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2014 Wiley India Pvt. Ltd.

JEE Knockout Improvement Paper

Physics

Question 1

Two large parallel copper plates are 5.0 cm apart and have a uniform electric field between them as depicted in figure given below.

An electron is released from the negative plate at the same time that a proton is released from the positive plate. Their distance from the positive plate when they pass each other is

(Take mp/me = 2000 and neglect the force of the particles on each other.)

(1) 56.7 10 m (2) 106.7 10 m

(3) 52.5 10 m (4) 53.7 10 m

Solution

We take the positive direction to be to the right in the figure. The acceleration of the proton is

p

p

eEa

m (1)

and the acceleration of the electron is

e

e

eEa

m

(2)

where E is the magnitude of the electric field, mp is the mass of the proton, and me is the mass of the electron. We take the origin to be at the initial position of the proton.

Then, the coordinate of the proton at time t is

2

p

1

2x a t (3)

and the coordinate of the electron is

2 2

e e

1 1( )

2 2x L a t x L a t (4)

where L is the separation between the plates.

They pass each other when their coordinates are the same, therefore from Eq. (3) and Eq. (4), we get

2 2

p e

2

p e

1 1

2 2

2(5)

( )

a t L a t

Lt

a a

Substituting value of t2 from Eq. (5) into Eq. (4) we get


p

p e

p e

e

e p

p

e

5

1

1

10.050m

1 2000

2.5 10 m

p

ax L

a a

eE mL

eE m eE m

mL

m m

Lm

m

Distance from the positive plate when electron and proton pass each other is 52.5 10 m.

Hence, the correct option is (3).

Question 2

If the electric field in a plane electromagnetic wave is given by 6 1

0 sin [(3 10 m ) ]E x t , the value of

is

(1) 10.01 rads (2)

110 rads

(3) 1100 rads (4)

14 19 10 rads

Solution

In the equation 6 1

0 sin [(3 10 m ) ]E x t ,

6 13 10 m is the wave number.

We know that,

kc

Then the value ofangular frequency () is

6 8 143 10 3 10 9 10


Question 3

The current in the circuit shown in the below figure is (1) zero (2) 0.024 A

(3) 0.03 A (4) 0.036 A


Solution In the given circuit, D1 is is forward biased and D2 is reverse biased. So, D1 acts as a short circuit and D2

an open circuit. Therefore, the equivalent circuit is as shown below

Now the total resistance of circuit:

100 150 250

6V (Given)

R

V

We have to find current I. We know that

60.024A

250

V R I

VI

R

I


Question 4 A uniform solid sphere, initially at rest, is pure rolled up an incline by a force F acting at the top of the

sphere parallel to the inclined plane as shown below. When the center of sphere moves a distance l up the

incline, then the work done by the

(1) force F is 2Fl

(2) force F is Fl


(3) force F is Fl sin

(4) gravity force is 2mgl sin

Solution

Since the sphere is purely rolling, the top point has a speed double that of the center so top points

cumulatively travel double the distance the center travels. Hence the work done by F is 2Fl. It gains

height of l sin . So work done by gravity is mgl sin.


Question 5

A particle of mass m is projected with velocity u at an angle with the horizontal. Its angular momentum about the point of projection when it is at the highest point of its trajectory is

(1) 3 2sin cos

2

mu

g

(2)

3 sin 2

2

mu

g

(3) 3 sin cos

2

mu

g

(4)

3 2sin

2 cos

mu

g

Solution

At the highest point, the particle has only horizontal velocity which remains constant, therefore

cosxv u u

Length of the perpendicular to the horizontal velocity from point O is the maximum height and is given

by the relation 2 2

max

sin

2

uH

g

Hence the required angular momentum is

max

3 2sin cos

2

x

L mvr

mu H

muL

g


Question 6


If the radius of the bubble on one side of tube is r and difference in height of liquid of density in

manometer is h, then surface tension of liquid used to make the bubble is

(1) 4

rh gT

(2) T = 2rhg

(3) 2

2

rh gT

(4)

2

rh gT

Solution

Pressure at point B

04

(1)BT

P Pr

Also,

B 0 (2)P P gh

From Eq. (1) and (2)

4

Trgh


Question 7

What is the equivalent capacitance of the capacitors in the figure given below? Given that: 1 6 3FC C

and 3 5 2 42 2 4FC C C C .


(1) 4.5F (2) 1.5F

(3) 2F (4) 3F

Solution

A little inspection of the given figure will show that the same can be redrawn as:

eq 4 3 5 3 5 2 6 1

1 1 1

[ / ( )]

1 1

2 2 2 3 3

1

3

C C C C C C C C C

Thus,

eq 3FC


Question 8

The amplitude of a lightly damped oscillator decreases by 4.0% during each cycle. What percentage of

mechanical energy of the oscillator is lost in each cycle?

(1) 4.0% (2) 6.0%

(3) 8.0% (4) 16%

Solution

The mechanical energy can be written as the maximum kinetic energy. Hence,

max

21

2

E K

mv

2 21

2mA (1)


On differentiating

(1) 2dE mA dA (2)

(2) Dividing Eq. (2) by Eq.(1), we get

2

2 21

2

2

2(4.0%)

8.0%

dE mA dA

EmA

dA

A


Question 9

The circuit shown below is equivalent to

(1) OR gate (2) AND gate (3) NAND gate (4) NOR gate

Solution

The truth table is

A B C X Y

0 0 1 0 1

0 1 0 1 0

1 0 0 1 0

1 1 0 1 0


Question 10

The smallest radius of an flat track around which a bicyclist can travel, if her speed is 36 km h1

and the s between tyres and track is 0.32, will be (Take g = 10 ms

2)

(1) 72 m (2) 24 m

(3) 31.25 m (4) 45 m

Solution


The maximum value of static friction is s FN = smg. If the bicycle does not slip, then f smg. This means

(3) 2 2

s

s

v vg R

R g

Consequently, the minimum radius with which a cyclist is moving at 36 km h1

= 10 m s1

can round the curve without slipping is

2 1 2

min

(10 ms )31.25 m

(0.32)(10 ms )s

vR

g


Question 11

A proton with a speed of 6 12.2 10 ms is shot into a region between two plates that are separated by a

distance of 0.18 m. A magnetic field exists between the plates, and it is perpendicular to the velocity of

the proton, as shown in the following figure.

What must be the magnitude of the field so that the proton just misses colliding with the opposite plate?

(Mass of proton = 271.67 10 kg .)

(1) 0.21 T (2) 0.19 T

(3) 0.13 T (4) 0.27 T

Solution

The proton will follow a circular path inside the magnetic field. To just avoid colliding with the opposite

plate, the radius of the circular path of the proton must be equal to 0.18 m. 2

27 6

19

(1.67 10 ) (2.2 10 )

(1.60 10 )(0.18)

0.13 T

mvqvB

r

mvB

qr

B


Question 12

The figure given below shows a ray of light whose angle of incidence is o

1 60 , travelling through two

solid materials and then undergoing total internal reflection at a solid-liquid interface.


What is the largest possible index of refraction for the liquid?

(1) 1.33 (2) 1.28

(3) 1.50 (4) 1.21

Solution

Applying Snells law at the solid-solid interface,

1 1 2 2sin sinn n

At the solid-liquid interface,

2 2 3sin sin(90 )on n

Thus, the largest possible index of refraction for the liquid will be


3 2 2

1 1

o

sin

sin

1.40 sin 60

1.21

n n

n


Question 13

If the maximum and the minimum voltage of amplitude modulated (AM) wave is 10 V and 2 V

respectively then the modulation index m is (1) 5 (2) 0.5

(3) 0.67 (4) 1.5

Solution Consider the following figure, Em is the amplitude of the audio signal (modulating) wave and EC is the

amplitude of the carrier signal.

From the figure, we have

max minc

max minm

2

2

V VE

V VE

Modulation index is given by

max minm

c max min

V VEm

E V V

max min

max min

10 2 80.67

10 2 12

V Vm

V V

If m > 1, it is an over-modulated wave and possesses detection problems. Hence, m < 1 is recommended.


Question 14 Monochromatic light of wavelength 700 nm is used in a Young's double slit experiment. One of the slits

is covered by a transparent sheet of thickness 51 4 10 m. made of a material of refractive index 1.5. How many fringes will shift due to the introduction of the sheet?

(1) 18 (2) 10

(3) 9 (4) 5


Solution The number of fringes that will shift can be calculated as

Total fringe

Number of frinshift

Fringe wges

idth

11

( )t( )t

Substituting values, we have5

9

1 5 1 1 4 10 m110

700 10 m

( . ) .( )t

Ten fringes will shift due to the introduction of the sheet in one of the interfering beams.


