Physics

1. A plane is inclined at an angle 30o = with respect to the horizontal. A particle is projected with

a speed u = 2 ms-1 , from the base of the plane, making an angle 15o = with respect to the plane

as shown in the figure. The distance from the base, at which the particle hits the plane is close to : (Take g = 10 ms-2

A. 26 cm B. 14 cm C. 20 cm D. 18 cm Ans. C Solution-

2 sin

tgcos

=

2 sin15 2 2 sin15

tgcos30 3

102

= =

t = 0.12 s

21s 2cos15 t gsin30 t

2= −

21 1s 2 cos15 0.12 10 (0.12)

2 2= −

s 9.195m 19.5 cm= =

2. Space between two concentric conducting spheres of radii a and b (b>a) is filled with a medium of resistivity p. the resistance between the two spheres will be:

A.

1 1

4 a b

+

B.

1 1

2 a b

+

C.

1 1

4 a b

−

D.

1 1

2 a b

−

Ans. C Solution-

for the resistance

dr

dRArea

=

2

drdR

4 r

=

R b

2

0 a

drdR

4 r

=

b

a

1R

4 r

− =

1 1

R4 a b

= −

−

1 1the resis tance is

4 a b

3. In the formula X = 5YZ2, X and Z have dimensions of capacitance and magnetic field, respectively. What are the dimensions of Y in SI units ? A. [M-2 L0 T-4 A-2] B. [M-2 L-2 T6 A3] C. [M-3 L-2 T8 A4]

D. [M-1 L-2 T4 A2] Ans. C

Solution- X = 5 YZ2

2

XY

5Z=

Now, for dimension

2

[X][Y]

[Z ]=

as [X] is capacitance and [Z] is magnetic field.

− −

− −

− −

= =

2 1 2 4

3 3 8 4

1 2

A M L T[Y] M L T A

MA T

[Y] = [M–3 L–2T8A4]

4. A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only

gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in 24 hours around the planet? [Given : Mass of planet = 8 x 1022 kg, Radius of planet = 2 x 106 m, Gravitational constant G = 6.67 x 10-11 Nm2/kg2] A. 17

B. 13 C. 11 D. 9 Ans. C

Solution- For gravitational force

2

g

mvF

r=

for speed of spaceship

11 22

6 3

GM 6.67 10 8 10V

r (2 10 20 10 )

− = =

+

V = 1.625 × 103 m/s

2 r

Tv

=

As nT = 24 × 60 × 60

=

6

3

2 (2 10 )n 24 3600

1.625 10

−

=

3

6

24 3600 1.625 10n

2 (2.0 10 )

n = 11

5. A square loop is carrying is steady current I and the magnitude of its magnetic dipole moment is

m. If this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be:

A.

2m

B.

4m

C.

3m

D.

m

Ans. B

Solution- m = NIA (magnetic moment)

m = 1 × I × a2

m = Ia2

Now, 4a = 2πr

2a

r =

for the circular Loop,

= =

2 22a 4Iam I

4m

m =

Hence, the new magnetic moment is 4m

6. A coil of self-inductance 10 mH and resistance 0.1 is connected through a switch to a battery

of internal resistance 0.9 . After the switch is closed, the time taken for the current to attain 80%

of the saturation value is: [take In5 = 1.6] A. 0.016 s B. 0.324 s

C. 0.103 s D. 0.002 s Ans. A

Solution-

eq

L 0.0100.01s

R 0.1 0.9 = = =

+

Now, for the current

I = I0(1 – –t/τ)

0.80I0 = I0 (1 – e–t/τ)

e–t/τ = 0.20

−

= = −t

In(0.20) 1.60.01

t 0.016 s=

7. The graph shows how the magnification m produced by a thin lens varies with image distance v. What is the focal length of the lens used?

A.

2b

ac

B.

a

c

C.

2b c

a

D.

b

c Ans. D Solution- For this lens,

1 1 1

1

1

f v u

v v vm

f v u

vm

f

= +

= + = −

= −

When v = a,

1 1a

mf

= − …. (1)

When v = a + b

( )2 1

a bm

f

+= − ….(2)

Equation (2) – (1)

( )

( )

( )

2 1 1 1

1 1

a b am m

f f

a b ac

f f

a bac

f f

bc

f

bc

f

bf

c

+ − = − − −

+= − − +

+= −

= −

=

=

8. In Li+ +, election in first Bohr orbit is excited to a level by a radiation of wave length . When the

ion gets de excited to the ground state in all possible ways (including intermediate emissions), a

total of six spectral lines are observed. What is the value of ?

(Given: h=6.63 x 10-34 Js; c=3 x 108 ms-1)

A. 10.8 nm B. 11.4 nm C. 9.4 nm

D. 12.3 nm Ans. A

Solution- hc

E

=

( )13.6 9 0.85 9

12.75 9

1240 .nm

12.75 9

10.8 nm

hceV

hc

eV

eV

eV

− =

=

=

=

9. A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass

of 7

8

M and is converted into a uniform disc of radius 2R. The second part is converted into a

uniform solid sphere. Let I1 be the moment of inertia of the disc about its axis and I2 be the moment of inertia of the new sphere about its axis. The ratio I1/I2 is given by :

A. 65 B. 285 C. 140

D. 185 Ans. C

Solution- for the first part

2 21 1 1

1 1 7MI M R (2R)

2 2 8

= =

2 21

7 7I 4MR MR

16 4= =

2 2 2

2

2 M R 2 M R MRI

5 8 2 5 8 4 80

= = =

2

1

22

7MR

I 74 80 1401I 4MR

80

= = =

1

2

I140

I=

10. The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m

which is carrying a current of 10 A is : [Take 7 24 10o NA − −= ]

A. 3

T

B. 1

T

C. 9

T

D. 18

T Ans. D Solution-

a a

x tan302 2 3

= =

For a single wire

0IB (sin60 sin60 )

4 (x)

= +

74 10 10 3 3

B2 21

42 3

− = +

B = 6 × 10–6T

Net magnetic field = 3 × B

= 18 × 10–6T

= 18μT

11. The correct figure that shows, schematically, the wave pattern produced by superposition of two waves of frequencies 9 Hz and 11 Hz, is:

A.

B.

C.

D. Ans. D

Solution- fbeat = f1 – f2

= 11 – 9

= 2 Hz

Now, for time period

beat

1 1T

f 2Hz= =

T = 0.5 s

12. The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit ?

A. 0.90 mm B. 1.00 mm C. 1.16 mm D. 1.36 mm Ans. C

Solution- Stress = 379 MPa

Force ⇒ F = 400 N

The minimum diameter is d

2

ForceStress

d

2

=

6

2

400379 10

d

2

=

d = 1.15 × 10–3 m

d 1.15 mm=

13. In free space, a particle A of charge 1 C is held fixed at a point P. Another particle B of the

same charge and mass 4 g is kept at a distance of 1 mm from P. If B is released, then is velocity

at a distance of 9 mm from P is :

[Take 9 2 21

9 104 o

NM C

−= ]

A. 3.0 x 104 m/s

B. 1.5 x 102 m/s C. 2.0 x 103 m/s D. 1.0 m/s

Ans. C

Solution- Using conservation of mechanical energy

ΔU + ΔK = 0

–ΔU = ΔK

21mv U

2= −

12 9 12

6 2

3 2

1 9 10 10 9 10 104 10 v

2 10 9 10

− − −−

− −

= −

3v 2 10 m / s=

14. A simple pendulum of length L is placed between the plates of a parallel plate capacitor having

electric field E, as shown in figure. Its bob has mass m and charge q. The time period of the pendulum is given by:

A.

