P2-16-3-6 PAPER-2 Code -...

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VMC/Paper-2 1 JEE Entrance Exam-2016/Advanced

JEE ENTRANCE EXAM-2016/ADVANCED

P2-16-3-6 PAPER-2 Code : 6 Time : 3 Hours Maximum Marks : 186

Read the following Instructions very carefully before you proceed.

GENERAL

1. This sealed booklet is your Question Paper. Do not break the seal till you are told to do so.

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

3. Use the Optical Response Sheet (ORS) provided separately for answering the questions.

4. The paper CODE is printed on the left part as well as the right part of the ORS. Ensure that both these codes are identical and same as that on the question paper booklet. If not, contact the invigilator for change of ORS.

5. Blank spaces are provided within this booklet for rough work.

6. Write your name, roll number and sign in the space provided on the back cover of this booklet.

7. After breaking the seal of the booklet at 2.00 pm, verify that the booklet contains 36 pages and that all the 54 questions along with the options are legible. If not, contact the invigilator for replacement of the booklet.

8. You are allowed to take away the Question Paper at the end of the examination.

OPTICAL RESPONSE SHEET

9. The ORS (top sheet) will be provided with an attached Candidate’s Sheet (bottom sheet).

The Candidate’s Sheet is a carbon-less copy of the ORS.

10. Darken the appropriate bubbles on the ORS by applying sufficient pressure. This will leave an impression at the corresponding place on the Candidate’s Sheet.

11. The ORS will be collected by the invigilator at the end of the examination.

12. Your will be allowed to take away the Candidate’s Sheet at the end of the examination.

13. Do not tamper with or mutilate the ORS. Do not use the ORS for rough work.

14. Write your name, roll number and code of the examination centre, and sign with pen in the space provided for this purpose on the ORS. Do not write any of these details anywhere else on the ORS. Darken the appropriate bubble under each digit of your roll number.

DARKENING THE BUBBLE ON THE ORS

15. Use a BLACK BALL POINT PEN to darken the bubble on the ORS.

16. Darken the bubble COMPLETELY.

17. The correct way of darkening a bubble is as:

18. The ORS is machine-gradable. Ensure that the bubbles are darkened in the correct way.

19. Darken the bubble ONLY IF you are sure of the answer. There is NO WAY to erase or “un-darken” a darkened bubble.

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PART-I PHYSICS

SECTION I (Maximum Marks : 18) This section contains SIX questions. Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories :

Full Marks : +3 If, only the bubble corresponding to the correct option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –1 In all other cases.

1. There are two Vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale.

The Vernier scale of one of the calipers 1( )C has 10 equal divisions that correspond to 9 main scale divisions.

The Vernier scale of the other caliper 2( )C has 10 equal divisions that correspond to 11 main scale divisions.

The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers 1C and 2 ,C

respectively, are :

(A) 2.87 and 2.86 (B) 2.85 and 2.82 (C) 2.87 and 2.87 (D) 2.87 and 2.83 1.(D) Reading of vernier scale C1 can be calculated by the usual method Reading M .S.R. L.C. V .S.R.

2 8 0 1 7. . = 2 87. cm In second scale C2, length of V .S.D is greater than M .S.D.

So, usual formula will not be applicable. Let the point where scales coincide be P.

Total reading 2 8. mm x

From figure, 8 1 7 1 1x mm . mm 0 3. mm

So, Reading 2 83. mm

2. The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius R is given by

2

0

3 ( 1)5 4

Z Z eER

The measured masses of the neutron, 1 151 7,H N and 15

8 O are 1.008665 u, 1.007825 u, 15.000109 u and

15.003065 u, respectively. Given that the radii of both the 157 N and 15

8 O nuclei are same, 1u = 931.5 2MeV / c

(c is the speed of light) and 20/(4 ) 1.44e MeV fm. Assuming that the difference between the binding

energies of 157 N and 15

8 O is purely due to the electrostatic energy, the radius of either of the nuclei is : (1 fm 1510 m)

(A) 2.85 fm (B) 3.03 fm (C) 3.42 fm (D) 3.80 fm

2.(C) 2 2

0 0

3 7 6 126.5 4 5 4

Ne eE

R R

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2 2

00 0

3 8 7 156.5 4 5 4

e eE

R R

2

00

425 4

NeE E E

R .................. (i)

Binding energy of 157 N : 1 7 1.007825 8 1.008665 15.000109 E

Binding energy of 158 O : 2 8 1.007825 7 1.008665 15.003065 E

Difference in B. E. = 1 2| |E E = –1.007825 + 1.008665 + 0.002956 = 0.003796 u

This is equal to E in equation (i)

42 10.003796 931.5 (1.44)5 R

3.42R fm

3. The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a

length of 1 m at 10 .C Now the end P is maintained at 10 ,C while the end S is heated and maintained at 400 .C The system is thermally insulated from its surroundings. If the thermal conductivity of wire PQ is twice

that of the wire RS and the coefficient of linear thermal expansion of PQ is 5 11.2 10 ,K the change in length of the wire PQ is :

(A) 0.78 mm (B) 0.90 mm (C) 1.56 mm (D) 2.34 mm

3.(A)

Since thermal resistance of RS is twice that of PQ, temperature of junction QR will be 140°C.

Variation of temperature from P to Q will be linear. Hence, is 0l l T T

Temperature of center of PQ can be taken as T. i.e. 51 1.2 10 75 10l = 1.2 × 65 × 10–5 = 0.78 mm

Hence (A)

4. A small object is placed 50 cm to the left of a thin convex lens of focal length 30 cm. A convex spherical mirror of radius of curvature 100 cm is placed to the right of the lens at a distance of 50 cm. The mirror is tilted such that the axis of the mirror is at an angle 30 to the axis of the lens, as shown in the figure.

