Physics

1. Two sources of sound S1 and S2 produce sound waves of same frequency 660 Hz. A listener

is moving source S1 towards S2 with a constant speed u m/s and he hears 10 beats /s. The

velocity of sound is 330 m/s. Then, u equals:

A. 2.5 m/s

B. 10.0 m/s

C. 15.0 m/s

D. 5.5 m/s

Ans. A

Solution- ⎯⎯→1 2uS S

For frequency from S1

1

vf f

− =

1

330f 660

330

− =

For frequency from S2

2

v 330f f 660

330

+ + = =

Beat frequency f = f2 – f1

10 = 2 (330 + μ) –2 (330 – μ)

μ = 2.5 m/s 2. The electron in a hydrogen atom first jumps from the third excited state to the second excited

state and subsequently to the first excited state. The ratio of the respective wavelengths, λ1

/ λ2, of the photons emitted in this process is :

A. 27 / 5

B. 20 / 7

C. 7 / 5

D. 9 / 7

Ans. B

Solution- For the wavelength

2 21 2

1 1 1R

n n

= −

From n2 = 4 to n1 = 3

1 1

1 1 1 1 7RR

9 16 9 16

= − → =

For the second jump from n = 3 to n = 2

2

1 1 1R

9 4

= −

2

1 5R

4 9=

1

2

5 9 16 20

36 7 7

= =

1

2

20

7

=

3. A small speaker delivers 2 W of audio output. At what distance from the speaker will one

detect 120 dB intensity sound ? [Given reference intensity of sound as 10-12 W / m2]

A. 10 cm

B. 30 cm

C. 20 cm

D. 40 cm

Ans. D

Solution- For 120 dB sound

Sound level = 0

I10log

I

12

I120 10 log

1 10−

=

I = 1 W/m2

Now, for distance

2 2

P 2I

4 r 4 r= =

2

21 r 0.40m

4 r= → =

r 40 cm=

4. In the given circuit, the charge on μF capacitor will be :

A. 24 μC

B. 5.4 μC

C. 9.6 μC

D. 13.4 μC

Ans. A

Solution- For the upper branch

4μF in series with (1μF + 5μF)

eq

4 6C 2.4 F

4 6

= =

+

Change on 4μF capacitor =Ceq × v = 2.4 × 10 μC = 24 μC

Charge on 4μF capacitor is 24μC

5. A smooth wire of length 2 r is bent into a circle and kept in a vertical plane. A bead can slide

smoothly on the wire. When the circle is rotating with angular speed ω about the vertical

diameter AB, as shown in figure, the bead is at rest with respect to the circular ring at position

P as shown. Then the value of ω2 is equal to :

A. ( )2 / 3g r

B. 2 /g r

C. 3

2

g

r

D. ( )3 /g r

Ans. A Solution-

r = R sin θ in the vertical direction

N cos θ – mg = 0

N cos θ = mg ….(1)

In horizontal direction

2 rN sin m 0

2− =

2 rNsin m

2= ….(2)

2 2m r r

tanmg 2g

→ = =

2r r

2g3r

=

=2 2g

r 3

6. Consider the LR circuit shown in the figure. If the switch S is closed at t = 0 then the amount

of charge that passes through the battery between t = 0 and L

tR

= is :

A. 22.7

EL

R

B. 27.3

EL

R

C. 2

7.3EL

R

D. 2

2.7EL

R

Ans. A

Solution- In an LR circuit

I = I0 (1 – e–t/τ)

L

R =

For charge passing

t/0

0 0

Q I dt I (1 e ) dt

− = = −

t/

0

EQ e

R

− = +

E

QR e

= + −

2

R E L ELQ

R e Re R 2.7R

= = =

2

ELQ

2.7R=

7. A tuning fork of frequency 480 Hz is used in an experiment for measuring speed of sound (𝒱)

in air by resonance tube method. Resonance lengths of the air column, l1 = 30 cm, l2 = 70

cm, Then, 𝒱 is equal to :

A. 384 ms-1

B. 332 ms-1

C. 379 ms-1

D. 338 ms-1

Ans. A

Solution- For the wavelength of sound

= −2 12( )

2(740 30) 80 cm = − =

0.8 cm =

Speed = f

Speed = 0.80 × 480

Speed = 384 m/s

8. A solid sphere, of radius R acquires a terminal velocity 𝒱1 when falling (due to gravity) through

a viscous fluid having a coefficient of viscosity . The sphere is broken into 27 identical solid

spheres. If each of these spheres acquires a terminal velocity, 𝒱 2, when falling through the

same fluid, the ratio (𝒱 1/ 𝒱 2) equals:

A. 1/9

B. 9

C. 27

D. 1/27

Ans. B

Solution- For terminal velocity

2

T 0

2 rV ( )g

9= −

Hence, VT α r2

For the ratio of radius

3 34 4R 27. r

3 3 =

R

r3

→ =

2 2

12 2

2

V R R9

V r R

9

= = =

1

2

V9

V=

9. Three particles of masses 50 g, 100 g and 150 g are placed at the vertices of an equilateral

triangle of side 1 m (as shown in the figure). The (𝒳,𝒴) coordinates of centre of mass will be

:

A. 3 7

,8 12

m m

B. 7 3

,12 8

m m

C. 3 5

,4 12

m m

D. 7 3

,12 4

m m

Ans. D

Solution- For position of center of mass

+ +

=+ +

1 1 2 2 3 3cm

1 2 3

m r m r m rr

m m m

+ + +

=+ +

oˆ ˆ ˆ50(0) 100 i 150(0.5 i 1.0 sin60 j)

50 150 100

cm

7 3 7 3ˆ ˆr i j , m12 4 12 4

= + =

10. A system of three polarizers P1, P2, P3 is set up such that the pass axis of P3 is crossed with

respect to that of P1. The pass axis of P2 is inclined at 60° to the pass axis of P3. When a beam

of unpolarized light of intensity I0 is incident on P1, the intensity of light transmitted by the

three polarizes is I. The ratio (I0/I) equals (nearly):

A. 10.67

B. 5.33

C. 1.80

D. 16.00

Ans. A

Solution- After the first polarizer

01

II

2=

Now, as second polarizer is at 30° with respect to first

I2 = I1 cos2θ

( )2

20 02

I I 3I cos 30

2 2 2

= =

02

3II

8=

Now, third polarizer is at 60° with respect P2

2

203 0

3I 3 1I cos 60 I

8 8 2

= =

03

3II I

32= =

0I 3210.67

I 3→ = =

11. The ratio of the weights of a body on the Earth’s surface to that on the surface of a planet is

9 : 4. The mass of the planet is 1

9th of that of the Earth. If ‘R’ is the radius of the Earth, what

is the radius of the planet? (Take the planets to have the same mass density).