Question 15

A cylindrical tube is partially filled with water such that air column length is 0.1 m. The air column

resonates in its fundamental mode with a tuning fork. The next resonance is observed at air column length of 0.35 m as the level of the water is lowered. The end correction is

(1) 2.5 cm (2) 20 cm

(3) 1.5 cm (4) 10 cm

Solution

Let is the frequency of the tuning fork, v is the speed of sound in air, e the end correction and l the length of air column then

Fundamental frequency

14

v

l e

(1)

First overtone frequency

2

3

4

v

l e

(2)

From Eqs. (1) and (2), we get,

2 1

3

4 4

v v

l e l e

1 23 l e l e

3 0.1 0.35e e 0.3 3 0.35e e 2 0.05me

e = 0.025 m = 2.5 cm


Question 16

If is the breaking stress of a material of density, then the longest wire of that material that can hang

freely without breaking is

(1) g

(2)

2 g


(3)2

g

(4)

2 g

Solution

Highest tension is at the top point is equal to the weight of the wire, therefore

Weight of wire = mg= Al g

Weight Stress ( ) =

Area

Al g

A

l g

lg


Question 17

The figure given below shows a circuit containing a battery, resistor R1, resistor R2 ( 2 200R ), and an

inductor ( 800HL ).

Switch S has been closed for a long time and the current through the inductor is 20.0 mA . At time 0t ,

the switch is opened. At what time t is the current through the inductor10.0 mA ?

(1) 2 (ln 2) s (2) 4 (ln 2) s

(3) 6 (ln 2) s (4) 8 (ln 2) s

Solution

The energy stored in the inductor will dissipate through the resistor 2R when the switch is opened. The

current will decay with time.

0 exp/

ti i

L R


0

6

6

ln

800 10 10ln

200 20

4 10 (ln 2)

4 (ln 2) s

L it

R i

t

s


Question 18

If one complete cycle of a reversible process is carried out on an ideal gas so that its final state is the same

as its initial state, which of the following quantities is the only one which can be non-zero?

(1) The change in the volume of the gas.

(2) The net heat absorbed by the gas.

(3) The change in the entropy of the gas.

(4) The change in the internal energy of the gas.

If one complete cycle of a reversible process is carried out on an ideal gas so that its final state is the same

as its initial state, which of the following quantities is the only one which can be non-zero?

(5) The change in the volume of the gas.

(6) The net heat absorbed by the gas.

(7) The change in the entropy of the gas.

(8) The change in the internal energy of the gas.

Solution

Volume, entropy and internal energy are state variables. So, in a reversible cyclic process these would

not change. However, heat is a path variable. In a reversible cyclic process there would be a finite

quantity of heat exchanged with the surrounding.

Hence, the correct option is (2)

Question 19

The stopping potential for electrons emitted from a surface illuminated by light of wavelength 491 nm is

0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. What is this new wavelength?

(1) 664 nm (2) 529 nm

(3) 382 nm (4) 276 nm

Solution

Photon energy is given by


34 8

9

19

(6.63 10 )(3 10 )

491 10

4.05 10 J

hcE

E

Stopping potential, V0 = 0.710 V Maximum kinetic energy will be

max 0

19

19 19

max

(1.6 10 C)(0.710V)

1.136 10 J 1.14 10 J

K eV

K

Therefore, work function

max

19

19

(4.05 1.14) 10 J

2.91 10 J

E K

For the changed value of incident wavelength,

max

19 19

19

'

2.91 10 J (1.43V)(1.6 10 C)

5.20 10 J

E K

E

Therefore the required wavelength is

34 8

19

''

(6.63 10 )(3 10 )

5.20 10

' 382 nm

hc

E


Question 20

A coil is formed by winding 250 turns of insulated 16-gauge copper wire (diameter = 1.3 mm) in a single layer on a cylindrical form of radius 12 cm. Neglecting the thickness of the insulation, the resistance of

the coil will be (Resistivity of copper is 81.69 10 m )

(1) 2.4 (2) 4.2

(3) 5.6 (4) 6.5

Solution The resistance of the coil is given by

LR

A

where L is the length of the wire, is the resistivity of copper, and A is the cross-sectional area of the wire. Since each turn of wire has length 2r, where r is the radius of the coil, then

L = (250)2r = (250) (2)(0.12 m) = 188.5 m. If rw is the radius of the wire itself, then its cross-sectional area is

2

wA r

m m

The resistivity of copper 81.69 10 m

Thus, the resistance will be


(4) 8

6 2

(1.69 10 m)(188.5m)2.4

1.33 10 m

LR

A


Question 21 A body of mass m is dropped from a height h = 10 m. Simultaneously another body of mass 2 m is thrown

up vertically with a speed 110 5 msv such that it collides with the first body at a height h/2 = 5 m. If

the collision is perfectly inelastic, the velocity of the combined mass as it returns to the ground is (Take g

= 10 m s

)

(1) 10 ms1

(2) 110 2 ms

(3) 15 ms1

(4) 15 5 ms

Solution In a perfectly inelastic collision, the colliding bodies combine after collision and move as a single system.

In an inelastic collision, law of conservation of momentum is also applicable.

Let v1 be velocity of m just before collision, v2 be velocity of 2m just before collision and v0 be combined

velocity just after collision

2 21 1 2

2

hv u g

2 21 0 v gh

2 1

1 10ms 10m 10 msv gh

and

2 22 2

2 22 2

2 22

22

21

22

(10 5

500 100

) 10 10

ms 20

hv u g

v u gh

v

v

v

Applying conservation of momentum to the above system, we get

2 1 02 3mv mv mv


2 1 02 3v v v

2 10

10

2

3

2 20 10

3

10 ms

v vv

v

Note that the system goes further up after collision but on the return journey, the velocity of the combined

system at h /2 is still v0. Applying conservation of energy, we get

2 20 f1 1

3 3 32 2 2

hmv mg m v

2 20 fv gh v (1)

where vf is the velocity of the combined mass as it strikes the ground. Substituting the values in Eq. (1)

we get

(10)2 + 100 = vf

2

1f 10 2 msv

Thus, the velocity of the combined mass as it returns to the ground will be 1f 10 2 ms .v


Question 22

The least count of a stop watch is 1/5 s. Two persons A and B use this watch to measure the time period of

an oscillating pendulum. Person A takes the time period of 30 oscillations and person B takes the time period of 50 oscillations. Neglecting all other source of error, we can say that

(1) Absolute error in measurement by A is greater than that of B (2) Absolute error in measurement by A is equal to that of B

(3) Accuracy in measurement by B is greater than that of A

(4) Accuracy in measurement by B is equal to that of A

Solution

(5)

1

2

1/ 5 1s

30 150

1/ 5 1s

50 250

T

T

2 1T T

So accuracy in measurement by person B is greater than that by person A.


Question 23

A charge of 9q is fixed to one corner of a square, while a charge of 8q is fixed to the opposite corner.

In terms of q , what charge should be fixed to the center of the square, so that the potential is zero at each

of the two empty corners?

(1) / 2q (2) 2q

(3) / 2q (4) 3 / 2q


Solution

Let the edges of square of length r.

The electric potential at an empty corner of the square

0 0 0

0

9 8

4 4 4 ( / 2)

1( 2)

4

q q QV

r r r

q Qr

Since this potential must be zero, we have

2

qQ


Question 24

The power radiated by a black body is P, and it radiates maximum energy around the wavelength 0 . If

the temperature of the black body is now changed so that it radiates maximum energy around a

wavelength 03 / 4 , the power radiated by it will increase by a factor of

(1) 4/3 (2) 16/9 (3) 64/27 (4) 256/81

Solution

From Wiens law, we have

1 1 2 2T T

Substituting values, we get


00

3'

4

4'

3

T T

T T

The power at a given temperatures is 4P Ae T

Substituting and solving for both the temperatures, we get 4 4

4 4

4

3

256

81

P T

P T

P P


Question 25

Two long, thin wires parallel to the z-axis carry currents in the positive z-direction, as shown in the figure

given below.

The wire carrying 50 A current is in the xz plane and is 5 m from the z-axis. The wire carrying 40 A current is in the yz plane and is 4 m from the z-axis. What is the magnitude of the magnetic field at the origin?

(1) 0 T (2) 61 10 T

(3) 62.83 10 T (4) 64 10 T

Solution

Magnetic field at the origin due to the two wires

0 0

0

(50) (40) ( ) ( )2 (5) 2 (4)

5 ( )

B j i

B i j

Thus, magnitude of magnetic field will be

0

7

6

52

5(4 10 )( 2)

2.83 10 T

B

B



Question 26

A shell is fired with a horizontal velocity in the positive x-direction from the top of an 80m high cliff.

