1

PHYSICS

1. The dimensions of 20e /4 hc , where 0e, ,h and c are electronic charge, electric permittivity, Planck’s

constant and velocity of light in vacuum respectively

(A) 0 0 0[M L T ] (B) 1 0 0[M L T ] (C) 0 1 0[M L T ] (D) 0 0 1[M L T ]

Solution (A):

[e] [AT], 1 3 4 20 [M L T A ], 2 1[h] [ML T ]

and 1[c] [LT ]

2 2 2

1 3 4 2 2 1 10

e A T

4 hc M L T A ML T LT

0 0 0[M L T ]

2. A particle is projected upwards. The times corresponding to height h while ascending and while

descending are t1 and t2 respectively. The velocity of projection will be

(A) 1gt (B) 2gt (C) 1 2g t t (D) 1 2g t t

2

Solution (D):

If 1t and are time of ascent and descent respectively then time of flight 1 22u

T t tg

1 2g(t t )u

2

3. A particle is moving on a circular path of radius r with uniform velocity v . The change in velocity when

the particle moves from P to Q is o( POQ 40 )

(A) o2vcos40

(B) o2vsin40

(C) o2vsin20

(D) o2vcos20

Solution (C): Change in velocity = 2vsin( /2) 2vsin20

4. The masses of 10 kg and 20 kg respectively are connected by a massless spring as shown in figure. A

force of 200 N acts on the 20 kg mass. At the instant shown, the 10 kg mass has acceleration 212m/sec . What is the acceleration of 20 kg mass

(A) 212m/sec

(B) 24m/sec

(C) 210m/sec

(D) Zero Solution (B): As the mass of 10 kg has acceleration 12m/s2 therefore it apply 120N force on mass 20kg in a backward direction.

Net forward force on 20 kg mass = 200 – 120 = 80N

Acceleration 2804m/s

20 .

2

5. A body of weight 50 N placed on a horizontal surface is just moved by a force of 28.2 N. The frictional force and the normal reaction are

(A) 10 N, 15 N (B) 20 N, 30 N

(C) 2 N, 3 N (D) 5 N, 6 N Solution (B):

Frictional force = f 28.2cos45 1

28.2 20 N2

Normal reaction R 50 28.2sin45 30 N

6. The kinetic energy of a body decreases by 36%. The decrease in its momentum is

(A) 36% (B) 20% (C) 8% (D) 6% Solution (B):

P 2mE P E

In given problem K.E. becomes 64% of the original value.

2 2

1 1

P E 64E0.8

P E 100E 2P 0.8P

2P 80% of the original value.

i.e. decrease in momentum is 20%. 7. Two identical spheres are placed in contact with each other. The force of gravitation between the

spheres will be proportional to (R = radius of each sphere)

(A) R (B) 2R (C) 4R (D) None of these Solution (C):

23

64

2 2 2

4G R

M M R3F G F R

R (2R) R

8. The diameter of rain-drop is 0.02 cm. If surface tension of water be 372 10 newton per metre, then the

pressure difference of external and internal surfaces of the drop will be

(A) 4 21.44 10 dyne cm (B) 4 21.44 10 newton m

(C) 3 21.44 10 dyne cm (D) 5 21.44 10 newton m

Solution (A): 3

2

2

2T 2 72 10P 1440 N/m

r 0.01 10

4 21.44 10 dyne/cm

9. If W be the weight of a body of density in vacuum then its apparent weight in air of density is

(A) W

(B) W 1

(C)

W

(D) W 1

Solution (D):

Apparent weight in air = W – upthrust = V g V g = V g 1

= W 1

3

10. Out of the following, in which vessel will the temperature of the solution be higher after the salt is completely dissolved. (A) A (B) B (C) Equal in both (D) Information is not sufficient Solution (B): When salt crystals dissolves, crystal lattice is destroyed. The process requires a certain amount of energy (latent heat) which is taken from the water. In vessel (B), a part of intermolecular bonds has already been destroyed in crushing the crystal. Hence less energy is require to dissolve the powder and the water will be at higher temperature.

11. If the r.m.s. velocity of a gas at a given temperature (Kelvin scale) is 300 m/sec. What will be the r.m.s.

velocity of a gas having twice the molecular weight and half the temperature on Kelvin scale = (A) 300 m/sec (B) 600 m/sec (C) 75 m/sec (D) 150 m/sec Solution (D):

rms rms

3RT Tv v

M M

2 21

1 2 1

v TM

v M T

2

1

2

1 1

2

v 300v

2 2 =150m/sec

12. An ideal monoatomic gas is taken round the cycle ABCDA shown in the PV diagram in the given fig. The

work done during the cycle is

(A) 1

PV2

(B) 2 PV

(C) PV (D) Zero Solution (C): Conceptual

13. Four massless springs whose force constants are 2k, 2k, k and 2k respectively are attached to a mass

M kept on a frictionless plane (as shown in figure). If the mass M is displaced in the horizontal direction, then the frequency of oscillation of the system is

(A) 1 k

2 4M (B)

1 4k

2 M (C)

1 k

2 7M (D)

1 7k

2 M

Solution (B): The two springs on left side having spring constant of 2k each are in series, equivalent constant is

1k

1 1

2k 2k

. The two springs on right hand side of mass M are in parallel. Their effective spring

constant is (k 2k) 3k . Equivalent spring constants of value k and 3k are in parallel and their net value

of spring constant of all the four springs is k 2k 3k

Frequency of mass is 1 4k

n2 M

4

14. A glass tube 1.5 m long and open at both ends, is immersed vertically in a water tank completely. A tuning fork of 660 Hz is vibrated and kept at the upper end of the tube and the tube is gradually raised out of water. The total number of resonances heard before the tube comes out of water, taking velocity of sound air 330 m/sec is

(A) 12 (B) 6 (C) 8 (D) 4 Solution (B): Suppose N resonance occurred before tube coming out.