2

2

2L

qEg

m

+

B.

22

L

qEg

m

−

C.

2 22

2

2L

q Eg

m

−

D.

2L

qEg

m

+

Ans. A Solution-

Now, for geff

2 2effmg (mg) (eE)= +

as 2

eff 2

T 2 2g qE

gm

= =

+

15. One mole of an ideal gas passes through a process where pressure and volume obey the relation

21

12

oo

VP P

V

= −

. Here Po and Vo are constants. Calculate the change in the temperature of the

gas if its volume change from Vo to 2Vo.

A.

1

2

o oPV

R

B.

5

4

o oPV

R

C.

1

4

o oPV

R

D.

3

4

o oPV

R Ans. B

Solution- As 2

00

V1P P 1

2 V

= −

at V = V0

01 0

P1P P 1

2 2

= − =

at V = 2V0

2

2 0 0

1 1 7P P 1 P

2 2 8

= − =

02

7PP

8=

For temperature at V0

00

0 02

7P2V

P P78TnR 4 nR

= =

Change in Temp = T2 – T1

0 0P V7 1

4 2 nR

= −

0 0P V5T

4 nR =

16. A source of sound S is moving with a velocity of 50 m/s towards a stationary observer. The observer measures the frequency of the source as 1000 Hz. What will be the apparent frequency of the source when it is moving away from the observer after crossing him? (Take velocity of sound in air is 350 m/s) A. 750 Hz

B. 857 Hz C. 807 Hz D. 1143 Hz Ans. A

Solution-

For apparent frequency

app 0

V 0f f

V 50

− =

−

0

3501000 f

350 50=

−

0

6f 1000 Hz

7=

0

Vf f

V 50

=

+

350 6

f 1000350 50 7

= +

7 6

f 1000 750Hz8 7

= =

17. When heat Q is supplied to a diatomic gas of rigid molecules, at constant volume its

temperature increases by T. The heat required to produce the same change in temperature, at a

constant pressure is :

A.

3

2Q

B.

7

5Q

C.

2

3Q

D.

5

3Q

Ans. B

Solution- for diatomic gas

p

7RC

2=

v

5RC

2=

For constant volume

Qv = nCv ΔT

For constant pressure

Qp = nCpΔT

p p p

v v v

Q nC T C 7R 2

Q nC T C 2 5R

→ = = =

p

v

Q 7

Q 5=

p v

7Q Q

5=

18. In an experiment, brass and steel wires of length 1 m each with areas of cross section 1 mm2

are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress required to produce a net elongation of 0.2 mm is, [Given, the Young’s Modulus for steel and brass are, respectively, 120x109 N/m2 and 60x109 N/m2]

A. 1.2X106 N/m2 B. 0.2X106 N/m2 C. 8.0X106 N/m2 D. 4.0X106 N/m2

Ans. C

Solution- A = 1 mm2

X = 0.2 mm = 2 × 10–4 m

1m=

Now, for spring constant

9

1 11

1

Y AYA 120 10 AK K

L L 1

= → = =

9

2 22

2

Y A 60 10 AK

L 1

= =

91 2eq

1 2

K K 120 60K 10 A

K K 120 60

= =

+ +

Keq = 40 × 109A

Now, for force

F = (40 × 109)A × (0.2 × 10–3)

6

2

F N8 10

A m=

19. Water from a tap emerges vertically downwards with an initial speed of 1.0 ms-1. The cross-sectional area of the tap is 10-4 m2. Assume that the pressure is constant throughout the stream of water and that the flow is streamlined. The cross-sectional area of the stream, 0.15 m below the tap would be : (Take g=10 ms-2)

A. 1x10-5 m2 B. 2x10-5 m2

C. 5x10-4 m2 D. 5x10-5 m2

Ans. D

Solution- u = 1 m/s

A = 10–4 m2

h = 0.15 m

using equation of continuity

A1V1 = A2V2

10—4 × 1 = A2V2 ….(1)

Using Bernoulli’s equation

( )2 21 2

1P V V gh P

2+ − + =

2 22 1V V 2gh→ = +

2 22V 1 2 9.8 0.15= +

V2 = 2 m/s

as A2 × 2 = 10—4

5 22A 5 10 m−=

20. A metal coin of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible mass as shown in the figure. The system is initially at rest. The constant torque, that will make the system rotate about AB at 25 rotations per second in 5s, is close to :

A. 4.0 x 10-6 Nm B. 1.6 x 10-5 Nm C. 2.0 x 10-5 Nm

D. 7.9 x 10-6 Nm Ans. C

Solution- m = 5 gm = 0.005 kg

r = 1 cm = 0.01 m

f

25 225rpm rod / s

1

= =

−

= = =

f 0 25 210 rad / s

t (1) 5

25MR I

4

= =

3 255 10 (10 ) 10

4− − =

51.963 10 N.m− =

τ = 2 × 10–5 N.m

21. The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of

the Zener diode is 6 V and the load resistance is, 4LR k= . The series resistance of the circuit is

1iR k= . If the battery voltage VB varies from 8 V to 16 V, what are the minimum and maximum

values of the current through Zener diode?

A. 0.5 mA; 8.5 mA

B. 1.5 mA; 8.5 mA

C. 0.5 mA; 6 mA D. 1 mA; 8.5 mA Ans. B

Solution- In breakdown, for zener

VB = Vz = 8V

3L 3

6i 1.5 10 A

4 10

−= =

3R 3

(8 6)i 2 10 A

1 10

−

−

−= =

IZenr = iR – iL = (2 – 1.5) × 10–3 = 0.5 × 10–3A

For VB = 16V

IL = 1.5 × 10–3A

3 3R

(16 6)i 10 A 10 10 A

1− −−

= =

IZ = (10 – 1.5) mA = 8.5mA

22. Two radioactive substances A and B have decay constants 5 and respectively. At t = 0, a

sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei

to become

21

e

will be :

A. 2/

B. 1/2

C. 1/4

D. 1/ Ans. B

Solution- for A

5 tA 0N N e− =

for B, tB 0N N e−=

5 t

0At

B 0

N eN

N N e

−

−→ =

4 tA2

B

N 1e

N e

− = =

2 4 te e− − =

2 4 t− =−

1

t2

=

23. A bullet of mass 20 g has an initial speed of 1 ms-1, just before it starts penetrating a mud wall of thickness 20 cm. IF the wall offers a mean resistance of 2.5 x 10-2 N, the speed of the bullet after emerging from the other side of the wall is close to :

A. 0.7 ms-1 B. 0.3 ms-1 C. 0.1 ms-1 D. 0.4 ms-1

Ans. A

Solution- mass, m = 20g = 0.020 kg

u = 1 m/s

V = ?

for 2

22.5 10a 1.25 m / s

0.020

−− = = −

using third equation

V2 – u2 = 2 as

V2 – 12 = 2 (–1.25) (0.20)

1

V 0.707 m / s2

= =

24. A cubical block of side 0.5 m floats on water with 30% of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water ?

