Download JEE Main 2014 1. The current voltage relation of diode is given by I

JEE Main 2014
PHYSICS
1. The current voltage relation of diode is given by I = (e1000V/T – 1) mA, where the applied V is in volts
and the temperature T is in degree kelvin. If a student makes an error measuring ± 0.01 V while
measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA?
(1) 0.2 mA
(2) 0.02 mA
(3) 0.5 mA
(4) 0.05 mA
Solution
The given current voltage relation is
I = (e1000V/T – 1) mA.
Now at I = 5 mA, Eq. (1) becomes
5 = (e1000V/T – 1)
or
e1000V/T = 6 mA
Differentiating Eq. (1) w.r.t V we get
(1)
dI
1000
 e1000V /T 
dV
T
1000
dI  e1000V /T 
 dV
T
1000
 6 mA 
 0.01
300
 0.2 mA
Hence, the correct option is (1).
2. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by
the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The
relation between H, u and n is:
(1) 2gH = n2u2
(2) gH = (n – 2)2u2
2
(3) 2gH = nu (n – 2)
(4) gH = (n – 2)u2
Solution
Time taken by the particle to reach highest point is t1 
u
g
1
sy  u yt  ayt 2
2
1
 H  ut  gt 2
2
t
u  u 2  2 gH
g
Using second kinematics relation we get nu u  u 2  2 gH

g
g
nu  u  u 2  2 gH  nu  u  u 2  2 gH
u(n  1)
2
 u 2  2 gH  u 2 (n  1)2  u 2  2 gH
u 2 (n 2  1  2n  1)  2 gH  u 2 (n 2  2n)  2 gH
2 gH  n(n  2)u 2
Hence, the correct option is (3).
3. A mass m is supported by a massless string wound around a uniform hollow cylinder of mass m and
radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?
2g
3
5g
6
(2)
g
2
(4) g
Solution
The various forces acting on the mass m are shown below in free body diagram
Let a be the acceleration with which the mass m will move downwards and  be the angular
acceleration produced in uniform hollow cylinder. Therefore,
a R
(1)
Now applying condition of translational motion for mass m we have
mg  T  ma
(2)
Applying the condition for rotational motion for uniform hollow cylinder we get
TR  mR 2
T  mR  T  mR
a
R
T  ma
(3)
From (2) and (3) we get
mg  ma  ma
mg  2ma  a 
g
2
Hence, the correct option is (2).
4. A block of mass m is placed on a surface with a vertical cross-section given by y = x3/6. If the
coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed
without slipping is
1
2
(1) m
(2) m
6
3
1
1
(3) m
(4) m
3
2
Solution
At limiting equilibrium
  tan 
(1)
Now
dy d  x3 
  
dx dx  6 
x2

2
Substituting Eq. (2) in Eq. (1) we get
tan  
(2)
x2
2
x2
0.5 
2
2
x 1

x  1
Now the maximum height above the ground at which the block can be placed without slipping is
x3 13 1
y   m
6 6 6
Hence, the correct option is (1).
5. When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx2
where a and b are constants. The work done in stretching the unstretched rubber-band by L is:
1
(1) aL2 + bL3
(2)  aL2  bL3 
2
2
3
1  aL2 bL3 
aL bL

(3)
(4) 


2 2
3 
2
3
Solution
The restoring force acting on the rubber band is
F = ax + bx2
Small amount of work done can be expressed as
dW = Fdx
The work done in stretching the unstretched rubber-band by L will be
L
W   (ax  bx 2 )dx
0

ax 2 bx 3

2
3
L
0
aL2 bL3


2
3
Hence, the correct option is (3).
6. A bob of mass m attached to an inextensible string of length l is suspended from a vertical support.
The bob rotates in a horizontal circle with an angular speed  rad/s about the vertical. About the point
of suspension
(1) Angular momentum is conserved.
(2) Angular momentum changes in magnitude but not in direction.
(3) Angular momentum changes in direction but not in magnitude.
(4) Angular momentum changes both in direction and magnitude.
Solution
The torque is given by relation
  mg  l sin  .
Direction of the torque is perpendicular to the axis of rotation.

As is perpendicular to L , direction of L changes but magnitude remains same.
Hence, the correct option is (3).
7. Four particles, each of mass M and equidistant from each other, move along a circle of radius R under
the action of their mutual gravitational attraction. The speed of each particle is
GM
GM
(1)
(2) 2 2
R
R
GM
1 GM
(3)
(4)
1 2 2
1 2 2
R
2
R




Solution
Let the speed of each particle be v. The situation can be express with help of diagram shown below
F
2
Substituting F 
GM
F

 F'
2
2
R 2
GM 2
, we get
4R2
and F ' 
2
GM 2
2
Mv 2
R
GM 2

R 2 R 2
2  GM 2

2 R 2

2

2

GM 2 Mv 2

4R2
R
GM 2 Mv 2

4R2
R
GM 2  1 1  Mv 2


R 2  2 4 
R
2
GM  1 1 

 Mv 2
R  2 4 
v
GM
R
 2  4  1 GM
(1  2 2)

 
 4 2  2 R
Hence, the correct option is (4).
8. The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length
constant when its temperature is raised by 100°C is : (For steel Young's modulus is 2 × 1011 Nm–2 and
coefficient of thermal expansion is 1.1 × 10–5 K–1)
(1) 2.2 × 108 Pa
(2) 2.2 × 109 Pa
7
(3) 2.2 × 10 Pa
(4) 2.2 × 106 Pa
Solution
As length is constant,
L
Strain=
  T
L
Now pressure = stress = Y × strain
= 2 × 1011 × 1.1 × 10–5 × 100
= 2.2 × 108 Pa
Hence, the correct option is (1).
9. There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1 and d2
are filled in the tube. Each liquid subtends 90° angle at center. Radius joining their interface makes an
angle  with vertical. Ratio d1/d2 is
1  sin 
1  sin 
1  tan 
(3)
1  tan 
1  cos 
1  cos 
1  sin 
(4)
1  cos 
(1)
(2)
Solution
Now equating pressure at A we get
 R cos  R sin   d 2 g   R cos  R sin   d1 g
 R cos  R sin   d 2   R cos   R sin   d1
 R cos   R sin  
d
 1 
d 2  R cos   R sin  
cos   sin 
cos   sin 
Dividing numerator and denominator by cos  we get

 cos   sin  

d1 
cos 


d 2  cos   sin  


cos 


1  tan 

1  tan 
Hence, the correct option is (3).
10. On heating water, bubbles being formed at the bottom of the vessel detach and rise. Take the
bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the
vessel. If r << R, and the surface tension of water is T, value of r just before bubbles detach is (Density
of water is  w )
R2
w g
(2) R 2
3T
(3) R 2
w g
(4) R 2
T
w g
6T
3 w g
T
Solution
When the bubble gets detached, then the surface tension will be equal to the buoyant force acting on the
bubble.
Buoyant force is
FB   T  dl sin 
(1)
Force due to surface tension is
4
FS   R 3  w g
3
(2)
Equating (1) and (2) we get
4
3
 T  dl sin   3  R 
w
g
4
T sin   dl   R 3  w g
3
r 4 3
T  2 r    R  w g
R 3
2
T  2 r
4
  R3  w g
R
3
2R4 w g
2w g
 r  R2
3T
3T
Hence, none of the given options is correct.
 r2 
11. Three rods of copper, brass and steel are welded together to form a Y-shaped structure. Area of
cross-section of each rod = 4 cm2. End of copper rod is maintained at 100°C whereas ends of brass and
steel are kept at 0°C. Lengths of the copper, brass and steel rods are 46 cm, 13 cm and 12 cm
respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities
of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through
copper rod is
(1) 1.2 cal/s
(2) 2.4 cal/s
(3) 4.8 cal/s
(4) 6.0 cal/s
Solution
Consider the figure given below
Now we know that rate of flow of heat through copper rod will be equal to sum of the rate of flow of
heat through brass rod and steel rod. Therefore, we have
dQc dQb dQs


dt
dt
dt
0.92  4 100  T  0.26  4  T  0  0.12  4  T  0 


46
13
12
3.68 100  T  1.04T 0.48T


46
13
12
o
 T  40 C
Rate of heat flow through copper rod will be
0.92  4(100  40)
Qc 
46
0.92  4  60

