AANNSSWWEERR,, HHIINNTTSS && SSOOLLUUTTIIOONN TEST 2 (ADVANCED...1 2 0 4 r 80 Heat t dt R3 Joule 11. The potential difference across a 10H inductor as a function of time is shown in

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CTS-PT-II(Advanced)-Paper–I&2-Solution Page: 1

CH

AM

P TEST SERIES for JEE, 2020

PART TEST – 2, JEE (ADVANCED)

PAPER – 1 & 2

TEST DATE: 24 November, 2019

PHYSICS, CHEMISTRY & MATHEMATICS

AANNSSWWEERR,, HHIINNTTSS && SSOOLLUUTTIIOONN



HINTS & SOLUTION

PAPER - 1

PHYSICS

SECTION – A One or More than one correct type

This section contains 5 questions. Each questions has FOUR options (A), (B), (C) and (D) . out of which ONLY ONE option is correct.

1. Magnetic field exist in the space and given as 20

2

B ˆB x k , where B0 and are positive constants. A

particle having positive charge ‘q’ and mass ‘m’ is projected with speed ‘v0’ along positive x-axis from the

origin. What is the maximum distance of the charged particle from the y-axis before it turns back due to the

magnetic field. (Ignore any intersection other than magnetic field)

(A)

1/ 32

0

0

m v

3qB

(B)

1/ 32

0

0

3m v

2qB

(C)

1/ 32

0

0

3m v

qB

(D)

1/ 32

0

0

m v

qB

Ans. C

Sol. F qv B

x yˆ ˆq(v i v j)B

20y x2

BF q x v

0 maxv x

20y 2

0 0

qBdv x dx

m

30 max

0 2

qB xv

3m

1/ 32

0max

0

3m vx

qB

Alternate solution Impluse momentum theorem along x-axis

maxx 20

x 02

0

qB xv dt mv

maxx 20

02

0

qB xdx mv

1/ 32

0max

0

3m vx

qB

2. A non-conducting hollow cone has charge density . A part ABP is

cut and removed from the cone. The potential due to the remaining

portion of the cone at point ‘P’ is

(A) 0

5 R

6

(B)

0

5 R

24

(C) 0

5 R

3

(D)

0

5 R

12

Ans. D

A

B 600

R

P

y

x v0

vx

vy v0

(q, m)



Sol. Potential due to circular non conducting disc, C

0

rv

2

(where r is

radius of disc)

So, P0

.V

2 2

0 0

5 R 1 5 R

2 3 ( ) 2 12

3. A charge particle ‘q’ lies at the centre of two concentric hollow

sphere of inner radii R and 3R and outer radii 2R and 4R

respectively. What amount of work has to be performed to slowly

transfer the charge ‘q’ from centre through the orifice to infinity.

(A)

2

0

5 q

96 R (B)

2

0

q

16 R

(C)

2

0

7 q

96 R (D)

2

0

q

96 R

Ans. C

Sol. Work performed

2R 4R 22 2

0 00R 3R

1 1 7 qE dV E dV

2 2 96 R

4. A point charge ‘q’ is placed at a distance ' ' (on the y-axis)

vertically above the surface which lies in the xz-plane as shown in

the figure. The flux of eclectic field passing through the plates is

(A) 0

q

6 (B)

0

q

18

(C) 0

q

24 (D)

0

q

48

Ans. C

Sol. OAB ABCD

0 0 0

q q q

48 48 24

5. A hollow non conducting sphere of radius R has a positive charge q uniformly

distributed on its surface. The sphere starts rotating with a constant angular

velocity ‘’ about an axis passing through center of sphere, as shown in the

figure. Then the net magnetic field at center of the sphere is

(A) 0q

2 R

(B)

0q

3 R

(C) 0q

6 R

(D)

0q

12 R

Ans. C

O

y

q

A

x

-

B C

D

P

5

R3

O R

q

z

O

y

q (0, ,0)

( ,0, )

x -

q R

2R 3R

4R

z



Sol. Magnetic field due to the ring on its axis is 2

0

2 2 3 / 2

r iB

2 (r x )

(due to ring)

2/ 2 0

3 / 22 2

/ 2

2 Rcos Rd(Rcos )

2 /B

2 (Rcos ) (Rsin )

Where 2

q

4 R

0qB6 R

One or More than one correct type

This section contains 8 questions. Each questions has FOUR options (A), (B), (C) and (D) . ONE OR MORE THAN ONE of these four options is (are) correct.

6. When the steady state is reached by keeping switch in position –1, then switch

‘S’ is transferred from position 1 to 2. Then

(A) Charge on 3F capacitor is 30 C after S is transferred from position 1 to 2.

(B) Charge on 3F capacitor is 650/19 C after S is transferred from position 1 to

2.

(C) Work done by the battery is 80 Joule after S is transferred from position 1

and 2

(D) Work done by the battery is 80 Joule after S is transferred from position 1

and 2

Ans. B,D

Sol. By applying Kirchhoff’s law charge transferred by the battery is 80

C19

7. A cubical box of side 5m contains helium gas at a pressure 320 N/m2. During an observation time of 1 second,

an travelling with root mean square speed parallel to one of the sides of the cube was found to make 1000

hits with a particular wall, without any collision with the other atoms. (take 25

R J/mol K3

). Then

(A)The temperature of gas is 16 103 K.

(B) The temperature of gas is 8 103 K.

(C) The total mass of helium gas in the box is 1.2 gm.

(D) The total mass of helium gas in the box is 2.4 gm.

Ans. A,C

Sol. Time between two consecutive collision 1

sec1000

So, rms

2 1

v 1000

rmsv 2000 5 10000m/s

rms3RT

v 10000M

So, 8 3

310 4 10T 16 10 K25

33

Mass of helium gas PV = nRT

m

PV RTM

O R

q

Rd

6F 19v

3F

10F

10F

20F

S

1 2



3

3

PVM 320 125 (4 10 )m 1.2gm

25RT16 10

3

8. In the given graph, an ideal gas can change its state from A to C by two

paths ABC or AC. Then

(A) If the internal energy of gas at ‘A’ is 10 J and the amount of heat supplied

in path AC is 200J, then the internal energy of gas at ‘C’ is 150 J.

(B) If the internal energy of gas at ‘A’ is 10 J and the amount of heat supplied

in path AC is 200 J; then the internal energy of gas at ‘C’ is 75 J.

(C) If the internal energy of gas at state B is 20 J. Then the amount of heat

supplied to the gas to go from A to B may be 5 Joule.

(D) If the internal energy of gas at state B is 20 J. Then the amount of heat

supplied to the gas to go from A to B must be 10 Joule.

Ans. A,D Sol. In path A C

1dW 4 10 10 4 40 20 60J

2

dQ dU dW

C A200 U U 60

C AU 140 U 140 10 150J

In path A B QAB = UB - UA

A AB B

B B

U TU 1 U 1

U T

120 1 10Joule

2

9. In the cylinder region of radius R = 2m, there exists a time varying magnetic field B

such that dB

2Tesla / sec.dt

A charged particle having charge q = 2C is placed at

the point P at a distance d from its center O. Now, the particle is moved in the

direction perpendicular to line OP by an external agent up to infinity so that there is

no gain in kinetic energy of charged particle. Then

(A) work done by external agent is 4 Joule when d = 4m

(B) work done by external agent is 4 Joule when d = 8m

(C) work done by external agent is independent of d.

(D) work done by external agent is positive.