Hence by using (2N 1)v

l4n

(2N 1) 3301.5

4 660

N 6 .

15. A point charge is surrounded symmetrically by six identical charges at distance r as shown in the figure.

How much work is done by the forces of electrostatic repulsion when the point charge q at the centre is removed at infinity (A) Zero

(B) 206q /4 r

(C) 20q /4 r

(D) 2012q /4 r

Solution (B):

Total potential at the centre 0

6qV

4 r

. Required work done

2

0

6qq.V

4 r

16. In the following circuit a 10 m long potentiometer wire with resistance 1.2 ohm/m, a resistance R1 and an

accumulator of emf 2 V are connected in series. When the emf of thermocouple is 2.4 mV then the deflection in galvanometer is zero. The current supplied by the accumulator will be

(A) 4 10–4 A

(B) 8 10–4 A

(C) 4 10–3 A

(D) 8 10–3 A

Solution (A):

E x l i l 3

4E E 2.4 10i 4 10 A

l l 1.2 5

.

17. The three resistances A, B and C have values 3R, 6R and R respectively. When some potential

difference is applied across the network, the thermal powers dissipated by A, B and C are in the ratio (A) 2 : 3 : 4 (B) 2 : 4 : 3 (C) 4 : 2 : 3 (D) 3 : 2 : 4 Solution (C):

Thermal power in 2

2A

2i 4A P 3R i R

3 3

Thermal power in 2

2B

i 2B P 6R i R

3 3

Thermal power in 2CC P i R

A B CP : P : P4 2

: :1 4 : 2 : 33 3

5

18. A particle having a charge of 10.0 C and mass 1 g moves in a circle of radius 10 cm under the

influence of a magnetic field of induction 0.1T . When the particle is at a point P , a uniform electric field

is switched on so that the particle starts moving along the tangent with a uniform velocity. The electric field is (A) 0.1 V/m (B) 1.0 V/m (C) 10.0 V/m (D) 100 V/m Solution (C): When the particle moves along a circle in the magnetic field B, the magnetic force is radially inward. If an electric field of proper magnitude is switched on which is directed radially outwards, the particle may experience no force. It will then move along a straight line with uniform velocity. This will be the case when qE vB E vB

also mv qBr

r vqB m

So 2qB r

Em

6 2 2

3 6

(10 10 ) (0.1) 10 1010 V /m

1 10 10

19. The magnetic field in the cylindrical region shown in figure increases at a constant rate of 20 mT/sec.

Each side of the square loop ABCD has a length of 1 cm and resistance of 4. Find the current in the wire AB if the switch S is closed

(A) 71.25 10 A , (anti-clockwise)

(B) 71.25 10 A (clockwise)

(C) 72.5 10 A (anti clockwise)

(D) 72.5 10 A (clockwise)

Solution (A): 2 2

3

7

e A dB (1 10 )i . 20 10

R R dt 16

1.25 10 A

(Anti-clockwise).

20. The instantaneous values of current and emf in an ac circuit are I 1/ 2sin314t amp and

E 2sin(314t /6)V respectively. The phase difference between E and I will be

(A) /6 rad (B) /3 rad (C) /6 rad (D) /3 rad

Solution (A):

Phase difference relative to the current (314t ) (314t)6 6

21. Two large parallel plates are connected with the terminal of 100 V power supply. These plates have a

fine hole at the centre. An electron having energy 200eV is so directed that it passes through the holes. When it comes out it's de-Broglie wavelength is (A) 1.22 Å

(B) 1.75 Å

(C) 2 Å

(D) None of these

6

Solution (A): Energy of the electron, when it comes out from the second plate = 200 eV – 100 eV = 100 eV Hence accelerating potential difference = 100 V

Electron12.27 12.27

1.23ÅV 100

22. Consider an initially pure M gm sample of AX, an isotope that has a half life of T hour, what is it’s initial

decay rate (NA= Avogrado No.)

(A) AMN

T (B) A0.693MN

T (C) A0.693MN

AT (D) A2.303MN

AT

Solution (C):

t0N N e t

0dN

N edt

Initially at t = 0, 0t 0

dNN

dt

where N0 = Initial number of undecayed atoms

Mass of the sample

Mass of a single atom of XA

A

MNM

A /N A

A A

t 0

MN 0.693MNdN

dt A AT

23. In the following circuit the equivalent resistance between A and B is

(A) 20

3 (B) 10 (C) 16 (D) 20

Solution (C): According to the given figure A is at lower potential w.r.t. B. Hence both diodes are in reverse biasing, so equivalent, circuit can be redrawn as follows.

