(A) It is vertical for both the cases (a) and (b)

(B) It is vertical for case (a) and is at 45° to the x-z plane and lies in the plane of the disc for case (b)

(C) It is horizontal for case (a) and is at 45° to the x-z plane and is normal to the plane of the disc for

case (b)

(D) It is vertical for case (a) and is at 45° to the x-z plane and is normal to the plane of the disc for case

(b)

Sol: [A] Factual

Paragraph for Question 11 to 12

The -decay process, discovered around 1900, is basically the decay of a neutron (n). In the laboratory,a proton (p) and an electron (e–) are observed as the decay products of the neutron. Therefore, consideringthe decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energyof the electron should be a constant. But experimentally, it was observed that the electron kinetic energyhas a continuous spectrum. Consider a three-body decay process, i.e. n p + e + ev , around 1930,Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino ( )ev to be masslessand possessing negligible energy, and the neutron to be at rest, momentum and energy conservationprinciples are applied. From this calculation, the maximum kinetic energy of the electron is 0.8 × 106 eV.The kinetic energy carried by the proton is only the recoil energy.

11. If the anti-neutrino has a mass of 3 eV/c2 (where c is the speed of light) instead of zero mass, what

should be the range of the kinetic energy, K, of the electron?

(A) 60 0.8 10 eVK (B) 63.0eV 0.8 10 eVK

(C) 63.0eV 0.8 10 eVK (D) 60 0.8 10 eVK

Sol: [C] Even if neutrino is at rest, its energy is moc2

Maximum kinetic energy of electron is < 0.8 × 106 eV

Also, antineutrino cannot take whole energy




12. What is the maximum energy of the anti-neutrino?

(A) zero (B) much less than 0.8 × 106 eV

(C) nearly 0.8 × 106 eV (D) much larger than 0.8 × 106 eV

Sol: [C]


Most materials have the refractive index, n > 1. So, when a light ray from air enters a naturally occurring

material, then by Snell’s law, 1 2

2 1

sinsin

nn

, it is understood that the refracted ray bends towards the

normal. But it never emerges on the same side of the normal as the incident ray. According to

electromagnetism, the refractive index of the medium is given by the relation r rcnv

, where

c is the speed of electromagnetic waves in vacuum, v its speed in the medium, r and r are the relativepermittivity and permeability of the medium respectively.

In normal materials, both r and r are positive, implying positive n for the medium. When both r

and r are negative, one must choose the negative root of n. Such negative refractive index materialscan now be artificially prepared and are called meta-materials. They exhibit significantly different opticalbehavior, without violating any physical laws. Since n is negative, it results in a change in the direction ofpropagation of the refracted light. However, similar to normal materials, the frequency of light remainsunchanged upon refraction even in meta-materials.

13. For light incident from air on a meta-material, the appropriate ray diagram is

(A)

Air

Meta-material (B)

Air

Meta-material (C)

Air

(D)Air

Meta-material

Sol: [C] Angle of refraction will be negative in case n is negative.

14. Choose the correct statement:

(A) The speed of light in the meta-material is | |v c n

(B) The speed of light in the meta-material is | |cvn

(C) The speed of light in the meta-material is v = c

(D) The wavelength of the light in the meta-material (m) is given by m = air | n |, where air is the

wavelength of the light in air.

Sol: [B] Factual




SECTION- III (Total Marks : 16)(Multiple Correct Answer (s) Types)

This Section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) outof which ONE or MORE may be correct.

15. A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touchingit. The correct statement (s) is (are)

(A) The emf induced in the loop is zero if the current is constant

(B) The emf induced in the loop is finite if the current is constant

(C) The emf induced in the loop is zero if the current decreases at a steady rate

(D) The emf induced in the loop is finite if the curent decreases at a steady rate.

Sol: [A, C] Faraday’s law

16. Six point charges are kept at the vertices of a regular hexagon of side L and centre O, as shown in the

figure. Given that 2

14 o

qKL

, which of the following statement (s) is (are) correct?