Ans. A,B,C,D Sol. dB

E d Adt

2 2 2E 2 x d R k, where dB

( k)dt

2

2 2

R kE

2 x d

0

W qE dx

E

d

r

O

dx

x

4Pa

8Pa

5m3

15m3

P

A

B

C

V

P

d

q

O

4Pa

8Pa

5m3

15m3

P

A

B

C

V



2 2

2 2 2 20 0

qR k d qR kqEcos dx dx

42 x d d x

Alternate solution By applying Faraday’s Law for an imaginary triangular loop having

vertex O, P and point at infinity q E d can be found out

10. A uniform conducting ring of mass m = 2kg, radius r = 2m and resistance 8 is kept on smooth, horizontal

surface. A time varying magnetic field 2ˆ ˆB (i t j) Tesla is present in the region, where t is time in second

and take vertical as y-axis. (Take 2 = 10). Then

(A) Time when ring starts toppling is 1 sec

(B) Time when ring starts toppling is ¾ sec

(C) Heat generated through the ring till the instant when the ring start toppling is 40/3 Joule

(D) Heat generated through the ring till the instant when the ring start toppling is 80/3 Joule

Ans. A,D

Sol. 2 2BA (t )( r )

2d

2 r tdt

22 r t

iR R

For toppling

| B | mgr

2(i. r ).1 mgr

2

22 r t r mgrR

2 3

mgrt 1sec

2 r r

22

2 2 r tHeat i Rdt dt.RR

12 22

0

4 r 80Heat t dt

R 3

Joule

11. The potential difference across a 10H inductor as a function of time is

shown in the figure. At t = 0, current is zero. Then

(A) The current in the inductor at t = 6 sec is 3 amp.

(B) The current in the inductor at t = 1 sec is ¼ amp

(C) The current in the inductor at t = 4 sec is 5/2 amp

(D) The variation of current versus time is parabola.

Ans. A,B,C,D

Sol. L Ldi

V L Ldi V dtdt

L1

L di V dt .6.10 302

f iL(l l ) 30Amp

f30

I 3amp10

for t 2,

V(volts)

10V

t(sec) 0 2 6



L10

V t 5t2

i 1

L

0 0

Ldi V dt 5 tdt

5

L(i 0)2

i 1/ 4

for t 2

L5

V 15 t2

i 4

t 2

5Ldi 15 t dt

2

i 5 / 2amp

12. Initially the switch ‘S’ is open for a long time. Now the switch ‘S’ is closed at

1 2 1t 0(R 2 ,R 2 ,C 1F and C2 = 1F). Then

(A) The current through the wire AB at t= 2 sec is 1/2e ampere.

(B) The current through the wire AB at t = 2 sec is 1/e ampere.

(C) The charge on the capacitor C2, which was initially uncharged , at t = 2

sec is 1

12e

coulombs

(D) The charge on the capacitor C1 at t = 2sec is 1

12e

coulombs

Ans. A,C,D Sol.

When the switch is closed then at t = 0 the distribution of the charge is shown in the figure.

By applying Kirchoff’s Law '1(1 q)

I2

q t

1/ 2 0

dq dt

1 q 2

1/ 21q 1 e2

' 1/ 21

dq 1I e

dt 4

13. An electric kettle has two coils of different power. When one coil is switched on, it takes 30 minutes to boil a

certain amount of water and when coil is switched on, it takes 45 minutes to boil same amount of water .Then

(A) It takes 150 minutes to boil same amount of water when both the coils are used in series combination.

(B) It takes 75 minutes to boil same amount of water when both the coils are used in series combination

(C) It takes 18 minutes to boil same amount of water when both the coils are used in parallel combination

(D) It takes 25 minutes to boil same amount of water when both the coils are used in parallel combination

Ans. B,C

B A R1

R2 C1 C2 2v

+1/2C

-1/2C

+1/2C

-1/2C

(at t = 0)

B A R1

R2 C1 C2 2v

'1 1I / 2 I

+q

-q -q

+q

'1 1I / 2 I

I1

I

(I – I1)

B A R1

R2 C1 C2 2v

S



Sol. 2 2

1 2 21

V VH (30 60),H (45 60)

R R

1 2 1 21 2

30 45H H 3R 2R

R R

2 2 2

series s s 11 2 1

1 1

V V VH t t H (30 60)

3(R R ) RR R

2

s5

t (30 60) 75min2

2 2 2 2

P P P1 2 1 1

V V V 2VH t t

R R R 3R

2 2

P1 1

5V Vt (30 60)

3R R

P3 30 60

t 18min5

SECTION – C (Single digit type)

This section contains FIVE questions. The answer to each question is a single Digit integer ranging from 0 to 9 both

inclusive

1. A ideal gas whose adiabatic exponent equals is expanded so that the amount of heat transferred to the gas

is equal to twice of decrease of its internal energy. The equation of the process is

1

KTV

= constant (where T

and V are absolute temperature and volume respectively). Then find the value of ‘K’.

Ans. 3

Sol. dQ dU dW

dQ 2dU

3dU dW

3nCvdT PdV

3nRdT nRT

dV1 V

From this TV3 = constant So, K = 3 2. When steady state is reached, switch S is opened. Then the amount of heat

rejected by the resistor is 800 K Joule. Then find the value of K.

Ans. 0 Sol. There is no transfer of charge through the battery

3. A coil of inductance L = 5H and resistance R = 55 is connected in series to the mains alternating voltage of

frequency f = 50 Hz in series. What can be the non-zero capacitance of the capacitor (in F) connected in

series with the coil if the power dissipated has to remain unchanged. (take 2 = 10)

Ans. 1

Sol. 2

av 2

V R V RP VIcos V

z z Z

12F

10

6F

20 volts

S



It means Z has to be same in both cases

2 2 2 21 2 L C LZ Z R X R (X X )

L C2X X

6 62

1C 10 1 10

2 L

4. A particle having charge ‘q’ and mass ‘m’ is projected with velocity ˆ ˆ ˆ(4i 6j 3k) m/sec from the origin in a

region occupied by electric field ‘E’ and magnetic field ‘B’ such that 0ˆB B j and 0

ˆE E j (take 0qE

2m

).

Find the time (in sec) when the magnitude of velocity of the charge particle becomes 5 5 m/sec. (neglect the

gravity)

Ans. 8

Sol. y y yv u a t

0 1qE

0 6 tm

1t 3sec

2

2 2 2x y zv v v 5 5

2 2 2y4 v 3 125

yv 10m/ sec

y y yv u a t

0 2qE

10 0 tm

2t 5sec

So, 1 2t t t 8sec

5. In the circuit the key (K) is closed at t = 0. Find the current in ampere

through the key(K) at the instant 3t 10 n2sec

Ans. 2

Sol. i1 is current in the first loop

4

t

5 101

40 3i 1 e amp

20 2

i2 is current in the second loop

3

1

102

40 1i e amp

40 2

So, total current through key (K)

1 2i i i

3 12amp

2 2

8 15

L = 10mH

210-6F

40V

(2)

12

25 (1)

8 15

L = 10mH

210-6F

40V

K

12

25



CHEMISTRY

SECTION – A

One or More than one correct type This section contains 5 multiple choice questions. Each questions has FOUR options (A), (B), (C) and (D) . out of which ONLY ONE option is correct.

1. Br

COOHCH3

CH3CH3

t - BuOK

ExcessX

Br

COOHCH3

CH3CH3

t - BuOK

ExcessY

Which of the following option is correct?

(A) CH3

CH3CH3

X is : Y isCH3

CH3CH3

COOK

(Major product)

(B)

X isCH3

CH3CH3

COOK

CH3

CH3CH3

Y is (Major product):

(C) CH3

CH3CH3

Both X & Y are (only)

(D) CH3

CH3CH3

COOK

Both X & Y are (only)

Ans. B



Sol.

COOH

Br

CH3

CH3

CH3

H

CH3 C

CH3

CH3

O. .

COO KCH3

CH3CH3

H

Br

CH3

CH3

CH3

COOH

CH3 C

CH3

CH3

O. .

COO KCH3

CH3CH3

COOH

Br

CH3

CH3

CH3

CO O

. .

2. The relative rates of solvolysis of following iodides are

I I

O

I

(x) (y) (z)

(A) x > y > z (B) y > x > z (C) x > z > y (D) z > x > y

Ans. D Sol. I +

2 cation

I

both SN2 and SN1 are slow

O

I

:NGP

:

3. The suitable reagent for the following conversion is

O

Me

O

O

Me

(A) m – CPBA (B) H2O2 / AcOH (C) BuOH/HCl (D) H2O2/NaOH

Ans. D



Sol. H2O2 + OH

. . H O O

. . + H2O

O

Me

OOH. .