The shell strikes the ground 1330m from the base of the cliff. What is the initial speed of the shell? (Take

g =10 ms2

.)

(1) 1332.5ms (2) 1170ms

(3) 182ms (4) 19.8ms

Solution

The initial velocity in the y-direction is zero.

2

0

2

1

2

1( 80) 0 (0) (10)

2

y yy y u t a t

t t

804s

5t

Horizontal displacement is1330m . If the initial speed is 0v , then

0

1

0

1330

1330332.5 ms

4

v t

v


Question 27

The nuclei of a radioactive element, with decay constant , start getting produced at a constant rate . Considering that before the decay began, there was no radioactive element present, the number of nuclei

at time t is given by

(1) (1 )te

(2)

te

(3) te

(4) (1 )te

Solution

The rate of radioactive decay is given by

d

d

NN

t

0 0

dd

N tNt

N

Solving the above relation, we get

(1 )tN e



Question 28

Each of the four corners of a square with edge a are occupied by a point masses having mass m each.

There is a fifth particle of mass m, which is placed at the center of the square. How much work must an

external agent do, to remove the mass from the center to a point far away?

(1) 24Gm

a (2)

2

2

4Gm

a

(3) 24 2 Gm

a (4)

24

2

Gm

a

Solution

Consider two configurations: (1) when the mass is at the center of the square, and (2) when the mass has been removed far away. The work done in moving from configuration (1) to (2) is equal to the difference

in the potential energy of the two configurations.

Potential energy of the mass at the center,

i

2

4/ 2

4 2

Gm mU

a

Gm

a

When the same mass is removed to a point far away, its potential energy becomes zero. Hence, f 0 JU .

Thus, the work done will be equal to

f i

2

2

4 20

4 2

W U U

Gm

a

Gm

a


Question 29

The rain-soaked shoes of a person may explode if ground current from nearby lightning vaporizes the

water. Water has density 1000 kg m3

and requires 2000 kJ kg1

to be vaporized. If horizontal current lasts 2.00 ms and encounters water with resistivity 150 m, length 12.0 cm, and vertical cross-sectional area 15 10

5 m

2, then average current required to vaporize the water is

(1) 130.0 A (2) 12.25 A

(3) 10.0 A (4) 31.0 A

Solution

The mass of the water over the length is

(6) w

3 5 2(1000 kgm )(15 10 m )(0.12 m)

0.018 kg

m d Al

,

and the energy required to vaporize the water is

(7) vap

1

4

(2000 kJkg )(0.018 kg)

3.6 10 J

Q L m

.


The thermal energy is supplied by joule heating of the resistor as

(8) 2Q P t I R t .

Since the resistance over the length of water is

w

5 2

5

150 m 0.120 m

15 10 m

1.2 10

lR

A

R

,

The average current required to vaporize water is

4

5 3

3.6 10 J

(1.2 10 )(2.0 10 s)

12.25 A

QI

R t

I


Question 30

Light of wavelength 625 nm passes through a single slit of width 0.320 mm and forms a diffraction

pattern on a flat screen located 8.00 m away. Determine the approximate distance between the middle of

the central bright fringe and the first dark fringe.

(1) 0.0156 cm (2) 0.516 cm

(3) 1.56 cm (4) 5.16 cm

Solution

For single slit diffraction,

1sinn

d

For the first dark fringe 1n , d is the width of the slit and is the wavelength of light.

1

91

3

1 3

3

sin

625 10sin

0.320 10

sin (1.95 10 )

1.95 10 rad

d

Thus, the distance between the middle of the central bright fringe and the first dark fringe is


3

3

tan [ For small angles tan ,

is the distance between the slit and screen]

8.00(1.95 10 )

15.6 10 m

1.56 cm

y L L

L

y


Chemistry

Question 31 Which of the following is not used as a chemical test to test for the presence of protein?

(1) Biuret test (2) Millons test (3) Ninhydrin test (4) Molisch test

Solution Biuret test is used to test the presence of protein. It is a reaction in which 1% solution of copper sulphate

reacts with the peptide bond and turns the solution violet in color.

Millons test is given by any compound that has phenolic hydroxyl group. So proteins containing amino acids with hydroxyl group will give a pink to dark red color as a positive result. The reagent ninhydrin is used to detect ammonia, primary and secondary amines. On treatment, deep

blue/purple coloration is observed.

The Molisch reagent is used to detect carbohydrates(monosaccharides, disaccharides and polysaccharides). On treatment, purple ring is formed at the interface between the layers.

So the presence of proteins can be detected using Biuret, Millon and Ninhydrin tests, out of which

Millons test is not a specific test for proteins. But proteins cannot be detected at all using Molisch test, which is used to detect carbohydrates


Question 32 In which of the following molecules, the surrounding atoms imposed minimum bond angle at the central

atom?

(1) Cl2O (2) H2Te (3) XeF2 (4) 3I

Solution The shape of Cl2O is based on the tetrahedral geometry, so the bond involved in it is around tetrahedral

angle. In the case of H2Te, the shape is angular with bond angle around 90 because the central atom is not participating in hybridization.

The shape of XeF2 and 3I

is based on trigonal bipyramidal geomtry, and both molecules are linear due to

which atoms impose maximum bond angle at the central atom, that is, 180. Hence, the correct option is (2).

Question 33

The vapor pressure of a dilute aqueous solution is 23.45 torr at 27C, whereas the vapor pressure of pure water at this temperature is 23.76 torr. If values of Kb and Kf for water are 0.51 K/m and 1.86 K/m, then

the incorrect statement is:


(1) The molality of the solution 7.3 101 molal

(2) The boiling point of the solution is 100.37C

(3) The freezing point of the solution is 1.36C

(4) The osmotic pressure of the solution is 17.86 atm at 27C

Solution From Raoults law equation, the mole fraction of the solute is given by

o

A BB Bo

A

23.76 23.450.013

23.76

p px x

p

1Moles of solute 0.013 molMolality( ) 0.73 7.3 10 molal(1 0.013) 18Mass of solvent (in kg)

kg1000

m

The colligative properties are determined as

b b b0.51 0.73 0.37 C 100.37 CT K m T

f f f1.86 0.73 1.36 C 1.36 CT K m T

nRT

V (Data is not sufficient to calculate the volume of the solution)


Question 34 Which of the following cations are paramagnetic?

(1) Yb2+

(2) Lr3+

(3) Th4+

(4) None of these [Atomic numbers are Yb = 70, Lr = 103, Th = 90]

Solution The electronic configuration of above cations is as follows:

Yb2+

: [Xe]4f14

is diamagnetic

Lr3+

: [Rn]5f14

is diamagnetic

Th4+

: [Rn] 6d0 4s

0 is diamagnetic

Hence, none of them is paramagnetic.


Question 35

Which of the following statements is not correct?

I. o-Nitrophenol has intramolecular hydrogen bonding

II. Dipole moment of p-dinitrobenzene has non-zero value III. Among lithium halides, LiI is more covalent than LiBr

IV. SF6 molecule obeys octet rule

(1) I, II and III (2) II and IV (3) I, III and IV (4) I, II and IV

Solution I: The nitro group in o-nitrophenol has an electronegative oxygen atom bonded with double bond, and

electropositive hydrogen atom bonded with oxygen in OH group. These groups are present at adjacent positions that make it suitable for intramolecular hydrogen bonding.


II: Dipole moment of p-dinitrobenzene is zero as the net dipole moment is the resultant of two equal and

opposite vectors.

NO2O2N

++

(III) According to Fajans rules, for the same cation, as the size of anion increases, covalent character of the molecule increases. Therefore, LiI will be more covalent than LiBr.

(IV) SF6 does not obey octet rule as the central atom sulphur in this molecule has 12 electrons in its outermost shell.

Thus, statements (II) and (IV) are incorrect.


Question 36

Consider the redox reaction in acidic medium: 2 33 4Mn BiO MnO Bi

. Choose the correct

statement from the following:

(1) The total electrons involved in the balanced chemical equation is 5.

(2) Two moles of electrons are donated per mole of 3BiO

(3) 2.8 mol of acid are required for the change from 3BiO to Bi

3+

(4) Acid is used as reducing agent in this reaction.

Solution:

In this reaction, 3BiO is reduced by accepting electrons which are donated by Mn

2+ that is getting

oxidized. Balancing the redox equation, we get: 2

2 4[Mn 4H O MnO 8H 5 ] 2e

3

3 2[BiO 6H 2 Bi 3H O] 5e

2 3

3 4 22Mn 5BiO 14H 2MnO 5Bi 7H O

From the balanced equation, it is clear that total of 10 electrons are involved in the reaction, and 2.8 mol

of acid are required per mole of 3BiO .