[Take, density of water = 103 kg/m3] A. 30.1 kg B. 65.4 kg C. 87.5 kg D. 46.3 kg Ans. C

Solution-

For the maximum weight

Mg = ρL (extra volume of liquid displaced) g

M = 103 (0.502 × 0.7 × 0.5)

M = 87.5 kg

maximum weight is 87.5 kg

25. A submarine experiences a pressure of 5.05 x 106 Pa at a depth of d1 in a sea. When it goes further to a depth of d2, it experiences a pressure of 8.08 x 106 Pa. Then d2 –d1 is approximately

(density of water = 103 kg/m3 and acceleration due to gravity = 10 ms-2): A. 400 m B. 500 m C. 300 m D. 600 m

Ans. C

Solution- as for the pressure

P1 = P0 + ρgd1

5.05 × 106 = P0 + 103 × 10 × d1 ….(1)

For the depth d2

8.08 × 106 = P0 + 103 × 10 × d2 ….(2)

→ (d2 – d1)

(8.08 – 5.05) × 106 = 103 × 10 × (d2 – d1)

2 1d d 300 m− =

26. Light is incident normally on a completely absorbing surface with an energy flux of 25 Wcm-2. If

the surface has an area of 25 cm2, the momentum transferred to the surface in 40 min time

duration will be : A. 5.0 x 10-3 Ns B. 3.5 x 10-6 Ns C. 6.3 x 10-4 Ns D. 1.4 x 10-6 Ns

Ans.

Solution- Intensity, 2

WI 25

cm=

2 2

WI 25

(10 m)−=

4

2

WI 25 10

m=

For momentum transferred

Force = Pressure × Area

Δ momentum = force × time

I

A timeC

=

4

4

8

25 1025 10 40 60

3 10

−=

− = 3momentum 5.0 10 N.s

27. In a Young’s double slit experiment, the ratio of the slit’s width is 4 : 1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be:

A. 25 : 9

B. ( )

4

3 1 :16+

C. 9 : 1 D. 4 : 1

Ans. C

Solution- for the ration of width

1

2

W 4

W 1=

01 1

2 2 0

4II W 4

I W 1 I→ = = =

( ) ( )2 2

max 1 2 0 0I I I 2 I I= + = + = 9 I0

For Imin

( )2

min 1 2I I I= −

( )2

0 02 I I= − = I0

max 0

min 0

I 9I 9

I I 1= =

28. The time dependence of the position of a particle of mass m = 2 is given by ( ) 2ˆ ˆ2 3r t ti t j− . Its

angular momentum, with respect to the origin, at time t = 1 is :

A. ( )ˆ ˆ34 k i− −

B. ( )ˆ ˆ48 i j+

C. 36 k̂

D. -48 k̂ Ans. D

Solution- m = 2 kg

2ˆ ˆr 2 i 3t j= + −

For the angular momentum

dr ˆ ˆV 2i 6tjdt

= = −

at t = 2s

ˆ ˆV 2i 12j= −

L m(r V)=

ˆ ˆ ˆ ˆ2(2i 12j) (2i 12j)= − −

ˆ24k 2= −

2ˆ48k kg.m / s=−

29. Two blocks A and B of masses mA = 1 kg and mB = 3 kg are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The maximum force F that can be applied on B horizontally, so that the block A does not slide over the block B is : [Take g = 10 m/s2]

A. 16 N B. 40 N

C. 8 N D. 12 N Ans. A

Solution- For the upper block

Using second law of motion

m × a = μmg

a = 0.2 × 10 = 2 m/s2

Now, F – 8 = 2 × 4 (f = 4 × 0.2 × 10 f = 8 N)

F = 16 N

The force is 16 N.

30. A 2 mW laser operates at a wavelength of 500 nm. The number of photons that will be emitted per second is : [Given Planck’s constant h = 6.6 x 10-34 Js, speed of light c = 3.0 x 108 m/s]

A. 5 x 1015

B. 1 x 1016 C. 1.5 x 1016 D. 2 x 1016

Ans. C

Solution- For energy of each photon

34 8

9

nc 6.62 10 3 10E

500 10

−

−

= =

E = 3.972 × 10–19 J

no. of photons 3

19

2 10

3.972 10

−

−

=

= 1.5 × 1016

Chemistry

1. 1 g of a non-volatile non – electrolyte solute is dissolved in 100 g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1 : 5. The ratio of the elevation in their boiling

points, ( )

( )

b

b

T A

T B, is :

A. 5 : 1 B. 1 : 5 C. 10 : 1 D. 1 : 0.2

Ans. B

Solution- Elevation in boiling point has the formula (represented by ΔT)

ΔT = Kb m, where Kb is the molal boiling point elevation constant and m is the concentration

of the solute in the solution.

Molality of solution A, mA = =

110

100

1000

[ divide by 1000 to convert in kg)

Similarly, molality of solution B, B

1m 10

100

1000

= =

= =

b A b A b A

B A b B b B

( T ) (K m) (K ) mA

( T ) (K m) (K ) mB

1 10

5 10=

1

5=

∴ 1 : 5

2. The crystal field stabilization energy (CFSE) of [Fe(H2O)6]Cl2 and K2[NiCl4],respectively, are:

A. 0.4 0.8o tand− −

B. 0.4 1.2o tand− −

C. 0.6 0.8o tand− −

D. 2.4 1.2o tand− −

Ans. A

Solution- CFSE energy depends on the kind of ligand strong or weak. In case of weak ligand, pairing

does not take place. In case of strong ligand, pairing happens.

22 6 2 2 6[Fe(H O) ]Cl [Fe(H O) ] 2Cl+ −+

+ +2 22 6ln[Fe(H O) ] , Fe oxidation state

Fe : [Ar] 3d6 4s2

Fe+2 : [Ar] 3d6 ⇒ t2g will have (2,1,1)

and eg will have (1,1)

∴ CFSE = –0.4 × 4Δ0 + 0.6 × 2Δ0

0 0 01.6 1.2 0.4= − + = −

K2 [NiCl4] 2 K+ + [NiCl4]2–

Ni is in +2 oxidation state in [Ni Cl4]2–

Ni+2 → [Ar] 3d8 and Cl– is a weak ligand ⇒ no pairing

∴ t2g → (2,1,1) and eg (2,2)

∴ CFSE = –0.6 × 4 Δt + 0.4 × Δt (as it has tetrahedral geometry)

t0.8= −

3. The correct match between Item – I and Item – II is :

Item – I Item – II

(a) High density

polythene

(I) Peroxide

catalyst

(b) Polyacrylonitrile (II) Condensation

at high

temperature

& pressure

(c) Novolac (III) Ziegler –

Natta Catalyst

(d) Nylon 6 (IV) Acid or base

catalyst

A. (a) → (III), (b) → (I), (c) → (IV), (d) → (II)

B. (a) → (IV), (b) → (II), (c) → (I), (d) → (III)

C. (a) → (III), (b) → (I), (c) → (II), (d) → (IV)

D. (a) → (II), (b) → (IV), (c) → (I), (d) → (III)

Ans. A

Solution- High density polythene (HDPE) is manufactured by Ziegler – natta catalyst

Polyacrylonitrile is an example of addition polymer.