Equivalent resistance between A and B

R = 8 + 2 + 6 = 16 . 24. In the given detector circuit, the suitable value of carrier frequency is

(A) << 109 Hz (B) << 105 Hz (C) >> 109 Hz (D) None of these Solution (A):

Using RCfcarrier

1

We get time constant, sRC 912 10101000

Now HzT

99

1010

11

Thus the value of carrier frequency should be much less than ,109Hz say 100 kHz.

7

25. PQR is a right angled prism with other angles as 60o and 30o. Refractive index of prism is 1.5. PQ has a thin layer of liquid. Light falls normally on the face PR. For total internal reflection, maximum refractive index of liquid is (A) 1.4 (B) 1.3 (C) 1.2 (D) 1.6 Solution (B): For TIR at PQ ; C

From geometry of figure 60 i.e. 60 C sin60 sinC

Liquid

Liquid Pr ismPr ism

3 3

2 2

Liquid Liquid3

1.5 1.32

.

26. If angular diameter of Sun is about 30 and it's distance from earth is 1.5 1011 m, then solar diameter is

(A) 1.1 107 m (B) 1.5 108 m (C) 1.4 109 m (D) 1.9 1011 m Solution (C): We know that

11 91D r 1.5 10 1.4 10 m

2 180

27. A uniform rod AB of mass m and length 2a is falling freely without rotation under gravity with AB

horizontal. Suddenly the end A is fixed when the speed of the rod is v. The angular speed with which the rod begins to rotate is

(A) v

2a (B)

4v

3a (C)

v

3a (D)

3v

4a

Solution (D): By conservation of Angular momentum about O

2L mL 3 v 3v

mv2 3 2 L 4a

28. The energy levels of the hydrogen spectrum is shown in figure. There are some transitions A, B, C, D

and E. Transition A, B and C respectively represent

(A) First member of Lyman series, third spectral line of Balmer series and the second spectral line of

Paschen series (B) Ionization potential of hydrogen, second spectral line of Balmer series and third spectral line of

Paschen series (C) Series limit of Lyman series, third spectral line of Balmer series and second spectral line of Paschen

series (D) Series limit of Lyman series, second spectral line of Balmer series and third spectral line of Paschen

series Solution (C): Conceptual

8

29. According to Bohr's theory, the expressions for the kinetic and potential energy of an electron revolving in an orbit is given respectively by

(A) 2

0

e

8 r

and

2

0

e

4 r

(B)

208 e

r

and

204 e

r

(C) 2

0

e

8 r

and

2

0

e

4 r

(D)

2

0

e

8 r

and

2

0

e

4 r

Solution (A):

P.E.2 2

0

ke e

r 4 r

; K.E.

2

0

1 e(P.E.)

2 8 r

30. The following figure shows a logic gate circuit with two inputs A and B and the output Y. The voltage

waveforms of A, B and Y are as shown below

The logic gate is (A) NOR gate (B) OR gate (C) AND gate (D) NAND gate

Solution (D): Conceptual

9

monochlorination

(1 mol) (1 mol X)

Mg/ether

Cl

(1 mol Y)

MgCl

2 2HO CH CH C CH

(Z) (0.1 mol)

MgCl monochlorination

(1 mol) (1 mol X)

Mg/ether

Cl

(1 mol Y)

MgCl

2 2HO CH CH C CH

(Z) (0.1 mol)

CHEMISTRY

31. LiAlH (excess)4

H O2

Product of the reaction is:

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

Solution (C):

4LiAlH reduces amides to amines.

32. HO CH CH C CHMonochlorination Mg 2 2

ether (2.9 gm)X Y Z salt,

the number of moles of Z

produced would be (assuming the yield to be 100% in each stage of reaction)

(A) 1 (B) 0.01 (C) 0.082 (D) 10

Solution (C):

2 2HO CH CH C CH 2

2.9moles

70

(a mole)

a 2.9/70 2.9a 2 0.082

2 1 70

33. 0.117 g of an organic compound on heating with conc. 3HNO and silver nitrate in carius furnace gave

0.42 g of AgCl. Find the percentage of chlorine in the compound (M.Mass, Ag=108, Cl=35.5) (A) 61.4 (B) 64 (C) 89 (D) 18

Solution (C):

(0.117g)Organiccompound AgCl(0.42g)

mass of chlorine =0.42

35.5 0.10g(108 35.5)

% of chlorine in the organic compound

0.10

100 88.8 890.117

10

O

O

Cl

(–C–(CH ) –C–NH–(CH ) –NH –2 4 2 6 n)O O

Nylon 6,6

n

H C OH H C O H

O :(XII)

: O :

(XIII)

H C O H H C O H

: O : : O :

(XIV) (XV)

2H C OH

2NH

2CHOH

2NH

(V) (VI)

2CHOH

2NH

(VII)

2CHOH

2NH

(VIII)

OH

2NH2CH

2H C OH

2NH

2CHOH

2NH

(V) (VI)

2CHOH

2NH

(VII)

2CHOH

2NH

(VIII)

2H C OH

2NH

2CHOH

2NH

(V) (VI)

2CHOH

2NH

(VII)

2CHOH

2NH

(VIII)

H C OH H C O H

O :(XII)

: O :

(XIII)

H C O H H C O H

: O : : O :

(XIV) (XV)

H C OH H C O H

O :(XII)

: O :

(XIII)

H C O H H C O H

: O : : O :

(XIV) (XV)

H C OH H C O H

O :(XII)

: O :

(XIII)

H C O H H C O H

: O : : O :

(XIV) (XV)

34. The resonating structures of the following compounds are arranged in order of decreasing stabilities. Which option is incorrect?