(A) The electric field at O is 6K along OD

(B) The potential at O is zero

(C) The potential at all points on the line PR is same

F E

A D

L

C

S T

P

R

O

–q+q

+2q –2q

B

+q

(D) The potential at all points on the line ST is same

Sol: [A, B, C] Enet = 6K along OD

F E

A D

C–q+q

2q –2q

B

–q+q

KK

2K2K

K

K

17. Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from

the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q

has most of its mass concentrated near the axis. Which statement (s) is (are) correct?

(A) Both cylinders P and Q reach the ground at the same time

(B) Cylinder P has larger linear acceleration than cylinder Q

(C) Both cylinders reach the ground with same translation kinetic energy

(D) Cylinder Q reaches the ground with larger angular speed.




Sol: [D]2

sin

1c

g Qa ImR

Given p QI I

P QC Ca a

cylinder P will reach ground first.

Velocity at ground 2 2

21 /

ghvK R

p Qv v

p Q

18. Two spherical planets P and Q have the same uniform density , masses MP and MQ, and surface areas

A and 4A, respectively. A spherical planet R also has uniform density and its mass is (MP + MQ). The

escape velocities from the planets P, Q and R, are VP, VQ and VR, respectively. Then

(A) VQ > VR > VP (B) VR > VQ > VP (C) VR / VP = 3 (D) VP / VQ = ½

Sol: [B, D] 283ev GR

( )ev R

24 PA R 24 4 QA R

2

4Q

P

RR

or 2Q

P

RR

12

P

Q

vv

Also, 3 3 34 4 43 3 3R P QR R R

3 3 1 PR Q

Q

RR RR

3

8 QR

R QR R

R Q PR R R

R Q Pv v v




19. The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on

a horizontal surface with angular speed and (ii) an inner disc of radius 2R rotating anti-clockwise with

angular speed /2. The ring and disc are separated by frictionless ball bearings. The system is in the x-z

plane. The point P on the inner disc is at a distance R from the origin, where OP makes an angle of 30°

with the horizontal. Then with respect to the horizontal surface

(A) the point O has a linear velocity ˆ3R i

(B) the point P has a linear velocity 11 3 ˆˆ4 4

R i R k

Ox

z

30°3R R

2R

P(C) the point P has a linear velocity

13 3 ˆˆ4 4

R i R k

(D) the point P has a linear velocity 3 1 ˆˆ3

4 4R i R k

Sol: [A, B]

x

z

ˆ3ov R i

ˆˆ ˆ3 cos30 sin302 2p

R Rv R i k i

11 3 ˆˆ4 4R Ri k

20. In the given circuit, the AC source has = 100 rad/s. Considering the inductor and capacitor to be ideal,

the correct choice (s) is (are)

(A) The current through the circuit I is 0.3 A

(B) The current through the circuit, I is 0.32 A

100µF

0.5H

20V

I(C) The voltage across 100 resistor = 102 V

(D) The voltage across 50 resistor = 10 V




Sol: [A, C]1 100CXL

50lX C

1 1 1100 100 50 50Z j j

100µF

0.5H

I2

Solving 20 10Z

20 0.320 10

i A

2 21 (100) (100) 100 2Z

120 0.1 2

100 2I A

Voltage across 100 = 102 volt

Z2 = 502 220 0.2 2

50 2I

Voltage across 50 = 0.22 × 50 = 102

PART II : CHEMISTRY

SECTION- I (Single Correct Answer Type)This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) outof which ONLY ONE is correct

21. In the cyanide extraction process of silver from argentite ore, the oxidizing and reducing agents used are

(A) O2 and CO respectively (B) O2 and Zn dust respectively

(C) HNO3 and Zn dust respectively (D) HNO3 and CO respectively

Sol: [B] 2 2 2Ag S 4NaCN 2Na[Ag(CN) ] Na S

2 2 2 4oxidizing agent

Na S 2O Na SO

2 2 4reducing agent2Na[Ag(CN) ] Zn Na [Zn(CN) ] 2Ag

22. The reaction of white phosphorus with aqueous NaOH gives phosphine along with another phosphorus

containing compound. The reaction type; the oxidation states of phosphorus in phosphine and the other

product are respectively

(A) redox reaction; – 3 and –5 (B) redox reaction; +3 and +5

(C) disproportionation reaction; –3 and +5 (D) disproportionation reaction; –3 and +3