If the origin of the coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of the point

(x, y) at which the image is formed are :

(A) (25,25 3) (B) (0, 0) (C) (125/ 3, 25/ 3) (D) (50 25 3,25)

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4.(A) Image formed by lens is at (75, 0). This acts as a virtual image for mirror. Suppose that mirror is not tilted Then, for mirror 25, ?, 50u V F

1 1 125 50v

50v cm ……………..(i)

This image is formed on x-axis Now, if the mirror is rotated clockwise by 30º, image rotates by 60º clockwise. Look at the ray striking the pole of mirror. This ray rotates by 60º. Image lies on this ray. Hence image rotates by 60º. New co-ordinates of image will be 50 50cos60º 25x cm

50 350sin 60º2

y cm

Hence (A) 5. A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure

510iP Pa and volume 3 310iV m changes to a final state at 5(1/ 32) 10fP Pa and 3 38 10fV m in an

adiabatic quasi-static process, such that 3 5P V constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps: an isobaric expansion at iP followed by

an isochoric (isovolumetric) process at volume .fV The amount of heat supplied to the system in the two-step

process is approximately : (A) 112 J (B) 294 J (C) 588 J (D) 813 J

5.(C) In the first process : i fi fPV P V

fi

f i

VPP V

32 8

53

............. (i)

For the two step process

( )i f iW P V V 5 310 (7 10 )

27 10 W J

( )2 f f i ifU P V PV 2 21 1 10 10

1 4

2 23 3 9. 10 102 4 8

U J

Q W U 2 2 29 477 10 10 108 8

Q J 588 J

6. An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of half-life

18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use?

(A) 64 (B) 90 (C) 108 (D) 120

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6.(C) 0 64dN N Xdt

where X is the permissible limit of Radioactivity

Laboratory can be considered safe when 0tdN X N e

dt

0064

tN N e

164

te ln 64t

1/ 2

ln 26 ln 2t

t

1/ 26 108t t days

SECTION II (Maximum Marks : 32) This section contains EIGHT questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is(are)

correct. For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened. Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened. Zero Marks : 0 If none of the bubbles is darkened. Negative Marks : –2 In all other cases. For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will result in +4

marks; darkening only (A) and (D) will result in +2 marks; and darkening (A) and (B) will result in –2 marks, as a wrong option is also darkened.

7. Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixed by a massless,

rigid rod of length 24l a through their centres. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is . The angular

momentum of the entire assembly about the point ‘O’ is L

(see the figure). Which of the following statement(s) is (are) true?

(A) The magnitude of angular momentum of the assembly about its center of mass is 217 / 2ma (B) The center of mass of the assembly rotates about the z-axis with an angular speed of / 5

(C) The magnitude of the z-component of L

is 255ma

(D) The magnitude of angular momentum of center of mass of the assembly about the point O is 281ma 7.(AB)

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For (A) For Angular momentum about CM, we will calculate L about axis marked as (1)

22

14 2

2 2m amaL

217

2ma w A is correct

For (B)

4 2 9

5 5CMm a m a aV

m

4 2 9

5 5CMm l m l ll

m

2CM

aboutCM

V al l

Component of Ω along z-axis cos 5

a ll a

5

B is correct

For (C) Moment of inertia about axis marked as (2)

2222

24 2

4 24 4

m amaI ml m l

2217

174ma ml

So, 2 2L I and net 2 1about z axisL L cos L sin 280 75. ma C is incorrect

For (D),

955CM about O CMlL m V

9 955 5

a lm 2 24

815

m a D is incorrect

8. A rigid wire loop of square shape having side of length L and resistance R is moving along the x-axis with a

constant velocity 0v in the plane of the paper. At 0,t the right edge of the loop enters a region of length 3L

where there is a uniform magnetic field 0B into the plane of the paper, as shown in the figure. For sufficiently

large 0v , the loop eventually crosses the region. Let x be the location of the right edge of the loop. Let ( ), ( )v x I x

and ( )F x represent the velocity of the loop, current in the loop, and force on the loop, respectively, as a function

of x. Counter-clockwise current is taken as positive.

Which of the following schematic plot(s) is(are) correct? (Ignore gravity)

(A) (B) (C) (D)

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8.(BD) At a general position x,

BLViR

2 2B L VFR

2 2dv B Lmv V

dx R

2 2

0B L xV V

mR

i.e. velocity will decrease linearly. When the loop is fully within field, force and current will be zero and velocity is constant. While coming out,

current will be clockwise, force backward and velocity will decrease linearly. Hence, (B), (D) 9. In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic

motion is 7( )2 .5R rT

g

The values of R and r are measured to be (60 1) mm and (10 1) mm,

respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01 s. Which of the following statement(s) is(are) true?

(A) The error in the measurement of r is 10% (B) The error in the measurement of T is 3.57% (C) The error in the measurement of T is 2% (D) The error in the determined value of g is 11%

9.(ABD) Mean time period 0.52 0.56 0.57 0.54 0.595

0.556 0.56sec as per significant figures Error in reading 1| | 0.04 meanT T

2| | 0.00meanT T

3| | 0.01meanT T

4| | 0.02meanT T

4| | 0.03meanT T

Mean error 0.1/ 5 0.02

% error in 0.02100 100 3.57%0.56

TTT

% error in 0.001 100 10%0.010

r

% error 0.001 100 1.67%0.060

R

% error in ( )100 100 2g R r Tg R r T

0.002 100 2 3.570.05

4% 7% 11%

10. Light of wavelength ph falls on a cathode plate inside a vacuum tube as shown in the figure. The work

function of the cathode surface is and the anode is a wire mesh of conducting material kept at a distance d

from the cathode. A potential difference V is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is e , which of the following statement(s) is(are) true?