(A) 2 2CH N N CH N N

2 2CH N N CH N N

(B) > > >

(C) > > >

(D) All of these

Solution (B): 35. Which of the following classification is wrong for the given polymers (A) Polyisoprene Natural rubber and homo-polymer

(B) Dacron Polyester and co-polymer

(C) Nylon-66 Polyamide and condensation polymer

(D) Buna-N Homopolymer and step growth polymer

Solution (D): (A) Polyisoprene is natural rubber. Its monomeric unit is isoprene.

n

(B) Terylene or Dacron is a condensation polymer of ethylene glycol and terephthalic acid

(C) Nylon-66 is a polyamide condensation polymer

2 4 2 2 6 2Adipic acid Hexamethylenediamine

HOOC (CH ) COOH H N(CH ) NH 553K

Highpressure

(D) Buna N is a copolymer

CN

Peroxide2 2 2

|nCH CH CH CH nCH CH

(Butadiene) (Acrylonitrile)

CN

2 2 2(Buna N)

|—(CH CH CH CH CH CH—)

36. Which of the following chemicals can be added for sweetening of food items at cooking temperature and does not provide calories?

(A) sucrose (B) glucose (C) aspartame (D) sucralose Solution (D):

Refer NCERT 12 2nd part page 450

Sucralose is trichloroderivative of sucrose. Its appearance and taste are like sugar. It is stable at cooking temperature. It does not provide calories.

37. Which of the following reactions will generate 3° alcohol?

(I) || +CH MgBr(excess) H3

3CH - C-Cl

O

(II)dil.H SO2 4

(III) +LiAlH H4 (IV)

+Mg H

ether

(A) I, II only (B) II, III only (C) III, IV only (D) All

11

H C–3 C–OH

CH3

CH3 CH3

CH –C–O¯3

CH3

..

H C—3 C—CH —CH2 3

CH3

OH

H C—3 C—CH —CH2 3

CH3

+

H C—3 C—CH—CH3

CH3

H

+H C—3 C—CH—CH2

CH3

—

H

O

LiAlH4

H¯ (S 2)N

OHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOHOH

OHO¯..O

CH MgX2

+

O

CH2

Cl

Solution (D):

(A)

O

2CH MgBr33

||CH C Cl

+H

(B) or H

hydride

Shift

H O2

(C) +H

(D) Mg

Ether

H

38. Which of the following will show electrophilic aromatic substitution with 2NaNO HCl

(I) Aniline

(II) Orthomethylaniline

(III) Phenol

(IV) N,N-dimethylaniline

(A) I, II only (B) II, III only (C) III, IV only (D) All of these Solution (C): I and II Diazonium salt will be formed, whereas in III & IV EAS will take place.

39. Identify the wrong statement in the following (A) chlorofluorocarbons are responsible for ozone layer depletion (B) greenhouse effect is responsible for global warming (C) acid rain is mostly because of oxides of nitrogen and sulphur

(D) ozone layer does not permit infrared radiation from the sun to reach the earth

Solution (D): (A) Chloroflurocarbon compounds (CFCs), also known as freons are responsible for ozone deplection.

UV2 2 2CF Cl (g) Cl CF Cl(g)

The chlorine radicals causes the breakdown of ozone layer.

3 2Cl(g) O (g) ClO(g) O (g)

(B) Green house gases such as carbondioxide, methane, water vapour, nitrous oxide, CFCs and ozone trap heat radiations and causes global warming. (C) When pH of the rain water drops below 5.6, it is called acid rain. SO2 and NO2 after oxidation and

reaction with water are major contributors to acid rain.

2 2 2 2 4

2 2 2 3

2SO (g) O (g) 2H O( ) 2H SO (aq)

4NO (g) O (g) 2H O( ) 4HNO (aq)

(D) The upper stratosphere consists of considerable amount of ozone (O3), which protects us from the

harmful ultraviolet (UV) radiations (l 255nm) coming from the Sun. These radiations cause skin

cancer (melanona) in humans.

12

COH

SOCl2(1)

CClO

NH3(2)

C 2NHO

NaOH Br2(3)

2NH

H C C O C CH3 3(4)

O O3HN C CH

O

(P)

COH

SOCl2(1)

CClO

NH3(2)

C 2NHO

NaOH Br2(3)

2NH

H C C O C CH3 3(4)

O O3HN C CH

O

(P)

40. The structure of product (P) is:

(CH CO) O3 2(4)

(P)

(A) (B) (C) (D) Solution (D):

41. Total number of lone pair of electron in 3I ion is:

(A) 12 (B) 3 (C) 6 (D) 9

Solution (4):

3I

I I I

.. .. ..

:: I I : I.. .. ..

No. of lone pairs =9

42. a bpOH 7 0.5 pK 0.5 pK is true for aqueous solution containing which pair of cation and anion:

(A) 6 5 4 3C N NH , CH COO (B) 4NH , F (C) Both (A) and (B) (D) None of these

Solution (C):

a b a bpOH 7 0.5 pK 0.5 pK or pH 7 0.5pK 0.5pK is valid for salt of wA and wB.