 4.8cal/s
46
Hence, the correct option is (3).
12. One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure.
The process BC is adiabatic. The temperatures at A, B, and C are 400 K, 800 K and 600 K respectively.
Choose the correct statement
(1) The change in internal energy in whole cyclic process is 250R.
(2) The change in internal energy in the process CA is 700R.
(3) The change in internal energy in the process AB is –350R.
(4) The change in internal energy in the process BC is – 500R.
Solution
The change in internal energy is given by relation
U CA  nCV T
 nCV  TA  TC 
(1)
Now temperature at point A is TA = 400 K and temperature at point C is TC= 600 K. Substituting the
values in Eq. (1) we get
5R
U CA  1 
 400  600 
2
5R

 ( 200  500 R
2
Hence, the correct option is (4).
13. An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the
mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by
additional 46 cm. What will be length of the air column above mercury in the tube now? (Atmospheric
pressure = 76 cm of Hg)
(1) 16 cm
(2) 22 cm
(3) 38 cm
(4) 6 cm
Solution
The figure given below shows an open tube immersed in mercury
Now we have
P + x = P0
P = P0 – x
[where P0 = 76 cm of Hg]
P = (76 – x)
Again we can write
8 A 76 = (76 – x) A (54 – x)
8 76 = (76 – x) 54 – x)
608 = 4104 – 76x – 54x + x2
x2 – 130x + 3496 = 0
x = 38
(neglecting the negative value of quadratic equation)
Length of air column will, be
L = 54 cm – x
= 54 cm – 38 cm
= 16 cm
Hence, the correct option is (1).
14. A particle moves with simple harmonic motion in a straight line. In first s, after starting from rest
it travels a distance a, and in next s it travels 2a in same direction then
(1) Amplitude of motion is 3a
(2) Time period of oscillations is 8
(3) Amplitude of motion is 4a
(4) Time period of oscillations is 6
Solution
As it starts from rest, we have x = A cos t.
At t = 0, Eq. (1) becomes x = A cos 0 = A
when t = , Eq. (1) becomes
x = A cos
A – a = A cos
Aa

 cos  
A
when t = 2, Eq. (1) becomes
x = Acos2
A – 3a = Acos2
A  3a
 cos 2 
A
Now we know that
cos 2= 2cos 2– 1
2
A  3a
 Aa
 2
 1
A
 A 
A  3a 2 A2  2a 2  4 Aa  A2

A
A2
A2  3aA  A2  2a 2  4 Aa
2a 2  aA  A  2a
Now
A  a  A cos   2a  a  2a cos 
a
1
cos  
 cos  
2a
2
2

cos   cos
T
3
2


   T  6
T
3
Hence, the correct option is (4).
15. A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of
air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.
(1) 12
(2) 8
(3) 6
(4) 4
Solution
The diagram showing a pipe closed from one end is shown below
Now the length of pipe l  0.85cm.
Speed of sound in air v = 340 m/s.
As we know in fundamental mode

4
l
  4l
  4  0.85
  3.4cm
Now the frequency can be calculated using
v  
v


340

3.4
 100 Hz
Therefore, the possible frequencies are 100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz, and 1100 Hz. The
number of possible natural oscillations of air column in the pipe is 6.
Hence, the correct option is (3).

16. Assume that an electric field E  30 x 2 i exists inspace. Then the potential difference VA – VO,
whereVO is the potential at the origin and VA the potentialat x = 2 m is
(1) 120 J
(2) –120 J
(3) – 80 J
(4) 80 J
Solution
We know that relation between electric field and potential is
dV
E
dx
dV   Edx
Taking definite integral both sides we get
VA
2
2
 dV   30 x dx
VO
0
2

x3 
VA  VO   30  
3 0

  10 x 3 
2
0
 80 J
Hence, the correct option is (3).
17. A parallel plate capacitor is made of two circular plates separated by a distance 5 mm and with a
dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 104
V/m, the charge density of the positive plate will be close to
(1) 6 × 10–7 C/m2
(2) 3 × 10–7 C/m2
4
2
(3) 3 × 10 C/m
(4) 6 × 104 C/m2
Solution
Dielectric constant between the two plates of parallel plate capacitor K = 2.2.
Electric field in the dielectric E = 3 × 104 V/m.
Let the charge density of the positive plate be  =?
Now using the relation for electric field

E
K 0
  K 0 E
 2.2  8.85  10 12  3  10 4
 6  10 7 C/m 2
Hence, the correct option is (1).
18. In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1
kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building
will be:
(1) 8 A
(2) 10 A
(3) 12 A
(4) 14 A
Solution
Power consumed by 15 bulbs of 40 W each = (15 40) W
Power consumed by 5 bulbs of 100 W each = (5 100) W
Power consumed by 5 fans of 80 W each = (5 80) W
Power consumed by heater = 1 kW = 1000 W
Voltage of electric mains V = 220 V.
Now the total power consumed can be expressed as
P total = V I
15 40 + 5 100 + 5 80 + 1000 = V I
600 + 500 + 400 + 1000 = 220 I
2500
I
220
 11.36 A
 12 A
Hence, the correct option is (3).
19. A conductor lies along the z-axis at –1.5 z < 1.5 m and carries a fixed current of 10.0 A in  a z

direction (see figure). For a field B  3.0  104 e  0.2 x a y T, find the power required to move the
conductor at constant speed to x = 2.0 m, y = 0 m in 5 × 10 –3 s. Assume parallel motion along the x-axis
(1) 1.57 W
(2) 2.97 W
(3) 14.85 W
(4) 29.7 W
Solution
Current I = 10.0 A.
Time t = 5 × 10 –3 s.
Length of the conductor l = 3.0 m.

Magnetic field B  3.0  104 e  0.2 x a y T.
The work done can be calculated as
2
W   F . dx
0
2
2
  IlB. dx   10  3  3.0  104 e 0.2 x dx
0
0
2
2
 9  10
3
e
0.2 x
0
 e0.2 x 
dx  9  10 

 0.2  0
3
3
9  10
9  103
 e0.22  1 
1  e  0.4 
0.2
0.2
 9  103  0.33

 2.97  103 J
Now average power can be calculated using
Work
Time
 2.97  103 


0.2
W

 
t
5  103
 2.97 W
P=
Hence, the correct option is (2).
20. The coercivity of a small magnet where the ferromagnet gets demagnetized is 3 103 Am–1. The
current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet
gets demagnetized when inside the solenoid, is
(1) 30 mA
(2) 60 mA
(3) 3 A
(4) 6 A
Solution
Coercivity of small magnet
B
 3  103 Am –1 .
0
Number of turns of solenoid N = 100.
Length of solenoid L = 10 cm = 10  102 m.
Let the required current be I =?
Now we know that magnetic field of solenoid is given by relation
B  0 nI
B
0

[ where n is the number of turns per unit length]
N
I
L
3  103 
NI
100i

L 10  102
I  3A
Hence, the correct option is (3).
21. In the circuit shown here, the point C is kept connected to point A till the current flowing through
the circuit becomes constant. Afterward, suddenly, point C is disconnected from point A and connected
to point B at time t = 0. Ratio of the voltage across resistance and the inductor at t = L/R will be equal to
(1)
e
1 e
(3) –1
(2) 1
(4)
1 e
e
Solution
Let the voltage drop across resistor be VR and voltage drop across inductor be VL.
Applying Kirchhoff's law in closed loop, we get
VR  VL  0
VR  VL
VR
 1
VL
Thus the ratio of the voltage across resistance and the inductor at t = L/R will be 1:1.
Hence, the correct option is (3).
22. During the propagation of electromagnetic waves in a medium
(1) Electric energy density is double of the magnetic energy density
(2) Electric energy density is half of the magnetic energy density
(3) Electric energy density is equal to the magnetic energy density
(4) Both electric and magnetic energy densities are zero
Solution
During the propagation of electromagnetic waves in a medium electric energy density is equal to the
magnetic energy density because energy is equally divided between electric and magnetic field, that is,
B2 1
 0E2
2 0 2
Hence, the correct option is (3).
3