Question 37

At 25C, 0.056 mol O2 and 0.02 mol N2O were placed in a 1 liter container where the following equilibrium was established:

2 2 22N O(g) 3O (g) 4NO (g).

At equilibrium, the NO2 concentration was 0.02 M. What is the value of KC for this reaction? (1) 23.2 (2) 48.78

(3) 34.33 (4) 0.043

Solution

The reaction involved is


2 2 22N O(g) 3O (g) 4NO (g)

Initial conc. 0.02 0.056 0

Final conc. 0.02 2 0.056 3 4x x x

Now, at equilibrium as [NO2] = 0.02 4x = 0.02 x = 0.005 M.

Therefore, [O2] = 0.056 3 (0.005) = 0.041 M, and [N2O] = 0.02 2(0.005) = 0.01 M 4 4

2

2 3 2 3

2 2

[NO ] (0.02)23.2

[N O] [O ] (0.01) (0.041)CK


Question 38 Consider the one molal aqueous solution of each of the coordination compounds given below?

(I) 2 6 3[Fe(H O) ]Cl (II) 2 5 2 2[Fe(H O) Cl]Cl H O

(III) 2 3 3 2[Fe(H O) Cl ] 3H O (IV) 2 4 2 2[Fe(H O) Cl ]Cl 2H O

The correct statement about them is (1) I will show maximum freezing point

(2) III will show maximum boiling point

(3) III will show minimum freezing point (4) I will show maximum boiling point

Solution The elevation in boiling point and depression in freezing point of a solvent depend on the number of

particles, or moles dissolved in it. Equal molal concentration indicates that equal moles of each compound

are dissolved.

But each compound may dissociate in different way in the solution. .

(I) 32 6 3 2 6[Fe(H O) ]Cl [Fe(H O) ] 3Cl

(II) 22 5 2 2 2 5[Fe(H O) Cl]Cl H O [Fe(H O) Cl] 2Cl

(III) 2 3 3 2[Fe(H O) Cl ] 3H O No dissociation

(IV) 2

2 4 2 2 2 4 2[Fe(H O) Cl ]Cl 2H O [Fe(H O) Cl ] Cl

In this way, the first compound gives maximum ions on dissociation, it causes maximum depression in freezing point and maximum elevation in boiling point. Therefore, solution (I) has maximum boiling

point and minimum freezing point.

In contrast, solution III has minimum boiling point and maximum freezing point since it does not dissociate.


Question 39 Which of the following electronic configuration is not possible?

(1) [Kr] 4d8 5s

1 (2) [Kr] 3d

7 4s

2 (3) [Xe] 4f

14 5d

7 6s

2 (4) [Ar] 3d

5 4s

1

Solution The electronic configuration is done by using Aufbaus law, that is, electron should be filled in increasing order of energy. In [Kr] 3d

7 4s

2, the configuration 3d

7 4s

2 is already included with [Kr], the correct one is

[Kr]4d75s

1 for Ru. All other configurations are written correctly.


Question 40

In the reaction, CH3CH2CHXCH3 CH3CH2CH=CH2


The percentage of 1-ene increases for given X in the order

(1) Br < SMe2

< NMe3

(2) Br < NMe3

< SMe2

(3) SMe2

< NMe3

< Br (4) SMe2

< Br < NMe3

Solution Hoffmann elimination follows anti-Saytzeff rule according to which hydrogen is extracted from the

carbon atom having more number of hydrogen atoms, and this takes place in the presence of bulkier and

sterically hindered base and bulkier leaving group. Among the given four groups, the increasing order of bulkiness of group is

Br < SMe2

< NMe3

.

The order of increasing reactivity towards Hofmann elimination is Br < SMe2

< NMe3

, so the

percentage of 1-ene increases in the same order.


Question 41 The concentration of a drug in the body is often expressed in units of milligrams per kilogram of body weight. The initial dose in an animal was 25.0 mg/kg body weight. After 2.00 h, 10.0 mg/kg body weight

is disappeared through metabolism. If metabolism is of first order then the approximate rate of

metabolism after 2 h is (1) 0.083 mg/min (2) 0.076 mg/min (3) 0.063 mg/min (4) 0.042 mg/min

.

Solution:

For first order reactions: lna

kta x

and Rate [Drug]k

0[Drug] 25 mg and [Drug] 25 10 15 mgta a x

Therefore, 11 1 25

ln ln 0.0042 min120 15

ak

t a x

1 1Rate 0.0042 min 15 mg 0.063 mg min

Hence, the correct option is (3) .

Question 42

In the following table, the first column lists some compounds of xenon with oxygen and fluorine and the second column lists the shapes. Then which of the following options is correct:

Column I Column II

(A) XeOF4 (p) Distorted octahedron

(B) XeF6 (q) Square pyramidal

(C) XeO2F2 (r) Octahedral

(D) [XeO6]4

(s) Trigonal bipyramid

(1) (A) (p); (B) (q); (C) (r); (D) (s) (2) (A) (q); (B) (p); (C) (s); (D) (r)

(3) (A) (q); (B) (s); (C) (p); (D) (r) (3) (A) (s); (B) (r); (C) (q); (D) (p)

Solution

Xenon reacts directly only with fluorine. However, oxygen compounds can be obtained from fluorides.

The structures of the given compounds are as follows:

XeOF4 Square pyramidal (that is octahedral with one

position unoccupied)

XeF6 Distorted octahedron


XeO2F2 Trigonal pyramidal (with one position unoccupied)

[XeO6]4

Octahedral


Question 43 In a compound, the anions (X) form hexagonal close packing and cations (Y) occupy only one-third of

octahedral voids. The general formula for the compound is

(1) YX (2) Y2X3 (3) YX3 (4) Y3 X

Solution

The atoms per unit cell 1 1

12 2 3 1 66 2

. So, the number of anions (X) per unit cell is 6.

Now since it is hexagonal close packing, the number of octahedral voids present is equal to the number of

anions present. As only 1/3 of the voids are occupied by cations,

Number of cations (Y)1

6 23

Therefore, formula is 2 6 3Y X YX


Question 44

Gas A on reaction with propene, followed by Zn/H2O forms acetaldehyde and formaldehyde. When this gas A is passed through KOH at low temperature, a deep red compound B and a gas are obtained. The

compounds A and B are, respectively,

(1) O3, KO2 (2) O2, KO2 (3) O3, KO3 (4) KOH, O2

Solution

When ozone reacts with an alkene, then aldehyde is formed (two molecules of same for symmetrical alkenes and mixture for unsymmetrical alkenes). This confirms that Gas A is O3.

3 3 2 3FormaldehydeAcetaldehyde

O CH CH CH CH CHO HCHO

The reaction of ozone with KOH is analyzed as follows:

3 3 2 2(Anothergas)(A) Red

(B)

KOH O 2KO 5O H O


Question 45 The correct order of increasing atomic radii among Se, O, S and As is

(1) As < S < Se < O (2) Se < O < S < As (3) O < S < Se < As (4) S < Se < As < O

Solution From O to Se group, the trend for atomic size is O < S < Se. Since As comes before Se in the period, so As is bigger in size than Se as in a period the size decreases from left to right. So, the correct trend is

given in option (3).


Question 46


Each of the following mixtures was added to a flask or a separation funnel that contained diethyl ether (as

an organic solvent) and mixed well. In which case, would the organic compound be present in the aqueous phase (soluble in aqueous NaHCO3) ?

(1) p-Cresol + aqueous NaHCO3 (2) p-Chlorophenol + aqueous NaHCO3

(3) p-Nitrophenol + aqueous NaHCO3 (4) Picric acid + aqueous NaHCO3

Solution

In general, phenols are more acidic than alcohols but less acidic than carboxylic acid. Carboxylic acids

can react with weak base NaHCO3, and are soluble in aqueous sodium bicarbonate. So, only those phenols can be dissolved in aqueous NaHCO3 which are sufficiently acidic to react with weak base,

NaHCO3.

Out of the given compounds, only picric acid is exceptionally acidic which can react with weak base NaHCO3. So, it dissolves in aqueous phase.

Hence, the correct option is (4). Quick Tip: Just look for the phenol that is as much as or more acidic than carboxylic acid. Picric acid (2,4,6-trinitrophenol) is very acidic due to three electron withdrawing groups on the benzene ring.