A. 2

R

B. 4

R

C. 9

R

D. 3

R

Ans. A

Solution- earth earth earth

planet planet planet

W mg g9

W 4 mg g= = =

as 2

GMg

R=

22p pe e

2p p ee

R RGM M9

4 GM M RR

= =

as p e

1M M

9=

2

p

e

R99

4 R

=

p

e

R 1 1

R 4 2= =

ep

R RR

2 2= =

12. In an amplitude modulator circuit, the carrier wave is given by,

C(t) =4 sin (20000 t)while modulating signal is given by, m(t) = 2 sin (2000 t) . The values of

modulation index and lower side band frequency are :

A. 0.5 and 10 kHz

B. 04 and 10 kHz

C. 0.3 and 9 kHz

D. 0.5 and 9 kHz

Ans. D

Solution- Modulation index m

c

A

A=

20.50

4= =

For carrier wave frequency

4cc

20000f 10 Hz 10KHz

2 2

= = = =

Message wave frequency

m

2000f 1000Hz 1KHz

2

= = =

Lower side bond frequency = fc – fm = 10 –1 = 9 KHz

Hence, the correct option is 0.5 and 9KHz

13. An electron, moving along the 𝒳 –axis with an initial energy of 100 eV, enters a region of

magnetic field B = (1.5 x 10-3T) k at S (see figure). The field extends between 𝒳 = 0 and

𝒳 =2 cm. The electron is detected at the point Q on a screen placed 8 cm away from the point

S. The distance between P and Q (on the screen) is :

(electron’s charge = 1.6 x 10-19C, mass of electron = 9.1 x 10-31 kg)

A. 1.22 cm

B. 11.65 cm

C. 2.25 cm

D. 12.87 cm

Ans. D

Solution- K = 100 ev

3 ˆB 1.5 10 Tk−=

For the radius of path

q

mvR

B=

q

2m(ke)R

B=

3.1 19

3 19

2 9.1 10 100 1.6 10R

1.5 10 1.6 10

− −

− −

=

R = 2.248 cm

In the figure

2

sin2.248

=

AB AB 2

tanBD 6 1.026

= = =

AB = 11.69 cm

AC = 11.69 + 2.248 (1 – cosθ)

AC 12.87 cm=

14. A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure.

What would be the heat energy absorbed by the gas, in this process ?.

A. 40 J

B. 30 J

C. 35 J

D. 25 J

Ans. C

Solution- For a diatomic gas, at constant pressure p

7RC

2=

For heat

ΔQ = nCp Δ J

W = 10 J

W = 10 = nR ΔT

ΔQ = nCpΔT = 7R

n T2

7 7

Q nR T 102 2

= =

Q 35J =

15. A carnot engine has an efficiency of 1/6, When the temperature of the sink is reduced by 62°

C, its efficiency is doubled. The temperatures of the source and the sink are, respectively,

A. 99° C, 37° C

B. 62° C, 124° C

C. 37° C, 99° C

D. 124° C, 62° C

Ans. A

Solution- Let temperature of sin k is Tc and source is Tn

Efficiency, c

n

T1

T= −

c c

n n

T T1 51

6 T T 6= − → =

When Tc is reduced by 62°C

c c

n n n

T 62 T2 1 621 1

6 T 3 T T

−= − → = − +

nn

1 5 621 T 372K 99 C

3 6 T= − + → = = 2

c

5T 372 310K 37 C

6= = =

(99 C, 37 C)→

16. A plane electromagnetic wave having a frequency 𝒱 = 23.9 GHz propagates along the positive

z – direction in free space. The peak value of the Electric Field is 60 C/m. Which among the

following is the acceptable magnetic field component in the electromagnetic wave ?

A. ( )7 2 112 10 sin 1.5 10 0.5 10B x t j−= +

B. ( )7 3 112 10 sin 0.5 10 1.5 10B z t i−= +

C. ( )7 3 112 10 sin 0.5 10 1.5 10B z t i−= −

D. ( )3 1160sin 0.5 10 1.5 10B x t i= +

Ans. C

Solution- 00 8

E 60B

c 3 10= =

B0 = 20 × 10–8 T = 2 × 10–7T

ω = 2πf = 2π × 23.9 × 109 = 1.5 × 1011 rod/s

ω = 1.5 × 1011 rod/s

=vk

11

1

8

1.5 10k 500 m

v 3 10

− = = =

−= − 7 3 11 ˆB 2 10 sin(0.5 10 z 1.5 10 t)i

17. A uniform cylindrical rod of length L and radius r, is made from a material whose Young’s

modulus of Elasticity equals Y. When this rod is heated by temperature T and simultaneously

subjected to a net longitudinal compressional force F, its length remains unchanged. The

coefficient of volume expansion, of the material of the rod, is nearly) equal to :

A. 6F / ( r2YT)

B. 9F / ( r2YT)

C. 3F / ( r2YT)

D. F / (3 r2YT)

Ans. C Solution-

For the length to remain unchanged

thermal stress

Compression strainY

=

2

F LT

LY r

= =

2

F

YT r =

For 3 =

2 2

3F

YT r =

18. Figure shows a DC voltage regulator circuit, with a Zener diode of breakdown voltage = 6V.

If the unregulated input voltage varies between 10 V to 16 V, then what is the maximum

Zener current ?

A. 1.5 mA

B. 7.5 mA

C. 2.5 mA

D. 3.5 mA

Ans. D

Solution- When in breakdown state

Vz = Breakdown voltage = 6V

Current in 4kΩ

zV 6I(4k ) 1.5mA

4 4 = = =

Now, when input is 16V

s

16 6I 5mA

2

−= =

Zener current = Is – I(4k Ω) = 5 – 1.5

Max. Zener current = 3.5 mA

19. One kg of water, at 20° C, is heated in an electric kettle whose heating element has a mean

(temperature averaged) resistance of 20 Ω. The rms voltage in the mains is 200 V. Ignoring

heat loss from the kettle, time taken for water to evaporate fully, is close to :

[Specific heat of water = 4200 J/kg °C), Latent heat of water = 2260 kJ/kg]

A. 10 minutes

B. 16 minutes

C. 3 minutes

D. 22 minutes

Ans. D

Solution- Heat supplied = Power x time = P × t

Heat needed = heat needed to heat water + evaporate water

P × t = msΔT + mLv

2

3200t 1 420080 2260 10

20 = +

1298

t 1298 & min60

= =

t 22min

20. Half lives of two radioactive nuclei A and B are 10 minutes and 20 minutes, respectively. If,

initially a sample has equal number of nuclei, then after 60 minutes, the ratio of decayed

numbers of nuclei A and B will be:

A. 3 : 8

B. 1 : 8

C. 9 : 8

D. 8 : 1

Ans. C

Solution- After a time t

N = N0 (0.5)t/τ

→ half life

Decayed atoms, 0N N N = −

t/0N N (1 (0.5) ) = −

Now, for A, after 60 min

60

610A 0 0N N (1 (0.5) ) N (1 0.5 ) = − = −

0A

63NN

64 =

For B

60

320B 0 0N N 1 0.50 N (1 0.5 )

= − = −

0B

7NN

8 =

Ratio = 0A

0B

63NN 8 9

64 7N 8N= =

21. Two particles are projected from the same point with the same speed u such that they have

the same range R, but different maximum heights, h1 and h2. Which of the following is correct

?