Sol: [D]1 3

4 2 2 2 3P 3NaOH 3H O 3NaH P O P H

23. The shape of XeO2F2 molecule is

(A) trigonal bipyramidal (B) square planar (C) tetrahedral (D) see-saw

Sol: [D]Xe

O

O

F

F

sp3d hybridised with one lone pair and shape see-saw

24. For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the

elevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than

the concentration of solvent, the vapour pressure (mm of Hg) of the solution is

(take Kb = 0.76 K kg mol–1)

(A) 724 (B) 740 (C) 736 (D) 718

Sol: [A] 2.5 gm of non-electrolyte

100 gm of H2O

Tb = 2 K

Tb = Kbm

A

2.5 10002 0.76M 100

MB = 9.50A A B B

0A A B A

p p n np n n n

(B = solute, A = solvent)

A760 p 2.5 18760 9.5 100

A760 p 36

Ap 760 36 724 mm of Hg

25. The compound that undergoes decarboxylation most readily under mild condition is

(A)

COOH

COOH (B)

COOH

O(C)

COOH

COOH(D)

CH2COOH

O

Sol: [B]

OHO

O due to keto group at -position




26. Usin the data provided, calculate the multiple bond energy (kg mol–1) of a C C bond in C2H2. That

energy is (take the bond energy of a C – H bond as 350 kJ mol–1)

2 2 22C(s) H (g) C H (g) H = 225 kJ mol–1

2C(s) 2C(g) H = 1410 kJ mol–1

2H (g) 2H(g) H = 330 kJ mol–1

(A) 1165 (B) 837 (C) 865 (D) 815

Sol: [D] 2 2 22C(s) H (g) C H (g) H = 225 kJ mol–1

2C(s) 2C(g) H = 1410 kJ mol–1

2H (g) 2H(g) H = 330 kJ mol–1

B.E. of C – H = 350

Hf (C2H2) = (B.E.Reactant – B.E.Product)

225 = 1410 + 330 – 350 × 2 – B.E. (C C)

B.E. (C C) = 1410 + 330 – 700 – 225 = 815

27. The major product H of the given reaction sequence is

2 4CN 95% H SO3 2 3 HeatCH CH CO CH

G H

(A)

CH3 CH

C COOH

CH3

(B)

CH3 CH

C CN

CH3

(C) CH3 CH2 C COOH

CH3

OH

(D)CH3 CH C C

CH3O

NH2

Sol: [D]CH3

O

CH3 CN

CH3

CH3

CN

OH 2H OCH3

CH3

OH

NH2 O

2 4conc. H SO (95%)

CH3 CH3

NH2

O

28. NiCl2[P(C2H5)2(C6H5)]2 exhibits temperature dependent magnetic behavior (paramagnetic /

diamagnetic). The coordination geometries of Ni2+ in the paramagnetic and diamagnetic states are

respectively

(A) tetrahedral and tetrahedral (B) square planar and square planar

(C) tetrahedral and square planar (D) square planar and tetrahedral

Sol: [C] 8 2Ni(28) : 3d .4s

2 8 0Ni : 3d .4s




If it is paramagnetic, then

sp3 tetrahedralIf it is diamagnetic, then

dsp2 square planar

SECTION- II (Paragraph Type)This Section contains 6 multiple choice questions relating to three paragraphs with two questions on eachparagraph. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE iscorrect.


In the following reaction sequence, the compound J is an intermediate.

3 2 2

3 23

(CH CO) O i) H ,Pd / CCH COONa ii) SOCl

iii) anhyd. AlCl I J K

J (C9H8O2) gives effervescence on treatment with NaHCO3 and a positive Baeyer’s test.