43. A sample of air is saturated with benzene (vapour pressure=100 mm Hg at 298 K) at 298K, 750mm Hg pressure. If it is isothermally compressed to one third of its initial volume, the final pressure of the system is: (A) 2250 torr (B) 2150 torr (C) 2050 torr (D) 1950 torr

Solution (C):

Pressure of air=750-100=650 mm of Hg

On compressing fP 650 3 mm of Hg 1950 mm of Hg

So, TP (1950 100) 2050 mm of Hg

SOCl NaOH BrNH2 23(1) (2) (3)

13

44. For a first order reaction, nA B whose concentration vs time curve is as shown in the figure. If half-

life for this reaction is 24 minutes. Find out the value of n.

(A) 1 (B) 2 (C) 3 (D) 4

Solution (B):

At t 48 min, [A] [B]

Also, 0[A][A] [B]

4

0[A] nx xnA B

0 0 00

[A] [A] [A]x [A] n n 3

4 4 4

45. 100 ml of 0.1M 2 3NaAl(OH) CO is neutralized by 0.25 N HCl to form 3 2NaCl, AlCl andCO . Volume of

HCl required is: (A) 10mL (B) 40mL (C) 100mL (D) 160mL

Solution (D):

0.01 0.04

2 3 3 2 2NaAl(OH) CO 4HCl NaCl AlCl 2H O CO

0.04 0.25 Y

0.04V 0.16

0.25 l

0.16 1000 160mL

46. In the process shown in the figure on an ideal diatomic gas, the value of q and H respectively is:

(A) 1 1 1 179.5 P V and 94.5 P V (B) 1 1 1 155.5 P V and 94.5 P V

(C) 1 112 P V and 0 (D) 1 179.5 P V and defined ( P varies)

Solution (A):

47. The percentage packing efficiency of the two dimensional arrangement of sphere for plane ABCDEF shown below is:

(A) 90.64% (B) 74.05% (C) 68.02% (D) 78.54%

Solution (A):

14

48. 4 2 3 2NH COONH (s) 2NH (g) CO (g) .

If equilibrium pressure is 6 atm for the above reaction; pK will be:

(A) 32 (B) 27 (C) 4/27 (D) 1/27

Solution (A):

4 2 3 22p p

NH COONH (s) 2NH (g) CO (g)

2p p 6 p 2 atm 3 2NH COP 4 atm; P 2 atm

3 2

2 2p NH COK [P ] [P ] 4 2 32

49. The critical volume of a gas is 0.072 lit. 1mol . The radius of the molecule will be, in cm

(A)

1

323310

4

(B)

1

323410

3

(C)

1

323310

4

(D)

1

38310

4

Solution (A):

3 23 3A c

16 16b r N ; V 3b 0.072 1000 6.023 10 r 3

3 3

50. A proton accelerated from rest through a potential difference of ‘V’ volts has a wavelength associated

with it. An alpha particle in order to have the same wavelength must be accelerated from rest through a potential difference of:

(A) V volt (B) 4V volt (C) 2V volt (D) V

volt8

Solution (D):

p

h....(1)

2m eV

h......(ii)

2m (2e)V

Equation (i) & (ii) we have p pV

2m eV 2 4 m 2eV V8

51. Following is the variation of physical adsorption with temperature :

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

Solution (B):

Rate of physical adsorption gradually decreases with increase the temperature.

52. The increasing order of acidic nature of 2 2 3Li O, BeO, B O ,CuO is:

(A) 2 2 3Li O BeO CuO B O (B) 2 3 2BeO CuO B O Li O

(C) 2 2 3Li O CuO BeO B O (D) 2 3 2B O CuO BeO Li O

15

Solution (A):

On moving left to right in periodic table, acidic nature of oxide increases. So, metallic oxides are basic,

non metallic oxides are acidic and BeO is amphoteric.

53. Which is the right structure of 4XeF ?

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

Solution (C):

Lone pair-lone pair repulsion is maximum, So to minimize the repulsion the lone pairs occupy the

positions which are trans to each other in the given octahedral geometry. Therefore, © option is correct.

54. Consider the following statements and arrange in the order of true/false as given in the codes.

1S : 2 4N H is pyramidal about each N atom

2 2S : NH OH is pyramidal about the N atom and bent about the O atom

3 3S : CH COCl is trigonal planar about the carbon atom (attached to O and Cl).

(A) FTT (B) FTF (C) TTTT (D) FFT

Solution (C):

55. 2 5 4 3 6 2 3[Fe(H O) NO]SO and [Cr(NH ) ](NO ) both are paramagnetic species with spin only magnetic

moment of 3.93 BM . The hybridization of central metal ions in these species respectively are:

(A) Both 3 2sp d (B) Both 2 3d sp

(C) 3 2 2 3sp d and d sp (D) 2 3 3 2d sp andsp d

Solution (C):

3.93 n(n 2); so n=3 (here n is number of unpaired electron(s))

56. An ore of tin containing 2 4FeCr O is concentrated by:

(A) Magnetic separation (B) Froth floatation

(C) Leaching method (D) Gravity separation

Solution (A):

An ore of tin containing 2 4FeCr O is concentrated by magnetic separation as 2 4FeCr O is ferromagnetic.