A. R2 = 4 h1h2

B. R2 = 16 h1h2

C. R2 = h1h2

D. R2 = 2 h1h2

Ans. B

Solution- For the particle at same range angle of projection = θ1 90 – θ

22sin cos

Rg

=

For max. height

2 2

1

sinh

2g

=

2 2 2 2

2

sin (90 ) cosh

2g 2g

− = =

4 2 2

1 2

sin cosh h

4g

=

2 2

1 2 2

(2 sin cos )h h

16 g

=

2

1 2

Rh h

16=

21 2R 16h h=

22. A particle is moving with speed v b x= along positive 𝒳 –axis. Calculate the speed of the

particle at time t = (assume that the particle is at origin at t = 0).

A. b2

B.

2

2

b

C.

2

2

b

D.

2

4

b

Ans. C

Solution- V b x=

Differentiating w.r.t. x

dv 1 dx

bdt dt2 x

=

dv bv b(b x)

dt 2 x 2 x= =

2dv b

dt 2=

2b

dv dt2

=

2b

v t2

=

Now, at t = τ

2b

v2

=

23. The number density of molecules of a gas depends on their distance r from the origin as, n(r)

= n0 4re −. Then the total number of molecules is proportional to :

A. 1

20n

B. 1

40n

C. 3

0n −

D. 3/4

0n −

Ans. D

Solution- 4r

0n(r) n e−=

For total no. of molecules = 0

n(r)dv

4r 4

0

0

N n e 4 r dr

−=

−→ = 3/40N n

24. A block of mass 5 kg is (i) pushed in case (A) and (ii) pulled in case (B), by a force F = 20 N,

making an angle of 30° with the horizontal, as shown in the figures. The coefficient of friction

between the block and floor is μ = 0.2 the difference between the acelerations of the block,

in case (B) and case (A) will be : (g = 10 ms-2)

A. 0.4 ms-2

B. 0.8 ms-2

C. 3.2 ms-2

D. 0 ms-2

Ans. B

Solution- For A

N = 50 + 20 sin30° = 60N

For acceleration

5 × aA = 20 cos30° – μ × N

A

35a 20 0.2 60

2= −

A

10 3 12a

5

−=

For B

N + 20 sin30° – 50 = 0

N = 40N

For acceleration

5 aB = 20 cos 30° – 0.2 × 40

B

10 3 8a

5

−=

2B A

4a a 0.8 m / s

5− = =

25. Let a total charge 2 Q be distributed in a sphere of radius R, with the charge density given by

p(r)=kr, where r is the distance from the centre. Two charges A and B, of – Q each, are placed

on diametrically opposite points, at equal distance , a from the centre, If A and B do not

experience any force then.

A. 1/42a R−=

B. 1

4

3

2

Ra =

C. 1/48a R−=

D. / 3a R=

Ans.

Solution- ρ(r) = kr

For charges to experience zero force, net electric field at distance = 0

Now, for field inside the shell using gauss law

a 2

2

00

kr4 r drE 4 a

a42

0 0

4 k rE 4 a

4

=

4

0

k aE

4

=

Now, for k in terms of Q

R

2

0

2Q kr4 r dr=

R4 4

0

r 4 kR2Q k4

4 4

= =

4

2Qk

R=

Now, for field to be zero

( )

− =

=

=

=

=

=

2

20

2

20

2

4 20 0

2

4 2

4 4

1/4

k a kQ0

4 2a

k a KQ

4 4a

2Q a 1 Q

4 4R 4a

2a 1

R 4a

R 8a

a 8 R

26. A spring whose unstretched length is l has a force constant k. The spring is cut into two pieces

of unstretched lengths l1 and l2 where, l1=nl2 and n is an integer. The ratio k1/k2 of the

corresponding force constants, k1 and k2 will be :

A. 2

1

n

B. n

C. n2

D. 1

n

Ans. D

Solution- For the spring constant

as Δx α l

For constant force

F = k Δx

1 1

kx

C

k =

Hence, 1 21 2

C Ck , k= =

1 22 2

C Ck , k

n= =

1 2

2 2

k C 1

k n C n= =

1

2

k 1

k n=

27. Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a

current of 5 A. (see figure) (7

0 4 10 −= N – A -2)

A. 1.5x10-5T

B. 2.5x10-5T

C. 2.0x10-5T

D. 3.0x10-5T

Ans. A Solution-

d 25 9 4cm= − =

0IB (sin sin )

4 d

= +

0IB (2sin )

4 d

=

74 10 5 3

B 24 0.04 5

− =

B = 1.5 × 10–5 T

28. Consider an electron in a hydrogen atom, revolving in its second excited state (having radius

4.65o

A ). The de-Broglie wavelength of this electron is :

A. 3.5o

A

B. 9.7o

A

C. 6.6o

A

D. 12.9o

A

Ans. B

Solution- For an electron rotating in orbit

n n2 r n =

For second excited state, n = 3

3 32 (r ) 3 =

10

3

2 (4.65) 10

3

− =

3 9.7Å =

29. A moving coil galvanometer, having a resistance G, produces full scale deflection when a

current Ig flows through it. This galvanometer can be converted into (i) an ammeter of range

0 to 0 0( )gI I I by connecting a shunt resistance RA to it and (ii) into a voltmeter of range O to

V (V=GI0) by connecting a series resistance RY to it. Then,

A. 2

A VR R G= and

2

0

gA

V g

IR

R I I

= −

B. 2

A VR R G=0

g

g

I

I I

−

and

A

V

R

R=

2

0 g

g

I I

I

−

C. 2

A VR R G= and A

V

R

R=

0

g

g

I

I I−

D. 2

A VR R G=0 g

g

I I

I

−

and

2

0

gA

V g

IR

R I I

= −

Ans. A

Solution- When a galvanometer is used ammeter

g G 0 g AI R (I I )R= −

g

A A0 g

IR R

I I

= −

For voltmeter

Ig = (RG + Rv) = v

For v = GI0 = RGI0

0 g Gv

g

(I I )RR

I

−=

2A v 2 A vG

R R R R R G= =

2

gA

v 0 g

IR

R I I

→ =

−

30. A transparent cube of side d, made of a material of refractive index μ2, is immersed in a liquid

of refractive index μ1 (μ1 < μ2). A ray is incident on the face AB at an angle (shown in the

figure). Total internal reflection takes place at point E on the face BC.

Then must satisfy :

A.

21 2

2

2

sin 1

− −

B. 1 1

2

sin

−

C. 1 1

2

sin

−

D.

21 2

2

2

sin 1

− −

Ans. A

Solution-

For TIR at BC

2 c 1sin sin90

2 c 1sin

1c

2

sin

Now, using shell’s Law at AB

1 2 csin sin(90 ) = −

1 2 csin cos( ) =

1 2 csin cos→ (for TIR)

2 221 2 1

2

sin

−

2 22 1

1

sin −

2221

sin 1

−

Chemistry

1. What will be the major product when m-cresol is reacted with propargyl bromide (HC = C —

CH2Br) in presence of K2CO3

A. B.

C. D.

Ans. C

Solution-

Above anion act as nucleophile for SN2 attack on CH C — CH2 — Br and acetone acting as polar

aprotic solvent.