Question 47

On heating of potassium permanganate and potassium dichromate, the number of gases evolved are respectively

(1) 1, 2 (2) 2, 1

(3) 1, 1 (4) 2, 2

The following reactions taking place on heating: Heat

4 2 4 2 22KMnO (aq) K MnO (s) MnO (s) O (g) (9)

Heat2 2 7 2 4 2 3 24K CrO 2C4K Cr O (aq) (s) (r O Os) (g)3


Question 48 Amine is basic in nature but after reaction with some reagents, its basicity is reduced to a great extent. This concept is used to distinguish between primary, secondary and tertiary amines. Which of the

following pairs can be distinguished by Hinsbergs test? (1) Benzylamine and benzamide (2) Allylamine and propylamine

(3) p-Toluidine and N-methylaniline (4) Cyclohexylamine and aniline

Solution

Hinsbergs test is very good method to distinguish between primary, secondary and tertiary amines. In this test, a mixture of amine and benzenesulphonyl chloride (C6H5SO2Cl) is treated with excess KOH and

then acidified after some time. Each type of amine gives different visible results.


Primary amines react with benzenesulphonyl chloride to form N-alkylbenzenesulphonyl amide

which is soluble in alkali. Secondary amines react with Hinsbergs reagent to give a sulphonamide which is insoluble in alkali. Tertiary amines, however, do not react with Hinsbergs reagent. The amines given in the option (3) are primary (p-Toluidine) and secondary (N-methylaniline), so

they can be distinguished by Hinsbergs test. Hence, the correct option is (3).

Question 49 Consider the reaction given below, the maximum possible deuterium atoms found in the product of the reaction is

(1) 3 (2) 5 (3) 7 (4) 8

Solution

Since -hydrogens of carbonyl group are acidic in nature and the reaction medium is basic in the given problem, so all hydrogen atoms alpha to carbonyl group will be substituted by deuterium atom. In addition, in this reaction, the hydrogen atoms at alpha position to the double bond are also substituted

because the double bond is in conjugation with the carbonyl group.

Overall reaction:

So, in the reactant, a total of eight hydrogen atoms can be substituted to form a product with eight deuterium atoms.


Question 50


Select the incorrect statement:

(1) The most probable speed increases with increase in temperature. (2) The fraction of gaseous molecules moving with most probable speed decreases with increase in

temperature.

(3) Fraction of gaseous molecules moving with average speed is less than the fraction of molecules

moving with root mean square speed. (4) In a mixture of gases, heavy molecules are unlikely to be found with very high speeds.

Solution

Since mp2RT

vM

, on increasing temperature vmp will increase.

On increasing temperature, the distribution graph tends to flatten, which indicates that fraction of

molecules with most probable velocity decreases.

From Maxwell distribution, it is clear that average speed fraction is greater than root mean square speed.

Since speed is inversely proportional to the root of molar mass, heavier molecules move relatively with

less speed.


Question 51 Which of the following options is correct when six ligands are approaching towards central metal ion along axis?

(1) Energy of t2g set of d-orbitals increases and that of eg set decreases as compared to that in free metal

ion. (2) Energy t2g set of d-orbitals decreases and that of eg set increases as compared to that in free metal ion.

(3) Energy of both sets of d-orbitals increases equally as compared to that in free metal ion.

(4) Energy of eg set of d-orbitals increases more than t2g set of d-orbitals, but the energy of both sets of d-orbitals increases as compared to that in free metal ion.

Solution

When ligands approach along the axes in octahedral fields, the electrons present in the eg set of d-orbitals

2 2 2( , )x y zd d faces more repulsion as compared to the electrons in t2g set (dxy, dyx, dzx) which point in

between the axes. But repulsion is faced by electrons from both sets of d-orbitals. Hence, the energy will

increase for both sets of d-orbitals.


Question 52 Predict the product when cis-3-methylcyclopentanol is treated with TsCl/pyridine, followed by sodium bromide.

(1) trans-1-Bromo-3-methylcyclopentane (2) cis-1-Bromo-3-methylcyclopentane


(3) 2-Methylcyclopentene (4) 1-Methylcyclopentene

Solution When TsCl/pyridine reacts with cis-3-methylcyclopentanol, it reacts with alcoholic group of the

compound and forms tosylate which on reaction with NaBr gives SN2 to produce trans-1-bromo-3-

methylcyclopentane.


Quick Tip: Tosylate on nucleophilic substitution undergoes SN2 pathway which is associated with

inversion of configuration

Question 53

A buffer solution is formed by adding 0.2 mol each of CH3COOH and CH3COONa in one liter solution. The pH upon adding 1.0 ml of 0.1 M HCl to 10 ml of this buffer [Given: pKa (CH3COOH) = 4.74]

(1) increases by 0.04. (2) decreases by 0.04.

(3) increases by 2.7. (4) decreases by 2.7.

Solution

Given that 33 3[CH COOH] [CH COONa] 0.2 10 10

Initial pH of the buffer solution is 3

a 3

[Salt] 0.2 10 10pH p log 4.74 log 4.74

[Acid] 0.2 10 10K

After addition of 0.1 M HCl to this buffer, few moles of CH3COONa convert to CH3COOH. Therefore,

Moles of salt, -3 -3 -33CH COONa = (210 0.110 ) =1.910

Moles of acid, 3 3 33CH COOH 2.0 10 0.1 10 2.1 10

Therefore, 3

3

3

3

1.9 10 1.9[CH COONa] = M and

11/1000 11

2.1 10 2.1[CH COOH] = M

11/1000 11

a

[Salt] 1.9 /11pH p log 4.74 log 4.7

[Acid] 2.1/11K

So, the pH decreases by 0.04.


Question 54 Which of the following monomers would be most reactive towards cationic polymerization?

(1) 2 3CH CH OCH (2) 2CH CH CN

(3) 2 2CH CH NO (4) 2 3CH CH CH

Solution

Cationic polymerization is the polymerization in which cations are the reacting species. So, carbocations

are the intermediates in the polymerization. The compound in which carbocation produced is very stable,

will react relatively faster than the others. The monomer given in the first option produces most stable carbocation because electron releasing group

OCH3 stabilize it by resonance. Hence, it is most reactive in cationic polymerization. Hence, the correct option is (1).


Question 55 Consider the transition from energy level n = 6 to n = 4 for He

+, the wavelength observed is the same as

the transition in [consider no reduced mass effect]

(1) Li2+

from n = 6 to n = 4 (2) B4+

from n = 6 to n = 4

(3) H from n = 3 to n = 2 (4) Be3+

from n = 12 to n = 8

Solution

Using Rydberg formula 2H 2 2

1 2

1 1 1R Z

n n

, we have:

For transition from energy level n = 6 to n = 4 for He+ is

2

H 2 2

He

1 1 12

4 6R

This can be expressed in terms of following transitions:

2H H2 2 2 2

He

1 1 1 1 12

4 6 2 3R R

same as in H from 3 2n

2 2

H H H2 2 2 2 2 2 2 2

He

2

1 1 1 1 1 1 13 3

2 3 3 2 3 3 6 9

same as in Li from 9 6

R R R

n

2 2

H H H2 2 2 2 2 2 2 2

He

4

1 1 1 1 1 1 14 4

2 3 4 2 4 3 8 12

same as in B from 12 8

R R R

n

2 2

H H H2 2 2 2 2 2 2 2

He

3

1 1 1 1 1 1 15 5

2 3 5 2 5 3 10 15

ame as in Be from 15 10

R R R

s n

Hence, the wavelength is the same in the case of H from n = 3 to 2.


Question 56 A sample of mercury (II) sulphide contains 82.6% mercury by mass. The mass of HgS that can be made

from 30 g of mercury is [At wt of Hg = 201, S = 32]

(1) 34.8 g (2) 36.3 g (3) 4.8 g (4) 24.8 g

Solution

The percentage of Hg in pure HgS

201100 86.2%

233

This is more than the given percentage of Hg, it implies that the sample is not pure. So, we cannot use

given percentage to calculate mass of pure HgS produced by 30 g Hg.


Consider x g of HgS can be obtained from 30 g of Hg. It indicates 30 g Hg is the 86.2% of x g HgS.

86.230 34.8

100x x g


Question 57 Consider the half-reactions of a galvanic cell given below:

2

2 o

2 2 MnOMnO H Mn 2H O; 1.23 VE

2

o

2 PbClPbCl Pb 2Cl ; 0.27 VE

The correct statement about the cell is (1) Standard cell potential is 0.95 V.

(2) During withdrawal of electricity from cell, lead is reduced.

(3) During withdrawal of electricity from cell, manganese is oxidized.

(4) During withdrawal of electricity from cell, two electrons are transferred from lead to manganese.

Solution

The reduction potential of MnO2 is more than that of PbCl2, so, Pb2+

cannot reduce manganese oxide. For

cell reaction to occur, redox reaction is necessary. Hence, if MnO2 is reduced, lead should be oxidized.

Reduction half reaction: 2

2 o

2 2 MnOMnO 4H 2 Mn 2H O; 1.23 Ve E

Oxidation half reaction: o

2 PbPb 2Cl PbCl 2 ; 0.27 Ve E

Cell reaction: + 2

2 2 2MnO Pb 2Cl 4H Mn PbCl 2H O ocell 1.23 0.27 1.5 VE


Question 58 Consider the compound shown in the figure.