O

Me

O H

O . . O

O

Me m-cPBA gives Baeyer – Villiger’s Oxidation with Ketone. 4. The intermediate in the reaction given below is:

N2

CH3CH3

h

H

CH3 CH3

H

(A) Free radical (B) Carbocation (C) Carbanion (D) Carbene Ans. D Sol.

N2CH3

CH3

H

CH3 CH3

H

h

Carbene 5. The correct combination of reagents required to effect the following conversion is:

H2C COOCH3

H2C COOCH3

O

OH

(A) (i) Na, xylene, heat (ii) H2O2, NaOH

(B) (i) NaOEt, EtOH (ii) Na, xylene, heat

(C) (i) Na, xylene, Me3SiCl, heat (ii) H3O+

(D) (i) TiCl3, Zn – Cu, (CH3)3SiCl, heat (ii) H3O+

Ans. C Sol. Acyloin condensation Sol.

H2C C

H2C

OCH3

O

C

O

OCH3

2Na 2Na- +2e H2C C

H2C

OCH3

O-

C

O-

OCH3

H2C C

O

CH2C

O

2Na 2Na +2e-H2C C

O-

CH2C

O-

H2C C

O-

CH2C

O-

H2C C

OH

CH2C

O-

H2C C

OH

CH2C

OH

H3O+

Tautomerism



One or More than one correct type

This section contains 8 questions. Each questions has FOUR options (A), (B), (C) and (D) . ONE OR MORE THAN ONE of these four options is (are) correct.

6.

+ (CH3)3CCOClAlCl3

C CH3CH3

CH3

+ CO + HCl

Which of the following statement(s) is/are correct?

(A) S for the rearrangement of 3 3(CH ) CCO

to 3 3(CH ) C

is positive.

(B) 3 3(CH ) C

is more stable than 3 3(CH ) CCO

(C) It’s an example of homogeneous catalysis

(D)

NO2

Solvent used for Friedel Craft acylation is

Ans. A,D

Sol. A evolution of CO results in S = +ve

7.

CN3

O

NH2

X Y

intermediate

X & Y are:

(A) X is acyl cation (B) X is nitrene

(C)

NH

O

(D)

N

C

N

O

H

H Ans. B,D

Sol.

CN3

O

NH2

-N2

CN

O

NH2

:.. N

NH2

C O

..

NH

NH

O

(nitrene)

8. In the reaction, 2 6 2 422x B H BH x BH

. The reactant(s) ‘x’ is (are)

(A) NH3 (B) CH3NH2 (C) (CH3)2NH (D) (CH)3N

Ans. A,B,C

Sol. 3 2 6 3 2 3 4H N: B H [H N BH NH ] [BH ]

3 2 2 6 3 2 2 2 3 4CH NH : B H [CH NH BH NH CH ] [BH ]



3 2 6 3 2 2 3 2 4(CH )NH B H (CH ) NH BH NH(CH ) BH

3 2 6 3 3 3(CH )N B H 2H B N CH

9. Which of the following are ture for Bakelite?

(A) Cross linked polymer (B) Condensation polymer

(C) Thermosetting polymer (D) Addition polymer

Ans. A,B,C Sol. Facts Bakelite is formed from monomers phenol and methanol

10. Which of the following statements about the peptide bond is(are) correct?

(A) The C – N bond of the peptide linkage is longer than that of an amide

(B) The

O||

C NH portion of the peptide linkage is planar

(C) The C – N bond length of peptide is shorter than that of an amine

(D) The rotation about CO – N bond of peptide is restricted.

Ans. B,C,D

Sol. Facts 11. In which of the following pairs at least one of the compounds give positive Tollens test?

(A) Glucose and sucrose (B) Glucose and fructose

(C) Fructose and sucrose (D) Maltose and sucrose

Ans. A,B,C,D Sol. Glucose , Fructose , Maltose are reducing sugars. 12. Which of the following is/are not the products in the following reaction?

O

O

H

H

i)DIBALH(1Eq)

ii) CH2=CHMgBr

iii)H2O weekly Acidic medium

iv) PCC

(A)

OO

CH2

H

(B)

OO

CHOH

(C)

O

H

H

CH3

OH

(D) O

CH2

CH2

H

H

Ans. B,C,D



Sol.

O

ODiBAL CH3

CHO

OH

CH2 = CHMgBr

OHOH

CH

CH2PCC

OO

CH2

13. Which of the following can give aldehyde by the oxidation with Jones reagent (CrO3/py)

(A)

CH2OH

OH

(B) 2 2CH CH CH OH

(C) CH3 C

CH3

CH3

CH2OH

(D)

CH2OH

NH2 Ans. A,B,C,D

Sol. A, B, D because hydrate is unstable, ‘C’ due to steric factor.



inclusive

1. Pure(+2) – chlorobutane gives how many different dichlorbutanes?

Ans. 5

Sol.

C C

Cl

Cl

C C C C

Cl

Cl

C C C C

Cl

C

Cl

C C C

Cl

C C

Cl

(1) (1) (2) (1) Configuration of Cl at C-2 is fixed.

2.

No. of aldol condensation productOH-

i) O3

ii) Zn - H2O

Ans. 0



Sol.

i)O3

ii)Zn - H2O

CHO

CHO

: OH

No aldol condensation

3. Number of amine isomer of the molecular formula C4H11N is/are structural isomers

Ans. 8

Sol.

C C C C NH2 C C C C CH3

NH2

C C

C

C NH2

C C

C

NH2

C C C N

H

C

C

C C N

H

C C C N

C

C C C C

C

N

H

C

4. Number of following substituent’s those are deactivating but ortho and para directing

CH3 , C R

O

CH3

, NH C R

O

, CH3 S C2H5

O

. . ,CH3 C NH2

O

, F

Ans. 3

5. Total number of chiral carbons in - D (+) glucose is:

Ans. 5



MATHEMATICS

SECTION – A Single Correct Type

This section contains 5 multiple choice questions. Each questions has FOUR options (A), (B), (C) and (D) . Only ONE

of these four options is correct.

1. Consider two parabola y = x2 – x + 1 and 2 1y x x

2 , the parabola

2 1y x x2

is fixed and

parabola y = x2 – x + 1 rolls without slipping around the fixed parabola, then the locus of the focus of the

moving parabola is

(A) 3

y4

(B) 3

y4

(C) 2y x (D) 2y x

Ans. A

Sol. Let F be the fixed focus and M be the moving focus and T be the varying points of mutual tangency. The

tangent line at T makes equal angle with FT and with a vertical line. This and congruence of the two parabola

imply that MT is vertical and FT = MT and must line on directrix of y = x2 – x + 1

So, 3

y4

2. Three circles C1, C2, C3 with radii r1 , r2, r3 (r1 < r2 < r3) respectively are given as r1 = 2, and r3 = 8 they are

placed such that C2 lies to the right of C1 and touches it externally, C3 lies to the right of C2 and touches it

externally. There exists two straight lines each of which is a direct common tangent simultaneously to all three

circles then r2 is equal to

(A) r2 = 4 (B) r2 = 5 (C) r2 = 10 (D) r2 = 16

Ans. A

Sol.