2. Which one of the following is likely to give a precipitate with AgNO3 solution?

A. CH2 = CH — Cl B. CHCl3 C. (CH3)3CCl D. CCl4

Ans. C

Solution- (CH3)3CCl will give Cl– readily forming tertiary carbocation. Hence (CH3)3CCl likely to give a

precipitate with AgNO3 Solution-

3. The INCORRECT match in the following is:

A. ΔG° < 0, K > 1 B. ΔG° < 0, K < 1 C. ΔG° = 0, K = 1 D. ΔG° > 0, K <

1

Ans. B

Solution- ΔG° = - 2.303 RT log Keq

If ΔG°<0

-2.303RT logK<0

logK > 0

K>1

4. Benzene diazonium chloride on reaction with aniline in the presence of dilute hydrochloric acid

gives :

A. B.

C. D.

Ans. B

Solution-

5. The INCORRECT statement is:

A. LiNO3 decomposes on heating to give LiNO2 and O2.

B. Lithium is least reactive with water among the alkali metals.

C. LiCl crystallises from aqueous solution as LiCl.2H2O.

D. Lithium is the strongest reducing agent among the alkali metals.

Ans. A

Solution- 3 2 2 2

12LiNO Li O 2NO O

2

⎯⎯→ + +

6. The C–C bond length is maximum in

A. graphite B. C70 C. diamond D. C60

Ans. C

Solution-Carbon in diamond is sp3 hybridized. In graphite and fullerene there is both C — C and C = C in

conjugation, hence there is partial double bond character between carbon atoms.

7. Consider the following reactions :

‘A’ is:

A. CH2 = CH2 B. CH3 – C CH

C. CH CH D. CH3 – C C – CH3

Ans. B

Solution-

8. In which one of the following equilibria, Kp Kc?

A. 2 22NO(g) N (g) O (g)+ B. + 22C(s) O (g) 2CO(g)

C. 2 2 3NO (g) SO (g) NO(g) SO (g)+ + D. 2 22Hl(g) H (g) l (g)+

Ans. B

Solution- Kp=Kc (RT) Δng If Δng = 0

Kp = Kc

If Δng 0, Kp Kc

Hence (B) is correct answer.

9. The coordination numbers of Co and Al in [Co(Cl)(en)2]Cl and K3[Al(C2O4)3], respectively, are

A. 6 and 6 B. 5 and 3 C. 3 and 3 D. 5 and 6

Ans. D

Solution- en and 22 4C C − are a bidentate ligand. So coordinate number of [Co(Cl)(en)2]Cl is 5 and

K3[Al(C2O4)3] is 6

10. Which of the given statements is INCORRECT about glycogen?

A. It is present in some yeast and fungi

B. It is present in animal cells

C. Only α-linkages are present in the molecule

D. It is a straight chain polymer similar to amylase

Ans. D

Solution- Glycogen is a multibranched polysaccharide.

11. The primary pollutant that leads to photochemical smog is:

A. acrolein B. nitrogen oxides

C. ozone D. sulphur dioxide

Ans. B

Solution- NO2 and Hydrocarbons are primary precursors of photochemical smog.

12. The ratio of number of atoms present in a simple cubic, body centered cubic and face centered

cubic structure are, respectively :

A. 1 : 2 : 4 B. 4 : 2 : 3 C. 4 : 2 : 1 D. 8 : 1 : 6

Ans. A

Solution- No. of atoms simple cubic = 1, bcc = 2 and fcc = 4

13. Thermal decomposition of a Mn compound (X) at 513 K results in compound Y, MnO2 and a

gaseous product. MnO2 reacts with NaCl and concentrated H2SO4 to give a pungent gas Z. X, Y

and Z, respectively.

A. K2MnO4, KMnO4 and SO2 B. K3MnO4, K2MnO4 and Cl2

C. K2MnO4, KMnO4 and Cl2 D. KMnO4, K2MnO4 and Cl2

Ans. D

Solution-

14. An ' Assertion' and a 'Reason' are given below. Choose the correct answer from the following

options.

Assertion (A): Vinyl halides do not undergo nucleophilic substitution easily.

Reason (R): Even though the intermediate carbocation is stabilized by loosely held p-electrons,

the cleavage is difficult because of strong bonding.

A. Both (A) and (R) are correct statements but (R) is not the correct explanation of (A)

B. Both (A) and (R) are wrong statements

C. Both (A) and (R) are correct statements and (R) is the correct explanation of (A)

D. (A) is a correct statement but (R) is a wrong statement.

Ans. D

Solution- Due to resonance there is partial double bond character between carbon and chlorine,

hence it does not undergoes nucleophilic substitution reaction.

15. The molar solubility of Cd(OH)2 is 1.84 × 10–5 M in water. The expected solubility of Cd(OH)2 in a

buffer solution of pH = 12 is :

A. 6.23 × 10–11 M B. 1.84 × 10–9M C. 92.4910 M

1.84− D. 2.49 × 10–

1M

Ans. D

Solution- Ksp of Cd(OH)2 = 4s3 = 4 × (1.84 × 10–5)3

If pH = 12

pOH = 2

[OH–] = 10–2 M

4 × (1.84 × 10–5)3 = [Cd2+] [OH–]2

3 15

2

4

4 (1.84) 10[CD ]

10

−+

−

=

Cd2+ = 4 × 6.22 × 10–11 = 2.49 × 10–10 M

16. The compound used in the treatment of lead poisoning is:

A. EDTA B. Cis-platin C. D-penicillamine D.

desferrioximeB

Ans. A

Solution- a. EDTA (ethylene diamine tetra acetate) is used for lead poisoning

b. Cis platin is used as an anti cancer drug

c. D-penicillamine is used for copper poisoning

d. desferrioxime B is used for iron poisoning

17. The pair that has similar atomic radii is:

A. Ti and Hf B. Mn and Re C. Sc and Ni D. Mo and W

Ans. D

Solution- Due to lanthanide contraction Mo and W have similar atomic radii.

18. 25 g of an unknown hydrocarbon upon burning produces 88g of CO2 and 9g of H2O.

This unknown hydrocarbon contains.

A. 24g of carbon and 1 g of hydrogen B. 22g of carbon and 3g of hydrogen

C. 18g of carbon and 7g of hydrogen D. 20g of carbon and 5g of hydrogen

Ans. A

Solution-

19. The decreasing order of electrical conductivity of the following aqueous solutions is:

0.1 M Formic acid (a)

0.1 M acetic acid (b)

0.1 M Benzoic acid (c)

A. a > c > b B. c > a > b C. c > b > a D. a > b > c

Ans. A

Solution- Stronger the acidic strength greater will be its electrical conductivity.

Ka value of formic acid > benzoic acid > acetic acid.