23. A thin convex lens made from crown glass     has focal length f . When it is measured in two
2

different liquids having refractive indices 4/3 and 5/3, it has the focal lengths f1 and f2 respectively. The
correct relation between the focal lengths is
(1) f1 f2 f
(2) f1 f and f2 becomes negative
(3) f2 f and f1 becomes negative
(4) f1 and f2 both become negative
Solution
3
Refractive index of crown glass   .
2
Refractive index of first liquid 1 = 4/3.
Refractive index of second liquid 2 = 5/3.
Applying Lens maker's formula we get
 1
1 
1 
   1  
f1  1  R1 R2 
1  3 / 2  1
1 

 1   
f1  4 / 3   R1 R2 
Similarly from lens maker formula, we get
 1
1 
1 
   1  
f 2  2
 R1 R2 
1  3 / 2  1
1 

 1   
f 2  5 / 3   R1 R2 
1  3  1
1 
   1   
f  2   R1 R2 
… (1)
….. (2)
…. (3)
Now dividing Eq. (1) and Eq. (3) we get
 1   3 / 2  1
1 
9 
 1   
  
  1
 f1    4 / 3   R1 R2   f   8   f  1  f  4 f
1
f1  3 
f1 4
1
1 
 3  1

1


 

  1  
2 
 2   R1 R2 
 f 
Again dividing Eq. (2) and Eq. (3) we get
 1   3 / 2  1
1 
 9

 1   
 1
  

f
5
/
3
R
R
f  10 
f
1
 1
 2
2 




 f 2  5 f
3
f
f
5
1


1 
 3  1
2
2
  1
 
  1   
2 
 2   R1 R2 
 f 
Hence, the correct option is (2).
24. A green light is incident from the water to the air–water interface at the critical angle (). Select the
correct statement
(1) The entire spectrum of visible light will come out of the water at an angle of 90° to the normal.
(2) The spectrum of visible light whose frequency is less than that of green light will come out to the air
medium.
(3) The spectrum of visible light whose frequency is more than that of green light will come out to the
air medium.
(4) The entire spectrum of visible light will come out of the water at various angles to the normal.
Solution
We know that as the frequency of visible light increases refractive index increases. The critical angle
are refractive index are related by relation given below
1
sin  c 

where  c is the critical angle and  is the refractive index. From the above relation we conclude that
refractive index is inversely related to critical angle. Therefore as the refractive index increases, the
critical angle decreases. Now light having frequency greater than green light will suffer total internal
reflection and light having frequency less than green light will suffer refraction and pass to air.
Hence, the correct option is (2).
25. Two beams, A and B, of plane polarized light with mutually perpendicular planes of polarization are
seen through a Polaroid. From the position when the beam A has maximum intensity (and beam B has
zero intensity), a rotation of Polaroid through 30° makes the two beams appear equally bright. If the
initial intensities of the two beams are IA and IB respectively, then IA / IB equals
(1) 3
(2) 3/2
(3) 1
(4) 1/3
Solution
Rotation of Polaroid is   30o.
Now according to law of Malus
I  I 0 cos 2 
For beam A, initial intensity is IA, the by Malus law the final intensity will be
I A '  I A cos 2 30
2
 3
 I A  

 2 
3I
 A
(1)
4
For beam B, initial intensity is IB, the by Malus law the final intensity will be
2
1
I B  I B cos 60  I B   
2
I
 B
(2)
4
Now as I A '  I B' from Eq. (1) and Eq. (2) we get
3I A I B
I
1

 3I A  I B  A 
4
4
IB 3
Hence, the correct option is (4).
2
26. The radiation corresponding to 32 transition of hydrogen atoms falls on a metal surface to
produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10–4 T. If the radius
of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is
close to
(1) 1.8 eV
(3) 0.8 eV
(2) 1.1 eV
(4) 1.6 eV
Solution
Magnetic field B = 3 × 10–4 T.
Radius of the largest circular wire r = 10.0 mm  10.0  103 m.
Charge of electron e  1.6  1019 C.
Mass of electron m  9.1  1031.
Now the stopping potential can be calculates using relation
mv
r
qB
2meV 1 2m

V
eB
B e
Squaring both sides we get

r2 
2mV
B2e
2
2
4
3
19
B 2 r 2 e  3  10  10.0  10  1.6  10 
V 

J
2m
2  9.1  10 31
 0.8 eV
Energy corresponding to 32 transition of hydrogen atoms will be
 1 1
1 1
 13.6  5 
E  13.6  2  2  eV  13.6    eV  
 eV
2 3 
4 9
 36 
 1.88eV
Therefore the work function will be = 1.88eV  0.8eV=1.08eV  1.1eV
Hence, the correct option is (2).
27. Hydrogen (1H1), Deuterium (1H2), singly ionized Helium (2He4)+ and doubly ionized lithium (3Li6)++
all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the
wave lengths of emitted radiation are 1, 2, 3 and 4 respectively then approximately which one of the
following is correct?
(1) 41 = 22 = 23 = 4
(2) 1 = 22 = 23 = 4
(3) 1 = 2 = 43 = 94
(4) 1 = 22 = 33 = 44
Solution
The wavelength can be calculates using relation
1
1 1 
 RZ 2  2  2 

1 2 
1 3RZ 2
 1
 RZ 2 1    

4
 4 
4
(1)

3RZ 2
1
For Hydrogen (1H1), Z =1 therefore the wavelength corresponding to hydrogen can be calculated using
Eq. (1) as
4
4
(2)
1 

2
3R  1 3R
For Deuterium (1H2), Z =1 therefore the wavelength corresponding to hydrogen can be calculated using
Eq. (1) as
4
4
(3)
2 

2
3R  1 3R
For singly ionized Helium (2He4)+, Z =2 therefore the wavelength corresponding to hydrogen can be
calculated using Eq. (1) as
4
4
(4)
3 

2
3R  2 12 R
For doubly ionized lithium (3Li6)++ , Z =3 therefore the wavelength corresponding to hydrogen can be
calculated using Eq. (1) as
4
4
(5)
4 

2
3R  3
27 R
Now from Eq. (2), Eq. (3), Eq. (4), and Eq. (5) we get
 1  2  43  94
Hence, the correct option is (3).
28. The forward biased diode connection is
(1)
(2)
(3)
Solution
(4)
In forward bias, the p-side of diode should be connected to positive terminal of battery or the terminal
at higher potential and n-side should be connected to negative terminal of battery (terminal to lower
potential). Therefore the configuration in which diode is forward biased is
Hence, the correct option is (1).
29. Match List-I (Electromagnetic wave type) with List - II (Its association/application) and select the
correct option from the choices given below the lists:
(a)
List-I
Infrared waves
(i)
List-II
To treat muscular strain
(b)
Radiowaves
(ii)
For broadcasting
(1)
(2)
(3)
(4)
(c)
X-rays
(iii)
To detect fracture of bones
(d)
Ultraviolet rays
(iv)
Absorbed by the ozone
layer of the atmosphere
(a)
(iv)
(i)
(iii)
(i)
(b)
(iii)
(ii)
(ii)
(ii)
(c)
(ii)
(iv)
(i)
(iii)
(d)
(i)
(iii)
(iv)
(iv)
Solution
Infrared rays are used to treat muscular strain. Thus the correct mapping is (a)  (i).
Radio waves are used for broadcasting. Thus the correct mapping is (b)  (ii).
Fractures in bones are detected using X-rays. Thus the correct mapping is (c)  (iii).
Ultraviolet rays are absorbed by ozone layer. Thus the correct mapping is (d)  (iv).
Hence, the correct option is (4).
30. A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to
measure it?
(1) A meter scale
(2) A Vernier caliper where the 10 divisions in Vernier scale matches with 9 division in main scale and
main scale has 10 divisions in 1 cm
(3) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm
(4) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm
Solution
Measured value = 3.50 cm therefore least count of measuring instrument must be 0.01 cm = 0.1 mm.
For Vernier caliper the least count is = 0.01 cm = 0.1 mm.
Thus the student used Vernier caliper to measure the length of the rod.
Hence, the correct option is (2).
CHEMISTRY
31. The correct set of four quantum numbers for the valence electrons of rubidium atom (Z = 37) is
(1) 5, 0, 0, +1/2
(2) 5, 1, 0, +1/2
(3) 5, 1, 1, +1/2
(4) 5, 0, 1, +1/2
Solution
For Rb (Z = 37), the electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s1
As the last electron enters 5s orbital, n = 5, l = 0 and ml = 0, and since only one electron is present in the s-
orbital, ms = +1/2.
Hence, the correct option is (1)
32. If Z is a compressibility factor, van der Waals equation at low pressure can be written as
(1) Z 1 RT/pb
(2) Z = 1  a/VRT
(3) Z = 1  pb/RT
(4) Z = 1 + pb/RT
Solution
We know that compressibility factor is given by Z 
pV
RT
For one mole of gas, we have
a 