2 2

2 1 2 1 1 2DE AK r r r r 2 r r

Similarly 2 3EF 2 r r

APD = and PD = x then

1rAD BE CFtanPD PE PF x

32

1 2 1 2 2 3

rr

x 2 r r x 2 2r r r r

Hence, 3 22 1

1 2 2 3

r rr r

2 r r 2 r r

2 1 3 2r r r r 2 8 4

3. Let A, B, C and D be four distinct point on a line in that order. The circles with diameters AC is

x2 + y2 + ax + c = 0 and BD is x2 + y2 – by = 0 intersect at X and Y the line XY meets BC at Z. Let P be a

point on XY other than Z, the line CP intersects the circle with diameter AC at C and M, and line BP intersects

the circle with diameter BD at B and N and the equation of line AM and DN are bx + cy + a = 0 and

cx + ay + b = 0 respectively, then which of the following is true (where is a cube root of unity)

(A) a b c 1 (B) 2a b c 0 (C) 2a b c 0 (D) None of these

Ans. C

P A B C

D E

F



Sol.

a b c

b c a 0

c a b

2 2 2a(bc a ) b(b ac) c(ab c ) 0

3 3 3abc a b abc abc c

3 3 33abc a b c 0

3 3 3 2 2a b c 3abc (a b c)(a b c )(a b c ) 0

4. Let A and B the ends of the diameter 4x – y – 15 = 0 of the circle x2 + y2 – 6x + 6y – 16 = 0. A and B lie on

tangents at the end points of the points of the major axis of an ellipse such that line joining A and B is a

tangent to the same ellipse at a point P. If the equation of major axis of the ellipse is y = x then the distance

between the foci is

(A) 2 2 (B) 4 2 (C) 8 (D) 2 3

Ans. C

Sol. Solve y = x with circle

Circle must passes through the foci of the ellipse 5. Two circle with radii r1 and r2 respectively touch each externally. Let r3 be the radius of a circle that touches

these two circles as well as a common tangents to two circles then which of the following relation is true

(A)

3 1 2

1 1 1

r r r (B)

3 1 2

1 1 1

r r r (C) 3 1 2r r r (D) 3 1 2r r r

Ans. A

Sol. Using length of direct common tangent

21 2

21 3

PQ 4r rPQ PR QR

QR 4r r

3 1 2

1 1 1

r r r

One or More than one correct type This section contains 8 questions. Each questions has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four options is (are) correct.

6. A point M divides A and B in the ratio 1 : 2 where A and B diametrically opposite ends of a circle

2 2x y 5x 9y 22 0 square AMCD and BMEF on the length AM and MB are constructed on the same

side of line AB if co-ordinates of A is (1, 3) then find the orthocenter of ABE

(A) (1, 6) (B) (1, 5) (C) (3, 3) (D) (4, 6)

Ans. B,C

Sol. Shown in the figure since C(1, 5) is the orthocenter of triangle AEB, similarly for the other side the coordinates of C(3, 3)

A B

P R Q

C

A M

D C

(2, 4)

E(0,6) F

B

(1, 3) (4, 6)

(1, 5)



7. If largest and smallest value of y 4

x 3

is p and q where (x, y) satisfy 2 2x y 2x 6y 9 0 then which of the

following is true

(A) 4

p q3

(B) q = 1 (C) 4

p3

(D) 4

pq3

Ans. A,C

Sol. Given circle is 2 2(x 1) (y 3) 1

Let of tangent to circle is 2y 3 m(x 1) 1 1 m

(3, 4) lies on axis

21 2m 1 m 2 2(1 2m) 1 m

2 24m 2m 1 1 m 23m 4m 0

4m 0,

3

y 4 4m 0.or

x 3 3

Smallest is 0 and tangent 4

3

8. A particle P moves on the line y = x, x 0 with constant speed u m/sec. Another particle Q revolves along the

circle 2 2 1x 1 y

9 with constant speed m/ sec.

3

in anticlockwise direction, let at t = 0, both P & Q

simultaneously started, P from the point (0, 0) and Q from the point 2

,03

. Further at any time instant t sec,

let the distance of their separation is r(t)m. Also it was observed that at time t0 sec, both P and Q are at their

closed distance of approach, denoted by rminimum, then choose the correct statements from among the

following

(A) [12 rminimum] = 4 (where [.] denotes the greatest integer function)

(B) 07

t sec4

(C) 0t sec2

(D) 27

t sec2

dr(t) 162r(t) m / sec

dt 3

Ans. A,B

Sol. Equation of common normal y 0 1(x 1)

2 2 1y x 1 ; (x 1) y9

2 12y9

1y

3 2

Clearly shortest distance of approach is along common normal whose equation is x y 1 0

So, when they are at their closet distance P must be at 1 1

,2 2

and Q must be at

(3, 4)

1 1P ,

2 2

2A ,0

3

Q



1 1

1 ,3 2 3 2

covering angle

7

4

on the circle

Since P and Q happened to be at these points at same time so we can have

07 1

t4w 2u

also at this instance

0 min min3 2

t ,r [12r ] [6 2 4] 43 2

Also at any time instant t1 co-ordinate of P, Q are ut ut

,2 2

at Qcos( t) t

1 sin3 3

2 2

2 ut cos( t) ut sin( t)r (t) 13 32 2

9. Let A, B and C be three distinct points on y2 = 8x such that normals at these points are concurrent at P. The

slope of AB is 2 and abscissa of centroid of ABC is 4/3. Which of the following is(are) correct?

(A) Area of ABC is 8sq. units.

(B) Coordinate of P is (6, 0)

(C) Angles between normals are 450, 450, 900

(D) Angles between normals are 300, 300, 1200

Ans. A,B,C

Sol. Let A = (2t12, 4t1), B = (2t22, 4t2) and C = (2t32, 4t3)

Slope of AB = 2 t1 + t2 = 1 and t1 + t2 + t3 = 0 So, t3 = - 1

Also, 2 2 21 2 3 2 2

1 2

2 t t t 4t t 1

3 3

t1 = 1, t2 = 0 A = (2, 4), B = (0, 0) and C = (2, – 4) Hence, P = (6, 0)

10. All x in the interval 0,2

such that 3 1 3 1

4 2sinx cosx

is

(A) 12

(B)

11

36

(C)

13

36

(D)

6

Ans. A,B

Sol.

3 1 3 1

2 2 2 22

sinx cosx

sin cosx cos sinx sin2x12 12

sin x sin2x12

11

n and12 36

11. In a triangle ABC, a

2 3b , and 0C 60 then angle A and B is

(A) A = 1050 (B) B = 150 (C) A = 1150 (D) B = 1050

Ans. A,B



Sol.

a1

A B CbSince tan tana 2 2

1b

0A B 90 and 0A B 120

0A 105 and B = 150

12. Let cosA cosB x;cos2A cos2B y;cos3A cos3B z, then which of the following is true

(A) 2 2y

cos A cos B 12

(B) 21

(2x y 2) cosAcosB4

(C) 22x z 3x(1 y) (D) xyz 0, A,B R

Ans. A,B,C

Sol. 2 2 2x cos A cos B 2cosA cosB

2 2y 2(cos A cos B) 2

2 2y

cos A cos B 12

21

cosA cosB (2x y 2)4

and 3z 2x 3xy 3x

32x z 3x(y 1)

xyz = 3x(y + 1)

xyz = 0 A and B is not true

13. Let A(4,3), B(-4, 3) and C(0, -5) be the vertices of triangle and P(5, 0) let L, M and N be the feet of the

perpendiculars draw from P onto the sides BC, CA and AB respectively, then

(A) area of LMN is 5sq. units (B) the centroid of LMN does not exist

(C) the orthocenter of LMN is the origin (D) L, M and N are collinear

Ans. B,D Sol. P lies on circumcircle So, L,M,N are collinear



inclusive

1. The number of solution in the interval [0, ] of the equation 3 3sin x.cox3x sin3x.cox x 0 is ________

Ans. 5

Sol. Use 33sinx sin3x

sin x4

and 3

3cosx cos3xcos x

4

Equation becomes sin4x = 0

n

x4

2. The number of values of in the range [0, 2] satisfying the equation 4 2 8cos 2 2sin 2 17(sin cos ) is

_________________.

Ans. 4

Sol. Put sin2 = x given equation becomes 4 2 2x 1 17(1 2x x )

Divide by 2x , we have 22

1 1x 17 x 2

xx



Now put 1 7

y x yx 4

or 5

2

Now 7

y4

have no solution and 5

y2

, we have 1

x , 22

So, there are (4 solution in [0, 2])

3. A line divides an equilateral triangle ABC of side 1 unit into two parts with same perimeter and different area

S1 and S2 sq. units. If maximum value of 1

2

S m

S n (where m and n are coprime), then |m - n| is __________

Ans. 2

Sol. x y 1 x 1 y 1

3x y

2

3 1 2S S S

0

1

3

1xysin60

S xy 3 32 xy x xS 23 3

4 4

2 3 1S S S

2 1

3 3

S S 31 1 xy 1 x x

S S 2

1

3

S 3(max)