20. Among the following, the energy of 2s orbital is lowest in:

A. K B. Na C. H D. Li

Ans. A

Solution- Greater the nuclear charge, stronger will be the attraction, hence lower will be the energy

of 2s

21. In the following skew conformation of ethane, H’—C—C—H” dihedral angle is:

A. 58° B. 120° C. 149° D. 151°

Ans. C

Solution- Dihedral angle is then angle between bond pairs present on adjacent atoms.

22. The correct statement is:

A. leaching of bauxite using concentrated NaOH solution gives sodium aluminate and sodium

silicate

B. the Hall-Heroult process is used for the production of aluminium and iron

C. the blistered appearance of copper during the metallurgical process is due to the evolution

of CO2

D. pig iron is obtained from cast iron

Ans. A

Solution- Since NaOH is a strong hence it reacts with Al2O3 and SiO2 to form salts.

23. NO2 required for a reaction is produced by the decomposition of N2O5 in CCl4 as per the equation

2 5 2 22N O (g) 4NO (g) O (g)→ +

The initial concentration of N2O5 is 3.00 mol L–1 and it is 2.75 mol L–1 after 30 minutes.

The rate of formation of NO2 is:

A. 1.667 × 10–2 mol L–1 min–1 B. 4.167 × 10–3 mol L–1 min–1

C. 8.333 × 10–3 mol L–1 min–1 D. 2.083 × 10–3 mol L–1 min–1

Ans. A

Solution- 2 5 2 22N O (g) 4NO (g) O (g)⎯⎯→ +

t = 0 3.0M

t = 30 2.75M

2 5[N O ] 0.25

t 30

−=

2 5 3[N O ] [NO ]1 1

2 t 4 t

− =

22[NO ] 0.252 1.66 10 M/min

t 30−

= =

24. The correct name of the following polymer is:

A. Polyisoprene B. Polytert-butylene

C. Polisobutane D. Polyisobutylene

Ans. D

Solution- Isobutylene on polymerization will form given polymer.

25. Heating of 2-chloro-1-phenylbutane with EtOK/EtOH gives X as the major product.

Reacting of X with Hg(OAc)2/H2O followed by NaBH4 gives Y as the major product. Y is

A. B.

C. D.

Ans. A

Solution-

26. The IUPAC name of the following compound is:

A. 3, 5-dimethyl-4-propylhept-1-en-6-yne

B. 3-methyl-4-4(1-methylprop-2-ynyl)-1-heptene

C. 3-methyl-4-(3-methylprop-1-enyl)-1-heptyne

d. 3, 5-dimethyl-4-propylhept-6-en-1-1yne

Ans. A

Solution- According to IUPAC rules, select the largest chain including functional group, if alkene and

alkyne are present at equivalent position then priority is given to alkene.

27. Among the following, the INCORRECT statement about colloids is:

A. The range of diameters of colloidal particles is between 1 and 100 nm

B. The osmotic pressure of a colloidal solution is of higher order than the true solution at the

same concentration

C. They can scatter light

D. They are larger than small molecules and have high molar mass

Ans. B

Solution- Osmotic pressure of colloidal solution is lower than true solution of same concentration

28. In comparison to boron, beryllium has:

A. greater nuclear charge and greater first ionization enthalpy

B. lesser nuclear charge and lesser first ionization enthalpy

C. greater nuclear charge and lesser first ionization enthalpy

D. lesser nuclear charge and greater first ionization enthalpy

Ans. D

Solution- Boron has higher nuclear charge because it has greater atomic number and lower 1st I.E. than

beryllium due to fully filled s-orbital.

29. A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g mol–1) and 1.8 g of glucose

(molar mass = 180 g mol–1) in 100 mL of water at 27°C. The osmotic pressure of the solution is:

(R = 0.8206 L atm K–1 mol–1)

A. 8.2 atm B. 1.64 atm C. 4.92 atm D. 2.46 atm

Ans. C

Solution- 6 18

CRT .0821 30060 180

= = +

= 0.2 × .082 × 300 = 4.926 atm.

30. The temporary hardness of a water sample is due to compound X. Boiling this sample converts

X to compound Y, X and Y, respectively, are:

A. Mg(HCO3)2 and MgCO3 B. Ca(HCO3)2 and CaO

C. Mg(HCO3)2 and Mg(OH)2 D. Ca(HCO3)2 and Ca(OH)2

Ans. C

Solution- ⎯⎯→ + +3 2 2 2 2Mg(HCO ) (aq) Mg(OH) 2CO H O

Mathematics

1. Let S be the set of all 𝔞 ϵ R such that the equation, cos2𝒳 + 𝔞Sin = 2𝔞 – 7 has a Solution-

Then S is equal to :

A. [2, 6]

B. [3, 7]

C. [1, 4]

D. R

Ans. A

Solution- Given Cos 2x + α Sin x = 2 α – 7

⇒ 1 – 2 Sin2x + α Sin x =2 α – 7

⇒ 2 Sin2 x – α Sin x + 2 α – 8 = 0

2 8(2 8)Sinx

4

− − =

( 8)Sinx

4

− =

8 8Sinx ,

4 4

+ − − + =

Sin x = 2 (Not possible)

for solution

2 81 1

4

−−

4 2 8 4− −

4 2 12 −

[2,6]

2. An ellipse, with foci at (0,2) and (0, -2) and minor axis if length 4, passes through which of

the following points?

A. ( 2 ,2)

B. (1, 2 2 )

C. (2, 2 2 )

D. (2, 2 )

Ans. A

Solution- . Here, Foci of ellipse = (0, ±2)

This mean, the minor axis is y – axis and C = 2

Length of minor axis = 4

⇒ 2a = 4 ⇒ a = 2

And 2bc = 4

→ a = 2, be = 2

⇒ b2 e2 = 4

⇒ b2 – a2 = 4

⇒ b2 = 8

⇒ equation of ellipse

2 2

2 2

x y1

a b+ =

2 2x y1

4 8+ =

Clearly option A is satisfy the given curve.

3. The derivative of 1 sin cos

tan ,sin cos

x x

x x

− −

+ with respect to

2

x, where 0,

2x

is :

A. 2

B. 1

C. 2

3

D. 1

2

Ans. 2

Solution- Given 1 sinx cos xy tan

sinx cos x

− − = +

1 tanx 1y tan

tanx 1

− − = +

1 1 tanxy tan

1 tanx

− − = +

1y tan tan x4

− = −

0 x2

x 02

− −

1y x tan tanx x x ,4 2 2

− − = − − + =

y x4

= −

dy 12

d(x / 2) (1 / 2)= =

4. If a1, a2 a3, …… are in A.P. such that a1 + a7 +a16 = 40, then the sum of the first 15 terms of

this A.P. is :

A. 120

B. 150

C. 200

D. 280

Ans. C

Solution- a1, a2, a3, …………, an are in A. P.