29. The compound K is

(A) (B)O O

(C)O

(D) O

Sol: [C]

30. The compound I is

(A)

HO

(B)

OH

(C)

CH3O

(D)

HH

H

Sol: [A]

29 & 30O

O

CH3

O

CH2 H3CH COO O

O

CH3

O

CH2-

enolate




HO

O

O

CH2-

O

CH3enolate

OH O

O

CH3

O2H O/

OH

O

2 3H O CH COOH

(I) (J)

OH

O2H ,Pd / C

OH

O2SOCl

Cl

O3anhy. AlCl

O(K)(J)


The electrochemical cell shown below is a concentration cell.

M | M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+(0.001 mol dm–3) | M

The emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The

emf of the cell at 298 K is 0.059 V.

31. The value of G (kJ mol–1) for the given cell is (take 1F = 96500 C mol–1)

(A) –5.7 (B) 5.7 (C) 11.4 (D) –11.4

Sol: [D] G = –nFE

= –2 × 96500 × 0.059

= –11400 J

= –11.4 kJ

32. The solubility product (Ksp mol3 dm–9) of MX2 at 298 K based on the information available for the given

concentration cell is (take 2.303 × R × 298/F = 0.059 V)

(A) 1 × 10–15 (B) 4 × 10–15 (C) 1 × 10–12 (D) 4 × 10–12

Sol: [B] M | M2+ (from sparingly soluble MX2) || M2+ (0.001 mol–1) | M

It is a kind concentration cell

So reaction is M + M2+ (0.01 ) M2+ (solubility) + M(s)

20 Productcell 2

Reactant

0.059 [M ]E E log2 [M ]




30.059 [solubility]E 0 log

2 (10 )

30.059 50.059 log

2 10 5s 10

3 15eq 2K (MX ) 4s 4 10


Bleaching powder and bleach solution are produced on a large scale and used in several householdproducts. The effectiveness of bleach solution is often measured by iodometry.

33. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid

is

(A) Cl2O (B) Cl2O7 (C) ClO2 (D) Cl2O6

Sol: [A] An bleaching powder is CaOCl2

Ca

Cl

O Cl

The oxoacid is HOCl and its anhydride is Cl2O

34. 25 mL of household bleach solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4 N acetic acid.

In the titration of the liberated iodine, 48 mL of 0.25 N Na2S2O3 was used to reach the end point. The

molarity of the household bleach solution is

(A) 0.48 M (B) 0.96 M (C) 0.24 M (D) 0.024 M

Sol: [C] 2 2 2 2CaOCl H O Ca(OH) Cl

2 2Cl 2KI I 2KCl

2 2 2 3 2 4 6I 2Na S O Na S O 2NaI

2 22 3 4 62S O S O 2e

n-factor for 22 3S O is 1 hence normality = molarity..

No. of millimoles of Na2S2O3 = 1212 millimoles of Na2S2O3 6 millimole of I2 6 millimole of Cl2 6 millimole of CaOCl2

Hence millimole = M × V in mL6 = M × 25

6M 0.24M25

SECTION- III ((Multiple Correct Answer (s) Types))This Section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) outof which ONE or MORE may be correct.




35. The given graphs / data I, II, II and IV represent general trends observed for different physisorptionand chemisorption processes under mild conditions of temperature and pressure which of the followingchoice(s) about I, II, III and IV is (are) correct?