S 16 as 2

3

S 16 71

S 9 9

4. If the tangent at the point (h, 2) on the hyperbola

2 2

2 2

x y1

a b cuts the circle

2 2 2x y a at the points

Q(x1y1) and R(x2y2), then the value of 1 2

1 1

y y is ___________

Ans. 1

Sol. Equation of tangent 2

2 2 2

xh 2y 2y a1 0;x 1

ha b b

By solving 2

2 2

2

2y a1 y a 0

hb

2 4

2 2

2 4 2

4y 4y a1 y a 0

b b h

4 4 4

2 2

4 2 2 2 2

4a 4 a ay 1 y a 0

b h b h h

1 2

1 2

y y1

y y

5. In a triangle ABC a = 3, b = 4, c = 5, then the value of 1 2 3r r r r

6 is _____________

Ans. 6

Sol. 21 2 3rr r r

B C

A

S1

S2 1 - x 1 - y

1

x y



Paper – 2

PART – A (PHYSICS)

SECTION – I (Single Correct Choice Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

1. A rod ACD bent in shape shown in is moving in the x – y plane with a

velocity ˆv i . The magnetic field in the region is 0ˆB x k as shown in the

figure. What will be the emf induced between points A and D (along the

path ACD) as a function of x coordinate of point A.

(A) 2

0B vL

2 (B)

20B vL

3

(C) 2

0B vL

4 (D)

202 B vL

3

1Sol. C Emf along the path ACD must be equal to rate of change of flux through

ACD, because no emf will be produced in the straight segment AD.

0

L cos / 4 L / 22 3

20 0

00

220

0

20

d B x r r dr

xr rB xr r dr B

2 3

B vLd L dxB

dt 4 dt 4

B vLE

4

2. On an x-y plane the region (l) (x < 0) and region ll (x > 0) are having surface charge

densities + and - respectively. If a charge + q is shifted from (0, 0, -d) to (0, 0, +d).

Then find the work done by electrostatic forces in the above shifting.

(A) Zero (B) 0

2 d

(C) 0

4 d

(D) can’t be determined.

2Sol. (A)

3. The resistivity of a certain metal is , using a wire drawing machine, a wire of uniform cross-section is made

out of a volume V of the metal. The wire is shaped in a square in a square loop, which is placed in a uniform

magnetic field B whose direction is parallel to the loop axis. Find the charge flown through any cross-section

of the wire, when the magnetic field is switched off.

(A) VB

16

(B) VB

4

(C) VB

(D) VB

4

3.Sol. (A)

y

dr

x

+ -

y

x

i

i H

y

x

45o D

C

B L

A



4 . N turns of wire having cross sectional diameter D are tightly wound completely

on a conical core having a semi vertex angle as shown in the figure. The

winding starts at the vertex and there is no space left between the turns If the

current through this coil is I, what will be the magnetic moment of the coil ?

(A) 2 2 2ID N sin

3

(B)

2 2ID N sin

3

(C) 2 3 2ID N sin

3

(D)

2 3ID N sin

3

4.Sol. C

2

ND2 2 3 22

0

dxdM I xsin

D

l sin ID N sinM x dx

D 3

SECTION – I (Multi Correct Choice Type)

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE OR MORE may be correct.

5. Two different coils have self inductances L1 = 4 mH and L2 = 1 mH. The current in one coil is increased at a

constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of

time, the power provided to the two coils is same. At that time the current, the induced emf and the energy

stored in the first coil are i1, 1 and 1 respectively corresponding values for the second coil at the same

instant i2, 2 and 2 respectively. Then choose the correct option.

(A) 1

2

i 1

i 4 (B) 1

2

i4

i

(C) 1

2

4

(D) 1

2

1

4

5.Sol. (A, C, D)

6. An inductor and two capacitors are connected in the circuit as shown in the

figure. Initially capacitor A has no charge and capacitor B has charge CV.

Assume circuit has no resistance at all. At t = 0 switch S is closed, then [given

2

2 4

2LC sec & CV 100mC

10

]

(A) when current in the circuit is maximum, charge on each capacitor is same

(B) when current in the circuit maximum charge on capacitor A is twice the

charge on capacitor B

(C) q = 50 (1 + cos 100 t) mC, where q is the charge on capacitor B at time t

(D) q = 50(1 – cos 100 t) mC, where q is the charge on capacitor B at time t.

6.Sol. (A, D)

7. In a certain series LCR A.C. circuit it is found that XL = 2 XC and phase difference between the current and

voltage is / 4 . If now capacitance is made 1/4th & the inductance and resistance are doubled then which are

not correct.

(A) phase different between current and voltage is zero (B) circuit is at resonance

(C) current becomes 1

2times

(D) current becomes 1 / 2 times

α x

dx

S

L

A C B +Q0=CV

-Q0=-CV

α



7.Sol. ABC

L C

C C

L L

L C L c L C

X 2X

X 4 X

X 2X

X X 2 X 4 X 2 X 2 X 0

8. An inductor L of inductance L and resistance R is a connected across a

source EMF ‘E’ at t = 0 switch K is closed. Mark the correct statement .

(A) The current in the ciruit first increases then becomes constant.

(B) The energy stored in the inductor increases with time and finally

becomes constant.

(C) The power dissipation across R first increases then becomes constant

(D) None of these

8.Sol. ABC

9. In the given circuit, initially the charge on capacitor is Q0. At t = 0, the switch S is closed.

Which of the following statement(s) is/are correct?

(A) Maximum current through the inductor is 0Q

iLC

(B) Charge on the capacitor is zero at T LC2

.

(C) Current through the inductor is maximum at T LC2

.

(D) Maximum magnetic energy can be stored in the inductor is

20Q

2C.

9.Sol. A, B, C, D

Charge on capacitor at time is

q = Q0 cost

where 1

LC

220max

0max

At t ,q 02

t LC2

by conservation of energy

Q 1Li

2C 2

Qi

LC

SECTION – II (Matrix Type)

This section contains 2 questions. Each question has four statements (A, B, C and D) given in Column I and five statements (P, Q, R, S and T) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) in Column II. For example, if for a given question, statement B matches with the statements given in Q and R, then for that particular question, against statement B, darken the bubbles corresponding to Q and R in the ORS.

R L

E

C

L S



1. A charged parallel plate capacitor is connected to a cell of emf through a

resistor as shown in the figure. Then choose the possible curves in Column – ll

for the physical quantities given in Column – 1

Column – l Column – ll

(A) Charge (q) on capacitor (vertical axis) vs time (t) (horizontal axis.)

(P)

(B) Current (l) through the resistor (vertical axis) vs time (t) (horizontal axis.)

(Q)

(C) Electrical energy (U) stored in the capacitor (vertical axis) vs time (t) (horizontal axis).

(R)

(D) Rate of heat generation (P) through the (s) resistor (vertical axis) vs time (t) (horizontal axis).

(S)

1.Sol. (A P, Q; B R, S; C P, Q; D R)

2. Three very large conducting plates A, B and C (having negligible thickness) are

placed along the plane x = -d, x = 0 and x = +d respectively. Charges + 4Q, -6Q and

+4Q are imparted to plates A, B and C respectively. Different regions separated by

the plate are marked as follows.

x < -d Region – l

-d < x < 0 Region – ll

0 < x < d Region – lll

X > d Region – lV

Match the Column-l with Column-ll.

Column – l Column – ll

(A) E is directed along ^

i direction (P) Region – l

(B) E is directed along ^

i direction (Q) Region – ll

(C) Equipotential surface are parallel to y-z plane with dV / dx positive

(R) Region – lll

(D) Equipotential surface are parallel to yz plane with dV/dx negative

(S) Region – IV

2.Sol. (A Q, S; B P, R; C P, R; D Q, S)

l ll lll lV

x

A B C

+4Q -6Q +4Q

S E

R

C



SECTION – III Integer Answer Type

This section contains 8 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9.

1. Two discs of radius R and 2R are made up of same insulating material both has uniform charge density on

one surface. Both the discs are rotated about centrodial axis with constant angular velocity in uniform

magnetic field B. If E1 and E2 are the work done by external agent to provide this energy to the discs.