 p  2  (V  b)  RT
V


At low pressure, volume is very high, so V >> b and thus V  b  b.
a 
a
pV
a
a

1
 Z 1
 p  2  (V )  RT  pV  RT  
V 
V
RT
VRT
VRT

Hence, the correct option is (2)
33. CsCl crystallizes in body centered cubic lattice. If ‘a’ is its edge length then which of the following
expressions is correct?
(1) rCs  rCl  3a
3a
2
3a
(3) rCs  rCl 
2
(4) rCs  rCl  3a
(2) rCs  rCl 
Solution
For body centered cubic lattice of CsCl shown below,
Two Cl and one Cs+ touch each other at body diagonal of cube. Thus,
3a
2rCs  2rCl  3a  rCs  rCl 
2
Hence, the correct option is (3)
34. For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and
the evolved ammonia was absorbed in 60 mL of M/10 sulphuric acid. The unreacted acid required 20
mL of M/10 sodium hydroxide for complete neutralization. The percentage of nitrogen in the
compound is
(1) 6%
(2) 10%
(3) 3%
(4) 5%
Solution
Total mmol of H2SO4 taken = 60  0.1 = 6 mmol
H 2SO4  2NaOH  Na 2SO4  2H 2 O
This implies that each mole of sulphuric acid is neutralized by two moles of NaOH. So, mmol of H2SO4
1
1
neutralized by NaOH   20   1 mmol
2
10
Therefore, mmol of H2SO4 neutralized by evolved ammonia = 6  1 = 5 mmol
H 2SO 4  2NH 3  (NH 3 ) 2 SO 4
This indicates that each mole of sulphuric acid is neutralized by two moles of ammonia, so, mmol of
ammonia evolved  2  5  10 mmol
mmol of NH3  mmol of N atoms  mass of nitrogen  10  14 mg
Mass of N
140
% of N 
 100 
 100  10%
Mass of organic compound
1000  1.4
Hence, the correct option is (2)
35. Resistance of 0.2 M solution of an electrolyte is 50 . The specific conductance of the solution is
1.4 S m1. The resistance of 0.5 M solution of the same electrolyte is 280 . The molar conductivity of
0.5 M solution of the electrolyte in S m2 mol1 is
(1) 5  104
(2) 5  103
3
(3) 5  10
(4) 5  102
Solution
Given that specific conductance of 0.2 M solution,   1.4 S m 1  1.4  102 S cm 1
1 1 l
l
Now, we know that       1.4  102  50  0.7 cm 1
 R A
A
1 l
1
 0.7  2.5  103 S cm1
For 0.5 M solution,    
R A 280
1000   1000  2.5  103

 5 S cm 2 mol1  5  104 S m 2 mol1
and its molar conductivity is  
M
0.5
Hence, the correct option is (1)
36. For complete combustion of ethanol,
C 2 H 5 OH(l)  3O 2 (g)  2CO 2 (g)  3H 2 O(l)
the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ mol1 at 25°C. Assuming
ideality the enthalpy of combustion, cH, for the reaction will be (R = 8.314 J K1mol1)
(1) 1366.95 kJ mol1
(2) 1361.95 kJ mol1
1
(3) 1460.50 kJ mol
(4) 1350.50 kJ mol1
Solution
For the given reaction, H  U  ng RT  1364.47 
(1 8.314  298)
 1366.95 kJ mol1
1000
Hence, the correct option is (1)
37. The equivalent conductance of NaCl at concentration C and at infinite dilution is C and ,
respectively. The correct relationship between C and is given as (where the constant B is positive)
(1) C    ( B )C
(2) C    ( B )C
(3) C    ( B) C
(4) C    ( B) C
Solution
According to Debye–Huckle–Onsager equation,
C    B C .
Hence, the correct option is (3)
38. Consider separate solutions of 0.500 M C2H5OH(aq), 0.100 M Mg3(PO4)2(aq), 0.250 M KBr(aq) and
0.125 M Na3PO4(aq) at 25°C. Which statement is true about these solutions, assuming all salts to be
strong electrolytes?
(1) They all have the same osmotic pressure.
(2) 0.100 M Mg3(PO4)2(aq) has the highest osmotic pressure.
(3) 0.125 M Na3PO4 (aq) has the highest osmotic pressure.
(4) 0.500 M C2H5OH (aq) has the highest osmotic pressure.
Solution
According to van’t Hoff equation,   iCRT .
For 0.500 M C2H5OH (aq),   1 0.5  RT  0.5RT
For 0.100 M Mg3(PO4)2 (aq),   5  0.1  RT  0.5RT
For 0.250 M KBr(aq),   2  0.25  RT  0.5RT
and for 0.125 M Na3PO4 (aq),   4  0.125  RT  0.5RT
So, all have the same osmotic pressure, and so they are isotonic solutions.
Hence, the correct option is (1)
SO3 (g) if Kp = KC(RT)x where the symbols have usual
meaning then the value of x is (assuming ideality)
(1) 1
(2) 1/2
(3) 1/2
(4) 1
39. For the reaction SO 2 (g)  12 O 2 (g)
Solution
For the reaction, x = ng = 1  (1 + 1/2) = 1/2.
Hence, the correct option is (2)
40. For the non-stoichiometric reaction 2A + B  C + D, the following kinetic data were obtained in
three separate experiments, all at 298 K.
Initial Concentration (A)
Initial Concentration (B)
Initial rate of formation of C
(mol L1 s1)
0.1 M
0.1 M
1.2  103
0.1 M
0.2 M
1.2  103
0.2 M
0.1 M
2.4  103
The rate law for the formation of C is
dC
 k[A][B]
dt
dC
 k[A][B]2
(3)
dt
(1)
dC
 k[A]2 [B]
dt
dC
 k[A]
(4)
dt
(2)
Solution
For the reaction 2A + B  C + D, the rate of formation of C is
d[C]
 k[A]x [B] y
dt
From the given values in the table, we have
1.2  10 3  k[0.1]x [0.1] y
(1)
3
x
y
(2)
3
x
y
(3)
1.2  10  k[0.1] [0.2]
2.4  10  k[0.2] [0.1]
Dividing Eq. (1) by Eq. (3), we get
x
1.2  10 3 k[0.1]x [0.1] y
1 1

     x 1
3
x
y
2.4  10
k[0.2] [0.1]
2 2
Dividing Eq. (1) by Eq. (2), we get
y
1.2  10 3 k[0.1]x [0.1] y
1