The addition polymerization of acrylonitrile in presence of a peroxide catalyst leads to the

formation of polyacrylonitrile

Novolac is formed with the reaction of phenol or substituted phenol with formaldehyde.

Nylon 6 is a polymer developed from caprolactum at very high temperature and pressure,

when condensation happens Nylon 6 is formed.

4. The increasing order of nucleophilicity of the following nucleophiles is:

(a) 3 2CH CO

(b) H2O

(c) 3 2CH SO

(d) O H

A. (a) < (d) < (c) < (b) B. (d) < (a) < (c) < (b)

C. (b) < (c) < (a) < (d) D. (b) < (c) < (d) < (a)

Ans. C

Solution-

OH is highly nucleophilic as the electron density is on oxygen atom and no resonance

Next would be 3CH COO because in 3 3CH SO negative charge is dispersed on 3 oxygen atoms

as a result of resonance, where as in 3CH COO , electron is dispersed on only two oxygen

atoms. Hence, the electron density would be more on 3CH COO and hence more

nucleophilicity

The least would be of H2O as there is no negative charge. Hence the order is

2 3 3 3H O CH SO CH COO OH

5. The correct option among the following is:

A. Colloidal particles in lyophobic sols can be precipitated by electrophoresis. B. Colloidal medicines are more effective because they have small surface area.

C. Brownian motion in colloidal solution is faster if the viscosity of the solution is very high. D. Addition of alum to water makes it unfit for drinking.

Ans. A

Solution- Alum is added in muddy water to purify it.

Alum (aluminium sulphate), when added to raw water reacts with bicarbonate alkalinities

present in water and forms a gelatinous precipitate. This floe attracts other fine particles and

suspended raw material in raw water and settles down at the bottom of the container and

water is purified.

Colloidal medicines are more effective because of large surface area .Large surface area

makes adsorption rate more.

Coagulation of lyophoboic sols con be done by different methods

— electrophoresis

— mixing two oppositely charged sols

— By boiling

— Dialysis

— addition of electrolyte

6. The noble gas that does NOT occur in the atmosphere is :

A. He B. Kr C. Ra

D. Ne

Ans. C

Solution- The radio active decay of radium (226) results in formation of Radon, hence Radon is not

naturally occurring. All other noble gas occur in atmosphere.

7. The highest possible oxidation states of uranium and plutonium, respectively, are: A. 6 and 7 B. 4 and 6 C. 6 and 4 D. 7 and 6

Ans. A

Solution- Uranium and plutonium have multiple oxidation states.

Uranium exists in aqueous solution in the +3, +4, +5, +6 oxidation states, +6 is the most stable

oxidation state of Uranium.

Plutonium has +3, +4, +5, +6, +7 oxidation states.

∴ maximum for Uranium → +6 and plutonium → +7

8. Compound A (C9H10O) shows positive iodoform test. Oxidation of A with KMnO4/KOH gives acid B(C8H6O4). Anhydride of B is used for the preparation of phenolphthalein. Compound A is:

A.

B.

C.

D.

Ans. A

Solution- The iodoform test is a test for the presence of carbonyl compounds with the structure of R

COCH3 and alcohols with the structure RCH(OH) CH3.

Hence (B) cannot be answer

9. The major product ‘Y’ in the following reaction is:

A.

B.

C.

D. Ans. B

Solution- NaOCl

3 H

O||

ph—C—CH ⎯⎯⎯→

PH CH3

O

X is

10. The number of pentagons in C60 and trigons (triangles) in white phosphorus, respectively, are:

A. 20 and 4 B. 12 and 4 C. 12 and 3 D. 20 and 3

Ans. B

Solution- White phosphorous → P4

C60 is Buckminster fullerene

It has 20 hexagons & 12 pentagons with a carbon atom which has one π-bond and two single

bonds at each vertex of each polygon and a bond along each polygon edge.

11. For the reaction of H2 with I2, the rate constant is 2.5 x 10-4dm3 mol-1 S-1 at 327o C and 1.0 dm3 mol-1 S-1at 527oC. The activation energy for the reaction, in kJ mol-1 is:

(R = 8.314 J K-1 mol-1)

A. 166 B. 150 C. 59 D. 72

Ans. A

Solution- Let’s assume the rate constant as K1 and K2

K1 = 2.5 × 10–4 dm3 mol–1 s–1

K2 = 1 dm3 mol–1 s–1

= −

2

1 1 2

K Ea 1 1log

K 2.303K T T

−

= −

4

1 Ea 1 1log

2.3 8.314 600 8002.5 10

=

10.4 Ea 1log

2.5 2.3 8.314 2400

=

310

Ea 1log 4 10

2.3 8.3 2400

⇒ Ea = 164.93 kg/mol

12. The correct statement is : A. Sodium cyanide cannot be used in the metallurgy of silver. B. Zincite is a carbonate ore. C. Zone refining process is used for the refining of titanium. D. Aniline is a froth stabilizer.

Ans. D

Solution- Zincite : ZnO is the mineral form of Zinc oxide. It is not a carbonate ore.

→ Sodium cyanide is used in metallurgy of silver. The process of extraction of silver is called

as cyanide process. The ore is crushed, concentrated and then treated with sodium cyanide

solution.

→ Aniline is a froth stabilizer, stabilizes the froth and enhance the non-wettability of the

mineral particles. Mineral particles become wet by oils while gangue particles by water.

Zone refining is a method employed for Ge, Si, B, Ga, it is not used for titanium.

13. The minimum amount of O2 (g) consumed per gram of reactant is for the reaction:

[Given atomic mass : Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1)

A. 3 8 2 2 2( ) 5O ( ) 3 ( ) 4 ( )C H g g CO g H O I+ → +

B. 2 2 34 ( ) 3 ( ) 2 ( )Fe s O g Fe O s+ →

C. 22 ( ) ( ) 2 ( )Mg s O g MgO s+ →

D. 4 2 4 10( ) 5 ( ) ( )P s O g PO s+ →

Ans. B

Solution- 2 2 34Fe(s) 3O (g) 2Fe O (s)+ →

4 × 56 = 3 × 32

2

3 321gm of Fe react with gm of O

4 56

=

= 0.42 gm

Similarly

4 2 4 10P 5O P O+ →

1 gm of P4 react with 5 32

a 131 5

=

3 8 2 3 2C H 5O 3CO 4H O+ → +

1 gm of C3H8 react with 5 32

b 1(12 3 8 1)

=

+

22Mg O MgO+ →

1 gm of Mg react with324

0.672 24

= =

0.42 is least

14. The major product obtained in the given reaction is :

A.

B.

C.

D.

Ans. D

Solution- AlCl3 is used in friedel craft alkylation.

15. The correct statements among (a) to (d) are:

(a) saline hydrides produce H2 gas when reacted with H2O. (b) reaction of LiAlH4 with BF3 leads to B2H6 (c) PH3 and CH4 are electron – rich and electron – precise hydrides, respectively.