a1 + a7 + a16 = 40

⇒ a + a + 6 d+ a + 15d = 40

⇒ 3a + 21d = 40

40a 7d

3 + =

15

15S 29 14d 15 a 2d

2= + = +

4015

3=

= 200

5. Let 𝔞 ϵ (0, /2) be fixed. If the integral tan tan

tan tan

xdx

x

+=

−

A (𝒳) cos2𝔞 + B(𝒳) sin2𝔞 + C, where C is a constant of integration, then the functions A(𝒳) and B(𝒳)

are respectively :

A. 𝒳 + 𝔞 and loge |sin(𝒳 + 𝔞 )|

B. 𝒳 - 𝔞 and loge |cos(𝒳 - 𝔞 )|

C. 𝒳 - 𝔞 and loge |sin(𝒳 - 𝔞 )|

D. 𝒳 + 𝔞 and loge |sin(𝒳 + 𝔞 )|

Ans. C

Solution- Integral

tanx tanxdx A(x)Cos2 B(x)Sin2 C

tanx tanx

+= + +

−

L.H.S.

tanx tanxdx

tanx tanx

+

−

SinxCos CosxSindx

SinxCos CosxSin

+=

−

Sin(x )dx

Sin(x )

+=

−

Sin(x 2 )dx

Sin(x )

− +=

−

Sin(x )Cos2 Cos(x )Sin2dx

Sin(x ) Sin(x )

− −= +

− −

(x )Cos2 Sin2 log|Sin(x )| C = − + − +

Compare with R.H.S. of the given integral equation

A(x) = x – α, B=(x0 = log |Sin (x – α)|

6. A value of θ ϵ (0, /3), for which

2 2

2 2

2 2

1 cos sin 4cos6

cos 1 sin 4cos6 0

cos sin 1 4cos6

+

+ =

+

, is :

A. 7

36

B. 18

C. 9

D. 7

24

Ans. C

Solution- A value of B 0,3

for which

2 2

2 2

2 2

1 cos Sin 4Cos6

Cos 1 Sin 4cos6 0

cos Sin 1 4Cos6

+

+ =

+

R2 → R2 – R1, R3 → R3 – R1

2 21 cos Sin 4Cos6

1 1 0 0

1 0 1

+

− =

−

C1 → C1 + C2

22 Sin 4Cos6

0 1 0 0

1 0 1

=

−

Expanding along first column

⇒ 2[1 – 0] – 1[-4 Cos 6 θ] =0

⇒ 2 + 4 Cos 6 θ = 0

1Cos6

2 = −

26

3 9

= =

7. The Boolean expression ~(p ⇒ (~q)) is equivalent to :

A. (~p)⇒ q

B. P v q

C. Q ⇒ ~p

D. P ∧ q

Ans. D

Solution- Boolean expression

(P ( q))is equivalent

(P q) P q− → =

8. Let ( ) 5 | 2 |f x x= − − and ( ) | 1|g x x= + , 𝒳 ϵ R. If ( )f x attains maximum value at 𝔞 and g(𝒳

) attains minimum value at β, then ( )( )2

2

1 5 6lim

6 8x

x x x

x x→−

− − +

− +is equal to :

A. 3/2

B. -1/2

C. 1/2

D. -3/2

Ans. C

Solution- Let f(x) = 5 - |x – 2| and g(x) = |x + 1|

F(x) attains maximum value when

|x – 2| = 0 ⇒ x = 2 = α

G(x) = |x + 1|

G(x) attains minimum value of x = -1 = B

2

2x

(x 1)(x 5x 6)lim

x 6x 8→

− − +

− +

x 2

(x 1)(x 2)(x 3)Lim

(x 2)(x 4)→

− − −=

− −

(2 1)(2 3)

(2 4)

− −=

−

1

2=

9. The term independent of 𝒳 in the expansion of

682

2

1 3. 2

60 81

xx

x

− −

is equal to :

A. -72

B. -36

C. -108

D. 36

Ans. B

Solution- Expansion

682

2

1 x 3. 2x

60 81 x

− −

Term independent of x will be

6

2

2

1 3term independent of x in 2x

60 x

−

6

8 2

2

1 3termof Z in 2x

8 x

− − −

6

2

r 1 2

3T in 2x will be

x+

−

r

6 2 6 r

r 1 r 2

3T C (2x )

x

−

+

= −

= 6Cr 26-r (-1)r × 3r × x12-2r-2r

Case I

For term independent of x

12 – 4r = 0 ⇒ r = 3

T4 = -6C3 × 23 × 33 x6 = -20 × 23 × 33

Case II

For term of x-8 12 – 4r = -8

⇒ 4r = 20 ⇒ r = 5

T6 = 6C5. 21 . (-1)35 . x-8

Required ans. 3 31( 20)2 3

60= −

516 2 ( 1) 3 72 36 36

81= − − = − + = −

10. Let A, B and C be sets such that A B C . Then which of the following statements is

not true ?

A. If ( ) ,A C B− then A B

B. B C

C. If (A – B) C, then A C

D. (C A) (C B)=C

Ans. A

Solution- Let A, B and C be the set that A B C

Let A = {1, 2, 3, 4}

B = {3, 4, 5, 6}

C = {1, 2, 3, 4, 7}

Here A B 3,4 C =

A C B− =

But A B

So not true statement A

If A C B then A B−

11. If [𝒳] denotes the greatest integer ≤𝒳, then the system of linear equations [sin θ]𝒳 +[-cos

θ] 𝒴=0 have infinitely many solutions if

A. 2

,2 3

and has a unique solution if 7

,6

have infinitely many solutions if

B. 2 7

, ,2 3 6

has a unique solution if

C. 2 7

, ,2 3 6

.

D. Has a unique solution if 2

,2 3

and have infinitely many solutions if 7

,6

Ans.

Solution- If [x] denotes the greatest Integer ≤ x,

Then the system of linear equations

[Sin θ]x + [Cos θ] y = 0, [Cot θ]x + y = 0

[Sin θ] x + [-Cos θ] y = 0 ………(i)

[Cot θ] x + y = 0 ………(ii)

Case I

When 2 3

, Sin ,12 3 2

1 1Cos ,0 Cos 0,

2 2

− −

1Cos ,0

3

−

[Sin ] 0,[ cos ] 0,[cot ] 1 = − = = −

equation (i) & (ii) will

0x 0y 0system will have inf initely many selection

x y 0

+ =

− + =

Case II

When 7 1

, Sin ,06 2

−

3Cos 1.