16

57. In manufacture of sodium carbonate from Solvary (or ammonia soda) process the raw material used is:

(A) NaOH (B) 2 4Na SO (C) NaCl (D) 3NaHCO

Solution (C):

3 2 2 4 3NH CO H O NH HCO

4 3 3 4NaCl NH HCO NaHCO NH Cl

o150 C3 2 3 2 22NaHCO Na CO CO H O

4 2 3 2 22NH Cl Ca(OH) 2NH H O CaCl (byproduct)

58. Ammonia gas is not prepared by: (A) reduction of sodium nitrite or sodium nitrate by the reaction of zinc dust and sodium hydroxide

(B) Hydrolysis of calcium cyanamide

(C)heating ammonium chloride with slaked lime

(D) heating of 4 2 2 7(NH ) Cr O

Solution (D):

(A) Zn/NaOH

2 3 2NO 7[H] NH 2H O; Zn/NaOH

3 3 2NO 9[H] NH 3H O

(B) 2 2 3 3CaCN 3H O CaCO 2NH

(C) 4 2 3 2 2NH Cl Ca(OH) 2NH CaCl 2H O

(D) 4 2 2 7 2 2 3 2(NH ) Cr O N Cr O 4H O

59. The yellow colour solution of 2 4Na CrO changes to orange red on passing 2CO gas due to the

formation of:

(A) 5CrO (B) 3CrO (C) 2 2 7Na Cr O (D) 2 3Cr O

Solution (C):

2 2 22 2 2 3 3 4 2 7

CO H O H CO 2H CO ; 2CrO 2H Cr O (orange red) + 2H O

60. The ion that cannot be precipitated by both HCl and 2H S is:

(A) 2Pb (B) Cu (C) Ag (D) 2Sn

Solution (D):

(A)(B)(C) can be precipitated by HCl as well as 2H S as their insoluble chlorides and sulphides

respectively.

(D) 22Sn 2HCl SnCl (soluble) 2H . It can be precipitated only by 2H S as its insoluble

sulphide (brown).

17

MATHEMATICS 61. If f(x 1) f(x 1) 2f(x) and f(0) = 0, then f(n), n N, is

(A) nf(1) (B) {f(1)}n (C) 0 (D) none of these Solution (A): f(x 1) f(x 1) 2f(x) , f(0) 0

Putting x = 1, f(2) + f(0) = 2f(1),

f(2) = 2f(1)

Let f(k) = kf(1) for k n

Then f(n + 1) + f(n 1) = 2f(n)

f(n 1) 2(nf(1)) f(n 1) 2nf(1) (n 1)f(1) (n 1)f(1)

f(n) = nf(1) for all n N.

62. If f 2 g 2 and f 2 a, g 2 2a, then

x 2

f x g xlim

x 2

is equal to

(A) 1 (B) a (C) a (D) none of these Solution (C):

x 2 x 2

f x g x f x g xlim lim

x 2 1

= f 2 g 2 a 2a a

63. If the normal to the curve y = f(x) of the point (3, 4) makes an angle 3

4

with the positive axis, then

f(3) =

(A) 1 (B) 3/4 (C) 4/3 (D) 1 Solution (D):

Slope of the normal 1

dy /dx

3tan 1

4

at x = 3

dy

1dx

at x = 3

f (3) 1

64. Let 2

2 2

x dxf(x)

(1 x )(1 1 x )

& f(0) 0 . Then f (1) is

(A) log(1 2) (B) log(1 2)4

(C) log(1 2)4

(D) none of these

Solution (B):

2 2 2

2 2 2 22

x ( 1 x 1) 1 x 1 dx dxf(x) dx dx

(1 x )(1 x 1) 1 x 1 x1 x

2 1log(x 1 x ) tan x k

1f(0) log1 tan 0 k k 0

2 1f(x) log(x 1 x ) tan x

f(1) log(1 2)4

18

65. The value of

/4n n 2

0

(tan (x [x]) tan (x [x])) dx

(where [.] denotes the greatest integer function) is

(A) 1

n 1 (B)

1

n 1 (C)

2

n 1 (D) none of these

Solution (A):

Let

/4n n 2

0

I (tan (x [x]) tan (x [x])) dx

0 x4

[x] = 0 then

/4n 2 2 n 2

0

I (tan x(sec x 1) tan x) dx

/4/4 n 1n 2 2

0 0

tan xtan xsec x dx

(n 1)

1 10

n 1 (n 1)

66. Let f(x) be a continuous function such that the area bounded by the curve y = f(x), the x-axis, and the

lines x = 0 and x = a is 2a

1 sina2

. Then

(A) 2

f 12 8

(B)

2af(a) 1 sina

2

(C) 21f(a) asina a cosa

2 (D) none of these

Solution (C):

Here

a 2

0

af(x)dx 1 sina

2 . Differentiating w.r.t. a,

2af(a) asina cosa

2 .

67. Solution of the differential equation dy

xcos x y(xsinx cos x) 1dx

is

(A) xy = sin x c cos x (B) xy sec x = tan2x + c (C) xy + sin x + c cos x = 0 (D) none of these

Solution (A): The given equation can be written as

dy xsinx cos x 1

ydx xcos x xcos x

which is L.D. equation in y.