1    y  0
3
x
y
1.2  10
k[0.1] [0.2]
2
dC
 k[A] .
Therefore, rate of formation of C is
dt
Hence, the correct option is (4)
41. Among the following oxoacids, the correct decreasing order of acid strength is
(1) HOCl > HClO2 > HClO3 > HClO4
(2) HClO4 > HOCl > HClO2 > HClO3
(3) HClO4 > HClO3 > HClO2 > HOCl
(4) HClO2 > HClO4 > HClO3 > HOCl
Solution
Let us consider the stabilities of the conjugate bases, ClO, ClO 2 , ClO3 and ClO 4 formed from the
acids, HClO, HClO2, HClO3 and HClO4, respectively. These anions are stabilized by the delocalization
of the charge over oxygen atoms. More oxygen atom causes more delocalization of charge on conjugate
base. So, the stability order of respective conjugate base is:
ClO < ClO 2 < ClO3 < ClO 4
Now from conjugate acid base pair definition, more stable anion implies weaker is the conjugate base
and stronger is the respective acid. Thus, the acidic strength decreases in the order
HClO4 > HClO3 > HClO2 > HOCl
Hence, the correct option is (3)
42. The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is
(1) Ag
(2) Ca
(3) Cu
(4) Cr
Solution
The reactions at cathode for the electrolysis of aqueous salt solution of Ag, Ca, Cu and Cr salts are as
follows:
Ag   2e   Ag
H 2 O  e   12 H 2  OH 
Cu 2  2e   Cu
Cr 3+  3e   Cr
As Ca lies in the bottom of electrochemical series, this implies very high negative reduction potential of
calcium ions. Water has less negative reduction potential, so, water is reduced instead of calcium from
its aqueous solution. Thus, Ca cannot be obtained by electrolysis of its aqueous salt solution, H2 is
evolved instead at the cathode.
Hence, the correct option is (2)
43. The octahedral complex of a metal ion M3+ with four monodentate ligands L1, L2, L3 and L4 absorb
wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand
strength of the four ligands is
(1) L4 < L3 < L2 < L1
(2) L1 < L3 < L2 < L4
(3) L3 < L2 < L4 < L1
(4) L1 < L2 < L4 < L3
Solution
Strong field ligands cause higher splitting in the d orbitals. This means that the absorption of energy for
d–d transition is more, and so transmitted energy is more. In case of L1, L2, L3 and L4, as the energy
decreases in the order
Violet > Indigo > Blue > Green > Yellow > Orange > Red
so, the order ligand strength is L1 < L3 < L2 < L4.
Hence, the correct option is (2)
44. Which one of the following properties is not shown by NO?
(1) It is diamagnetic in gaseous state.
(2) It is a neutral oxide.
(3) It combines with oxygen to form nitrogen dioxide.
(4) Its bond order is 2.5.
Solution
The MO configuration of NO is  1s 2  *1s 2  2s 2  * 2s 2 2pz2  2px2  2p 2y  * 2p1x
As can be seen it is an odd electron species and so paramagnetic in gaseous state with one unpaired
electron.
Bond order = (10  5)/2 = 2.5.
Hence, the correct option is (1)
45. In which of the following reactions H2O2 acts as a reducing agent?
(i) H 2 O 2  2H   2e   2H 2 O
(ii) H 2 O 2  2e   O 2  2H 
(iii) H 2 O 2  2e   2OH 
(iv) H 2 O 2  2OH   2e   O 2  2H 2 O
(1) (i), (ii)
(2) (iii), (iv)
(3) (i), (iii)
(4) (ii), (iv)
Solution
The oxidation states of the species are as follows:
1
2
(i) H 2 O 2  2H   2e   2H 2 O
1
0
(ii) H 2 O 2  2e  O 2  2H 
1
2
(iii) H 2 O 2  2e   2O H 
1
2
0
2
(iv) H 2 O 2  2 O H   2e   O 2  2H 2 O
In (ii) and (iv) H2O2 acts as a reducing agent as it loses electrons to reduces the other species and itself
undergoes oxidation.
Hence, the correct option is (4)
46. The correct statement for the molecule, CsI3, is
(1) It is a covalent molecule
(2) It contains Cs+ and I3 ions
(3) It contains Cs3+ and I ions
(4) It contains Cs+, I and lattice I2 molecule.
Solution
CsI3 contains Cs+ and I3 ions because Cs cannot show +3 oxidation states and it is difficult to
accommodate I2 molecules into the lattice because of their large size.
Hence, the correct option is (2)
47. The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1:4. The ratio of
number of their molecule is
(1) 1:4
(2) 7:32
(3) 1:8
(4) 3:16
Solution
Let the mass of oxygen be x and that of nitrogen be 4x.
So, the number of moles of oxygen is x/32 and that of nitrogen is 4x/28 = x/7
x 7 7
 
The required ratio is
32 x 32
Hence, the correct option is (2)
48. Given below are the half-cell reactions
Mn 2   2e   Mn;
E o  1.18 V
2(Mn 3  e   Mn 2+ ); E o  1.51 V
The Eo for 3Mn2+  Mn + 2Mn3+ will be
(1) 2.69 V; the reaction will not occur.
(2) 2.69 V; the reaction will occur.
(3) 0.33 V; the reaction will not occur.
(4) 0.33 V; the reaction will occur.
Solution
Mn2+ + 2e Mn; Eo = 1.18V
[Mn3+ + e Mn2+]  2; E° = +1.51 V
[Mn2+ Mn3+ + e] 2; E° = 1.51 V
Adding Eq. (1) by Eq. (3), we get
(1)
(2)
(3)
3Mn2+  Mn + 2Mn3+
o
for which E  1.18  1.51  2.69 V
Negative electrode potential implies that disproportion is not spontaneous for Mn with +2 oxidation
state to 0 and +3 oxidation states.
Hence, the correct option is (1)
49. Which series of reactions correctly represents chemical reactions related to iron and its compound?
dil. H 2SO 4
H 2SO 4 , O 2
Heat
(1) Fe 
 FeSO 4 
 Fe2 (SO 4 )3 
 Fe
O 2 , heat
dil. H 2SO 4
Heat
(2) Fe 
 FeO 
 FeSO 4 
 Fe
Cl2 , heat
Heat, air
Zn
(3) Fe 
 FeCl3 