(A) For large potential difference eV / e , is

approximately halved if V is made four times

(B) e decreases with increase in and ph

(C) e increases at the same rate as ph for ph < hc /

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(D) e is approximately halved, if d is doubled

10.(A) maxK.E of e just after ejection

iph

hcK .E

maxK.E e reaching anode fph

hcK .E eV

For Ve

fK .E. eV

12

heVm eV

(A) is correct If & ph increase

fK .E decreases e increases

(B) is incorrect

2 f

hem K.E

phdd edt dt

(C) is incorrect

e is independent of d (D) is incorrect

11. Consider two identical galvanometers and two identical resistors with resistance R. If the internal resistance of

the galvanometers 2CR R / , which of the following statement(s) about any one of the galvanometers is(are)

true? (A) The maximum voltage range is obtained when all the components are connected in series

(B) The maximum voltage range is obtained when the two resistors and one galvanometer are connected in series, and the second galvanometer is connected in parallel to the first galvanometer

(C) The maximum current range is obtained when all the components are connected in parallel (D) The maximum current range is obtained when the two galvanometers are connected in series and the

combination is connected in parallel with both the resistors 11.(BC) For (A),

1 2 2g g CV i R i R

For (B),

2 4 g g CV i R i R 2 2 2g g C g g Ci R i R i R i R 1 2g CV i R R

2 1V V as

2 CR R , Hence 2 CR R

B is correct For (C)

1 2 2gI i i 2 2 Cg g

Ri iR

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For (D)

2 2gI i i 2

2 g Cg

i Ri

R

2 2 2C Cg g g g

R Ri i i iR R

12

2g

Ci RI RR

2 1I I as

2CRR

So ‘C’ is correct 12. While conducting the Young’s double slit experiment a student replaced the two slits with a large opaque plate in

the x-y plane containing two small holes that act as two coherent point sources 1 2S ,S emitting light of

wavelength 600 nm. The student mistakenly placed the screen parallel to the x-z plane (for z > 0) at a distance D = 3m from the mid-point of 1 2S S , as shown schematically in the figure. The distance between the sources

d = 0.6003 mm. The origin O is at the intersection of the screen and the line joining 1 2S S . Which of the

following is(are) true of the intensity pattern on the screen?

(A) Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction (B) Straight bright and dark bands parallel to the x-axis (C) Semi circular bright and dark bands centered at point O (D) The region very close to the point O will be dark 12.(CD) At any point on screen, which is at a distance r from O, in the plane of the screen path difference will depend only on r.

all such points have same path difference. bands will have circular shape. At O, path difference = 0 6003d . mm

2 1

100 52

n.

Dark band at O 13. A block with mass M is connected by a massless spring with stiffness constant k to a rigid wall and moves

without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position 0x ; Consider two cases: (i) when the block is at 0;x and (ii) when the block is at 0x x A . In both the

cases, a particle with mass m(<M) is softly placed on the block after which they stick to each other. Which of the following statement(s) is(are) true about the motion after the mass m is placed on the mass M?

(A) The amplitude of oscillation in the first case changes by a factor of Mm M

, whereas in the second

case it remains unchanged (B) The final time period of oscillation in both the cases is same (C) The total energy decreases in both the cases (D) The instantaneous speed at 0x of the combined masses decreases in both the cases

O

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13.(ABD) ' kM m

0kM

When m placed at mean position, Let velocity of M & m be 'v just after placing the block By momentum conservation, 0 ( ) ' M v M m v ( ) ' ' M A M m A

''

A M MA M m M m

When m placed at extreme position, Mv before placing = 0 M mv after placing = 0

extreme position and mean positions remain unchanged ' A A

2''

T

, which is same in both cases.

Energy decreases in 1st case, but not in 2nd case. Velocity at mean position = A which decreases in both cases. 14. In the circuit shown below, the key is pressed at time t = 0. Which of the following statement(s) is (are) true? (A) The voltmeter displays 5 V as soon as the key is pressed, and displays +5V after a long time (B) The voltmeter will display 0 V at time t = In 2 seconds (C) The current in the ammeter becomes 1/e of the initial value after 1 second (D) The current in the ammeter becomes zero after a long time 14.(ABCD)

Just after pressing key, 15 25000 0i

25 50000 0i

(As charge in both cap = 0) 1 0.2i mA 2 0.1i mA

And 125000B AV i V 5B AV V V

After a long time, 1 2& 0i i (steady state)

115 0 200

40q q C

And 225 0 100

20

q q C

220B AqV V 5B AV V V (A) is correct.

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For capacitor 1, 11 200 1

tq e C

11

15

ti e mA

For capacitor 2, 12 100 1 tq e C

12

110

ti e mA

21 25

20B AqV i V

5 1 5t tB AV V e e 5 1 2 te

At 2 5 1 1 0B At n , V V

(B) is correct.

At t = 1, 1 11 2

1 1 3 15 10 10

i i i e e .e

At t = 0, 1 21 1 35 10 10

i i i

(C) is correct. After a long time, 1 2 0i i (D) is correct.

SECTION III (Maximum Marks : 12) This section contains TWO paragraph. Based on each paragraph, there are TWO questions. Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct integer in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : +3 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases.

Paragraph for Questions 15 - 16

A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity is

an example of a non-inertial frame of reference. The relationship between the force rotF

experienced by a particle of

mass m moving on the rotating disc and the force inF

experienced by the particle in an inertial frame of reference is

2 rotrot in mF F v m r

Where rotv

is the velocity of the particle in the rotating frame of reference and r

is the position vector of the particle

with respect to the centre of the disc.