(d) HF and CH4 are called as molecular hydrides. A. (c) and (d) only B. (a), (b), (c) and (d). C. (a), (c) and (d) only. D. (a), (b) and (c) only

Ans. B

Solution-→ Saline hydrides such as NaH, LiH etc. react with water to form a base and hydrogen gas.

2 2MH H O MOH H (g)+ → +

The reaction is violent and produce fire

4 3 3 2 63Li AlH 4BF 3LiF 4AlF 2B H .→ + → + +

→ hydrides is an anion of hydrogen (H–) H is formed when one atom is hydrogen bonded to

another electropositive element.

∴ HF and CH4 are called as molecular hydrides

a,b and d correct

(C) is also correct as PH3 and CH4 are electron rich and electron precise hydrides.

Note that PH3 has no hybridization and CH4 has perfectly filled octet

16. The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be

about 9. The spectral series are:

A. Balmer and Brackett B. Paschen and Pfund

C. Brackett and Pfund

D. Lyman and Paschen

Ans. D

Solution- Lyman is when n1 = 2

Paschen is when n1 = 3

For shortest wavelength, →2n

−

=

−

2

2

1 113.6 2

( E lyman)max 2

1 1( E paschen)max13.6 2

4

9

1=

17. Points I, II and III in the following plot respectively correspond to

(Vmp : most probable velocity)

A. Vmp of N2 (300 K); Vmp of O2 (400K); Vmp of H2 (300 K)

B. Vmp of O2 (400 K); Vmp of N2 (300K); Vmp of H2 (300 K) C. Vmp of N2 (300 K); Vmp of H2 (300K); Vmp of O2 (400 K) D. Vmp of H2 (300 K); Vmp of N2 (300K); Vmp of O2 (400 K)

Ans. A

Solution- We know that

Most probable velocity

mp

2RTV

M=

R and T is constant

mp

1V

M , where M is the molar mass

Now,

( )

⎯⎯⎯⎯→

⎯⎯⎯⎯⎯→

H ,N ,O2 2 22 14 13

M increases

V decreasesmp

M:

when R & T is constant

Here T is different consider T

Vmp of O2, Vmp of N2, Vmp of H2

2R 400 2R 300 2R 300

32 28 2

Vmp of N2 < Vmp of O2 < Vmp of H2

18. The correct order of the first ionization enthalpies is :

A. Mn < Ti < Zn < Ni B. Zn < Ni < Mn < Ti C. Ti < Mn < Ni < Zn D. Ti < Mn < Zn < Ni

Ans. C

Solution- First ionization enthalpy is the energy required to remove an electron from the outer most

orbit.

As we go from left to right across a period, internuclear force increases, hence the attraction

between outer most electron and nucleus increases. As a result, more energy is required to

remove an electron from the atom which lies on right side of periodic table.

That is, ionization enthalpy increases as we go from left to right.

Position of Mn, Ti, Zn and Ni in periodic table is

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn

IE1 order Zn > Ni > Mn > Ti

19. For the reaction,

2 2 3

1

16

2 ( ) ( ) 2 ( ),

57.2 mol and

K 1.7 10c

SO g O g SO g

H kJ −

+

= −

=

Which of the following statement is INCORRECT? A. The equilibrium will shift in forward direction as the pressure increases. B. The equilibrium constant is large suggestive of reaction going to completion and so no

catalyst is required. C. The equilibrium constant decreases as the temperature increases.

D. The addition of inert gas at constant volume will not affect the equilibrium constant

Ans. B

Solution- 2So2(g) + O2(g) 2SO2(g)

All the statements can be explained by le chatelier’s principle.

It state that if constrain (P,T, or concentration) is applied to a system in equilibrium, the

equilibrium will shift and as to tend to counteract the effect of the constraint.

A) When pressure is volume

Hence, the number of moles per unit volume of various reactants and product will increase

Hence, the equilibrium will shift in the direction where there is decrease in number of mole

of gases

B) Incorrect as catalyst is required in the reaction.

V O2 5

2 2 22SO O 2SO (g)+

Here V2O5 acts as a catalyst.

D) The equilibrium constant remains unchanged with addition of inert gas

C) The equilibrium constant will decrease with increase in temperature because high

temperature will help in achieving equilibrium easily.

20. Number of stereo centres present in linear and cyclic structures of glucose are respectively:

A. 5 & 4 B. 5 & 5 C. 4 & 4

D. 4 & 5

Ans. D

Solution- Glucose: C6H12O6

Stereo center 4 different atoms or group of atoms present and sp3 carbon C1,2,3 and 4 are

stereocentre.

Cyclic

In cyclic, there are 5 stereocentre as marked above

21. Which of the following is NOT a correct method of the preparation of benzylamine from

cyanobenzene ? A. H2 /Ni B. (i) SnCl2 + HCl(gas) (ii) NaBH4 C. (i) HCl/H2O (ii) NaBH4 D. (i) LiAlH4 (ii) H3O+

Ans. B

Solution-

Sn/HCl + cyanide is Stephen’s reduction

R—CN SnCl /HCl23NaBH4

O||

R—CHR—C—H⎯⎯⎯→⎯⎯⎯⎯→

In this case Ph—CH2—NH2 won’t form

22. In chromatography, which of the following statements is INCORRECT for Rf ? A. The value of Rf can not be more than one.

B. Higher Rf value means higher adsorption. C. Higher Rf value depends on the type of chromatography D. Rf value is dependent on the mobile phase.

Ans. B

Solution- Rf in chromatography is retardation factor.

Rf values are strongly dependent on the nature of the adsorbent and solvent system.

Low polarity compounds have higher Rf values than high polarity compounds, because low

polarity means low adsorption and hence more mobility.

23. The major product ‘Y’ in the following reaction is :

A.

B.

C.

D.

Ans. B

Solution-

Major product is the result of stable carbocation.

24. Air pollution that occurs in sunlight is :

A. Frog B. Reducing smog C. Acid rain

D. Oxidising smog

Ans. D

Solution- Sunlight will oxidise. Oxidizing smog will form when the primary pollutants are transformed

through photochemical reaction into secondary pollutants (oxidant gases, ozone and

peroxyacetal nitrate)

Acid rain is the result of rain

Fog happens in winter

Smog → winter.

25. Which of the following graphs between molar conductivity ( m ) versus C is correct?

A.

B.

C.

D.

Ans. D

Solution- m

1

c

Both NaCl and KCl are strong electrolytes.

Less the size more hydration

hydration of Na+ > hydration of K+

conductance of Na+ < conductance of K+

26. The difference between Δ H and Δ U (Δ H - Δ U), when the combustion of one mole of heptanes

(1) is carried out at a temperature T, is equal to :

A. -3 RT B. 4 RT C. 3 RT D. -4 RT

Ans. D

Solution- Heptane → C7H16

C7H16(l) + 11 O2(g) → 7CO2(g) + 8H2O(l)

Δng is calculated only for gas

Δng = 7 – 11 = – 4

ΔH = ΔU + Δng RT

− = −H U (7 11)RT

= – 4RT

27. Which of these factors does not govern the stability of a conformation in acyclic compounds ?

A. Torsional strain

B. Angle strain C. Steric interactions D. Electrostatic forces of interaction

Ans. B

Solution- In acyclic, there is no angle strain.