 FeCl2 
 Fe
O 2 , heat
CO, 600  C
CO, 700  C
(4) Fe 
 Fe3O 4 
 FeO 
 Fe
Solution
The reaction in option (4) is correct.
In option (1), Fe2(SO4)3 produces oxide(s) of iron, not Fe.
In option (2) FeSO4 cannot be converted to Fe on heating, in this reaction, oxide(s) will be formed.
In option (3), FeCl3 cannot reduced to FeCl2 in the presence of air.
Hence, the correct option is (4)
50. The equation which is balanced and represents the correct product(s) is:
(1) LiO  2KCl  2LiCl  K 2 O
(2) [CoCl(NH 3 )5 ]  5H   Co 2   5NH 4  Cl
excess NaOH
(3) [Mg(H 2 O)6 ]2   (EDTA) 4  
[Mg(EDTA)]2  6H 2 O
(4) CuSO 4  4KCN  K 2 [Cu(CN) 4 ]  K 2SO 4
Solution
The equation in option (2) is correct since both charges as well as atoms are balanced.
For the rest,
(1) LiO  2KCl  2LiCl  K 2 O ; Li atoms are not balanced.
excess NaOH
(3) [Mg(H 2 O)6 ]2   (EDTA) 4 
[Mg(EDTA)]2  6H 2 O ; charge is not balanced.
(4) CuSO 4  4KCN  K 2 [Cu(CN) 4 ]  K 2SO 4 ; products are not correctly written.
The correct equations for other reactions are:
excess NaOH
(1) [Mg(H 2 O)6 ]2   (EDTA) 4  
[Mg(EDTA)]2  6H 2 O
(3) LiO  2KCl  2LiCl  K 2 O is unfavorable in forward reaction as LiO is stable, while K2O is unstable.
(4) 2CuSO 4  10KCN  2K 3 [Cu(CN) 4 ]  2K 2SO 4  (CN) 2
Hence, the correct option is (2)
51. In SN2 reactions, the correct order of reactivity for the following compounds
CH3Cl, CH3CH2Cl, (CH3)2CHCl and (CH3)3CCl is
(1) CH3Cl > (CH3)2CHCl > CH3CH2Cl > (CH3)3CCl
(2) CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl
(3) CH3CH2Cl > CH3Cl > (CH3)2CHCl > (CH3)3CCl
(4) (CH3)2CHCl > CH3CH2Cl > CH3Cl > (CH3)3CCl
Solution
Rate of SN2 reaction depends on stearic hindrance on carbon atom which bears leaving group.
Increasing stearic hindrance decreases possibility of back side attack by nucleophile. This decreases the
rate of SN2 reaction. Order of steric hindrance is (CH3)3CCl > (CH3)2CHCl > CH3CH2Cl > CH3Cl , so
the order of reactivity towards SN2 reaction is CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl.
Hence, the correct option is (2)
52. On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the
organic compound formed is
(1) an alkanol
(2) an alkanediol
(3) an alkyl cyanide
(4) an alkyl isocyanide
Solution
The reaction involved is
RNH 2  CHCl3  3KOH(alc.)  RNC   3KCl  3H 2 O
Alkyl isocyanide
Hence, the correct option is (4)
53. The most suitable reagent for the conversion of R–CH2–OH  R–CHO is:
(1) KMnO4
(3) CrO3
(2) K2Cr2O7
(4) PCC (Pyridinium Chlorochromate)
Solution
All others except PCC are stronger oxidizing agents and will convert R–CH2–OH  R–COOH. PCC is
a mild oxidizing agent and will convert R–CH2–OH  R–CHO.
Hence, the correct option is (4)
54. The major organic compound formed by the reaction of 1,1,1-trichloroethane with silver powder is
(1) Acetylene
(2) Ethene
(3) 2-Butyne
(4) 2-Butene
Solution
The reaction involved is
H 3C  C  Cl3  6Ag  Cl3  C  CH3  H3C  C  C  CH3
2-Butyne
Hence, the correct option is (3)
55. Sodium phenoxide when heated with CO2 under pressure at 125°C yields a product which on
acetylation produces C.
The major product C is
Solution
The reaction is
Hence, the correct option is (1)
56. Considering the basic strength of amines in aqueous solution, which one has the smallest pKb value?
(1) (CH3)2NH
(2) CH3NH2
(3) (CH3)3N
(4) C6H5NH2
Solution
C6H5NH2 is least basic as the lone pair is involved in resonance with benzene ring. In the case of three
aliphatic amines in aqueous solutions, the basicity depends on two factors: +I effect of alkyl group and
solvation of conjugate acid (aminium ion) of the respective base in the aqueous solution. +I effect as
well as basicity due to the group decreases as (CH3)3N > (CH3)2NH > CH3NH2 > NH3. However,
solvation of conjugate acids occurs by hydrogen bonding which depends on the number of hydrogen
atoms attached with nitrogen atom. The order of solvation of conjugate base is (CH3)3N < (CH3)2NH <
CH3NH2 < NH3. In this way, both factors are against each other and the order of resultant basic strength
becomes
(CH3)2NH > CH3NH2 > (CH3)3N > NH3
Hence the amine that has minimum pKb (maximum basic) is (CH3)2NH.
Hence, the correct option is (1)
57. For which of the following molecule significant   0?
(1) Only (a)
(2) (a) and (b)
(3) Only (c)
(4) (c) and (d)
Solution
In (a) and (b),  = 0 as the dipole moments cancel each other. However, in (c) and (d) more than one
conformation is possible, so, the dipole moment is not zero in those cases. For example, possible
conformations in (c) are shown below.
Similarly for structure (d), two conformations are possible.
Hence, the correct option is (4)
58. Which one is classified as a condensation polymer?
(1) Dacron
(2) Neoprene
(3) Teflon
(4) Acrylonitrile
Solution
Condensation polymer: Dacron is formed by condensation of ethylene glycol and terephthalic acid at
420–460 K using zinc acetate and antimony oxide as catalyst.
Addition polymers: Teflon formed by addition of tetrafluoroethylene. Acrylonitrile is a monomer to
form addition polymer “orlon”. Neoprene is a rubber formed by addition polymerization of
chloroprene, 2-chloro-1,3-butadiene.
Hence, the correct option is (1)
59. Which one of the following bases is not present in DNA?
(1) Quinoline
(2) Adenine
(3) Cytosine
(4) Thymine
Solution
The bases in DNA are adenine, cytosine, thymine and guanine. Quinoline is not present in DNA.
Hence, the correct option is (1)
LiAlH
PCl
alc. KOH
5
4
60. In the reaction, CH 3COOH 
 A 
 B 
 C , the product C is
(1) Acetaldehyde
(2) Acetylene
(3) Ethylene
(4)Acetyl chloride
Solution
The reaction is
Hence, the correct option is (3)
MATHEMATICS
61. If X = {4n– 3n – 1: n } and Y = {9(n – 1) : n }, where N is the set of natural numbers, then
X Y is equal to
(1) X
(3) N
(2) Y
(4) Y – X
Solution
X = {4n  3n  1: n  N}
4n  3n  1 = (1 + 3)n  3n  1
 nC0 30  nC1 31  nC2 32   nCn 3n  3n  1
 9(n C2  n Cn 3n 2 )  9k , where k is an integer, when n  2
When n = 1, 4n  3 n  1 = 4  3  1 = 0 = 9 k
Therefore, X contains multiples of 9.
Now Y = {9 (n  1): n  N} = {0, 9, 18,} = all multiples of 9
Thus,
X Y  Y
Hence, the correct option is (2)
62. If z is a complex number such that |z| 2, then the minimum value of z  1
2
5
(1) is strictly greater
2
3
5
(2) is strictly greater than but less than
2
2
5
(3) is equal to
2
(4) lies in the interval (1, 2)
Solution
Min z 
1
is when z is at B.
2
Therefore, Min z 
1 3
1
1
lies in the interval (1, 2).
 BC  1   . Hence, Min z 
2 2
2
2
Hence, the correct option is (4)
63. If a R and the equation, –3(x – [x])2 + 2(x – [x]) + a2 = 0 (where [x] denotes the greatest integer
x) has no integral solution, then all possible values of a lie in the interval:
(1) (–2, –1)
(2) (–, – 2) (2, )
(3) (–1, 0) (0, 1)
(4) (1, 2)
Solution
–3(x – [x])2 + 2(x – [x]) + a2 = 0  3{x}2 + 2{x} + a2 = 0
2
2
2

2
1 1
1 1 

 a 2  3{x}2  2{x}  3 {x}2  {x}         3 {x}   
3
3 3

 3   3  

Now we know that 0  {x} < 1.
2
2
1
4
1
4
1
1 2


Therefore,   {x}    0  {x}     0  3 {x}   
3
3 3
3
9
3
3


2
1
1 1

   3 {x}     1
3
3 3

 1  a2 < 1
Only possibility for non-integral solution is 0 < a2 < 1
Thus, a2  0, but when a = 0, there is integral solution for {x} = 0
Therefore, ( 1, 0)  (0, 1)
Hence, the correct option is (3)
64. Let and be the roots of the equation px2 + qx + r = 0, p 0. If p, q, r are in A.P. and
1


1

 4,
then the value of    is

34

9



2 13
9
(3)
61

9



2 17
9 
Solution
Given that,
1


2q = p + r
1
(1)
4

(2)
Since and are roots of equation px2 + qx + r = 0, we have:     
 
r
p
q
p
(3)
(4)
From eq (1), we have
 
4

From eq (2) and (3), we have 2(   ) 
Therefore, 2
(   )


1

 1  2  4 
Therefore, equation having roots
1

and
1

2q p  r
r

 1   1  
p
p
p
1

 1
1

 9
is x2  4x  9 = 0
1 4
  9  0  9x2 + 4x  1 = 0
x2 x
16
16  36
52 2 13
2
 1 
Therefore,         4 
 4  


81
81
9
9
 9 
Hence, the correct option is (2)
Therefore, equation having roots  and  is
3
1  f (1) 1  f (2)
65. If , β  0, and f(n) =  +  and 1  f (1) 1  f (2) 1  f (3)  K (1   )2 (1   )2 (   )2 , then K is
n
n
1  f (2) 1  f (3) 1  f (4)
equal to
(1) 1
(2) –1
1
(3) β
Solution
(4)  
111
1    1  2   2 1 1
1    1  2   2 1  3   3  1 
1  2   2 1  3   3 1  4   4 1  2
1 1 1 1
 1   2 = {(1 ) (1 ) (  )2)}
2 1  2
On comparison with the given equation, we get K = 1.
Hence, the correct option is (1)
66. If A is a 3 × 3 non-singular matrix such that AAAA and B A–1A, then BBequals
(1) B–1
(2) (B–1)
(3) I + B
(4) I
Solution
BB = A1A(A1A) = A1AA(A1) = A1AA(A1) = IA(A1) = I(A1A) = I.I = I2 = I
Hence, the correct option is (4)
67. If the coefficients of x3 and x4 in the expansion of (1 + ax + bx2)(1 – 2x)18 in powers of x are both
zero, then (a, b) is equal to
  14, 272   