Now consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a constant angular speed about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the x-axis along the slot, the y-axis perpendicular to the slot and the

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z-axis along the rotation axis k

. A small block of mass m is gently

placed in the slot at 2 ˆr R / i

at t = 0 and is constrained to move only along

the slot.

15. The distance r of the block at time t is :

(A) 2 24

t tR e e (B) 4t tR e e

(C) 22R cos t (D)

2R cos t

15.(B) In the expression 2 ( ) ( ) rot in rotF F m V m r

The third term is radially outward i.e. this goes towards increasing rotV .

2nd term is perpendicular to edge of slot and is cancelled by Normal reaction from edge of slot. So, for radial direction:

2F m r

2

22

m d r m rdt

i.e. 2

22

d r rdt

Solution to this differential equation is t tr Ae Be Where A and B are constants * for detailed solution of above differential equation see the end of this solution .

At 0,2

Rt r

2

R A B

Also, rotdrVdt

t tA e B e

At 0, 0 rott V , so,

0 ( ) A B 4

RA B

So, [ ]4

t tRr e e

Hence (B) * Detailed solution of differential equation

2 dva r vdr

2

0 / 2

v r

R

v dv r dr

2 2 2

22 2 8

v r R 2

24

R drv r

dt

2 2

/ 2 0/ 4

r t

R

dr dtr R

2 2

/ 2ln / 4

R

Rr r R

2 2 / 4ln

/ 2

r r R tR

2 2 / 42

tRr r R e

22

24 2

tR Rr e R

22 2 .

4 2

t tR Rr e r R r e

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2

214

t tRr R e e 4

t tRr e e

16. The net reaction of the disc on the block is :

(A) 2 ˆˆm Rcos t j mgk (B) 2 2 212

t t ˆˆm R e e j mgk

(C) 2 ˆˆm R sin t j mgk (D) 212

t t ˆˆm R e e j mgk

16.(D) From previous question, term 2 rotm V

gives normal reaction from edge ( )eN

4

t t

rotdr RV e edt

( )4

t tR e e

So, 2 . .4

t te

RN m e e

2

ˆ2

t t

eMRN e e j

Normal reaction from bottom of slot ˆ mgk

Net reaction from slot 2

ˆˆ2

t tMRR e e j mgk

Hence (D) Paragraph for Questions 17 - 18

Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r << h . Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at 0V and the top plate at 0V .

Due to their conducting surface the balls will get charged, will become

equipotential with the plate and are repelled by it. The balls will

eventually collide with the top plate, where the coefficient of

restitution can be taken to be zero due to the soft nature of the material

of the balls. The electric field in the chamber can be considered to be

that of a parallel plate capacitor. Assume that there are no collisions

between the balls and the interaction between them is negligible.

(Ignore gravity)

17. Which one of the following statements is correct ? (A) The balls will bounce back to the bottom plate carrying the opposite charge they went up with (B) The balls will stick to the top plate and remain there (C) The balls will execute simple harmonic motion between the two plates (D) The balls will bounce back to the bottom plate carrying the same charge they went up with 17. (A) Balls will gain + ve charge and hence move towards – ve plate.

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On reaching – ve plate, balls will attain – ve charge and come back to + ve plate. So on, balls will keep oscillating. But oscillation is not S.H.M., As fore on balls is not x. (A) is correct. 18. The average current in the steady state registered by the ammeter in the circuit will be :

(A) proportional to 1/ 20V (B) zero

(C) proportional to 20V (D) proportional to the potential 0V

18.(C) As the balls keep on carrying charge from one plate to another, current will keep on flowing even in steady state. When at bottom plate, if all balls attain charge q,

0k q Vr

0V rqk

Inside cylinder, electric field 0 0E V V h

02V h .

Acceleration of each ball, 20

2qE hra .Vm k m

Time taken by balls to reach other plate, 200

2 2 12

h h .k m k mta V rhrV

If there are n balls,

Average current, 00 VavV rn q ri n

t k k m 2

0avi V

PART-II CHEMISTRY



19. The correct order of acidity for the following compounds is

(A) I > II > III > IV (B) III > I > II > IV (C) III > IV > II > I (D) I > III > IV > II

19.(A) According to ortho effect ortho substituted benzoic acid is strongest acid among o, m and p substituted benzoic acid. +M effect of OH decreases acidic strength of benzoic acid. The correct order of acidity is I > II > III > IV.

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20. The major product of the following reaction sequence is

(A) (B) (C) (D)

21. In the following reaction sequence in aqueous solution, the species X, Y and Z, respectively, are :

Ag Ag with time22 3

clear solution white precipitate black precipitateS O X Y Z

(A) 32 3 2 2 2 3 2[Ag(S O ) ] , Ag S O , Ag S (B) 5

2 3 3 2 3 2[Ag(S O ) ] , Ag SO , Ag S

(C) 33 2 2 2 3[Ag(SO ) ] , Ag S O , Ag (D) 3

3 3 2 4[Ag(SO ) ] , Ag SO , Ag

21.(A) Ag Ag2 32 3 2 3 2 2 2 3 2

X Y ZClearsolution White black

precipitate precipitate

S O [Ag(S O ) ] Ag S O (s) Ag S s

22. The qualitative sketches I, II and III given below show the variation of surface tension with molar concentration

of three different aqueous solutions of KCl, 3CH OH and 3 2 11 3CH (CH ) OSO Na at room temperature. The

correct assignment of the sketches is :

(A) I : KCl II : 3CH OH III : 3 2 11 3CH (CH ) OSO Na

(B) I : 3 2 11 3CH (CH ) OSO Na II : 3CH OH III : KCl

(C) I : KCl II : 3 2 11 3CH (CH ) OSO Na III : 3CH OH

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(D) I : 3CH OH II : KCl III :

3 2 11 3CH (CH ) OSO Na

22.(D)