There can definitely be torsional strain which is caused because of repulsion of electrons,

steric interactions and electrostatic force of attraction will be present.

28. A hydrated solid X on heating initially gives a monohydrated compound Y. Y upon heating above

373 K leads to an anhydrous white powder Z. X and Z, respectively, are :

A. Washing soda and soda ash B. Baking soda and dead burnt plaster. C. Washing soda and dead burnt plaster. D. Baking soda and soda ash.

Ans. A

Solution-

Soda ash is Na2CO3

Both are white in colour

Now, based on Z, X can be either

It is given that Y is monohydrated

Na2CO3.H2O is Y

X = NaCO3.10H2O (washing soda)

Z = Na2CO3 (soda ash)

29. The INCORRECT statement is:- A. The spin –only magnetic moment of [Ni (NH3)4 (H2O)2]2 is 283 BM. B. The spin –only magnetic moments of [Fe(H2O)6]2+ and [Cr(H2O)6]2+ are nearly similar.

C. The color of [CoCl(NH3)5]2+ is violet as it absorbs the yellow light. D. The gemstone, ruby, has Cr3+ ions occupying the octahedral sites of beryl.

Ans. D

Solution- 2

2 6Fe(H O)+

Fe+2

Fe : [Ar] 3d6 4s2

Fe+2: [Ar] 3d6 (H2O is a weak ligand no pairing)

Hybridization: sp3d2

n = 4

4(4 2)= +

24 2 6 B.M.= =

[Cr(H2O)6]2+

Cr+2 : [Ar] 3d4

n = 4

correct statement

For [Ni(NH3)4 (H2O)2]2+

n = 2 2(2 2) 2.83 BM= + =

In gemstone, ruby has Cr+3 ion occupying the octahedral sites of Aluminium oxide (Al2O3)

normally occupied by Al+3 ion and not Cr+3ions

This is incorrect.

30. The pH of a 0.02M (NH4OH)=10-5 and log 2 = 0.301

A. 2.65 B. 4.65 C. 4.35 D. 5.35

Ans. D

Solution- NH4Cl → strong acid

NH4OH → weak base

For strong acid and weak base salt

K C

Hkb

+ =

14 2

5

10 2 10[H ]

10

− −+

−

=

( )1/2

1110 2−=

111log[H ] log2 10

2+ − =

2log[H ] log2 11+ = −

= 0.3 – 11 = –10.7

log[H ] 3.35+ =−

pH = – log [H+] =5.35

Mathematics

1. The smallest naturel number n, such that the coefficient of x in the expansion of n

2

2

1x

x

+

9x

is nC23, is:

A. 38 B. 23 C. 58 D. 35

Ans. A

Sol. Given that

The coefficient of x in the expansion of n

2

3

1x

x

+

is nC23. Find the smallest hadural number n =

?

General term

Tr+1 = nCr X2n–2r X–3r

2n – 5r = 1 2n = 5t + 1

2n 1

r5

− =

n n232n 1

5

coeff of x C C−

= =

2n 1 2n 1

23 or n 235 5

− − = − =

2n – 1 = 115 n = 38

n = 58 n = 38 is smallest

2. If 5x + 9 = 0 is the directrix of the hyperbola 162 – 9y2 = 144, then the corresponding focus is

A. 5

,03

B. (–5, 0) C. 5

,03

−

D. (5, 0)

Ans. B

Sol. The given eq- of hyperbola

2 2x y

19 16− =

Compare the origin eq- of hyperbola and find out a,b and e

a = 3, b = 4

2

2

2

be 1

a= +

2 16e 1

9= +

5

e3

=

Focus is (–3e, 0) = (–5 , 0)

3. The angles A, B and C of a triangle ABC are in A.P and a : b = 1 : 3. If c = 4 cm, then the area C

in eq. of this triangle is

A. 4 3 B. 4 / 3 C. 2 / 3 D. 2 3

Ans. D

Sol. 2B = A + C

and A + B + C = π

3B = π

B3

=

2

A C3

+ =

a 1

b 3=

2RsinA 1

2RsinB 3=

sinA 1

sinB 3=

1

sinA2

=

A = 30

a 2,b 2 3,c 4 = = =

1

2 3 2 2 32

= =

4. If the tangent to the curve = −2

xy ; x R;

x 3 (x 3), at a point (α, β) (0, 0) on it is parallel

to the line 2x + 6y – 11 = 0, then:

A. |6α + 2β| = 19 B. |2α + 6β| = 19 C. |2α + 6β| = 11 D. |6α + 2β| =

9

Ans. A

Sol. 2

xy

x 3=

−

Differentiable with respected to X in the pint (α, β) 2

2 2( , )

dy 3

dx ( 3)

− −= −

Given

2

2 2

3 1

3( 3)

− − −=

−

0, 3( 0)=

α = 3 and β = 1

2

Hence |6α + 2β| =19

5. Minimum number of times a fair coin be tossed so that the probability of getting at least one

head is more than 90% is:

A. 8 B. 6 C. 5 D. 7

Ans. D

Sol. Pr (at least on head in n tosses)

= 1 – Pr (no head in n tosses)

n

11

2

= −

So, Just we solved

n

1 991

2 100

−

n

1 1

2 100

n 7 =

6. The negation of the Boolean expression ~SV(~r s) is equivalent to:

A. ~ s ^ ~ r B. r C. s ^ r D. s v r

Ans. C

Sol. ~ (~ S (~r s))

S (r ~ s)

(S r) (s ~ s)

(S r) (s)

(s r)

7. The number of real roots of the equation 5 + |2x - 1| = 2x (2x – 2) is:

A. 3 B. 1 C. 2 D. 4

Ans. B

Sol. 5 + 2x – 1 = 2x (2x – 2)

If x2 1

5 + 2x – 1 = 2x (2x – 2)

Let 2x = t

5 + t – 1 = t (t – 2)

t = 4, – 1

2x = 4

x = 2

only solution

If 2x < 1

5 + 1 – 2x = 2x (2x – 2)

Let 2x = t

5 + 1 – t = t (t – 2)

6 – t = t2 – 2t

t2 – 3t – 6 = 0

(t – 3) (t – 2) = 0

t = 2, 3

2x = 3, 2x = – 2

Rejected

8. The tangent and normal to the ellipse 3x2 + 5y2 = 32 at the point P(2, 2) meet the x-axis at Q and

R, respectively. Then the area (in sq. units) of the triangle PQR is:

A.

68

15

B.

34

15

C.

16

3

D.