Which of the following statements is not correct?

(1) Both protons are more acidic than acetic acid.

(2) All CC bonds in the dianion, 2

4 4C O

are equal.

(3) All CO bonds in the dianion, 2

4 4C O

are equal.

(4) The pKa1 and pKa2 values are equal for both deprotonations.

Solution

After deprotonation, the dianion obtained is stabilized by resonance and four membered ring becomes aromatic due to which it is more acidic than acetic acid. In general, first deprotonation is easier than the


second, and pKa1 < pKa2. Thus, all four resonance structures are equally stable, which implies that all CC and CO bonds are equal. Hence, the correct option is (4).

Question 59

For the reaction A B, H = 4 kcal mol1, S = 10 cal mol1 K1. The reaction is spontaneous when the temperature is (1) 400 K (2) 300 K

(3) 500 K (4) None of these.

Solution

AsG = ve for spontaneous change and G = H TS, we have 1

1 1

4000 cal mol400 K

10 cal mol K

HT S H T T T

S


Question 60 When the alkyl bromides (listed here) were subjected to hydrolysis in a mixture of ethanol and water

(80% EtOH/20% H2O) at 55C, the rates of the reaction showed the order:

I. CH3Br

II. CH3CH2Br III. (CH3)2CHBr

IV. (CH3)3CBr

(1) I > II > III > IV (2) IV > III > II > I (3) IV > I > II > III (4) IV > III > I > II

Solution Two different mechanisms are involved in the hydrolysis of these compounds.

(CH3)3CBr reacts by SN1 mechanism only and apparently this reaction takes place faster. The other three

alkyl halides react by SN2 mechanism, and their reactions are slower because the nucleophile H2O is

weak. The reaction rates are affected by the steric hindrance and thus their order of reactivity is I > II > III

(as order followed by alkyl halides towards SN2 is methyl > 1 > 2). Hence, the correct option is (3).

Mathematics

Question 61 The following question consists of two statements, 1 and 2. Of the four choices given, choose the one that

best describes the two statements.

Statement 1: Contrapositive of If xy is even, then either x or y is even is If neither x nor y is even, then xy is not even Statement 2: Contrapositive of p q is ~ q ~ p (1) Statement 1 is True; Statement 2 is True; Statement 2 is a correct explanation for Statement 1. (2) Statement 1 is True; Statement 2 is True; Statement 2 is not a correct explanation for Statement 1.

(3) Statement 1 is True; Statement 2 is False.

(4) Statement 1 is False; Statement 2 is True.

Solution

Contrapositive of p q q p~ ~

r: xy is even


s: Either x or y is even

Contrapositive of r s is s r~ ~

This implies that if neither x nor y is even, then xy is not even.


Question 62

Let X be a binomial variate with parameters n and p. If the mean is 20 and variance is 15, then p is equal

to

(1) 1

3 (2)

1

4

(3) 1

2 (4)

3

4

Solution

Mean np and variance 2 npq . Therefore,

20np

and 15 = npq

Solving, we get 15 20

3

4

3 11 1

4 4

q

q

p q


Question 63

If the area bounded between two circles 2 2 2( 5) ( 5) 5x y and 2 2 2( 1) ( 1) 5x y is

1 1339 sin5

a b c

sq. unit, find .a b c

(1) 79 (2) 29

(3) 29

4 (4) 54

Solution

The area bounded between circles 2 2 25x y and 2 2 2( 13) 52 yx is same as the area bounded

between the given two circles (see the following figure).


We have

2 2 2 2 25 5 ( 2 )13y x x 2 2( 2 ) 013x x

( 2 )(13 12 ) 03x x x x

2 2 1 03x

13x Dotted area is

5

1313

1

52 2 2 1

1

25 13 25 130 12 sin

2 2 2 2 5

25 25 1339

255 25 sin

2 2

sin4 2

5

5

x xx dx x

Therefore, the required area is

1

1

1

25 25 134 Dotted area 4 39 sin

4 2 5

1325 4 39 50sin

5

1339 sin

5a b c

where 25 4 50 29a b c . Hence, the correct option is (2).

Question 64

Let 1 be a cube root of unity and2 2 2 22(1 )(1 ) 3(2 1)(2 1) 4(3 1)(3 1) ( 1)( 1)( 1)E n n n

Then E is equal to

(1) 2 2( 1)

4

n n (2)

2 2( 1)

4

n nn


(3) 2 2( 1)

4

n nn

(4)

2 2( 1)1

4

n n

Solution The k

th term of the given summation is

2 2 3( 1)( 1)( 1) ( 1)( 1) 1k k k k k k k .

Therefore using the above result, we can rewrite the summation as 2 2

3 3

1 1

( 1)( 1)

4

n n

k k

n nE k k n n


Question 65

Four natural numbers are chosen (with repetition) and multiplied together. The probability that unit digit

of the product is odd other than 5 is

(1) 2

5 (2)

1

210

(3) 16

625 (4)

609

625

Solution Possible unit digit of any natural number is 0, 1, 2, , 9. That is, 10 possibilities. For four natural numbers with repetition, we have 10

4 choices.

For product to have odd unit digit other than 5, we must have four natural numbers have unit digit from 1,

3, 7, 9. Hence, there are 44 favourable cases.

Thus, the probability is 4 4

4 4

4 2 16

62510 5


Question 66

Let A, B and C be finite sets such that A B C and each one of the sets ,A B B C and

C A has 100 elements. Given that ( ) ( )A B A B A B , the number of elements in A B C

is

(1) 250 (2) 200

(3) 150 (4) 300

Solution

Let ( )n X denote the number of elements in X.

Then,

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ).n A B C n A n B n C n A B n B C n C A n A B C

( ) ( )n A n A B (since A B C ) Now,

( ) ( ) ( ) ( ).A B A B B A A B A B

Therefore,

( ) ( ) ( )

( ) ( ) 2 ( )

n A B n A B n A B

n A n B n A B


and

300 ( ) [ ( ) ( ) 2 ( )]

2 ( ) ( )

n A B n A n B n A B

n A n A B

Therefore

300( ) ( ) ( ) 150

2n A B C n A n A B


Question 67

Normal is drawn at point P to the ellipse 2

2 19

xy as shown in the following figure. The equation of

the normal to the ellipse at point P is

(1) 3 3 4 3x y (2) 3 3 4 3x y

(3) 3 3 5 3x y (4) 3 3 7 3x y

Solution The equation of line OP is y = mx.

1

3y x

Equation of the ellipse is 2

2 19

xy . To get the coordinates of point P, we get

2 2

2

2

19 3

41

9

9

4

3

2

x x

x

x

x

Therefore, for point P, we consider

3

2x

and


1 3 3

2 23y

That is,

1 1

3 3( , ) ,

2 2x y

Equation of normal at point P is

1 1

2 21 1/ /

x x y y

x a y b

or 2

21 1

1 1

( )a x x y yb

x y

That is,

9[ (3 / 2)] 1[ ( 3 / 2)]

3 / 2 3 / 2

2 29 1 1

3 3

26 9 1

3

26 8

3

3 43

3 3 4 3

x y

xy

x y

x y

yx

x y


Question 68

If 1 cos

tansin

BA

B

, then tan 3A is equal to

(1) 3

tan2

B

(2) tan 2B

(3) tan B (4) tan 2B

Solution We have

1 costan tan

sin 2

B BA

B

Therefore,

2 2

2tan 2 tan( / 2)tan 2 tan

1 tan 1 tan ( / 2)

tan tan 2 tan( / 2) tan 3tan3 tan

1 tan tan 2 1 tan ( / 2) tan 2

A BA B

A B

A A B B BA

A A B B



Question 69

From a well shuffled pack of 52 cards, five cards are drawn at random. The probability of getting at least

two ace cards and at least two queen cards is

(1) 52

5

1782

C (2)

52

5

270

C

(3) 52

5

1632

C (4)

52

5

1000

C

Solution Five cards are drawn; sample space is

52C5. Probability of drawing at least two queen cards and two ace

cards are as follows: Two queen cards and two ace cards and one other:

4C2

4C2

44C1 = 6 6 44 = 1584.

Two queen cards and three ace cards: 4C2

4C3 = 6 4 = 24.