2

− −

( )Cos 3,

[Sin θ] = -1, [Cos θ0 = -1

[Cot θ] = {1, 2,3 ……}

-x – y = 0

Ix + y = 0

I = {1, 2, …..}

It will have unique solution is all cases x = 0, y = 0

12. The equation of a common tangent to the curves, 𝒴 2 = 16𝒳 and 𝒳𝒴 = -4, is:

A. 2𝒳 - 𝒴 + 2 = 0

B. 𝒳 - 2𝒴 + 16 = 0

C. 𝒳 + 𝒴 + 4 = 0

D. x - 𝒴 + 4 = 0

Ans. D

Solution- 4

y mxm

= + is always tangent to y2 = 16x ……(i)

If it is tangent to the xy = -4

4x mx 4

m

+ = −

m2 x2 +4x = -4m

m2 x2 + 4x + 4m = 0

for tangent D = 0

16 – 16m2 = 0 ⇒ m = 1 Put in equation (i)

y = x + 4

x – y + 4 = 0

13. A triangle has a vertex at (1,2) and the mid points of the two sides through it are (-1,1) and

(2,3).Then the centroid of this triangle is :

A. 7

1,3

B. 1 5

,3 3

C. 1

,23

D. 1

,13

Ans. C

Solution-

2 2

2 2

x 1 y 21, 1

2 2

x 3,y 0

B( 3,0)

+ += − =

= − =

−

3 3

3 3

x 1 y 22, 3

2 2

x 3,y 4

C(3,4)

+ += =

= − =

Centroid

1 2 3 1 2 3x x x y y y

,3 3

+ + + +

1 3 3 2 0 4,

3 3

− + + +

1,2

3

14. The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed

to be 45° from a point A on the plane. Let B be the point 30 m vertically above the point A. If

the angle of elevation of the top of the tower from B be 30°, then the distance (in m) of the

foot of the tower from the point A is :

A. 15 (3 - 3 )

B. 15 (1 + 3 )

C. 15 (3 + 3 )

D. 15 (5 - 3 )

Ans. C Solution-

Then, AB = 30m = NP

In ΔANM

MNtan45 1

AN = =

⇒ MN = AN

PM = MN – 30

= AN – 30

In ΔBPM

PM AN 30tan30

PB AN

− = =

1 AN 30

AN3

−=

BN 3AN 30 3= −

( )( )

30 3 3 130 3AN 15 3 3

23 1

+= = = +

−

15. Let 𝔞 ϵ R and the three vectors 3a i j k= + + , 2b i j k= + − and 2 3c i j k= + + . Then

the set : , and are coplanarS a b c=

A. Contains exactly two numbers only one of which is positive

B. Is empty

C. Is singleton

D. Contains exactly two positive numbers

Ans. B

Solution- Let R and three vectors

a ai j 3k

b 2i j k

= + +

= + −

And

c i 2j 3k= − +

[abc] 0=

1 3

2 1 0

2 3

− =

−

2(3 2 ) 1( 6) 3( 4 ) 0 − + − − + − − =

2 23 2 6 12 3 0 0 − − − − − = =

α2 – 18 = 0

α2 + 6 = 0

not possible for real α

S is empty set

16. A value of 𝔞 such that

19

log( )( 1) 8

e

dx

x x

+

= + + +

is:

A. 1

2

B. -2

C. 1

2−

D. 2

Ans. B

Solution-

1

e

dx 9log

(x )(x 1) 8

+

= + + +

1

e

(x 1) (x ) 9dx log

(x )(x 1) 8

++ + − +

= + + +

1 1

e

dx dx 9log

x x 1 8

+ +

− = + + +

1

e e

x 9log log

x 1 8

++

= + +

e e

2 1 2 9log log log

2 2 2 1 8

+ − = + +

e

2 1 2 1 9log log

2 2 2 8

+ + = +

2(2 1) 9

4 ( 1) 8

+ =

+

⇒ 8[4α2 + 4 α + 1] = 9[4α2 + 4α]

⇒ 32 α2 + 32 α + 8 = 36 α2 + 36 α

⇒ 4 α2 + 4 α – 8 = 0

⇒ α2 + α – 2 = 0

⇒ (α + 2) (α – 1) = 0

⇒ α = 1, -2

17. A plane which bisects the angle between the two given planes 2𝒳 - 𝒴 + 2z – 4 = 0 and 𝒳 +

2𝒴 + 2z – 2 = 0, passes through the point :

A. (1, 4, -1)

B. (2, -4, 1)

C. (1, -4, 1)

D. (2, 4, 1)

Ans. B

Solution- The given planes

2x – y + 2z – 4 = 0

And x + 2y + 2z – 2 = 0

Equation of angle bisectors are

2 2 2 2 2 2

2x y 2z 4 x 2y 2z 2.........(i)

2 ( 1) 2 1 2 2

− + − + + −=

+ − + + +

Case I

Take positive sign

2x – y + 2z – 4 = x + 2y + 2z – 2

X – 3y – z = 0 …….(ii)

Case II take negative sign

2x – y + 2z – 4 = -(x + 2y + 2z – 2)

2x – y + 2z – 4 = -x – 2y – 2z + 2

3x + y + 4z – 6 = 0 ………..(iii)

Option (2) satisfy equation (iii)

⇒ (2, -4, 1)

18. The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the

lines ( ) ( )2r i j i j k= + + − + − and ( ) ( )2r i j i j k= + + − + − is :

A. 1

3

B. 1

3

C. 3

D. 3

Ans. D

Solution- Perpendicular vector to the plane

i j k

n 1 2 1 3i 3j 3k

1 1 2

= − = − + +

− −

Equation of plane

-3 (x – 1) + 3(y – 1) + 3z = 0

⇒ x – y – z = 0

2 2 2

|2 1 4|d(2,1,4) 3

1 1 1

− −= =

+ +

19. 2 20

2sinlim

2sin 1 sin 1x

x x

x x x x→

+

+ + − − +is :

A. 3

B. 1

C. 2

D. 6

Ans. C

Solution- 2 2x 0

x 2sinxlim

x 2sinx 1 sin x x 1→

+

+ + − − +

Rationalize

( ) ( )2 2

2 2x 0

x 2sinx x 2sinx 1 sin x x 1lim

x 2sinx 1 sin x x 1→

+ + + + − +

+ + − − +

( )2 2x 0

x 2sinx 2lim

x 2sinx sin x x→

+

+ − +

0

0 form using L’hospital Rle

( )x 0

1 2cosx 2Lim

2x 2cosx 2sinx cosx 1→

+

+ − +

2 32

(2 1)

=

+

20. A circle touching the 𝒳 –axis and making an intercept of length 8 on the 𝒴 –axis passes

through the point :

A. (2,3)

B. (1,5)

C. (3,5)

D. (3, 10)

Ans. D

Solution- Equation of circles are

2 2 2(x 3) (y r) r− + =

2 2(x 3) (y 5) 25− + − =

2 2(x 3) (y 5) 25− + + =

2 2x y 6x 10y 9 0 ........(i)+ − − + =

2 2x y 6x 10y 9 0 ........(ii)+ − + + =

Length of y intercept

22 f c= −

3

5 = r

4

f r=

2

2

8 2 r 9

16 r 9

r 5

= −

= −

=

Option D (3, 10) satisfy equation (i)

21. A person throws two fair dice. He wins Rs. 15 for throwing a doublet (same numbers on the

two dice), wins Rs. 12 when the throw results in the sum of 9, and loses Rs. 6 for any other

outcome on the throw. Then the expected gain/loss (in Rs.) of the person is :