14

3

Ans. A

Sol. The equation

3x2 + 5y2 = 32

d.w.r. to x

6x + 5y y = 0

3x

y5y

− =

at the point P(3, 2)

3

y (2,2)5

− =

Tangent (y –2) = 3 16

(x 2) Q ,05 3

− −

Normal 5 4

(y 2) (x 2) R ,03 5

− = −

Area 21 68(QR)x QR

2 15= = =

9. Let λ be a real number for which the system of linear equations

6x y z+ + =

4 2x y z + − = −

3 2 4 5x y z+ − = −

has infinitely many solutions. Then λ is a root of the quadratic equation :

A. 2 6 0 + − =

B. 2 6 0 − − =

C. 2 3 4 0 − − =

D. 2 3 4 0 + − =

Ans. B

Sol. Δ = 0

1 1 1

4 0

3 2 4

= − =

−

1[ 4 ) 1[ 16 3 ) 1[8 3 ) 0− + − − + + + =

on solving we get 3 =

go by the option

Only α option will be satisfy by 3.=

Hence, Answer will be 2 6 0. −− =

10. A perpendicular is drawn from a point on the line 1 1

2 1 1

x y z− += =

− to the plane 3x y z+ + =

such that the foot of the perpendicular Q also lies on the plane 3x y z− + = . Then the co-

ordinates of Q are : A. (2,0,1) B. (4,o,-1) C. (1,0,2) D. (-1,0,4)

Ans. A

Sol. x 1 y 1 z

2 1 1

− += = =

−

Let a point P on the line is

( 1, 1, )+ −− +

Foot of ⊥ Q is given by

x 2 1 y 1 z ( 3)

1 1 1 3

− − ++ − − −= = =

Q lies on x + y + z = 3 and x – y + z = 3

x + z = 3 and y = 0

2 3

y 0 1 03

− + = + = =

Q is (2,0,1)

11. Let a, b and c be in G.P. with common ratio r, where 0a and 1

02

r . If 3a,7b and 15c are

the first three terms of an A.P., then the 4th term of this A.P. is :

A.

2

3a

B.

7

3a

C. a

D. 5a

Ans. C

Sol. Let’s Find GP’s terms of common ration r.

a = 0, b = ar, c = ar2

where a 0 and 1

a0 r .2

Given

3a, 7b, 15c → in A.P

14b = 3a + 15c

Put the value of a,b,c in above equation

14(ar) = 3a + 15(ar2)

15r2 – 14r + 3 = 0

1 3

r , (rejected)2 5

=

Common difference = 7b – 3a

= 7ar – 3a

7a

3a3

= −

2

a3

−=

4th terms is 2 15 2

15C a a a3 a 3

− = −

= a

12. The area (in sq. units) of the region bounded by the curves 2xy = and 1y x= + , in the first

quadrant is :

A.

3log 2

2e +

B.

3 1

2 log 2e

−

C.

1

2

D.

3

2

Ans. B

Sol. y = 2x and y = |x + 1|

Required Area 1 1 1x x

0 0 0((x 1) 2 )dx (x 1) x 2 dx+ − = + + −

12 x

2e 0

x 2x

2 log

− + −

2e

3 1

2 log= −

13. A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts

at a rate of 50 cm3/min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of the ice decreases, is :

A.

5

6

B.

1

18

C.

1

9

D.

1

36

Ans. B

Sol.

Volume of spherical iron ball

3 34V ((10 h) 10 )

3= + −

Volume differentiate with respect to time

2 3dv dh dv4 (10 5) Give 50 cm /m

dt dt dt

= + =

dv 1 cm

dt 18 min

− =

14. If 2 25 ( ) ,x xx e dx g x e c− −= + where c is a constant of integration, then g (-1) is equal to :

A. 1

B.

5

2−

C.

1

2−

D. -1

Ans. B

Sol. If 2 25 x xx e dx g(x)e c− −= + , c- integration constant

Let x2 = t

2x dx = dt

2 t1t e dt

2−

2 t t1[ t e 2e dt]

2− − +

4

22 xxx 1 e c

2−

= − − − +

4

2xg(x) x 1

2= − − −

= − − −

= −

1g(x) 1 1

25

2

15. Suppose that 20 pillars of the same height have been erected along the boundary of a circular

stadium. If the top of each pillar has been connected by beams with the top of all its non-adjacent pillars, then the total number of beams is: A. 180 B. 210 C. 170

D. 190

Ans. C

Sol. Any two non-adjacent pillars are joined by beam

no. of beams = no. of diagonals

C220 20= −

= 20 × 19 – 20

= 190 – 20

= 170

16. If 2lim

5,1 1

x ax b

x x

− +=

→ −then a+b is equal to :

A. -7 B. 1 C. 5

D. -4

Ans. A

Sol. If 2

x 1

x ax blim 5

x 1→

− +=

−

(1) 2 – a (1) + b = 0

1 – a + b = 0

a – b = 1 …..(1)

Now,

‘L’ hospital rule

2x – a = 5

2 – a = 5 ( x = 1)

a = – 3 …..(2)

Put in (1)

b = – 4

a + b = – 7

17. Let 1 2 3,a ,.....a a be an A.P. when a6=2. Then the common difference of this A.P., which

maximises the product a1a4a5, is :

A.

8

5

B.

2

3

C.

6

5

D.

3

2

Ans. A

Sol. Let a1, a2, a3……….be an A.P

With a6 = 2.

First term = 9

Common difference = d

a + 5d = 2

a1, a4.a5 = a(a + 3d) (a + 4d)

f(d) = (2 – 5d) (2 – 2d) (2 – d)

2 8

f (d) 0 d ,3 5

= =

8

f (d) 0 at d5

=

8

d5

=

18. Let (x)y y= be the solution of the differential equation,

2tan 2 tan , ,2 2

dyy x x x x xe

dx

+ = + −

, such that y(0)=1. Then :

A.

2

24 4 2

y y

+ − = +

B.

24 4

y y

− − =

C.

24 4

y y

+ − = −

D.

24 4

y y

− − = −

Ans. D

Sol. Differential equation,

dy

dx + y tan x = 2x + x2 tan

I.F = tanx dx n secxe e secx = =

y sec x = 2(2x x tanx)secx dx+ +

y sec x = x2 sec x + sec x [y] = 2x secxsecx

+

2y x cosx = +

y(0) 0 1 1= + = =

y = x2 + cos x

y 2x sinx = −

1

y4 2 2

= −

−

− = +

1y

4 2 2

Correct Answer is

− − = −

y y 24 4

19. The integral 2 4

3 3 3

6

sec cosx ec xdx

is equal to:

A.

5 13 33 3−

B.

7 56 63 3−

C.

4 13 33 3−

D.

5 26 33 3−

Ans. B

Sol. Integral 2

/3

4/3/6

sec xdx

tan x

2

/3

4/3/6

sec xdx

tan x

=

Let 2

4/3

sec xI dx

tan x=

Let tan x = t

Sec2x dx = dt

4/3

dt

t=

I = –3 e–1/3

( )/3

1/3

/63(tanx)

−

= −

1/3

1/6

13 3

3

= −

= 37/6 – 35/6

20. The distance of the point having position vector 2 6i j k− + + from the straight line passing

through the point (2, 3, -4) and parallel to the vector, 6 3 4i j k+ − is :

A. 7 B. 6

C. 2 13

D. 4 3

Ans. A

Sol.

AP

AD 61|n|

= =

2 2 2AP PD AP AD = − [By Pythagoras theorem]

2 2PD AP AD= −

110 61= −

= 7

21. Line are drawn parallel to the line 4 3 2 0,x y− + = at a distance3

5 from the origin. Then which

one of the following points lies on any of these lines?

A.

1 1,

4 3

B.

1 1,

4 3

−

C.

1 2,

4 3

−

D.