Three queen cards and two ace cards: 4C3

4C2 = 4 6 = 24

The above mentioned ways show that the favourable cases are:

1584 + 24 + 24 = 1632

Hence, the required probability is

52

5

1632

C


Question 70

If a1, a2, , a2014 are in arithmetic progression such that 1 2 2 3 2013 2014

1 1 13

a a a a a a

and a2 +

a2013 = 72, then find the value of 2014 1 .a a

(1) 20 (2) 30 (3) 40 (4) 50

Solution It is given that a1, a2,, a2014 are in arithmetic progression, such that

1 2

1

a a+

2 3

1

a a + . +

2013 2014

1

.a a = 3

3 2 2014 20132 1

1 2 2 3 2013 2014

1

.

a a a aa a

d a a a a a a

= 3

1 2 2 3 2013 2014

1 1 1 1 1 1 13

d a a a a a a

1 2014

1 1 1 1 20133

( 2013

a d a

d a a d a a d

2 2013 1 201472 2 2013a a a a a d

3 = 2013

( 72 2 )a a a

72a a2 = 671


a2 72a + 671 = 0

a = 11 or 61

d = 50 50

or2013 2013

|a2014 a1| = 2013 |d| = 50


Question 71

The number of five-digit numbers formed using the digits 0, 1, 2, 3, 4, and 8 without repetition and

divisible by 6 is (1) 98 (2) 116

(3) 150 (4) 216

Solution Out of six digits 0, 1, 2, 3, 4, 8, five digit numbers have to be formed without repetition; hence, one digit

has to be left out. For the number to be divisible by 6, number has to be even and divisible by 3, that is,

the sum of digits to be divisible by 3, current sum of all digits is 18. Case 1: 0 left out: The total number of five digit even numbers are

? ? ? ? 2 / 4 / 8

1 2 3 4 3 = 72 numbers

Case 2: 3 left out: The total number of five digit numbers are

? ? ? ? 0

4 3 2 1 1 = 24 numbers

? ? ? ? 2 / 4 / 8

3 3 2 1 3 = 54 numbers

The total number of five digit numbers = 72 + 24 + 54 = 150 numbers.


Question 72

For the system of equations ax + 4y + z = 0, 2y + 3z = 1 and 3x bz = 2, which of the following is not true? (1) Unique solution if ab 15 (2) Infinitely many solutions if a = 3 and b = 5 (3) No solution if ab = 15, a 3 (4) No solution if ab = 15, a 5

Solution We have

4 0

2 3 1

3 2

ax y z

y z

x bz

Here,

4 1

0 2 3

3 0

( 2 ) 3(12 2)

2 30

2( 15)

a

D

b

a b

ab

ab


0 4 1

1 2 3

2 0

4 24 4

4( 5)

xD

b

b

b

0 1

0 1 3

3 2

(6 ) 3

y

a

D

b

a b

4 0

0 2 1

3 0 2

4 12

4( 3)

z

a

D

a

a

Now the system equations will have a unique solution if,

0

15

D

ab

System of equations will have infinitely many solution if,

0

3, 5

x y zD D D

a b

System of equations will have no solution if,

0 and 0, 0, 0

15, 3, 5

x y zD D D D

ab a b


Question 73

The equation of the plane containing the line 1, 03 5

x yz and parallel to the straight line

1 2 3

3 1 1

x y z is 5x + 3y +18z = c. Find the value of c.

(1) 0 (2) 5

(3) 10 (4) 15

Solution

Since the plane contains the line 5 3 15x y and 0z , each and every point on the line also satisfies

equation of the plane.

Let say (p, q, r) is a point on the line 1, 03 5

x yz

So we have,

1, 0 .....(1)3 5

p qr


Also as (p, q, r) is a point on the plane 5 3 18x y z c , we get,

5 3 18 ....(2)p q r c

Comparing both the equations we have 15c .


Question 74 2

2

3( ) .

3

x x cf x

x x c

If the range of f (x) is

1,7

7

, then find c.

(1) 6 (2) 5

(3) 4 (4) 3

Solution We have

2

2

3( )

3

x x cy f x

x x c

2( 1) 3( 1) ( 1) 0y x y x c y

Therefore,

2 23( 1) 9( 1) 4 ( 1)

2( 1)

y y c yx

y

For real x 2 2

2 2

2

9( 1) 4 ( 1) 0

9 9 18 4 4 8 0

(9 4 ) 2(9 4 ) (9 4 ) 0

y c y

y y cy c cy

c y c y c

Given that

17 (7 1)(7 ) 0

7y y y

27 50 7 0y y

Therefore, 9 4 7 4 and 2(9 4 ) 50 4c c c c


Question 75

Evaluate: 2lim ( ).

xx x x

(1) (2)

(3) 1

2 (4)

1

2

Solution


2 22

2

2 2

2

2

2 2 2

( ) ( )lim ( ) lim

( )

lim( )

/lim

[ ( ) / ( / )]

1lim

[ ( / ) ( / ) 1]

1lim

[ (1 (1 / ) 1]

1

2

x x

x

x

x

x

x x x x x xx x x

x x x

x x x

x x x

x x

x x x x x

x x x x

x


Question 76 The area bounded by the curve y = f(x), x-axis and the ordinates x = 2 and x = k is (k + 1) cos (2k + 3).

Find the value of f(x).

(1) cos(2 3) 2( 1)sin(2 3)x x x (2) cos(2 3) 2( 1)sin(2 3)x x x

(3) sin(2 3) 2( 1)sin(2 3)x x x (4) ( 1)(cos(2 3))x x

Solution Area bounded by y = f(x), x-axis and the ordinate x = 2 and x = k is

2

( )

k

f x dx

Therefore,

2

( ) ( 1)cos(2 3)

k

f x dx k k (1)

Differentiating Eq. (1) with respect to k on both sides, we get

( ) 1 cos(2 3) ( 1)sin(2 3)(2)f k k k k

Therefore,

( ) cos(2 3) ( 1)(2)sin(2 3)f x x x x


Question 77 ABCD is a trapezium. If AB is parallel to CD; BC is perpendicular to AB and CD; BC = CD = 1; and

60 ,ADB find AB.

(1) 2 3

3 1 (2) 3 3

(3) 3 3 (4) 5 3 1

Solution


The following figure depicts the trapezium with given measurements.

Therefore,

2 21 1 2BD

And in ABD

2(sine rule)

sin60 sin75 sin75

AB BD

Thus,

sin60 3 / 2 2 32 2

sin75 ( 3 1) / (2 2) 3 1AB

3( 3 1) 3 3


Question 78 If a, b, c, are distinct positive real numbers such that a > b > c and 2ln(a c), ln(a2 c2), ln(a2 + 2b2 + c2) are in arithmetic progression, then

(1) , ,a b c are in arithmetic progression (2) a, b, c are in arithmetic progression

(3) a, b, c are in geometric progression (4) a, b, c are in harmonic progression

Solution We have that a > b > c and 2ln(a c), ln(a2 c2), ln(a2 + 2b2 + c2) are in arithmetic progression. These imply that

2ln(a2 c2) = 2ln(a c) + ln(a2 + 2b2 + c2) (a

2 c2)2 = (a c )2(a2 + 2b2 + c2)

(a + c)2 = a

2 + 2b

2 + c

2

a2 + c

2 + 2ac = a

2 + 2b

2 + c

2

b2 = ac a, b, c are in geometric progression.


Question 79

Two vectors r and s are non-collinear. The volume of parallelopiped whose co-terminal edges are p , q

and s is v1 and the volume of parallelopiped whose edges are p , q , r is v2. If

( ) ( ) log logb ap q r s ar bs (where a, b > 1), then the least value of v1 + v2 is

(1) 2 (2) 3

(3) 4 (4) 12


Solution We have

1 2

b a

( ) ( )

log log

p q r s p q s r p q r s v r v s

ar bs

1 b 2 alog ; logv a v b

Now, a, b > 1 v1, v2 > 0

1 21 2

1 2 1 2

(AM GM)2

2 ( 1)

v vv v

v v v v

Therefore, the least value of v1 + v2 is 2.


Question 80

Consider y x for 0 x 9. P(, ) is any point on y x and Q (2, 0) is a point not on the curve.

Then

(1) Point (0, 0) is closest to Q and point (9, 3) is farthest from Q.

(2) Point (0, 0) is closest to Q and 3 3

point ,2 2

is farthest from Q.

(3) Point 3 3

,2 2

is closest to Q and Point (9, 3) is farthest from Q.

(4) Point (0, 0) is closest to Q and point (3, 3) is farthest from Q.

Solution The graph is depicted in the following figure. We have OQ = 2. Therefore,

2 2 1/2((9 2) (3 0) ) 49 9 58AQ

Given that y x and P(, ) is any point on this curve, therefore


2 2 2

2

(Distance ) ( 2) ( 0)

( 2) ( )

PQ

Let us consider that 2( ) ( 2)f

Therefore,

( ) 2( 2) 1f

and

( ) 2 which i s posit( ive)f

For critical points,

( ) 0f

Therefore,

2( 2) 1 0

3 3

2 2

The point on the curve is 3 3

,2 2

and this point is closest to the point Q and its distance from the point

Q is

223 3 1 3

2 02 2 4 2

72 ( 2)

2OQ

which is equal to OQ. Note: Maximum occurs at end point and minima occurs at a point in (0, 9), that is, 0 < x < 9.