A. 1

2gain

B. 1

4loss

C. 2 gain

D. 1

2loss

Ans. D

Solution-

Win +15 +12 -6

Prob. 6/36 4/36 26/36

Probability of doublet = 6/36

Probability of sum of 9 = 4/36

Other Probability = 26/36

Expected gain/loss

6 4 2615 12 6

36 36 36= + −

90 48 156

36 36 36= + −

1 1

2 2= − −

So, 1

2 loss

22. Let z ϵ C with Im(z) = 10 and it satisfies 2

2 12

z ni

z n

−= −

+for some natural number n. Then :

A. n = 40 and Re(z)=-10

B. n = 20 and Re(z) = -10

C. n = 40 and Re(z) = 10

D. n =20 and Re(z) =10

Ans. A

Solution- Let z = x + 10i

Given 2z n

2i 12z n

−= −

+

2(x 10i) n2i 1

2(x 10i) n

+ − = −

+ +

⇒ (2x –n) + 20i = 92i -1) [(2x+n) + 20i]

Comparing real and imaginary part

⇒ 2x – n = 2(-20) – (2x +n) and 20 = 2(2x + n) – 20

⇒ 2x – n = -40 – 2x – n and 20 = 4x + 2n – 20

⇒ 4x = -40 and 4x + 2n = 40

⇒ n = ±40

⇒ n = 40 and Re(z) = -10

23. for an initial screening of an admission test, a candidate is given fifty problems to solve. If the

probability that the candidate can solve any problem is 4

5, then the probability that he is

unable to solve less than two problems is :

A.

4954 4

5 5

B.

49201 1

5 5

C.

48316 4

25 5

D.

48164 1

25 5

Ans. A

Solution- Total Problems = 50

4P(solving)

5=

1P(not solving)

5=

P (unable to solve less than two problems)

= p(not solving one Problem) + P(not solving zero Problem)

0 50 1 49

50 50

0 1

1 4 1 4C C

5 5 5 5

= +

50 49

50 59

4 450,

5 5.5= +

50 494 4

10.5 5

= +

494 4

105 5

= +

4954 4

.5 5

=

24. If 𝔞, β and are three consecutive terms of a non-constant G.P. such that the equations 𝔞𝓍2

+ 2β𝓍 + =0 and 𝓍2 + 𝓍 -1 =0 have a common root, the 𝔞(β + ) is equal to :

A. 𝔞β

B. 0

C. 𝔞

D. β

Ans. D

Solution- , , are in G.P.

2 2x 2 x 0 & x x 1 0 + + = + − =

Have a common roots. Both roots will be common

1 1 1

= = =

−

, ,2

= = = −

( )2

2 2

+ = − = − =

25. If the area (in sq. units) bounded by the parabola 𝒴2 = 4λ𝓍 and the line 𝒴 = λ𝒳, λ > 0, is 1

9, then λ is equal to :

A. 48

B. 4 3

C. 2 6

D. 24

Ans. D

Solution- . Parabola y2 = 4 λx

Line y = λx, λ > 0

y2 = 4 λx & y = λx

λ2 x2 = 4 λx

x = 0 & 4

x

=

( )4/

0

1Area 4 x x dx

9

= − =

4/

4/3/2 2

0

0

x x 12

3 2 9

2

− =

( )3/2

2

3/2 2

2 x4 16 1

3 2 9

− =

32 8 1

3 9 − =

y

8 124

3 9

= =

26. A group of students comprises of 5 boys and n girls. If the number of ways, in which a team

of 3 students can randomly be selected from this group such that there is at least one boy

and at least one girl in each team, is 1750, then n is equal to :

A. 27

B. 24

C. 28

D. 25

Ans. D

Solution- Given 5 boys and n girls

Total ways of forming team of 3 member under given condition

=5C1 . nC2 + 5C2 . nC1

⇒ 5C1 . nC2 + 5C2 . nC1 = 1750

5n(n 1)10n 1750

2

− + =

n(n 1)2n 350

2

− + =

⇒ n2 + 3n = 700

⇒ n2 + n – 700 = 0

⇒ n = 25

27. The general solution of the differential equation (𝒴2 – 𝒳3) d𝒳 – 𝒳𝒴d𝒴 =0 (𝒳 ≠ 0) is : (where

c is a constant of integration)

A. 𝒴2 – 2𝒳3 +c𝒳2 =0

B. 𝒴2 + 2𝒳2 +c𝒳3 =0

C. 𝒴2 – 2𝒳2 +c𝒳3 =0

D. 𝒴2 + 2𝒳3 +c𝒳2 =0

Ans. D

Solution- . Differential equation

(y2 – x3) dx – xy dy = 0 (x ≠ 0)

2 3 dyy x xy 0

dx− − =

2 3dyxy y x

dx− = −

22dy y

y x .......(i)dx x

− = −

Let y2 = v

dy dv2y

dx dx=

Put in equation (i)

21 dv 1v x

2 dx x− = −

2dv 2v 2x ........(ii)

dx x

− + = −

I.F. = 2 nx

2

e

2 1dx e

x x

−−= =

Solution of equation (ii)

2

2 2

1 1V 2x dx c

x x = − −

2

v2x c

x== −

Put v = y2

y2 = -2x3 – cx2

y2 + 2x3 + cx2 = 0

28. A straight line L at a distance of 4 units from the origin makes positive intercepts on the

coordinate axes and the perpendicular from the origin to this line makes an angle of 60° with

the line 𝒳 + 𝒴 = 0. Then an equation of the line L is :

A. ( ) ( )3 1 3 1 8 2x y+ + − =

B. 3 8x y+ =

C. 3 8x y+ =

D. ( ) ( )3 1 3 1 8 2x y− + − =

Ans. A and B Solution-

OP = 4

Given OP makes 60° with x + y = 0

Let slope of OP = m

m 1tan60

1 m

+ =

−

m 13 or 3

m 1

+ = −

−

m 1 3m 3 or m 1 3 3r + = − + = −

( )m 3 1 3 1 or m(1 3) 3 − = + + =

m 3 1 3 1or m

3 1 3 1

− + − =

− +

3 1 3 1tan or tan

3 1 3 1

+ − = =

− +

⇒ equation of line x cos α + y sin α = P

( ) ( )3 1 x 3 1 y 8 2 + + − =

( ) ( )or 3 1 x 3 1 y 8 2− + + =

29. The tangents to the curve 𝒴 = (𝒳 – 2)2 – 1 at its points of intersection with the line 𝒳 - 𝒴 =3,

intersect at the point :

A. 5

,12

B. 5

, 12

−

C. 5

,12

−

D. 5

, 12

− −

Ans. B Solution- Y=(x-2)2-1 y=x2-4x+3

Equation of chord of contact at (h,k) is

T=0 (y+k)/2=hx-2(x+h)+6 (2h-4)x-y= (4h+k-6) Comparing with 𝒳 - 𝒴 =3

h=5/2 and k=-1 So (h,k)=(5/2,-1)

30. If 20C1 + (22) 20C2 + (32) 20C3 + ………. + (202) 20C20 = A(2β ), then the ordered pair (A, β) is

equal to :

A. (380, 18)

B. (420, 18)

C. (380, 19)

D. (420, 19)

Ans. B Solution-

A = 420, B = 18