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PAPER – 1 : PHYSICS, CHEMISTRY & MATHEMATICS
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1
JEE-MAIN-2015 SOLUTIONS
PHYSICS
1.
Two stones are thrown up simultaneously form the edge of a cliff 240 m high with initial
speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents
the time variation of relative position of the second stone with respect to the first
(Assume stones do not rebound after hitting the ground and neglect air resistance, take g
= 10 m/s2)
The figure are schematic and not draw to scale)
(1)
Ans.
Sol:
2.
Ans.
Sol:
(2)
(3)
(4)
(3)
y1 = 10t – 5t2
y2  40t  5t 2
And y2 – y1 = 30 t for t ≤ 8 s
At t = 8s, first stone is on the ground and the second stone is at the edge of the diff &
moving down.
So at t = 8 s, y2 – y1 = 240 m
1
For t > 8
y2 – y1 = 240 – 40 t g t2
2
Hence option (3)
L
The period of oscillation of simple pendulum is T= 2 . Measured value of L is 20.0
g
cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be
90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is
(1) 2%
(2) 3%
(3) 1%
(4) 5%
(2)
4 2 L
g 2
T
g L 2T


g
L
T
2
1
21

 2.72%  3%
200 90
 correct option (2)
=
3.
Ans.
Sol.
Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively.
These are being pressed against a wall by a force F as shown. If the coefficient of friction
between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force
applied by the wall on block B is
(1) 100 N
(2) 80 N
(3) 120 N
(4) 150 N
(3)
The solution is based on the assumption that the blocks are in equilibrium.
20
A
f
B
20 20 100
f = 120 N
 Correct option (3)
4.
Ans.
Sol.
5.
A particle of mass m moving in the x direction with speed 2v is hit by another particle of
mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the
percentage loss in the energy during the collision is close to :
(1) 44%
(2) 50%
(3) 56%
(4) 62%
(3)
From law of conversation of linear momentum
2
Centre of Mass(x-direction)  m.2v = 3mvx  vx = v
3
2
(y-direction)  2m.v = 3mvy  vy = v
3
1
1
1
4 
2
2
4
m  2v   .2m  v    3m  .  v2  v2 
2
2
2
9 
9
 % Loss =
= 56%
1

2
2
m  2m   .2m  v 
2
2
 correct option (3)
Distance the centre of mass of a solid uniform cone form its vertex is Z0. If the radius of
its base is R and its height is h then Z0 is equal to:
3
(1)
Ans.
h2
4R
(2)
3h
4
(3)
5h
8
(4)
3h 2
8R
(2)
y
r
dy
R


 M  r 2 ydy
Centre of mass   

1
M
2
O
R H 
3

H
2
3
3 R
 2  r 2 y.dy  2  2 y3 .dy
R HO
R H H
H
3
H
4
 correct option (2)

6.
Ans.
Sol.
From a solid sphere of mass M and radius R cube of maximum possible volume is cut.
Moment of inertia of cube about an axis passing through its centre and perpendicular to
one of its faces is:
MR 2
MR 2
4MR 2
4MR 2
(1)
(2)
(3)
(4)
23 2
16 2
9 3
3 3
(3)
For maximum volume of cube, body diagonal of cube = Diameter of sphere
a 3  2R
a = side of cube
 2R 
Volume of cube = V = 

 3
3
2M
 2R 



4 3  3   3
 R 
3

m
4 MR 2
2
Moment of Inertia =  side  
6
9 3 
 correct option (3)
Mass of cube m 
M
3
4
7.
R
is removed,
2
as shown in the figure. Taking gravitational potential V = 0 at r = , the potential at the
centre of the cavity thus formed is: (G = gravitational constant)
From a solid sphere of mass M and radius R, a spherical portion of radius
(1)
Ans.
Sol:
GM
2R
(2)
GM
R
(3)
2GM
3R
(4)
2GM
R
(2)
Mass of complete sphere = M
M
“–ve” mass of cavity M1 =
8
Potential due to complete sphere V1  
r 
R
2
 v1  
GM
 3R 2  r 2 
2R 3
11GM
8R
M
3G  
3GM
8
  
Potential due to cavity V2 =
1
2R
R
3GM
V2  
8R
Now net potential at the given point
 11GM   3GM 
V  V1  V2  