For KCl curve - Increase of surface tension for inorganic salts

For CH3OH curve - Decrease of surface tension progressively for alcohols

For 3 2 311CH CH OSO Na curve - Decrease of surface tension before CMC (Critical Micelle Concentration)

and then almost unchanged

23. The geometries of the ammonia complexes of 2 2Ni , Pt and 2Zn , respectively, are :

(A) octahedral, square planar and tetrahedral (B) square planar, octahedral and tetrahedral (C) tetrahedral, square planar and octahedral (D) octahedral, tetrahedral and square planar

23.(A) 2 3 23 6[Ni(NH ) ] sp d Octahedral 2 2

3 4[Pt(NH ) ] dsp Square planar

2 33 4[Zn(NH ) ] sp Tetrahedral

24. For the following electrochemical cell at 298 K,

4 22Pt( ) | H ( , 1bar) | H ( , 1M) || M ( ), M ( ) | Pt(s) s g aq aq aq

cellE 0.092 V when 2

4

M ( )10

M ( )

xaq

aq. Given : 4 2

0M / M

RTE 0.151 V; 2.303 0.059 V

F

The value of x is : (A) –2 (B) –1 (C) 1 (D) 2

24.(D) 0 0 0cell c AE E E

24 2

4 22

Anode H 2H 2e

Cathode M 2e M

H M 2H M

0cellE 0.151– 0 0.151

2

220

cell cell 4H

M H0.059E E log

2 M p

0.059 10 10.092 0.151 log2 1

x

0.059 = 0.059 log102

x

log 10x = 2 x = 2

SECTION II (Maximum Marks : 32)

Vidyamandir Classes


This section contains EIGHT questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is(are)



25. According to Molecular Orbital Theory, (A) 2

2C is expected to be diamagnetic

(B) 22O is expected to have a longer bond length than O2

(C) 2N and 2N have the same bond order

(D) 2He has the same energy as two isolated He atoms 25.(AC) (A) is true because all electrons are paired.

(B) is false 22 2

6 0. . 2, . 32O O

B O B O

222

(Bond length) Bond length OO

(C) Both have B.O. = 2.5 (True) (D) No it will have lesser energy. (False) 26. The CORRECT statement(s) for cubic close packed (ccp) three dimensional structure is(are) : (A) The number of the nearest neighbours of an atom present in the topmost layer is 12 (B) The efficiency of atom packing is 74% (C) The number of octahedral and tetrahedral voids per atom are 1 and 2, respectively.

(D) The unit cell edge length is 2 2 times the radius of the atom 26.(BCD) CCP has a packing efficiency of 74%. The nearest neighbours in the top most layer is 9. The number of tetrahedral and octahedral voids are 2 and 1 respectively per atom.

The edge length is 2 2 times the radius of atom ( 2 4 )a r

27. Reagent(s) which can be used to bring about the following transformation is(are) :

(A) LiAlH4 in (C2H5)2O (B) BH3 in THF (C) NaBH4 in C2H5OH (D) Raney Ni/H2 in THF 27.(C) 4LiAlH in 2 3Et O, BH in THF & Raney 2Ni / H in THF reduces aldehydic group, carboxylic acid and ester

groups. 4NaBH in EtOH reduces only aldehyde groups.

28. Extraction of copper from copper pyrite (CuFeS2) involves (A) crushing followed by concentration of the ore by froth-flotation

Vidyamandir Classes


(B) removal of iron as slag (C) self-reduction step to produce ‘blister copper’ following evolution of SO2 (D) refining of ‘blister copper’ by carbon reduction 28.(ABC) CuFeS2

(A) The ore contains sulphide ions so in the purification, concentration is done by froth-floatation process.

(B) 2 2 2 22CuFeS O Cu S 2FeS SO

2 22FeS 3O 2FeO 2SO

2 3

slagFeO SiO FeSiO

(C) 2 2 2 22Cu S 3O 2Cu O 2SO Partialoxidation

2 2 2Blister copper

2Cu O Cu S 6Cu SO Self Reduction

(D) Copper is refined by electrolytic refining 29. The nitrogen containing compound produced in the reaction of HNO3 with P4O10 (A) can also be prepared by reaction of P4 and HNO3 (B) is diamagnetic (C) contains one N-N bond (D) reacts with Na metal producing a brown gas 29.(BD) 1. 4 3 3 4 2 2P 20 HNO 4H PO 20 NO 4 H O Hence, 2 5N O cannot be prepared using 4 3P and HNO 2. 4 10 3 2 2 5 3 4P O 2HNO 5H O N O 4H PO 3. 2 5N O is formed which is diamagnetic and does not have N–N bond. 4. 2 5 3 2

Brown coloured gasN O Na NaNO NO ( )

30. Mixture(s) showing positive deviation from Raoult’s law at 35°C is(are) : (A) carbon tetrachloride + methanol (B) carbon disulphide + acetone (C) benzene + toluene (D) phenol + aniline 30.(AB) Positive deviation is observed when the interaction between A and B in mixture is weaker than the interaction between A and B alone.

(Interaction between A–A + interaction between B–B) > (2×interaction between A–B). Benzene and toluene forms an ideal solution while phenol and aniline form a solution showing –ve deviation from Raoult's law.