I. II.

III. IV.

(A) I is physisorption and II is chemisorption (B) I is physisorption and III is chemisorption(C) IV is chemisorption and II is chemisorption (D) IV is chemisorption and III is chemisorption

Sol: [A, B, C, D] Self explanatory

36. For the given aqueous reactions, which of the statement(s) is (are) true?2 4dilute H SO

3 6excess KI K [Fe(CN) ] brownish-yellow solution

white precipitate + brownish-yellow filtrate

colourless solution

2 2 3Na S O

4ZnSO

(A) The first reaction is redox reaction(B) White precipitate is Zn3[Fe(CN)6]2

(C) Addition of filtrate to starch solution gives blue colour

(D) White precipitate is soluble in NaOH solution

Sol: [A, C, D] 2 4dilute H SO3 6 4 6 2

brownish yellow solutionKI K [Fe(CN) ] K [Fe(CN) ] I

4ZnSO

NaOH

starch2 4 2

white ppt. brownish yellow pptZn [Fe(CN) ] I blue colour

2 4

soluble due to formationNa [Zn(OH) ]

2 4 6colourless solution

Na S O 2NaI

2 2

orNa ZnO

2 2 3Na S O




37. With respect to graphite and diamond, which of the statement(s) given below is (are) correct?(A) Graphite is harder than diamond(B) Graphite has higher electrical conductivity than diamond(C) Graphite has higher thermal conductivity than diamond(D) Graphite has higher C – C bond order than diamond

Sol: [B, C, D] Self explanatory38. With reference to the scheme given, which of the given statement(s) about T, U, V and W is (are)

correct?

O

O

CH3 T

3 3 2CrO / H excess (CH CO) O W

V U

4LiAlH

(A) T is soluble in hot aqueous NaOH(B) U is optically active(C) Molecular formula of W is C10H18O4

(D) V gives effervescence on treatment with aqueous NaHCO3

Sol: [A, C]

O

O

CH3

CHO

CHO

CH3

3CrO / H

CH3

OH

OH

4LiAlH

3 2

excess(CH CO)

(T)

(V) (U)CH3

O

O

CH3

O O

CH3

39. Which of the given statement(s) about N, O, P and Q with respect to M is (are) correct?

CH3

HOH

Cl

HOH

Cl

OHH

CH3

HOH

CH3

HOH

Cl

H OH

Cl

OHH

CH3

OH OH

Cl

HOH

CH3

OH H

M N O P Q

(A) M and N are non-mirror image stereoisomers (B) M and O are identical

(C) M and P are enantiomers (D) M and Q are identical




Sol: [B, C]

CH3

HOH

Cl

HOH

Cl

OHH

CH3

HOH

CH3

HOH

Cl

H OH

M N O P Q

CH3

HOH

Cl

OHH

CH3

HOH

Cl

HOHR

S S

R R

S R

S

S

R

40. The reversible expansion of an ideal gas under adiabatic and

isothermal conditions is shown in the figure. Which of the

following statement(s) is (are) correct?

(A) T1 = T2

(B) T3 > T1

(C) wisothermal > wadiabatic

(D) Uisothermal > Uadiabatic

Sol: [A,C,D] For isothermal process temperature remains hence T1 = T2

During adiabatic expansion temperature decreases hence T3 < T1

For isothermal process work done is greater than the work done during adiabatic process

For isothermal process U = 0 but for adiabatic expansion internal energy decreases.

PART III : MATHEMATICS

SECTION- I : Single Correct Answer TypeThis section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) outof which ONLY ONE is correct

41. Let PQR be a triangle of area with 72,2

a b and 52

c , where a, b and c are the lengths of the

sides of the triangle opposite to the angles at P, Q and R respectively. Then 2sin sin 22sin sin 2

P PP P

equals

(A)3

4(B)

454

(C)23

4

(D)245

4

Sol: [C]2

22

2sin sin 2 3tan2sin sin 2 64 32

P P PP P

6 again 23 3

4 32




42. If a and b

are vectors such that 29a b

and ˆ ˆˆ ˆ ˆ ˆ(2 3 4 ) (2 3 4 ) ,a i j k i j k b then a

possible value of ˆˆ ˆ( ).( 7 2 3 )a b i j k is

(A) 0 (B) 3 (C) 4 (D) 8

Sol: [C] ˆˆ ˆ(2 3 4 ) (2 3 4 )a i j k i j k b

ˆˆ ˆ( ) (2 3 4 ) 0a b i j k

ˆˆ ˆ( ) (2 3 4 )a b i j k but | | 29 1a b

ˆˆ ˆ( ) ( 7 2 3 ) 4a b i j k

43. If P is a 3 × 3 matrix such that PT = 2P + I, where PT is the transpose of P and I is the 3 × 3 identity

matrix, then there exists a column matrix 000

xX y

z

such that

(A)000

PX

(B) PX = X (C) PX = 2X (D) PX = –X

Sol: [D] Let 1 2

1 2

1 2

11

1

a aP b b

c c

By PT =2P + I we observe that all non diagonal elements are zero.