 8R   8R 
GM
V
R
 Correct option (2)
A pendulum made of a uniform wire of cross sectional area A has time period T. When
an additional mass M is added to its bob, the time period changes to TM. If the Young’s
1
modulus of the material of the wire is Y then
is equal to :
Y
(g = gravitational acceleration)
 TM  2  A
 TM  2  Mg
(1) 
(2) 
  1 Mg
  1 A
 T 

 T 

1
8.
Ans.
  TM 2  A
(3) 1  
 
  T   Mg
(1)
  T 2  A
(4) 1  
 
  TM   Mg
5
Sol.
T  2

g





Mg 
A. 

 mg 


A.y 
Y
T2  l
Tm2  1   

 1  
T2

Tm2
Mg
 1
2
T
AY
2
 Tm
 Mg
 2  1 
T
 AY
 A
1  Tm2
  2  1
Y T
 Mg
9.
Ans.
Sol:
Consider a spherical shell of radius R at temperature T. The black body radiation inside it
can be considered as an ideal gas of photons with internal energy per unit volume
1 U
U
u   T 4 and pressure. p    . If the shell now undergoes an adiabatic expansion
3 V 
V
the relation between T and R is :
1
1
(1) T  e  R
(2) T  e3R
(3) T 
(4) T  3
R
R
(3)
U  T4V
dU  T 4 dV  4T 3V.dT
U
T4
dW  P.dV 
dV 
dV
3V
3
dQ  dW  dU  0
T4
.dV  T 4 .dV  4T3 V.dT  0
3
4 dV
dT
4
0
3 V
T
1
ln V  ln T  Constant
3
TV1/3  Constant
T  V 1/3
T  R 1
10.
A solid body of constant heat capacity 1 J/oC is being heated by keeping it in contact with
reservoirs in two ways:
6
Ans.
Sol.
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies
same amount of heat.
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies
same amount of heat.
In both the cases body is brought form initial temperature 100oC to final temperature
200oC. Entropy change of the body in the two cases respectively is
(1) ln2, 4ln2
(2) ln2, ln2
(3) ln2, 2ln2
(4) 2ln2, 8ln2
(No Answer)
T
T
S  C ln 1  c ln f
Ti
T1
T
473
s1  C ln f  ln
 ln1.27
Ti
373
T
T
T
s 2  c ln 1  C ln 2  ....c ln f
Ti
T1
T7
T
473
 c ln f  ln
 ln1.27
Ti
373
None of the given options is correct.
11.
Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an
adiabatic expansion, the average time of collision between molecules increases as Vq,

Cp 
where V is the volume of the gas. The value of q is:   

Cv 

3  5
3  5
 1
 1
(1)
(2)
(3)
(4)
6
6
2
2
Ans. (3)
mean free path   
Sol: t 
v
T
   and v  T
P
T
 t
P
1/ 2 
T V
V
12.
1 
2
For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d. Which one of the following represents these
correctly?
(graphs are schematic and not drawn to scale)
7
E
(1)
PE
E
(3)
KE
E
PE
(2)
KE
d
d
E
KE
(4)
d
PE
KE
PE
Ans. (2)
Sol: In SHM
1
1
KE = mv 2  m2 (A 2  d 2 );
2
2
1
1
PE = Kd 2  m2 d 2 ;
2
2

'KE' Vs 'd' is a parabola opening downward

'PE' Vs 'd' is a parabola opening upward

option (2) is correct.
A train is moving on a straight track with speed 20 ms-1. It is blowing its whistle at the
frequency of 1000 Hz. The percentage change in the frequency heard by a person
standing near the track as the train passes him is (speed of sound = 320 ms-1) close to:
(1) 6 %
(2) 12 %
(3) 18 %
(4) 24 %
Ans. (2)
 V 
Sol: When the train is approaching f a  f 0 

 V  Vs 
13.
When the train crosses the person f a' 