31. For ‘invert sugar’, the correct statement(s) is(are) : (Given: specific rotations of (+) –sucrose, (+) –maltose, L-(–)-glucose and L-(+)-fructose in aqueous solution are +66°, + 140°, –52° and + 92°, respectively) (A) ‘invert sugar’ is prepared by acid catalyzed hydrolysis of maltose (B) ‘invert sugar’ is an equimolar mixture of D-(+)-glucose and D-(–)-fructose (C) specific rotation of ‘invert sugar’ is – 20° (D) on reaction with Br2 water, ‘invert sugar’ forms saccharic acid as one of the products

31.(BC) 2H OMaltose 2D-( )-glucose

2H O

Invert sugarSucrose D-( )-glucose + L-(–) - fructose

Specific rotation of invert sugar is –20º (Observed rotation is 52 92 40 , which is for 2 moles of

mixture). On reaction with 2Br water, invert sugar forms gluconic acid as one of the product 4 2[HOOC(CHOH) CH OH] 32. Among the following, reaction(s) which gives(give) tert-butyl benzene as the major product is(are) :

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(A) (B) (C) (D)

SECTION III (Maximum Marks : 12) This section contains TWO questions. Based on each paragraph, there are TWO questions. Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct option in the ORS. For each question, marks will be awarded in one of the following categories: Full Marks : +3 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases.


Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equation:

2 ( ) 2 ( )X g X g

The standard reaction Gibbs energy, ,rG of this reaction is positive. At the start of the reaction, there is one

mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by . Thus,

equilibrium is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total

pressure of 2 bar. Consider the gases to behave ideally. (Given : R = 0.083 L bar K–1 mol–1)

33. The equilibrium constant KP for this reaction at 298 K, in terms of ,equilibrium is :

(A) 28

2 equilibrium

equilibrium

(B)

2

28

4equilibrium

equilibrium

(C) 24

2equilibrium

equilibrium

(D)

2

24

4equilibrium

equilibrium

33.(B)

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2

eqeq eq

X (g) 2X(g)

t 0 1

t t 12

2

eqeq

x xeq eq

12p 2 p 2

1 12 2

2

eq

eq

peq

eq

2

12K

122

12

2 2eq eq2 2eq eq

2 8

41

4

34. The INCORRECT statement among the following, for this reaction, is (A) Decrease in the total pressure will result in formation of more moles of gaseous X (B) At the start of the reaction, dissociation of gaseous X2 takes place spontaneously (C) 0 7equilibrium .

(D) 1cK

34.(C) (A) It is true statement 2X (g) 2X(g)

If P is decreased then reaction will move in forward direction according to Le-Chatelier's principle. (B) It is true statement At start of reaction Q = 0

0G G RT ln Q If Q = 0 Then G ve, Reaction is spontaneous

(D) It is true statement

0G ve

0PRT ln K G

pK 1 since c p CK K K is also less than 1.


Treatment of compound O with KMnO4/H+ gave P, which on heating with ammonia

gave Q. The compound Q on treatment with Br2/NaOH produced R. On strong heating, Q gave S, which on further treatment with ethyl 2-bromopropanoate in the presence of KOH followed by acidification, gave a compound T.

35. The compound R is :

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(A) (B) (C) (D) 35.(A) 36. The compound T is : (A) glycine (B) alanine (C) valine (D) serine

PART-III MATHEMATICS



37. Let 1 0 04 1 0

16 4 1P

and I be the identity matrix of order 3. If ijQ q is a matrix such that 50P Q I ,

then 31 32

21

q qq

equals :

(A) 52 (B) 103 (C) 201 (D) 205

37.(B) 1 0 04 1 0

16 4 1P

; 21 0 08 1 048 8 1

P

; 31 0 0

12 1 096 12 1

P

; 41 0 0

16 1 0160 16 1

P

Pattern of element 31P is 16 1,3,6,10,......

50th term is 16 1275

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[By observing that nT of S = 1 + 3 + 6 + 10 + ….. is 2

2n n ]

501 0 0

200 1 016 1275 200 1

P

; 500 0 0

200 0 016 1275 200 0

Q P I

31 32

21

16 1275 200 102 1 103200

q qq

38. The value of

13

1

11

sin sin4 6 4 6

k k k

is equal to :

(A) 3 3 (B) 2 3 3 (C) 2 3 1 (D) 2 2 3

38.(C)

136

16

sin 14 6 4

2sin 1 . sin

4 4 6k

k k

kk

= 13

612 cot 1 cot

4 4 6kk k

= 132 cot cot4 4 6

= 1 cot4 6

= 3 12 13 1

= 4 2 32 1 2 1 2 32

= 2 3 1

39. Let 1ib for i = 1, 2, …… , 101. Suppose 1 2 101log log ........ loge e eb b b are in Arithmetic Progression (A.P.)

with the common difference log 2e . Suppose 1 2 101......a a a are in A.P. such that 1 1a b and 51 51a b . If 1 2 51.....t b b b and 1 2 51.....s a a a , then

(A) 101 101ands t a b (B) 101 101ands t a b (B) 101 101ands t a b (D) 101 101ands t a b

39.(B) 22 1

1log log (2) 2e e

b b bb

b1 b2 ………are in G.P. with common ratio, r = 2. Let 1 1 1a b y y (given)

and 51 51a b x

1 5151 51( )2 2

s a a x y

51 511 1 1[2 1] 2 .t b b b = 512 . y y

5151 51 2 .2 2

ys t x y y

Also 50 5051 1 .2 .2b b x y

51 53 2.2 2

ys t x x = 47 53 02 2

x y s > t

Also 101 51 12a a a 101 2a x y

2 2

51101

1

( )b xbb y

Vidyamandir Classes


22 2 2

101 10122 0

x yx x xy yb a x yy y y

101 101b a

40. The value of 2 2

2

cos1 xx x dx

e

is equal to :

(A) 2

24

(B) 2

24

(C) 2 2e

(D) 2 2e

40.(A) / 2 2

/ 2

cos1 x

x xI dxe

/ 2

2

0

1 1cos1 1x xI x x dx dx

e e

/ 2 / 2/ 22 2

00 0

cos sin 2 sinI x x dx x x x x dx

/ 22 2 2

0/ 2

0

2 cos cos 2 1 24 4 4

x x x dx

41. Let P be the image of the point 3, 1, 7 with respect to the plane 3x y z . Then the equation of the plane

passing through P and containing the straight line 1 2 1x y z is :