Calculate E2/2E1.

1.Sol. 8

1

2 2 221 1 1 2 2

1 2 21 2

44

42 2

41 1

q

L 2m

q m R q R q m 4R, q R

2m 2 4 2m 2

R4 R

4

W 4 R 48

2W 2 2R

2. A long wire has uniform charge density = 3 C/m is kept along

z-axis. Another similar wire 2 is lying on x – y plane such that the

minimum separation between them is d = 2m. Calculate the work

done (in Joule) in moving is wire-2 upto origin. (approximately)

2.Sol. 1

y

x

z

Wire-1

Wire-2

=3C/m

=3C/m

2m

B B

+ + + + + + + + + + + +

+ + + + + + +

+ + + + + + +

B B

+ + + + + + + + +

+ + + + + + +

+ + + + + + +



3. Consider the circuit shown in the figure. Capacitors A and B, each have

capacitance C = 2 F. The plates of capacitor A are shorted using a wire of

resistance R 1 while the left plate of capacitor B is given an initial

charge Q = + 4C. The switch is closed at a time t = 0. What will be the initial

current (in ampere) drawn from the battery immediately after the switch is

closed ?

3.Sol. 7 The resistance shorting the capacitor plates may be considered in parallel with the capacitor , and at t = 0,

the capacitor behaves as a short circuit.

Q Q

7Q 7 4C 2C7

R 2CR 2 2 1

4. Given figure is the part of some circuit. Charge on the capacitor in

the circuit is given by tq 3 3 e , (coulomb) where t is time in

seconds. Let VA, VB and VC be the potentials of points A, B and C,

respectively. At t = 0 if VA = 5 V and a constant current i0 =1 Amp

flows through R 1 , then find the VP ( in volt)

4.Sol. 6

10

dqi i 1 3e

dt

P A

diV V L 5 V 1 V 6 V

dt

5. A conducting rod of length 1 m aligned along the unit vector 3 4ˆ ˆj k5 5

, is moving with a velocity

ˆ2i m/ s . The motion take place in a uniform magnetic field ˆ ˆB j 2k tesla. Find the magnitude of

induced emf across the rod in volt.

5.Sol. 4

1Ω

1A

VB

VP

+q

10 V VC

L=1 / 3 H

A

− 10 V

i

Q

2

Q

2

Q

2

Q2

R = 1Ω

I0 = 1A

B VB

VP

q

- q C = 1F

C VC

L=1 / 3 H VA = 5 V

A

A

C C

B +Q

R

R

E = 4Q / C



6. A block A of mass M kept on a rough horizontal surface is connected to a dielectric slab of mass M / 6 and

dielectric constant k by means of a light and inextensible string passing over a fixed pulley as shown in the

figure. The dielectric can completely fill the space between the parallel plate capacitor of plate area and

separation between the plates d kept in vertical position. Initially switch S is open and length of the dielectric

inside the capacitor is b. The coefficient of friction between the block A and the surface is 1/ 4 . Find the

minimum value of the emf (= V) of the battery so that after closing the switch the block A will move. Ignore any

other friction, Given 0

Mgd24

k 1

6.Sol. 2

7. The capacitor C1 in the figure initially carries a charge q0. When the switch S1

and S2 are shut, capacitor C1 is connected in series to a resistor R and a

second capacitor C2, which initially does not carry any charge.

If the heat lost in the resistor after a long time of closing the switch is

20 2

1 1 2

q C

kC C C, then Find the value of k.

7.Sol. 2

8. In the figure shown, the switch is closed at t = 0. The potential difference

across the resistance R 20 , 1 millisecond after closing the switch is K / e.

Find the value of K.

8.Sol. 8

d

V S

b

A

r = 10Ω

R2 =10Ω

R1 = 20 Ω

40 µf

E = 20 V

V

+q0

-q0 C1

S1

S2

C2

R



PART – B (CHEMISTRY)



1. The E2 reaction is carried out with different halogen leaving groups and percentage yield of two products is

given below:

CH3

CH3

+ CH3 - OE2

CH3

CH3+CH2

CH3X

Leaving group Conjugate acid(pka) (% yield of 2-

hexane) (% yield of 1-hexene)

I -10 81% 19%

Br -9 72% 28%

Cl -7 67% 33%

F 3.2 30% 70%

Which of the following statement is/are true in context of above reaction.

(A) Based on pKa’s of conjugate acid l- is best leaving group and F- is poorest leaving group.

(B) When l-, Br-, Cl- are used as leaving group. Zaitsev’s rule is followed.

(C) F- is strongest base (therefore poorest leaving group) and the transition state for reaction with F- as the

leaving group has the least double bong character.

(D) All of the avove. 1.Sol. (D)

2.

(CH3)2SO + BrCH2COOC2H5RLi

III

O

IVIII

The major product (IV) is:

(A)

O COOC2H5

(B)

O

COOC2H5

(C)

CHCOOC2H5

(D)

O

O

2.Sol. B



O OH

Me2 S+

C-HCOOC2H5

O-

COOC2H5

SMe2

O

(IV) 3. Which of the following sequence of reagents can be best utilized for the given transformation

C

O

H

R C

O

D

R

(A) Cl2, AcOH ; D2O ; liAlH4 / H2O

(B) D2O / D3O+ , Cl2, AcOH ; NaBH4 / H2O

(C)

C SH

C SH

H2

H2

, base ; D2O and then HgCl2 / CdCO3 – H2O

(D) Base ;

C SH

C SH

H2

H2

HgCl2 / CdCO3 – H2O and then D2O

3.Sol. C Aldehyde is first converted into cyclic thioacetal and then st. base like Ph – Li is used to generate carbanion and then

treated with D2O. Finally, the aldehyde is regenerated

4. Which of the following is the most likely product from the reaction illustrated by the curved arrows in the

formula ?

SCl

CH3

CH3

CH3

O O

(A)

CH3

CH3

CH3

Cl

+ SO2

(B)

CH3

CH3

CH3

SO

Cl

O

(C) CH3

CH3

CH3

ClO

+ SO

(D) CH3

CH3

Cl

+ SO2

4.Sol. A



SECTION – I (Multi Correct Choice Type) This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE OR MORE may be correct.

5.

R C NH2

O

+ H3 CHD CO

NH2C

Br2 + OH

Mixture of 1° amines

Which of the following statements is / are correct for above reaction

(A) A mixture of four different amines are formed confirming that a free carbonylnitrene intermediate is formed

(B) A mixture of two amines are formed confirming that rearrangement is strictly intramolecular

(C) If R is chiral, it migrates with retention of configuration

(D) A free carbonylnitrene C C

O

NH3

intermediate is formed and only two amines are formed

5.Sol. BC

6. Which combination of reactants will not give the species shown as reactive intermediate

N

O

O Br

HO

(A) 1- Bromo - 4 –nitrobenzene and NaOH (B) Bromobenzene and HONO2

(C) Nitrobenzene, Br2 and water (D) 4- Nitrophenol, Br2 and FeBr3

6.Sol. BCD 1 – Bromo–4–nitrophenol undergoes aromatic nucleophilic substitution by addition–elimination mechanism.

7. In which of the following cases, protonation of nitrogen will change its hybridization :

(A) N

H

(B) N

H

(C) N

H

(D) N

H 7.Sol. (A) is actually aromatic with no lone pair on N and hybridization being sp2. Protonation causes loss of aromaticity and

conversion to sp3 hybridization.



8.

(X)

O

N

H(i)

CH2

Br

excess

(i i)

H3O+

(Y)(i)

OH

OH ; BF3

(ii) KOH/ (iii)Ozonolysis

(iv) Hydrolysis, H3O+

(Z)

Select the correct Statements:

(A)

OCH2CH2

(Y) is

(B) (Z) has three carbonyl groups

(C)

O

OHC CHO(Z) is

(D) Formation of (X) involves equilibrium mixture of the iminium salt and the acylated enamine.