Question 81

Find the equation of the locus of a point P which moves such that its distance from the origin is twice its

distance from the point A(1, 2).

(1) 3x2 + 3y

2 8x 16y + 20 = 0 (2) 3x2 3y2 8x 16y + 20 = 0

(3) x2 + y

2 8x 16y 20 = 0 (4) x2 + y2 8x 16y + 20 = 0

Solution We have

OP = 2PA (OP)2 = 4(PA)2

x2 + y2 = 4[(x 1)2 + (y 2)2]

x2 + y2 = 4x2 + 4y2 8x 16y + 20

3x2 + 3y2 8x 16y + 20 = 0 Hence, the equation of the locus is 3x

2 + 3y

2 8x 16y + 20 = 0.


Question 82

Evaluate: 4 42

xdx

x .


(1) 2

1sin16

xc (2)

211 sin

2 2

xc

(3) 2

11 sin2 16

xc

(4) 2

11 sin2 4

xc

Solution

2 1Numerator2

dx

dx

Substituting x2 = t, we get

2 2

1

2 4

dtI

t

[ 2

4 = (2

2)

2 = 4

2]

11 sin

2 4

tc

2

11 sin2 4

xc


Question 83

If 1 2 3 sin ,2

y x x x x x R

is not differentiable for n values of x, find the value of n.

(1) 1 (2) 2

(3) 3 (4) infinite

Solution We have,

( ) 1 2 3 sin ,2

f x x x x x x R

Then,

(6 3 )sin , 12

(4 )sin , 1 22

( )

sin , 2 32

(3 6)sin , 32

xx x

xx x

f xx

x x

xx x

Therefore,


(6 3 ) cos 3sin , 12 2 2

(4 ) cos sin , 1 22 2 2

( )

cos sin , 2 32 2 2

(3 6) cos 3sin , 32 2 2

xx x x

xx x x

f xx

x x x

xx x x

For this function the left hand and the right hand derivatives of f(x) can be obtained by finding the left

hand and the right hand limits of f (x) at any point. Now at 1x ,

L F (1) 3, R F (1) 1

L F (1) R F (1)

At 2x ,

L F (2) , R F (2)

L F (2) R F (2)

At 3x

L F (3) , R F (3) 3

L F (3) R F (3)

Therefore f (x) is not differentiable at exactly 2 points x = 1 and 3.


Question 84

Find the value of 6[( 2 1) ]. ([] denotes greatest integer function.)

(1) 196 (2) 197

(3) 198 (4) 199

Solution 6( 2 1) I f

6( 2 1) f

6 6( 2 1) ( 2 1)I f f

= 2(T1 + T3 + T5 + T7) 6 6 4 6 2 6 0

2 4 62[( 2) ( 2) ( 2) ( 2) ]C C C

= 2(8 + 15.4 + 15.2 + 1)

= 198

1 197f f I

6[( 2 1) ] 197


Question 85

Straight lines are drawn from the point A(3, 2) to meet the line 6 7 30 0x y at point P. Then, the

locus of the mid-points of the segment AP is

(1) 2 2 30x y (2) 6 7 31x y


(3) 2 2 2(6 3) (7 2) 30x y (4) 6 7 32x y

Solution Let the line that passes through A (3, 2) be

3 2(say)

cos sin

x yr

Every point on this line is of the form (3 cos ,2 sin )P r r and this lies on the line

6 7 30 (6cos 7sin ) 2x y r (1)

Suppose M(h, k) be the mid-point of AP. Therefore,

2 6 cosh r and 2 4 sink r 6(2 6) 7(2 4) 2h k [from Eq. (1)]

12 14 62 0h k

6 7 31 0h k

Hence, the locus of M(h, k) is the line 6 7 31 0x y .


Question 86

Let S be the set of all 3 3 symmetric matrices whose entries are either 0 or 1. Five of these entries are 1 and four of these entries are 0. If a matrix is selected from set S, what is the probability that the selected

matrix will be a non-singular matrix?

(1) 1

4 (2)

1

3

(3) 1

2 (4)

2

3

Solution Symmetric matrix is the one that exists in the form

a d e

d b f

e f c

Case 1: All three diagonal elements are 1.

1 2 3

1 1 0 1 0 0 1 0 1

1 1 0 , 0 1 1 , 0 1 0

0 0 1 0 1 1 1 0 1

A A A

Case 2: Two diagonal entries are 0 and one entry is 1.

4 5 6

0 0 1 0 1 0 0 1 1

0 0 1 , 1 0 1 , 1 0 0 ,

1 1 1 0 1 1 1 0 1

A A A


7 8 9

0 0 1 0 1 0 0 1 1

0 1 1 , 1 1 1 , 1 1 0 ,

1 1 0 0 1 0 1 0 0

A A A

10 11 12

1 0 1 1 1 0 1 1 1

0 0 1 , 1 0 1 , 1 0 0 .

1 1 0 0 1 0 1 0 0

A A A

Hence, there exist a total of 12 symmetric matrices.

From these, the determinants of those matrices which have identical 2 rows or 2 columns are zero. Hence, A1, A2, A3, A4, A8, A12 are singular.

5 6

7 9 10

11

1, 1

1, 1, 1

1

A A

A A A

A

Hence, there are six non-singular matrices. Hence, the probability of selection of non-singular matrix is

6 1

12 2


Question 87

The locus of the foot of the perpendicular drawn from the centre of the hyperbola xy = c2, where c 0,

onto any tangent of the same curve is (x2 + y

2)

2 = kc

2xy. Then, find k.

(1) 1 (2) 2

(3) 4 (4) 8

Solution

Equation of any tangent to a rectangular hyperbola at a point , isct c t ty x t c .

Foot of the perpendicular drawn from the vertex (0, 0) to ty x t c can be found as,

2 21 1

x y c

t t t t

2 2 2 2,

1 1

c t ctx y

t t t t

Replacing these values of x and y in the equation under examination we get,

22 2

2

2 2 2 2 2 2 2 2

2

2 24 4

2 2 2 2 2 22 2 2 2

2 22 2 2 2

1 1 1 1

1 1

1 11 1

1 1

1 1

1

c t ct c t ctkc

t t t t t t t t

t t t tc kc

t t t tt t t t

kt t t t

k


Question 88


Two circles c1 and c2 touch each other externally and both circles also touch the coordinate axes. If c1 is

smaller than c2 and its radius is 2 units, then the radius of c2 is

(1) 6 4 2 (2) 2 2 2

(3) 3 2 2 (4) 6 4 2

Solution The following figure depicts the two circles c1 and c2 touching each other externally and also touching the

coordinate axes.

From the figure, we can write as OQ = OR + RQ

2

2 2

2

2 ( diagonal of a square is 2 side)

2 2 2 2

( 2 1) 2( 2 1)

r OP PR RQ

r r

r

Rationalizing, we get

2

2( 2 1) 2 1

2 1 2 1

6 4 2

r


Question 89

The direction cosines of two lines satisfy the equations 2 2 0l m n and 0mn nl lm . The angle

between two lines is

(1) 30 (2) 60

(3) 90 (4) 120

Solution Given that

2 2 0l m n (1)

0mn nl lm (2) From Eq. (1), we get

2( )n l m

Substituting n in Eq. (2), we get

(2)( ) 2( ) 0m l m l m l lm

2 22 3 2 2 0m ml l lm 2 2

2 2

2 5 2 0

2 4 2 0

(2 ) 2 (2 )

m ml l

m ml ml l

m m l l m l

(2 )( 2 ) 0l m l m

2 or2

lm l m

When 2m l , we get

[using 2( )]1 2 2

l m nn l m

Therefore, direction ratios of one line are 1: 2 : 2 .


When ,2

lm we get

2 1 2

l m n

[using 2( )n l m ]

which implies that the direction ratios of another line are 2: 1: 2.

Taking dot product of the above two direction ratios, we get 2 2 4 0 , which implies that the two

lines are inclined at 90 to each other.


Question 90

Solve: 2 2

x dy y dxx dy y dx

x y

.

(1) logy

xy cex

(2) 2log

xxy ce

y

(3) 1tanx

xy cy

(4) 1tany

xy cx

Solution We know that

1tan2 2

y x dy y dxd

x x y

Also, we know that

( )d xy x dy y dx

Given that

2 2

x dy y dxx dy y dx

x y

1

1tan ( )

tan ( ) (by integrating on both sides)

yd d xy

x

yd d xy c

x

1tany

xy cx


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