(A) 3 0x y z (B) 3 0x z (C) 4 7 0x y z (D) 2 0x y

41.(C) Let image be ( , , )P h k

3 1 7 2[6] ( 4)1 1 1 3

h k

1 5 3h k

( 1, 5, 3)P

Other point 0(0, 0, 0)

ˆˆˆ ˆ1 5 3 4 7

1 2 1

i j kn i j k

Plane is 4 7 0x y z

42. Area of the region 2, : 3 , 5 9 15x y R y x y x is equal to :

(A) 16

(B) 43

(C) 32

(D) 53

42.(C) Shaded area = Area of trapezium – Area of AED – Area of EBC

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3 1

4 3

1 3 5 3 32

x dx x dx

4 1

3/ 2 3/ 2

3 3

15 2 23 32 3 3

x x

15 2 162 3 3

15 362 2

SECTION II (Maximum Marks : 32) This section contains EIGHT questions. Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is(are)



43. Let P be the point on the parabola 2 4y x which is at the shortest distance from the center S of the circle

2 2 4 16 64 0x y x y . Let Q be the point on the circle dividing the line segment SP internally. Then :

(A) 2 5SP

(B) : 5 1 : 2SQ QP

(C) the x-intercept of the normal to the parabola at P is 6

(D) the slope of the tangent to the circle at Q is 12

43.(ACD) (2, 8), 4 64 64 2S r

22 8

2t t

t

32 8 2t t t

t = 2 P (4, 4)

4 16 20 2 5PS

SQ = 2

: 2 : 2 5 2 1 : 5 1SQ QP = 5 1 : 4

Normal at point 2x + y = 12 x intercept = 6

Vidyamandir Classes


Slope of tangent at Q is 12

.

44. Let 1 2 3ˆˆ ˆu u i u j u k be a unit vector in 3R and 1 ˆˆ ˆˆ 2

6w i j k . Given that there exists a vector v in

3R such that ˆ 1u v and ˆ ˆ 1w u v . Which of the following statement(s) is (are) correct ?

(A) There is exactly one choice for such v (B) There are infinitely many choices for such v

(C) If u lies in the xy-plane then 1 2u u

(D) If u lies in the xz-plane then 1 32 u u

44.(BC) 1 | | | | cos 1w u v w u v

cos 1 0

isw to plane of &u v

isw u and isw v

1 2 3is to 2 0w u u u u

If u in xy plane 3 0u 1 2 1 2| | | |u u u u

If u in xz plane 2 0u 1 3 1 32 0 | | 2 | |u u u u

For v w v

and | | 1 | | sin 1u v v

is angle b/w andu v

So v

can take many values where | | 1v

and 0w v

45. Let ,a b and :f be defined by 3 3cos(| |) | | sin(| |)f x a x x b x x x . Then f is

(A) differentiable at x = 0 if a = 0 and b = 1 (B) differentiable at x = 1 if a = 1 and b = 0 (C) NOT differentiable at x = 0 if a = 1 and b = 0 (D) NOT differentiable at x = 1 if a = 1 and b = 1

45.(AB) 3 3cos | | | | sin | |f x a x x b x x x

3 3

3 3

cos sin 2 0

cos sin 0

a x x bx x x h

a x x bx x x h x

3 2 3 3 2

3 2 3 3 2

sin 1 3 sin cos 3 1

sin 3 1 sin cos 3 1

a x x x b x x bx x x xf x

a x x x b x x bx x x x

Differentiable at x = 0

3 3

3 3

cos sin 1 1

cos sin 1 1

a x x bx x x x hf x

a x x bx x x h x

Differentiable at x = 1

46. Let

2 2

2 2 2 22

...2lim ,

! ...4

xn

n

n

n nn x n x xnf x

n nn x n x xn

for all x > 0. Then

Vidyamandir Classes


(A) 1 12

f f

(B) 1 23 3

f f

(C) ' 2 0f (D)

' 3 ' 23 2

f ff f

46.(BC)

2 22 2 2 2

2

...2lim , 0

!( ) ...4

xn

n

n

n nn x n x xnf x x

n nn x n x xn

2 2 2 22 2

2

1 / 2 /ln lim ln ln ln ... ln!

4

n

n

n x n x n x n nf x xn n x n n nx x

n

1lim ln ln ... ln .....1 2n

n n nxn n

1 1

20 0 2

11ln ln

1

xtx dt dt

t xt

1

2 20

11ln1

t txx dt

t t x

1

2 20

1ln ln1

txg x f x x dtt x

1

2 20

1ln1

txx dtt xf x e tx y

2

0

1

1

xyn dy

ye

21

1xf x f x n

x

0f x if 21 1

1x

x

21 1x x

2x x 0 1x (B) correct

3 4 23 10 5

fn n

f

2 32 5

fn

f

3 23 2

f ff f

(D) incorrect

32 2 05

f f . n

(C) correct.

47. Let : (0, )f and :g be twice differentiable functions such that ''f and ' 'g are continuous

functions on . Suppose ' 2 2 0, '' 2 0f g f and ' 2 0g . If 2

lim 1,' 'x

f x g xf x g x

then

(A) f has a local minimum at x = 2 (B) f has a local maximum at x = 2 (C) '' 2 2f f (D) '' 0f x f x for at least one x

Vidyamandir Classes


47.(AD) 2

( ) ( ) 1'( ) '( )x

f x g xLtf g g x

2

( ) ( ) ( ) ( ) 1( ) ( ) ( ) ( )x

f x g x f x g xLtf x g x f x g x

'(2) (2) (2) '(2) 1

'(2) "(2) "(2) '(2)f g f g

f g f g

0 (2) '(2) 10 "(2) '(2)

f gf g

(2) "(2)f f

Option C is incorrect and option D is correct. Since '(2) (2) 0f f

f has local minima at x = 2 Option A is correct.