So PX = –X

44. Let (a) and (a) be the roots of the equation 23 61 1 1 1 1 1 0a x a x a where a

> – 1. Then 0lim ( )a a and 0lim ( )a a are

(A)52

and 1 (B)12

and –1 (C)72

and 2 (D)92

and 3

Sol: [B] Let (1 + a)1/6 = t then

(t2 – 1) x2 + (t3 – 1) x + (t – 1) = 0

2 2 2( 1) ( 1) 4( 1)2( 1)

t t t t tx

t

1limt

3 1 14

x x and

12

x .

45. Four fair dice D1, D2, D3 and D4, each having six faces numbered 1, 2, 3, 4, 5 and 6, are rolled

simultaneously. The probability that D4 shows a number appearing on one of D1, D2 and D3 is

(A)91

216(B)

108216

(C)125216

(D)127216




Sol: [A] The probability = 5 5 5 616 6 6 6

= 91

216

46. The value of the integral/ 2

2

/ 2

ln cosxx x dxx

is

(A) 0 (B)2

42

(C)2

42

(D)2

2

Sol: [B]/ 2 2

/ 2ln cosxx x dx

x

= / 2 2

02 cosx x dx

= 2

2 / 20[2 sin 4 cos 4sin ] 4

2x x x x x

47. Let a1, a2, a3 .... be in harmonic progression with a1 = 5 and a20 = 25. The least positive integer n for

which an < 0 is

(A) 22 (B) 23 (C) 24 (D) 25

Sol: [D]1 1 ( 1) 1 1.

5 19 25 5n

na

99 4 125 19 n

na

990 254na n n

48. The equation of a plane passing through the line of intersection of the planes x + 2y + 3z = 2 and

x – y + z = 3 and at a distance 23 from the point (3, 1 –1) is

(A) 5x – 11y + z = 17 (B) 2 3 2 1x y (C) 3x y z (D) 2 1 2x y

Sol: [A] (x + 2y + 3z –2) + (x – y + z – 3) = 0

(1 ) (2 ) (3 )2 2 3 0x y

Ditance = 2

2 233 4 14

5 11 17x y z is plane.

SECTION- II : Paragraph TypeThis Section contains 6 multiple choice questions relating to three paragraphs with two questions on eachparagraph. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE iscorrect.





A tangent PT is drawn to the circle x2 + y2 = 4 at the point 3,1P . A straight line L, perpendicular to

PT is a tangent to the circle (x – 3)2 + y2 = 1.

49. A common tangent of the two circles is

(A) x = 4 (B) y = 2 (C) 3 4x y (D) 2 2 6x y

Sol: [D] Eequation for tangent of x2 + y2 = 4 is

22 1y mx m

22 1 0y mx m

2

2

0 3 2 1 11

m mm

on solving 12 2

m

equation of common tangent is 2 2 6x y

50. A possible equation of L is

(A) 3 1x y (B) 3 1x y (C) 3 1x y (D) 3 5x y

Sol: [A] Equation of tangent of P,

3 4x y

1 3m

slope of L is = 13

Let equation of tangent, to (x – 3)2 + y2 = 1 is

1 1( 3) 1 133

y x

3 ( 3) 2y x

3 1x y


Let f(x) = (1 – x)2 sin2x + x2 for all x IR and let g(x) =

1

2 1ln ( )

1

x tt f t dt

t

for all (1, )x .