% change in frequency =
=
 1
1
f0 V 

 V  Vs V  Vs
f0 V
V  Vs
f0 V
V  Vs
f f'
f
100  a a 100
fa
fa


  100
2Vs
2  20
 100  12% (approx.)
 100 =
340
V  Vs
A long cylindrical shell carries positive surface charge  in the upper half and negative
surface charge  in the lower half. The electric field lines around the cylinder will look
like figure given in :
(figure are schematic and not drawn to scale)
=
14.
8
(1)
(2)
(3)
(4)
Ans. (1)
Sol: The shell can be considered as a dipole as shown
+
P
E
E
-
So, at equatorial points
figure 1.
15.

the electric lines has to opposite to P . This satisfied only in
A uniformly charged solid sphere of radius R has potential V0 (measured with respect to
 ) on its surface. For this sphere the equipotential surfaces with potentials
3V0 5V0 3V0
V
,
,
and 0 have radius R1 , R 2 , R 3 and R 4 respectively. Then
2
4
4
4
(1) R1  0 and R 2  (R 4  R 3 )
(2) R1  0 and (R 2  R1 )  (R 4  R 3 )
(3) R1  0 and R 2  (R 4  R 3 )
(4) 2R  R 4
Ans. (3), (4)
Sol: For uniformly charged solid sphere at internal points the potential is greater than that on
KQ
the surface Vsurface = V0 
R
KQ
3R 2  r 2 
Vint (r  R) =
3 
2R
3 KQ 3
At center (r = 0); V 
 V0
2 R
2
3V
R1  0 , V1  0
 at
2
At R2;
KQ
5
5 KQ
3R 2  R 22  V0 
3
2R
4
4 R


9
  3R 2  R 22  
5 2
R
2
R2
R
 R 22  R 2 

2
2
3
At R3; V  V0 (which is less then V0) So R3 should be outside the sphere.
4
KQ 3 KQ
4R

 R3 
R3 4 R
3
KQ 1 KQ
Similarly at R4;

 R 4  4R
R4 4 R
4R 8R
Now, R 4  R 3  4R 

3
3
So, R 2   R 4  R 3 
So, option (3) is correct. 2R < R4 hence option (4) is also correct.
16.
In the given circuit, charge Q2 on the
2F capacitor changes as C is varied
from 1F to 3F . Q2 as a function of
'C' is given properly by : (figures are
drawn schematically and are not to
scale)
11FF
C
2F
E
(1)
(2)
Charge
Q2
Q2
1F
(3)
3F
C
1F
(4)
Charge
3F
C
Charge
Q2
Q2
1F
Ans.
Charge
3F
C
1F
3F
C
(2)
10
Sol:
1 F
C
C
3 F
2 F
E
Ceq 
E
3C
C3
3CE
C3
Q  2 2CE
 Q2 

3
C3
dQ 2   C  3 2E  2CE 
6E


2
2
dC 
 C  3
  C  3
dQ 2
Since
is decreasing with increase in value of C
dC
 option (2) is correct.
Q  Ceq .E 
17.
Ans.
Sol:
When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of
electrons is 2.5 104 ms 1 . If the electron density in the wire is 8  1028 m 3 , the
resistivity of the material is closed to:
(1) 1.6 108 m
(2) 1.6 107 m
(3)1.6 106 m
(4) 1.6 105 m
(4)
j = E
I
V

(1) where I = neAVd,
A
l
I
Now,
 neVd
A
1V
V
From (1), neVd  
 neVd 
l
 l
V
5


28
19
neVd l 8 10  1.6  10  2.5  104  0.1
 = 1.6  10-5 m .
11
18.
6V
P 2
E
1
E
9V
Q 3
3
In theEcircuit shown,Ethe current in the 1 resistor is:
(1) 1.3 A, from P to Q
(3) 0.13 A, from Q to P
Ans. (3)
Sol:
(2) 0 A
(4) 0.13 A, from P to Q
9V
i 3 P 5
x i1
3
9V
1
(–6V)
i2
6V
Q
i1  i2  i3
9x x6 x


5
3
1
27  3x  5x  30  15x
3
x
23
Current is (3/23) Amp from Q to P.
19.
Ans.
Sol:

Two coaxial solenoids of different radii carry current I in the same direction. Let F1 be

the magnetic force on the inner solenoid due to the outer one and F2 be the magnetic
force on the outer solenoid due to the inner one. Then :
 
(1) F1  F2  0


(2) F1 is radially inwards and F2 is radially outwards


(3) F1 is radially inwards and F2 = 0


(4) F1 is radially outwards and F2 = 0
(1)
Magnetic field due to the inner coil at external points will be zero.
So magnetic force (iLB) on the outer coil will be zero.
The magnetic field at any current element of inner coil due to outer coil is perpendicular
to it.
 