  16, 272 
3
3 



(3)  16, 251   
3 


  14, 251 
3 

Solution
(1 + x + bx2)(1  2 x)18
 (1  ax  bx 2 ){18 C0 (2 x)0 18 C1 (2 x)1 18 C2 (2 x)2 18 C3 (2 x)3 18 C4 (2 x)4 }
Coefficient of x3  18C3 (2)3  a  18C2 (2)2  b  18C1 (2)
3
2
8  18  17  16
18  17

 a  4
 2b  18
3  2 1
2 1
=  4 {2  3  17  16  9  17 a + a b} = 0
 153a  ab = 1632  51a  3b = 544
 272 
Only  16,
 satisfies it
3 

Hence, the correct option is (2)
68. If (10)9 + 2(11)1 (10)8 + 3(11)2 (10)7+ … + 10(11)9 = k (10)9, then k is equal to
(1) 100
(2) 110
121
441
(3) 10 
(4) 100

Solution
Let S = 1  109 + 2  111  108 + 3  112  107 +  9  10  118 + 10  119
(1)
11
S = 11  108 + 2  112  107 + 3  113  106 +  9  119
10
On subtracting (2) from (1), we get
1
S = 11  108  112  107   119 + 1110  109
10
=  109  11  108  112  107   119 + 1110
Therefore,
11 / 10  1  11
 10
10
9
10
(11 / 10)  1


10
  11
109 11 / 10   1
1 / 10
(2)
10
10
 11

1
S  1010  10  1  1110   1110  1010  1110
10
10

11
9
S = 10 = 100  10
Therefore, k = 100
Hence, the correct option is (1)

69. Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new
numbers are in A.P. Then the common ratio of the G.P. is
(1) 2  3 

 2  3 
(3)
2 3

 3  2 
Solution
Let the numbers are a, a r, a r2.
Now 2[2ar] = a + a r2 4 a r = a + a r2  r2  4 r + 1 = 0
r
4  16  4
2 3
2
Since the numbers form an increasing G.P. so r  1 .
Therefore, r  2  3
Hence, the correct option is (2)
sin( cos 2 x)
is equal to
x 0
x2
(1) –π
(2) π
70. lim

(3) 2  


Solution
2
sin( cos 2 x) sin  (1  sin x)  sin(   sin 2 x)


x 0
x2
x2
x2
lim
x 0

sin( sin 2 x) sin( sin 2 x)  sin 2 x


x2
 sin 2 x
x2
x0
x 0
=1=
Hence, the correct option is (2)
71. If g is the inverse of a function f and f '( x) 
(1)
1

1  {g ( x)}5
x

1
, then gʹ(x) is equal to
1  x5

 1  {g ( x)}5 

x
Solution
Given g (x) = f1 (x)
Therefore, f (g(x)) = x
f(g(x)) g(x) = 1
Thus, g '( x) 
Thus,
1

f '  g ( x) 
1
1
1  {g ( x )}5
g (x) = 1+ {g(x)}5
Hence, the correct option is (2)
72. If f and g are differentiable functions in [0, 1] satisfying f(0) = 2 = g(1), g(0) = 0 and f(1) = 6, then
for some c ]0, 1[
(1) f (c) = g(c)
(2) f (c) = 2g(c)
(3) 2f (c) = g(c)
(4) 2f (c) = 3g(c)
Solution
Let h (x) = f (x)  2g(x)
So that h (0) = f (x)  2 g(x) = 2  2  0 = 2
And h (1) = f (1)  2 g(1) = 6  2  2 = 2
Now h(x) is a differentiable function in [0, 1] and h(0) = h(1), so by Rolle’s theorem h(c) = 0 for some
c  (0, 1)
Therefore,
0 = f (c)  2 g (c)  f (c) = 2 g (c)
Hence, the correct option is (2)
73. If x = –1 and x = 2 are extreme points of f (x) log |x|  x2 x, then
1
1
(1)   2,    

   2,   
2
2
(3)   6,  
1

2
   6,   

1
2
Solution
f (x) =  log(x) +  x2 + x

 2 x  1
x
Now f( 1) =    2 + 1 = 0
f '( x) 

 4 x  1  0
2
Thus, the equations are
  2   1  0 and   8  2  0
On solving the above equations, we get
1
   ,  2
2
Hence, the correct option is (1)
f '(2) 
1
1  x

74. The integral   1  x   e x dx is equal to
x

(1) ( x  1)e
 ( x  1)e
x
x
1
x
1
x
c 
c

  xe

 xe
x
x
1
x
1
x
c
c

Solution
1
1
1
x
 x  1x
1  x x
1  x x 


x
1

x

e
dx

e
dx

x
1

e
dx

xe
c



2 
 
 
x
 x 

  xf '( x )  f ( x )dx  xf  x   c 
 

Hence, the correct option is (4)

75. The integral

1  4sin 2
0
x
x
 4sin dx equals
2
2
(1) 4 3  4 


 4 3  4 
   4 




3

2
44 3 
3
Solution


0

1  4sin 2
x
x
x

 4sin dx    1  2sin 
2
2
2

0 
2


2 
x
x
x

   1  2sin  dx   3  1  2sin  dx    1  2sin  dx
0
0
2
2
2



3


x 3 
x

2 cos 
2 cos 


2
2
 x 
  x 

(1
/
2)
(1 / 2) 




0 

3





   2 3   (4)    4(0)     2 3  
3

3





3
 2 3  4  

3
2 3
2

  4 3  4  4 3   4
3
3
Hence, the correct option is (2)

76. The area of the region described by A = {(x, y): x2 + y2 1 and y2 1 – x} is
 2
 2
  

 
2 3
2 3


2

4

3



2

4
3
Solution

y2 = 1  x  x = 1  y2
1
1


y3  
 1    4
1
Required Area is  (  12 )  2  (1  y 2 )dy  2  y     2  1    0   
0
2
3 0 2
2
 3   2 3

Hence, the correct option is (3)
77. Let the population of rabbits surviving at a time t be governed by the differential equation
dp(t ) 1
 p(t )  200 If p(0) = 100, then p(t) equals
dt
2
(1) 600 – 500 et/2
(2) 400 – 300 e–t/2
t/2
(3) 400 – 300 e
(4) 300 – 200 e–t/2
Solution
dp (t ) 1
dp (t )
 p(t )  200 
 dt
1
dt
2
p (t )  200
2
On integrating we get
p(t )
log
 200
log p(t )  400 1
c
2
 t  c1 
(1)
 t 1
(1/ 2)
2
2
2
Using initial Conditions
log 100  400 c1
(2)
  2log150  c1
2
2
From eqn. (1) and eqn(2) we get
p (t )  400 1
p (t )  400 1
log
 t  log150  log
 t
2
2
300
2
p (t )  400
 et / 2 or p (t )  400  300et / 2
300
Since initially number of rabbits is 100 and is decreasing
( p(t )  400)  300t / 2 or p(t )  400  300et /2
Hence, the correct option is (3)
78. Let PS be the median of the triangle with vertices P(2, 2), Q(6, –1) and R (7, 3). The equation of the
line passing through (1, –1) and parallel to PS is
(1) 4x + 7y + 3 = 0
(2) 2x – 9y – 11 = 0
(3) 4x – 7y – 11 = 0
(4) 2x + 9y + 7 = 0
Solution
Slope of required line is
1
2

( 9 / 2)
9
2
Equation is y  1   ( x  1) 9 y + 9 =  2 x + 2 or 2 x + 9 y + 7 = 0
9
Hence, the correct option is (4)
79. Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax + 2ay + c = 0 and
5bx + 2by + d = 0 lies in the fourth quadrant and is equidistant from the two axes then
(1) 3bc – 2ad = 0
(2) 3bc + 2ad = 0
(3) 2bc – 3ad = 0
(4) 2bc + 3ad = 0
Solution
4ax  2ay  c  0
5bx  2by  d  0
x
y
1