I. F. =

xsinx cos xdx

(tanx 1/x)dx logsec x log x log(xsec x)x cos xe e e e xsec x

Multiplying the given equation by I.F. = x sec x, and integrating, we get

xsec x

y xsec x c dxxcos x

= c + sec2 x dx

Multiplying by cos x, this result can also be written as xy = sin x + c cos x which is given in (A). Hence the correct answers are (A).

68. If cos6x 6cos4x 15cos2x 10

ycos5x 5cos3x 10cos x

, then

dy

dx is equal to

(A) 2sinx cosx (B) 2sinx (C) cos2x (D) sin2x

Solution (B):

Numerator = cos6x cos4x 5 cos4x cos2x 10 cos2x 1

= 2cosx cos5x 5cos3x 10cosx

Nr dyy 2cos x, 2sinx

Dr dx

19

69. The roots of the cubic equation (z + )3 = 3, 0 represent the vertices of an equilateral triangle of sides of length

(A) 1

| |3

(B) 3| | (C) 3| | (D)1

| |3

Solution (B):

We have, (z + )3 = 3

z = , z = w , z = w2 .

Thus, the vertices A, B and C of ABC are respectively, , w and w2 .

Clearly, AB = BC = AC = | | |1 w | = 3 ||.

70. The number of permutations of letters of word HINDUSTAN such that neither the pattern ‘HIN’ nor ‘DUS’

nor ‘TAN’ appears are (A) 12246 (B) 169194 (C) 166680 (D) 181434

Solution (B):

Let 1A be the property that HIN appears

Let 2A be the property that DUS appears

Let 3A be the property that TAN appears

1n(A ) 7! 5040

27!

n(A ) 25202!

3n(A ) 7! 5040

1 2 2 3 3 1n(A A ) n(A A ) n(A A ) 5! 120

1 2 3n(A A A ) 3! 6

1 2 3 i i j 1 2 3i j

n(A A A ) n(A ) n(A A ) n(A A A )

= 12246

Total number of ways to arrange the letters of the word HINDUSTAN9!

2! = 181440

Number of ways in which neither the pattern HIN or DUS and TAN appears = 181440 – 12246 = 169194 71. A is a 3 × 3 matrix with entries from the set {–1, 0, 1}. Then the probability that A is neither symmetric

nor skew-symmetric is

(A) 9 6 3

9

3 3 3 1

3

(B)

9 6 3

9

3 3 3 1

3

(C)

9 6 3

9

3 3 3 1

3

(D)

1

2

Solution (A): Total number of matrices = 39. Number of symmetric matrices = 36 Number of skew symmetric matrices = 33

The zero matrix is both symmetric and skew-symmetric the required probability is = 9 6 3

9

3 3 3 1

3

72. The value of

1 1 1log ....

5 4 8 16(0.2)

is

(A) 2 (B) 4 (C) 8 (D) 16 Solution (B):

1 1 1 1 4 1log ...to log log

5 5 54 8 16 1 1 2 20.2 0.2 0.2

2log 2 log 4 log 45 5 50.2 0.2 5 4

20

73. If the angles of elevation of the top of a tower from three collinear points A, B and C on a line leading to

the foot of the tower are o o o30 ,45 & 60 respectively, then the ratio, AB: BC , is

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

Solution (B):

o xtan30

a b c

a b c3

x x x

o xtan45

b c

b c1

x x

o xtan60

c

c 1

x 3

c 1 b 1 a

, 1 , 3 1x x x3 3

a 3 13

1b1

3

74. The coefficient of x5 in the expansion of 2 5 2(1 x ) (1 x) is

(A) 10 (B) 15 (C) 20 (D) 25 Solution (C):

Coefficient of x5 in 2 5 2(1 x ) (1 x) = Coefficient of x5 in 2 4 2(1 5x 10x )(1 2x x ) = 20

75. If , , are the roots of x3 + ax2 + b = 0, then the value of

is

(A) a3 (B) a3 3b (C) a3 (D) a2 3b Solution (C):

Since , , are the roots of the given equations, therefore + + = a, + + = 0 & = b.

Now, 2 2 2( ) ( )

= ( + + ) {( + + )2 3 ( + + )

= (a) {(a2 0)} = a3.

76. (A) Symmetric (B) Skew-symmetric (C) Data insufficient (D) None of these

Solution (A):

(ABBA) = (BA) (AB) = (AB) . (BA) = ABBA So given matrix is symmetric.

77. The set of exhaustive values of ‘’ for which the equation 2| x 1| x x 2 has at least one

positive root, is

(A) (–, –1] [1, ) (B) R (C) (–, –1] (D) [1, ) Solution (C):

2x x 2 x 1 , (Let x 1 0)

2x 3 0 , (No solution)

Now 2x x 2 1 x , (Let x 1 0)

21

y-axis

O 30°

C

M B P

45° A

(h, o) x-axis

xy 3

xy

2x 2 x 1 0

Let 2f(x) x 2 x 1

D 0

24 4 0 1, 1

f(0) 0 and b

02a

, (both roots –ve)

f(0) 0 , for all values of .

( 2 )

02

0

So for 1 both roots of this equation are –ve and 1 atleast one root is +ve and for –ve values of

, x 1 0 for positive values of x.

78. If a > 0, b > 0, c > 0 and the minimum value of 2 2 2 2 2 2a(b c ) b(c a ) c(a b ) is abc

then is (A) 2 (B) 1 (C) 6 (D) 3 Solution (C):

Use A G for the numbers ab2, ac2, bc2, ba2, ca2, cb2.