51. Consider the statements :

P : There exists some x IR such that f(x) + 2x + 2(1 + x2)

Q : There exists some x IR such that 2f(x) + 1 = 2x(1 + x)

Then

(A) both P and Q are true (B) P is true and Q is false

(C) P is false and Q is true (D) both P and Q are false

Sol: [C] P : (1 – x)2 sin2x + x2 + 2x = 2 + 2x2

22

2

2 2sin 12 1

x xxx x

Not possible.

Q : 2 2 2 22(1 ) sin 2 1 2 2x x x x x

22

2 1sin2(1 )

xxx

Which can be true for some values of x.

52. Which of the following is true ?

(A) g is increasing on (1, )

(B) g is decreasing on (1, )

(C) g is increasing on (1, 2) and decreasing on (2, )

(D) g is decreasing on (1, 2) and increasing on (2, )

Sol: [B] 2 2 22( 1)(́ ) ln 1 sin1

xg x x x x xx

...(i)

Let 2( 1)( ) ln

1xh x x

x

2

2

1´( ) 0 (1, )

( 1)x

h x xx x

( )h x is decreasing in (1, )

h(1) > h(x)

h(1) < 0

(́ ) 0g x

( )g x is decreasing in (1, )





Let an denote the number of all n-digit positive integers formed by the digits 0, 1 or both such that noconsecutive digits in them are 0, Let bn = the number of such n-digit integers ending with digit 1 and cn =the number of such n-digit integers ending with digit 0.

53. The value of b6 is

(A) 7 (B) 8 (C) 9 (D) 11

Sol: [B] One such number is 111111.

The numbers containing 51’s and 10’s is 4

The numbers containing 41’s and 2 0’s is 32 3C

Total number = 8.

54. Which of the following is correct ?

(A) a17 = a16 + a15 (B) a17 a16 + a15 (C) a17 a16 + a16 (D) a17 = a17 + a16

Sol: [A] bn ends with 1.

if we place 0 or 1 extra in them,

we get 2bn numbers.

cn ends with 0. We can only place 1 after them to have (n + 1) digit numbers.

Obviously, an + 1 = 2bn + cn

= bn + (bn + cn)

= bn + an

Now if we remove 1 from the last place of bn numbers, we will get al such (n – 1) digit numbers.

bn = an–1

So, an + 1 = an–1 + an

a17 = a16 + a15

SECTION- III (Multiple Correct Answer (s) Types)This Section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) outof which ONE or MORE may be correct.

55. If 2

0( ) ( 2)( 3)

x tf x e t t dt for all (0, )x , then

(A) f has a local maximum at x = 2

(B) f is decreasing on (2, 3)

(C) there exists some (0, )c such that f´́ (c) = 0

(D) f has a local minimum at x = 3




Sol: [A,B,C,D]

f´(x) = 2

( 2)( 3)xe x x

for maximum and minimum f (́x) = 0

x = 2, 3

Now f´́ (x) = 2 2 2

( 2)( 3)(2 ) ( 3) ( 2)x x xe x x x e x e x

f (́x) < 0 2

( 2)( 3) 0xe x x

( 2)( 3) 0x x

2 3x

f´ (́2) < 0 f(x) has a local maximum at x = 2f´́ (3) > 0 f(x) has a local minimum at x = 3

f´́ (x) = 0

2 3 2(2 10 14 5) 0xe x x x

3 22 10 14 5 0x x x

Let g(x) = 2x2 – 10x2 + 14x – 5g(0) = –5g(1) = 1

g(0) g(1) < 0

f´́ (x) has one root between 0 and 1.

56. If the straight lines 1 1

2 2x y z

k

and 1 1

5 2x y z

k

are coplanar, then the plane(s) containing

these two lines is (are)

(A) y +2z = –1 (B) y + z = –1 (C) y – z = –1 (D) y – 2z = –1

Sol: [B,C]

2 0 02 2 0 25 2

k kk

1 12 2 2 0 15 2 2

x y zy z

and

1 12 2 2 0 15 2 2

x y zy z




57. Let X and Y be two events such that 1 1( | ) , ( | )2 3

P X Y P Y X and 1( )6

P X Y . Which of the

following is (are) correct ?