2ad  abc 5bc  4ad 8ab  10ab
2 ( ad  bc ) bc  ad
x

2 ( ab)
ab
5bc  4ad
y
2ab
Now according to question point being in 4th quadrant and equidistance from axes, lies on y =  x
5bc  4ad
(bc  ad )
Therefore,

2 ab
ab
5bc  4ad = 2(bc  ad)  5bc  4ad = 2bc  2ad
 3bc  2ad = 0
Hence, the correct option is (1)
80. The locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any
tangent to it is
(1) (x2 + y2)2 = 6x2 + 2y2
(2) (x2 + y2)2 = 6x2 – 2y2
(3) (x2 – y2)2 = 6x2 + 2y2
(4) (x2 – y2)2 = 6x2 – 2y2
Solution
x2
y2
1
( 2)2
x cos  y sin 
Eqn. of tangent at P is

1
6
2
Ellipse is
( 6)2

Equation of tangent at (, ) is
(1)

( x   ) or  y  2 =   x + 2

or  x +  y = 2 + 2
(2)
Comparing Eq (1) and Eq (2), we get
cos  sin 
1

 2
2
6
2   
y 
cos 2   sin 2 
1
1

 2
2
2
2
6  2 
6 2  2  2   
 6 2 +2 2 = (2 + 2)
Therefore, locus is (x2 + y2)2 = 6 x2 + 2 y2
Hence, the correct option is (1)
or
81. Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle centred at (0, y), passing
through origin and touching the circle C externally, then the radius of T is equal to
1
1
(1) 2
(2) 4
3
2
Solution
(3)
(4)
3
2
AB =1 + y
BC = 1  y
Now by Pythagoras theorem (1 + y) 2 = 12 + (1  y) 2 4 y = 1
Therefore, y 
1
4
Hence, the correct option is (2)
82. The slope of the line touching both the parabolas y2 = 4x and x2 = –32y is
1
2
(1)
(2)
8
3
1
3
(3)
(4)
2
2
Solution
Let a Parametric point on y2 = 4x be (t2, 2 t)
Therefore, Equation of tangent is yt = x + t2
or
x = yt  t2
(1)
Now solving
x = yt  t2 and x2 =  32y for points of intersection.
Therefore,
(yt  t2)2 = 32y
or
y2t2 + t4  2yt3 =  32y
or
y2t2 + y(32  2t3) + t4 = 0
For line to be tangent, discriminate = 0
Therefore,
(32  3t3)2  4t6 = 0 or 4(16  t3)2  4 (t3)2 = 0
3
3
 (16  t  t )(16  t 3  t 3 )  0  d 3  8  t  2
1 1
Now slope 
(from 1)
t 2
Hence, the correct option is (3)
83. The image of the line
x3

3
x3

(3)
3
(1)
y 5

1
y 5

1
z2

5
z2

5
x 1 y  3 z  4


in the plane 2x – y + z + 3 = 0 is the line
3
1
5
x3 y 5 z 2





3
1
5
x3 y 5 z  2




3
1
5
Solution
Since,
3 (2) + 1 (1) + (5) (1) = 6  1  5 = 0
Therefore, Line is parallel to plane
Now finding image of point (1, 3, 4) in the plane. Let the image is (x1, y1, z1)
x1  1 y1  3 z1  4 2  2(1)  (3)  (4)  3 2  6
Therefore,




2
1
1
22  12  12
6
Therefore,
x1 =  3, y1 = 5, z1 = 2
x3 y 5 z 2
Therefore, required image line is


3
1
5
Hence, the correct option is (3)
84. The angle between the lines whose direction cosines satisfy the equations l + m + n = 0 and l2 =
m2 + n2 is
(1)
(3)

6

(2)

2

(4)
3
4
Solution
Since,
l=mn
Therefore, ( m  n)2 = m2 + n2 or 2 m n = 0 m = 0 or n = 0
If m = 0, then l2 = n2l = n
1
1
Therefore,
n2 + 0 2 + n 2 = 1  n 2   n  
2
2
1 
 1
,0,
Therefore, directions are 

2
 2
If n = 0, then l =  m m2 + m2 + 0 = 1
1
1
Thus, m 2   m  
2
2
1 
 1
,
,0 
Therefore, directions are 
2 
 2

1
cos    0  0   
2
3
Hence, the correct option is (3)
     
 
85. If [a  b b  c c  a ]  [a b c ]2 , then  is equal to
(1) 0
(2) 1
(3) 2
(3) 3
Solution
    
   
 
     
[a  b.b  cc  a ]  (a  b).[(b  c)  (c  a)]  (a  b).[(b  c)  c  a]
         
 (a  b).[(b  c  a)c  (b  c  c)a ]
    
 (a  b).[(b  c.a )c  0]
   
 
 
 (a  b.c)(a  b.c)  (a  b.c)2  [ab.c]2
Therefore,  = 1
Hence, the correct option is (2)
1
1
1
86. Let A and B be two events such that P( A  B)  , P( A  B)  and P( A)  , where A stands for
6
4
4
the complement of the event A. Then the events A and B are
(1) Independent but not equally likely
(2) Independent and equally likely
(3) Mutually exclusive and independent
(4) Equally likely but not independent
Solution
1
1 5
 P ( A  B)  1  
C
6 6
1
(1)
P( A  B) 
4
1
1 3
P ( A)   P ( A)  1  
4
4 4
Now, P (A  B) = P (A) + P (B)  P (A  B)
5 3
1
5 3 1 10  9  3 4 1
  P ( B )  or P ( B )    
 
6 4
4
6 4 4
12
12 3
3 1 1
P( A)  P( B)   
(2)
4 3 4
Therefore, from (1) and (2), A and B are independent.
Since P (A)  P (B), therefore not equally likely
Hence, the correct option is (1)
P( A  B) 
87. The variance of first 50 even natural numbers is
(1) 437
(3)
(2)
833
4
437
4
(4) 833
Solution
Variance = 2 Mean of squares  Square of Mean
50
 50

(2r )   (2r ) 


= r 1
  r 1
n
 n 




2
2
4(12  22  32   502 )  2(1  2  3   50) 



50
50


 50  51 
50 ( 51)(101)
 4
 4

6  50
 2  56 
= 3434  512 = 833
Hence, the correct option is (4)
2
2
1
88. Let f k ( x)  (sin k x  cos k x) where x  R and k ≥ 1. Then, f 4 ( x)  f 6 ( x) equals
k
1
1
(1)
(2)
4
12
1
1
(3)
(4)
6
3
Solution
1
1
f 4 ( x)  f 6 ( x)  (sin 4 x  cos 4 x)  (cos 6 x  sin 6 x)
4
6
1
1
 [(sin 2 x  cos 2 x) 2  2sin 2 x cos 2 x]  [(cos 2 x  sin 2 x)3  3cos 2 x sin 2 x(cot 2 x  sin 2 x)]
4
6
1
1
 [1  2sin 2 x cos 2 x]  [1  3cos 2 x sin 2 x]
4
6
1 1
1 1
3  12 1
2
2
  sin x cos x   cos 2 x sin 2 x 

4 2
6 2
12
12
Hence, the correct option is (2)
89. A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the
ground is 45°. It flies off horizontally straight away from the point O. After one second, the elevation of
the bird from O is reduced to 30°. Then, the speed (in m/s) of the bird is
(1) 20 2
(2) 20( 3  1)
(3) 40( 2  1)
(4) 40( 3  2)
Solution
d  20
 cot 30  3  d  20 3  20  20( 3  1)
20
Hence, the correct option is (2)
90. The statement ~(p ~q)is
(1) A tautology
(2) A fallacy
(3) Equivalent to p q
(4) Equivalent to ~ p q
Solution
p
q
q
p  q
 (p  q)
pq
T
T
F
F
T
T
T
F
T
T
F
F
F
T
F
T
F
F
F
F
T
F
T
T

Bi-conditional
Hence, the correct option is (3)