79. The equation of the circle, which is touched by the line y = x, has its centre on the positive direction of

the x-axis and cut off a chord of length 2 units on the line 3 y x, is

(A) 2 2x y 4x 2 0 (B) 2 2x y 4x 1 0

(C) 2 2x y 8x 8 0 (D) 2 2x y 4y 2 0

Solution (A):

hOP CP AC

2

h

CM hsin302

From ACM

2 2 2AC AM CM

2 2h h

12 4

2h

14

h 2 , radius 2

2 2(x 2) y 2 ; 2 2x y 4x 2 0

80. The locus of the middle points of chords of the parabola 2y 8x drawn through the vertex is a parabola

whose (A) focus is (2, 0) (B) latusrectum = 8 (C) focus is (0, 2) (D) latusrectum = 4 Solution (D):

If the middle point of a chord is (, ), then 22t 0 4t 0

,2 2

Eliminating t, 2

2

. So, the locus is 2y 4x .

22

81. The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is:

(A) 14.0 (B) 16.8 (C) 16.0 (D) 15.8 Solution (A):

Let 16 numbers are 1 2 15x ,x ,....,x ,16

1 2 15x x .... x 16 16 16 1 2 15x x x 16 16 16

1 2 15x x x 3 4 5 16 15 12 required mean 16 15 12

1418

82. An arc of a bridge is semi-elliptical with major axis horizontal. If the length of the base is 9 meter and the

highest part of the bridge is 3 meter from the horizontal; the best approximation of the height of the arch, 2 meter from the centre of the base is

(A) 11

4 m (B)

8

3m (C)

7

2m (D) 2m

Solution (B):

Let the equation of the semi elliptical are be, 2 2

2 2

x y1(y 0)

a b

Length of the major axis = 2a = 9 a = 9/2 Length of the semi minor axis = b = 3.

So the equation of the becomes 2 24x y

181 9

If x = 2, then 2 65y

9

1 8y 65

3 3 approximately.

83. If hyperbola 2 2 2x y a and 2xy c are of same size, then

(A) 2 2c 2a (B) c 2a (C) 2 22c a (D) none of these

Solution (C): Distance between their vertices will remain same (length of transverse axis) 2a is distance between vertices is 1st hyperbola (c, c) and (–c, –c) are vertices of 2nd .

2 22a c c c c 2a 2 2c 2 2a 2c

84. If sec and cosec are the roots of x2 px + q = 0 then

(A) p2 = q (q 2) (B) p2 = q (q + 2) (C) p2 + q2 = 2q (D) none of these Solution (B):

sec + cosec = p, sec . cosec = q

sin + cos = psin . cos, sin. cos = 1

q

sin + cos = p

q

2

2

p 21 2sin .cos 1

qq .

85. If 1 lies between the roots of the quadratic equation 2 23x sin x 2cos 0 . Then

(A) 5

3 6

(B) n 2n

(C) 5

2n 2n6 6

(D) none of these

23

Solution (D):

2 2f 1 0 3 3sin 2cos 0 2sin 3sin 1 0

1 5

sin 1 2n 2n2 6 6

and 2n , n I

2

86. The area of a ABC is a2 (b c)2. Then tanA is equal to

(A) 4

3 (B)

3

4 (C)

8

15 (D) none of these

Solution (C):

Here = a2 (b c)2

= (2s 2c) (2s 2b) = 4(s b) (s c)

1 (s b) (s c) A

tan4 2

22

A 12tan 2

2 4tanAA 11 tan 12 4

87. 1 1cos [cos (2cot ( 2 1)] is equal to

(A) 2 1 (B) 1 2 (C) 4

(D)

4

Solution (D):

Let 1cot ( 2 1) ( 2 1) cot

Therefore 2

1 12

1 tancos [cos2 ] cos

1 tan

2

1 12

cot 1 2 2 2cos cos

4 2 2cot 1

1 1 3cos

42

88. Equation of plane which contains the line x 1 y 2 z 3

1 3 2

and which is perpendicular to the plane

2x 7y 5z 2 is

(A) x y z 6 (B) x y z 2 (C) 2x y z 3 (D) x y z 2

Solution (D): Plane contains line (1), so it passes through point (1, 2, 3)

a x 1 b y 2 c z 3 0

Now a 1 b 3 c 2 0 …(i)

As well plane is perpendicular i.e., 2a 7b 5c 0 …(ii)

From equation (i) and (ii) a : b : c is (1, –1, 1)

or equation of plane is x 1 y 2 z 3 0 x y z 2

89. The value of 2 2 2ˆ ˆ ˆ|a i| |a j| |a k| is

(A) 2a (B) 2 2a (C) 3 2a (D) none of these

24

Solution (B):

Putting 1 2 3ˆ ˆ ˆa a i a j a k , we have 2 3

ˆ ˆ ˆa i a k a j , whence 2 2 2

2 3ˆa i a a .

Similarly , 2 2 2

1 3ˆa j a a and

2 2 21 2

ˆa k a a .

Therefore 2 2 2 2 2 2

1 2 3ˆ ˆa i a k 2(a a a ) 2a .

90. The negation of s r s is equivalent to:

(A) s r (B) s r (C) s r s (D) s r s

Solution (A):

S R S

S R S