8.Sol. A, B, D

9. Select the correct order with respect to the mentioned properties:

(A) CH3CH2OH > (CH3)2 CHOH > (CH3)2 CDOH > (CH3)3COH (Ease of oxidation)

(B) CH3

CH3

CH2> > > (Heat of hydrogenation)

(C) > > >

C

CH3 CH3

O-

N

C

CH3 CH3

O

N

C

CH3 CH3

O

N

-

C

CH3 CH3

O

N

-

(stabil ity of resonating structures)

(D) PhSO3H > PhCOOH > PhOH > PhCH2NH2 (acidic strength)

9.Sol. A, B, C, D





1. Match the following Column I with Column II.

Column – I ( Reaction) Column – II ( Products)

(A)

OCH3

(i) Na / liq. NH3

(ii) H3O+ / H2O

(P)

O

(B)

CH3(i) O3 / Zn

(ii) OH- /

(Q)

O

(C) C CH CH CH2H2

(i) CH2 = CBr2 /

(ii) aq. KOH

(R) Diel’s Alder reaction

(D)

OH

OH

H+

Possible products

(S) Birch reduction

(T) CH3OH

1.Sol. A P,S,T ; B Q ; C P,R ; D P,Q

(A) OCH3

Na in liq. NH3

OCH3

H3O+

H2O

O

+ CH3OH

(B)

CH3

OO

CH3

OH-

O

OH

O

(i) O3 / Zn

(C)

CH2

CH2

+

Br

CH2

Br

Br

Br

aq. KOHO



(D) OH

OH

H+OH OH

O

OH OHO

2. Match the column

Column l Column ll

(A) CHO2KCN

(P) Cabanion formation

(B) CHO+CH3CHONaOH

(Q) Easy dehydration

(C)

O CH CH2

CH3CH3

(R) Acidic H removal

(D) O

Cl

EtO-

(S) Para substitution

2.Sol. (A P,R; B P, Q, R ; C S; D P, R) A – Benzoin condensation B – Aldol condensation C – Claisen rearrangement D - Favorskii reaction

SECTION – III

Integer Answer Type


1. Reaction of 3, 3, 6, 6 – tetramethyl -1,4 – cyclohexadiene, first with excess aqueous mercuric acetate, then

followed by sodium borohydride reduction, produces a mixture of isomeric C10H20O2 alcohols. Excluding

enantiomers, how many isomeric products may be formed in this reaction ?

1.Sol. 4



CH3CH3

CH3 CH3

Hg (OAc)2 / H2OOH

CH3CH3

CH3 CH3

Hg

OH

HgOAc

+

OH

HgOAcCH3 CH3

CH3 CH3

OH

HgAcO

NaBH4

OHHO

CH3 CH3

CH3 CH3

+

CH3CH3

HO

OHCH3 CH3

cis + trans cis + trans

AcO

2. How many stereo isomers are possible for 1, 3 – dichloro cyclopentane.

2.Sol. 3

(i) cis 1,3-dichloro cyclopentane (meso)

(ii) trans 1,3 – dichloro cyclopentane (Enatiomeric pair)

3. One hydrocarbon ‘P’ having molecular formula C8H10 can form only three monochloroderivative. ‘P’ on

oxidation with KMnO4 yields a compound, which on heating with aq. NH3 yields ‘Q’. ‘Q’ on treatment with

NaOH followed by NaOCl yields a salt which on acidification yields a compound ‘R’ having molecular formula

C7H7O2N. The compound ‘R’ is further treated with (NaNO2 + HCl) and then with base to give dipolar ion ‘S’

which readily losses two gases to give an intermediate. ‘T’ is finally converted into hydrocarbon ‘U’ on

dimerisation. Degree of unsaturation in the hydrocarbon ‘U’ will be

3.Sol. 9



CH3

CH3

(P)

COOH

COOH

C

O

C

O

NH

(Q)

NaOH

COOH

CONH2

NaOClCO2Na

NH2HCl

CO2H

NH2

(R)

NaNO2 + HClCO2H

N2

OH CO2

N2(S)

Benzyne

+N2+CO2

(T)

Dimensation

Biphenylene

(U)

KMnO4 NH3

4. Total number of stereoisomers of the compound will be :

COOH

PhHOOC

Ph

4.Sol. 5

5. Select the total number of methods to convert cyclohexanone into 1, 2-cyclohexanedione.

(I) NaBH4, C. H2SO4 / ∆, cold KMnO4, H2CrO4

(II) Br2 + H3O+; aqueous KOH; oxidation by CrO3 in acetic acid

(III) aqueous NaOH + ; ozonolysis

(IV) SOCl2, NaNH2, cold KMnO4

(V) NaCN + dilute H2SO4; H3O+;

(VI) Br2 + KOH (aqueous);

(VII) Br2 + H3O+; N2H4; KOH + ; cold alkaline KMnO4; HIO4

5.Sol. 3 - (I), (II), (III)



6. How many of the following compounds would undergo intermolecular aldol condensation when treated with

dilute aqueous caustic soda under suitable condition?

O

O

OHCCHO

O

OCH3CHO, PhCOCH3, CH3COCH3, , ,

OH3CO OCH3

,

6.Sol. 5

7. Find the total number of compounds which give yellow precipitate of iodoform when heated with I2 and alkali?

,

O

O

PhCOCH3, CH2 = CH - CHO,

OH

OH CH3CH(OH)CH2CH(OH)CH3, ICH2COCH3, CH3CONH2, CH3COCH2COCH3

7.Sol. 7

8. How many geometrical isomers (excluding enantiomers) are possible for the compound given below?

CH3

CH2

8.Sol. 8



PART – C (MATHEMATICS)



1. Value of 50 50

0 0 0 0

n 1 n 1

cos1 1 sin n sin1 cosn

is equal to

(A) sin 50° (B) sin50° - sin 1° (C) sin 51° - sin 1° (D) 1 / 2

1.Sol. C

0 0

n

0 0

n

0 0

50

a sin n 1 sin n

S sin n 1 sin1

S sin 51 sin 1

2. In ABC, A B C 1

cos cos cos2 2 2 4

then greatest angle of triangle

(A) lies in 0,2

(B) 2 5

lies in ,3 6

(C) 5

lies in ,6

(D)

2lies in ,

2 3

2.Sol. C

Let us assume that A B C; as b c a sinB sinC sinA;

Again A B C

4cos cos cos 12 2 2

1 sinA sinB sinC 2sinA

1sinA and as A is greatest

2

5A

6

3. Sum of series 1 1 1 15 9 15 23

cot cot cot cot ...3 3 3 3

upto infinite terms is equal to

(A) 4

(B)

3

(C)

6

(D)

12

3.Sol. B

4. The value of 8

r 1

tanrA tan r 1 A

, where A = 36° is

(A) -10 – tan A (B) – 10 (C) 10 + tan A (D) – 10 – 2 tan A 4.Sol. B

We have tan rA tan r 1 A cot A tan r 1 A tan A tanA

cot A tan r 1 A tanr A tan A

Now,

8 8

r 1 r 1

tan r A tan r 1 A cot A tan r 1 A tan rA tan A

tan9A 9 tanA cot A 10

SECTION – I (Multi Correct Choice Type)

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE OR MORE may be correct.