(A)2( )3

P X Y (B) X and Y are independent

(C) X and Y are not independent (D)1( )3

cP X Y

Sol. [A,B]( )( / )

( )P X YP X Y

P Y

1 1( ) 26 3

P Y

Similarly, 1 1( ) 36 2

P X

( ) ( ) ( ) ( )P X Y P X P Y P X Y = 1 1 1 22 3 6 3 .... (A)

1( ) ( ) ( )6

P X Y P X P Y

X, Y are independent ... (B)

1 1 1( ) ( ) ( )3 6 6

CP X Y P Y P X Y

58. If the adjoint of a 3 × 3 matrix P is 1 4 42 1 71 1 3

, then the possible value(s) of the determinant of P is (are)

(A) –2 (B) –1 (C) 1 (D) 2

Sol. [A,D] |adj P| = |P|3–1 = |P|2

|adj P| = 2

1 4 42 1 7 4 | |1 1 3

P

|P| = ± 2

59. Let f : (–1, 1) IR be such that 2

2(cos 4 )2 sec

f

for 0, ,4 4 2

. Then the values(s) of

13

f

is (are)

(A)312

(B)312

(C)213

(D)213




Sol. [A,B] f(cos 4) = 2

22 sec

= cos 2 1

cos2

2 1(2cos 2 1) 1cos2

f

Now, 2 12cos 1 cos 2 2 / 33

(1 / 3) 1 3 / 2f and 1 3 / 2

60. For every integer n, let an and bn be real numbers. Let function f : IR IR be given by

sin , [2 ,2 1]( )

cos , [2 1,2 ]n

n

a x for x n nf x

b x for x n n

, for all integers n

If f is continouous, then which of the following hold(s) for all n ?

(A) 1 1 0n na b (B) 1n na b (C) 1 1n na b (D) 1 1n na b

Sol: [B,D] f(x) = an + sin x , x [2n, 2n + 1]

bn + cos x , x (2n–1, 2n)f is continnous for all n

So f(2n + 0) = f(2n – 0)

f(2n + 0) = 0lim (2 )h

f n h

= 0lim sin (2 )nh

a n h

= an

f(2n – 0) = 0lim (2 )h

f n h

= 0lim cos (2 )nh

b n h

= bn +1

an – bn =1

again f(x) = an–1 + sinx, x [2n – 2, 2n – 1]bn + cosx, x (2n – 1, 2n)

Now f(2n –1 – 0) = f(2n –1 + 0)

0(2 1 0) lim (2 1 )

hf n f n h

= 10

lim sin (2 1 )nha n h

= an–1

0(2 1 0) lim (2 1 )

hf n f n h

= 0

lim cos (2 1 )nhb n h

= bn – 1

an–1 – bn = –1




AnswersChemistry

1. [C] 2. [D] 3. [C] 4. [D] 5. [C]6. [B] 7. [A] 8. [B] 9. [D] 10. [A]11. [C] 12. [C] 13. [C] 14. [B] 15. [A,C]16. [A,B,C] 17. [D] 18. [B,D] 19. [A,B] 20. [A,C]

Physics21. [B] 22. [D] 23. [D] 24. [A] 25. [B]26. [D] 27. [A] 28. [C] 29. [C] 30. [A]31. [D] 32. [B] 33. [A] 34. [C] 35. [B,D]36. [A,C,D] 37. [B,C,D] 38. [A,C] 39. [B,C] 40. [A,C,D]

Mathematics41. [C] 42. [C] 43. [D] 44. [B] 45. [A]46. [B] 47. [D] 48. [A] 49. [D] 50. [A]51. [C] 52. [D] 53. [B] 54. [A] 55. [A,B,C,D]56. [B,C] 57. [A,B] 58. [A,D] 59. [A,B] 60. [B,D]

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