48. Let ,a b and 2 2 0.a b Suppose 1: , , 0 ,S z z t ta ibt

where 1.i If z x iy and

,z S then (x ,y) lies on

(A) the circle with radius 12a

and centre 1 ,02a

for 0, 0a b

(B) the circle with radius 12a

and centre 1 ,02a

for 0, 0a b

(C) the x-axis for 0, 0a b

(D) the y-axis for 0, 0a b

48.(ACD)

2 2 21, , , 0 , , , 0a ibtS z c z t R t z c z t R t

a ibt a b t

As andz x iy z S

2 2 2 2 2 2

a btx ya b t a b t

0 , 0ayt bx ay t bbx

2

2 2 2 2 2 22 2

2 2

a axx xa y x a a ya bb x

2 2 2ax x y x 2 2 0xx ya

circle centre 1 , 02a

, radius 1

2a for 0; 0a b

For 0, 0b a equation (i) becomes, 0y lies on x-axis

For 0, 0a b equation (i) becomes, 0x lies on y-axis

49. Let , ,a R . Consider the system of linear equations 2ax y 3 2x y

Which of the following statement(s) is(are) correct ? (A) If 3a , then the system has infinitely many solutions for all values of and

(B) If 3a , then the system has a unique solution for all values of and

(C) If 0 , then the system has infinitely many solutions for 3a

(D) If 0 , then the system has no solution for 3a

49.(BCD) 2| | 2 6 2 3

3 2a

A a a

Vidyamandir Classes


12

| | 22

A

2| | 33a

A a

For 3a | | 0A system as infinite many solutions and if 1| | 0A and 2| | 0A

0 Option A is incorrect

If and 3a

0A , 2 1| | 0 | | 0A A System has infinite solutions

For 0 and 2 13 | | 0 | | 0 | | 0a A A A

System has no solution Option D in correct

50. Let 1: , 22

f R and 1: , 2

2g R

be functions defined by 2( ) [ 3]f x x and

( ) | | ( ) | 4 7 | ( )g x x f x x f x , where [y] denotes the greatest integer less than or equal to y for y R . Then :

(A) f is discontinuous exactly at three points in 1 , 22

(B) f is discontinuous exactly at four points in 1 , 22

(C) g is NOT differentiable exactly at four points in 1 , 22

(D) g is NOT differentiable exactly at five points in 1 , 22

50.(BC) 2 3f x x

24 7 4 7 3g x x f x x f x x x x

f x discontinuous at 1, 2, 3, 2

17 5 3 , 02

7 3 3 0, 1

7 3 2 1, 2

7 3 1 2, 3

77 3 0 3,4

75 7 0 , 24

3 2

x

x

x

x

x

x

x

Discontinuous at 1, 2, 3, 2

Non differentiable 0, 1, 2, 3, 2

SECTION III (Maximum Marks : 15) This section contains TWO paragraph. Based on each paragraph, there are TWO questions. Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct. For each question, darken the bubble corresponding to the correct integer in the ORS. For each question, marks will be awarded in one of the following categories:

Vidyamandir Classes


Full Marks : +3 If only the bubble corresponding to the correct answer is darkened. Zero Marks : 0 In all other cases.


Football teams 1 2andT T have to play two games against each other. It is assumed that the outcomes of the two games

are independent. The probabilities of 1T winning, drawing and losing a game against 2T are 1 1 1, and2 6 3

, respectively.

Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams 1T and 2T , respectively , after two games.

51. P (X > Y) is :

(A) 14

(B) 512

(C) 12

(D) 712

51.(B) For (X > Y) Match -1 Match – 2 P(X > Y)

T1 wins T1 wins 1 12 2

T1 wins T1 draws 1 12 6

T1 draw T1 wins 1 16 2

1 1 1 54 12 12 12

P X Y

52. P (X = Y) is :

(A) 1136

(B) 13

(C) 1336

(D) 12

52.(C) For (X = Y) Match -1 Match – 2 P(X > Y)

T1 wins T1 Lose 1 12 3

T1 Lose T1 win 1 13 2

T1 draw T1 draw 1 16 6

1 1 1 136 6 36 36

P X Y


Let 1 1 2 2, 0 and , 0F x F x , for 1 20 and 0x x , be the foci of the ellipse 2 2

19 8x y

. Suppose a parabola having

vertex at the origin and focus at 2F intersects the ellipse at point M in the first quadrant and at point N in the fourth

quadrant. 53. The orthocentre of the triangle F1MN is :

(A) 9 , 010

(B) 2 , 0

3

(C) 9 , 010

(D) 2 , 63

53.(A) Solving 2 4y x

2 24 1

3 8x

Vidyamandir Classes


3 36 62 2

M N

Equation of altitude through M

5 3622 6

y x

According to figure, orthocentre lies on x = axis

910

x orthocentre 9 0

10

54. If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the x-axis at Q, then the ratio of the area of the triangle MQR to area of the quadrilateral MF1NF2 is :

(A) 3 : 4 (B) 4 : 5 (C) 5 : 8 (D) 2 : 3 54.(C) Equation of normal at point M

32y mx m m ; where 36

m t

7 02

Q

Equation of tangent at point M is given by

3 . 6 19 2 8x y

3 6 19 2 8x y

6 0x y

3 62

M 6 0R

7 02

Q

5 6Area of4

MQR

Area of quadrilateral 1 2MF NF 12 2 62

2 6

1 2. 5. . 8

ar of MQR MF NFar of quad

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