5. The value of

20 20

1

p 1q 1

p2tan

q is equal to

(A) 200 (B) 220



(C)

20 20

1 1

p 1q 1

p qtan tan

q p (D) 400

5.Sol. A, C

1 1p qS 2 tan 2 tan

q p

20

r 1

400S 20 200

2 2 2

6. If in a triangle ABC circumradius is 8 times the inradius and B – C = 2

3; then

(A) A 1

sin is equal to2 4

(B) b : c :: 3 5 : 4

(C) b : c :: 3 5 : 2 (D) A B C 1

sin sin sin2 2 2 32

6.Sol. A, C, D

R = 8r R = 32R A B C A B C 1

sin sin sin 2sin sin sin2 2 2 2 2 2 16

or B C B C A 1

cos cos sin2 2 2 16

1 A a 1 A 1sin sin sin

2 2 2 16 2 4

Again

B C b c Atan cot

2 b c 2

b c3 15

b c

b : c :: 3 5 : 2

7. Let z satisfies arg z 3 4i arg z 1 2i arg iz arg z and z1 is a fixed complex number such that

1z z is constant then

(A) z lies on a semi – circle (B) 8

1z is equal to 625

(C) argument of z1 is 4

(D) 1arg of z 2 is

2

7.Sol. ABD

c 1

8

1 1

z 3 4iarg

z 1 2i 2

3 4i 1 2iso z 2 i z

2

z 625 arg z 22

8. A(2,0) is a point on the circle (x + 2)2 + (y - 3)2 = 25, A line through A(2, 0) making an angle of 450 with the

tangents to the circle at A is drawn. Then the equation of the circles with the centres on these lines are at a

distance of 5 2 units from point A and of radius 3 are

(A) 2 2(x 1) (y 7) 9 (B) 2 2(x 3) (y 7) 9

(C) 2 2(x 9) (y 1) 9 (D) 2 2(x 9) (y 1) 9

8Ans. A,B,C,D



Sol. Equation of tangents at A is 4x 3y 8 0

Let y = m(x- 2) a line thro(2,0) and m = -7 or 1

7

We have x 2 y

5 21 7

50 50

Centre of circle are (1,7),(3, 7)(9,1) and (-5, -1)

9. Let x and y represent the sum and product of two sides of a triangle such that 2 2x y z where z is the third

side, then (A) Triangle is acute angle triangle

(B) Inradius of the triangle is y 3

x z

(C) Area of the triangle is y 3

4

(D) All the above.

9.Sol. BC

(A) Let b c x and bc y and a z

Then b c a b c a bc

s s a 1 A

cos cos A 120bc 2 2 3

(B)

0y1 bc sin A sin120y 32 2r

a b c x zs 2 x z

2 2

(C) 1 1 3 y 3

bc sinA y2 2 2 4



1. Match the conics in Column I with the statements/ expression in Column - II

Column – I Column – II

(A) Circle (P) The locus of the point (h, k) for which the line hx + ky = 1 touches the circle x2 + y2 = 4

(B) Parabola (Q) Points z in the complex plane satisfying |z + 2| - |z – 2| = 3

(C) Ellipse (R) Pints of the conic have parametric representation 2

2 2

1 t 2tx 3 ,y

1 t 1 t

(D) Hyperbola (S) The eccentricity of the conic lies in the interval 1 x <

(T) Points z in the complex plane satisfying Re (z + 1)2 = |z|2 + 1

ANS. A - P, B - ST, C - R, D- QS



2. Match the following

Column – I Column – II

(A) The maximum value of

2

1

2

7 5 x 3sec

2 x 2

is (P) 2

3

(B) In minimum value of 1 2 1 25 1cosec 3x sec 3x

4 4

is (Q) 6

(C)

In a ABC , B is acute and c > b > c sin B , then ratio of areas of

circumcircles of triangle ABC1 and ABC2 ( ambiguous case) is [x], then x can be (where [.] denotes the greatest integer function)

(R) 5

12

(D)

Radius of largest circle which passes through the focus of the parabola 2y 4x 0 and is contained in is [2x], then x can be (where [.] denotes

the greatest integer function)

(S) 1

2.Sol. A P; B Q; C R,S; D P

(A)

2

1 1

22

7 5 x 3 1 5sec sec

2x 22 x 2

2 2

1 1 1 5 22

2 2 3x 2 x 2

(B) minimum value 1 1 2 1cosec 2 sec 1 where 3x 1

6 4

(C) 1 2AC AC

R2sinB 2sinB

(D) Equation of circle is 2 2 2x r 1 y r

Putting y2 = 4x and taking discriminant D = 0

r 4 SECTION – III

Integer Answer Type


1. ABC is a right-angled triangle with AC = 65 cm and B = 900. If r = 7 cm, if area of triangle ABC is abc (abc is three digit number) then (a – c) is ____________

1.Sol. 1

r = 7cm and R = 32.5 cm; Again using r = A B C

4Rsin sin sin2 2 2

0 0

2 2

2

7r R[cosA cosB cosC 1] cosA cos90 cos 90 A 1

32.5

39.5 79cosA sinA

32.5 65

79 65sin2A

65

D C2 C1 B

A



Now = 2R2 sinA sinC = R2sin2A =

2 2 2

2

65 79 65 144 1472 7sq.units.

2 465

2. Let r1 and r2 represent the radii of the smallest and largest circles passing through 1

a,a

and touching the

circle 4x2 + 4y2 – 4x – 8y – 31 = 0, where a is the least possible value of ratio of the sides of the regular n

sided (n 3) polygons inscribed and circumscribed to a unit circle, then r1 + r2 is ____________

2.Sol. 3

We have ratio of n sided regular polygon inside and outside unit circle =

2sinn cos ;

n2tan

n

which is least

when n = 3 (as n 3) 1

a2

Now

1,2

2 lies inside the circle with centre

1,1

2 and radius 3; Sum of radii of smallest and largest circle is R

Also, 2(r1 + r2) = 2R r1 + r2 = 3

3. Chord AB of the circle x2 + y2 = 100 passes through the point (7, 1) and subtends an angle of 600 at the

circumference of the circle. If m1 and m2 are the slopes of two such chords, then the value of |m1m2| is

Ans. 1

Sol.

Let the slope of the chord through point (7, 1) be m.

The equation of the line: y – 1 = m(x – 7)

or mx – y + 1 = 7m = 0

Perpendicular distance from (0, 0) = r

2

2

7m 1 r5

21 m

(7m – 1)2 = 25(1 + m2)

49m2 – 14m + 1 = 25 + 25m2

24m2 – 14m – 24 = 0

m1m2 = - 1

4. If a tangents of slope 4 of the ellipse 2 2

2 2

x y1

a b is normal to the circle 2 2x y 4x 1 0 then the

maximum value of ab is

Ans. 8

Sol .Equation of tangent is 2 2y 4x 16a b this normal passes through (-2,0)

2 2 2 20 8 16a b ;264 16a b

Giving AM GM

2 2

2 216a b 16a b2

2 264

16a b2

4ab 32

ab 8 5. The line through P , perpendicular to the chord of contact of the tangents drawn from the point P to the

parabola y2 = 16x touches the parabola x2 = 12y, then the locus of P is 2ax + 3y + 4a2 = 0 then a is ______



Ans. 4

Sol. Let P(x1, y1) equation of chord of contact yy1 = 8(x + x1)

Equation of line perpendicular to this and passes through P is 11 1y

y y (x x )8

This touches x2 = 12y

Locus is 8x 3y 64 0

6. Find the length of focal chord of the parabola x2 = 4y which touches the hyperbola 2x 4y 1 __________

Ans. 9

Sol. Let point 21 1(2t ,t ) and 2

2 2(2t ,t ) be the extremities of focal chord

t1t2 = -1

Equation of focal chord 1 2x(t t ) 2y 2 0 and line 2 1y mx m

4 is tangent to hyperbola passes

through (0, a)

1 21

m 1 t t4

Length of focal chord = 9

7. Let x, y R. If the maximum and minimum value of expression 2 2

2 2

x yE

x xy 4y

are M and m, then the

value of 3

M m2

is _____________.

7.Sol. 2

Let x = r cos and y = r sin

2 2

1E

4sin sin cos cos

2

5 sin2 3cos2

Now 5 10 5 sin2 3cos2 5 10

2 2

M and m5 10 5 10

2 2M 5 10 and m 5 1015 15

3 3 2

M m 10 22 2 15

8. If 2 22

2xtan x y cot x y 1

1 x

where x, y R+ and if the least possible value of y satisfying

the above is k, then the value of 8k is ___________. 8.Sol. 2

LHS 2; RHS 2 Equality is possible. Then for RHS x = 1

for LHS tan2 (1 + y) = 1

tan2y = 1

Least value of y is 1

4

8k 2

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AANNSSWWEERR,, HHIINNTTSS && SSOOLLUUTTIIOONN TEST 2 (ADVANCED...1 2 0 4 r 80 Heat t dt R3 Joule 11. The potential difference across a 10H inductor as a function of time is shown in