1 2,

4 3

− −

Ans. C

Sol. Line parallel to 4x – 3y + 2 = 0

Is given as 4x – 3y + = 0

Distance given from origin

3

5 5

=

3 =

Required lines are 4x – 3y + 3 = 0 and 4x – 3y – 3 = 0

By check option

Option C 1 2

,4 3

−

4x = 3y + 3 = 0 , 4x – 3y – 3 = 0

1 2

4 3 3 04 3

− − + =

1 24 3 3

4 3

− − −

0 = 0 –1 – 2 –3 0

Option is satisfy the equation.

Hence, option (C) is correct.

Option (D) 1 2

,4 3

− −

1 2

4 3 3 02 3

− − + =

1 24 3 3 0

4 3

− − − − =

1 2 3 0− + + 1 2 3 0− + −

Option D is wrong

Option (A) 1 1

,4 3

1 1

4 3 3 04 3

− + 1 1

4 3 3 04 3

− −

Option A is wrong

Option B 1 1

,4 3

−

1 1

4 3 34 3

− − + ,

1 14 3 3

4 3

+ − − −

1 1 3 0+ + 1 3 3 0+ −

22. Let ( ) log (sin ),(0 )ef x x x = and 1( ) sin ( ),( 0)xg x e x− −= . If α is a positive real number

such that a = (fog)’ (α) and b = (fog) (α), then:

A. 2 0a b a + + =

B. 2 22a b a + − = −

C. 2 1a b a − − =

D. 2 0a b a − − =

Ans. C

Sol. Let f(x) = loge(sin x), (0 < x < π)

g(x) = sin–1 (e–x), (x 0)

f(g(x)) = –x (f(g(x))1 = –1

f(g(x)) = –α = b (f(g(x)))1 = –1 = a

b = –α

a = –1

Now check from option.

Option (1) aα2 + bα – a = – 2α2

b = –α

a = –1

LHS. –α2 – (–α)α – (1) = –1

Option (2)

aα2 + bα + a = 0

b = –α, a = –1

L.H.S = –1α2 + (–α) (+α) – 1

= –2α2–1

Option (3)

aα2 – bα – a = + 1

b = –α, a = –1

L.H.S = –α2 – (α) α –

= + 1

Option (3) is correct.

Option (4) aα2 – bα – a = 0

L.H.S = –α2 – (α) α –(1) = 1

23. If the line ax + y =c, touches both the curves x2 +y2 =1 and y2 =4 2x , then |c| is equal to :

A.

1

2

B.

1

2

C. 2 D. 2

Ans. C

Sol. Tangent to the curve 2y 4 2x= is 2

y mxm

= +

It is tangent to the circle x2 + y2 = 1

2

2 / m1 m 1

1 m = =

+

tangent are y x 2 and y x 2= + =− −

Compare with y = ax + c

a 1 and c 2 = =

24. The sum

3 3 3 3 31 2 1 2 31 ...

1 2 1 2 3

+ + ++ + +

+ + +

3 3 3 31 2 3 ... 15 1(1 2 3 ... 15)

1 2 3 ... 15 2

+ + + ++ − + + + +

+ + + + is equal to

:

A. 1240 B. 620 C. 660 D. 1860

Ans. B

Sol. 2 2 2 2 2 2 2 2 21 2 1 2 3 1 2 3 .... 15

1 .....1 2 1 2 3 1 2 3 .... 15

+ + + + + + ++ + + +

+ + + + + + +

1

(1 2 3 ..... 15)2

−+ + + +

Sum 3 3 315

n 1

1 2 ....n 1 15 16

1 2 ..... n 2 2=

+ + = −

+ + +

15

n 1

n(n 1)60

2=

+= −

15 15

2

n 1 n 1

1 1n n 60

2 2= =

= + −

1 15 16 31 1 15 16

602 6 2 2

= + −

= 620

25. The sum of the real roots of the equation

6 1

2 3 3 0

3 2 2

x

x x

x x

− −

− − =

− +

, is equal to:

Ans. B

Sol. The equation

x 6 1

3x x 3 0

3 2x x 2

− −

− − =

− +

Expand x(–3x × (x+ 2) – 2x(x – 3)) + (–6) [(x + 2) 2 – 2(x – 3) (–3) + (–1) [4x + 9x]

25x 30x 30 5x 0 + − + =

x3 – 7x + 6 = 0

Sum of roots = 0

26. If both the mean and the standard deviation of 50 observations x1, x2, …., x50 are equal to 16,

then the mean of (x1 – 4)2, (x2 – 4)2, ….., (x50 – 4)2, is : A. 525

B. 480 C. 400 D. 380

Ans. C

Sol. Given mean and standard deviation of 50 observation x1, x2,….x50 are equal to 16.

Mean ix( ) 16

50 = =

ix 16 50 =

S.D. 2

ix ( )( ) 16

50

− = =

2i

x256 2

50 =

Required mean 2

i(x 4)

50

−=

2 ii

x 16 50 8 x

50

+ −=

= 256x2 + 16 – 8 × 16

= 400

27. If 1 1cos cos ,2

yx − −− = where 1 1,x− 2 2,

2

yy x− , then for all

2 2, ,4 4 cosx y x xy y− + is equal to :

A. 2 2 24sin 2x y −

B. 2 2 24cos 2x y +

C. 22sin

D. 24sin

Ans. C

Sol. If 1 1 ycos x cos ,

2− −− = where

y1 x 1, y 2, x

2− −

1 1 ycos x cos

2− − − =

1 1 ycos cos x cos cos

2− −

− =

2

2y yx 1 x 1 cos

2 4+ − − =

2

2xy ycos 1 x 1

2 4

− = − −

Squaring of both sides

2

2 2 2yx xycos 1 cos sin

4+ − = − =

4x2 + y2 – 4xy cos α = 4sin2α

28. If z and w are two complex numbers such that |zw| = 1 and arg(z)-arg(w)=2

, then :

A.

1

2

izw

− +=

B.

1

2

izw

−=

C. zw i= −

D. zw i=

Ans. C

Sol. |Zω| = 1

|Z| |ω| = 1

Let i1e

r=

Then i

2Z re

+

=

i

i i( /2)2Z e .e e i

− + = = = −

And i

i i /22z e .e e i

+ = = =

29. If the plane 2x – y + 2z +3 = 0 has the distances 1

3and

2

3units from the planes

4 2 4 0x y z − + + = and 2 2 0x y z − + + = respectively, then the maximum value of λ + μ

is equal to :

A. 13 B. 9 C. 5 D. 15

Ans. A

Sol. Distance formula

(i) | 6| 6 1

6 316 4 16

− −= =

+ +

| 6| 2− =

8, 4 =

(ii) | 3| 2

34 4 1

−=

+ +

|μ – 3| = 2

μ = 5, 1

max( ) 13 + =

30. The locus of the centers of the circles, which touch the circle, x2 + y2 = 1 externally, also touch

the y – axis and lie in the first quadrant, is:

A. 1 4 , 0x y y= +

B. 1 4 , 0y x x= +

C. 1 2 , 0x y y= +

D. 1 2 , 0y x x= +

Ans. D

Sol.

2 2k 1 | | + = +

2 2 2h k 1 h |h|+ = + +

2k 1 |h|= +

2y 1 x= +

y 1 x, x 0= +