As per Fleming’s Left Hand rule (or dF  I dl  B the force will be radially.


12
So net force on inner coil is also zero. (Every loop of inner coil is a uniform magnetic
field)
So. F = 0
20.
Two long current carrying thin wires, both with
current I, are held by insulating threads of length
L and are in equilibrium as shown in the figure,
with threads making an angle '  ' with the
vertical. If wires have mass  per unit length then
the value of I is:
(g = gravitational acceleration)
L

E
I
I
(1) sin 
(3) 2
Ans.
Sol:
gL
 0 cos 
gL
tan 
0
(2) 2sin 
(4)
gL
 0 cos 
gL
tan 
0
(2)
Assume length of the wire is 1 m
Since the wire is in equilibrium ,
 o I2
T sin  =
(1)
4L sin 
T cos  = g
(2)
On dividing (1) and (2),
oI2
tan  =
4L sin . g
 
T
T cos 
L

T sin 
2
oI /2(2Lsin) I
L sin 
L
L sin  I
g
Hence, I =
4Lg sin  tan 
4Lg sin 2 

o
o cos 
Lg
 o cos 
A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in
different orientations as shown in the figures below:
Hence, I = 2sin 
21.
13
z
z
I
B
B
I
I
(a)
(b)
y
I
x
z
B
(d)
y
I
I
I
I
22.
y
I
x
Key:
Sol.
z
B
I
(c)
y
I
I
x
I
I
I
x
I
If there is a uniform magnetic field of 0.3 T in the positive z direction, in which
orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium?
(1) (a) and (b), respectively
(2) (a) and (c), respectively
(3) (b) and (d), respectively
(4) (b) and (c), respectively
(3)
  MB sin 
U   MB cos 
In (a)  = 90
(b)  = 00
(c)  = 90
(d)  180
So, (b) is stable equilibrium
& (d) is unstable
An inductor (L = 0.03H) and a resistor (R = 0.15 k) are connected in series to a battery
of 15V EMF in a circuit shown below. The key K1 has been kept closed for a long time.
Then at t = 0, K1 is opened and key K2 is closed simultaneously. At t = 1ms, is closed in
the circuit will be  e5  150 
0.03H
0.15k
K2
K1
15V
Key:
Sol.
(1) 100 mA
(2) 67 mA
(4)
At t=  when K1 is closed
15
1
I0 

A.
150 10
When K1 is opened & K2 is closed.
(3) 6.7 mA
(4) 0.67 mA
14
I  I 0 e t / &  
L
0.03

 2 104
3
R 0.15  10
103
1  4
 I  e 210
10
1 5
 e
10
1
 e 5
 0.67mA
10
23.
Key:
Sol.
24.
A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field
of the light at a distance of 1m from the diode is:
(1) 1.73 V/m
(2) 2.45 V/m
(3) 5.48 V/m
(4) 7.75 V/m
(2)
1

P    0 E02   A  C
2

1

P    0 E02   A  C
2

2  0.1

12
8.85 10  4 12  3  108
 2.45 V / m
Monochromatic light is incident on a glass prism of angle A. If the refractive index of the
material of the prism is , a ray, incident at an angle , on the face AB would get
transmitted through the face AC of the prism provided:
A

B
Key:
Sol.


 1  
(1)   sin 1   sin  A  sin 1    

    



 1  
(3)   cos 1   sin  A  sin 1    
    


(1)
C


 1  
(2)   sin 1   sin  A  sin 1    

    



 1  
(4)   cos 1   sin  A  sin 1    
    


15
A

r1
c
B
A  r1  C
r1   A  C 
Using snell’s law on first face
sin    sin r1
  sin  A  C 

1
  sin  A  sin 1 




 1  
   sin 1   sin  A  sin 1    

    

For transmission through face AC


 1  
  sin 1   sin  A  sin 1    

    

25.
Key:
Sol.
26.
Key:
Sol.
On a hot summer night, the refractive index of air smallest near the ground and increases
with height from the ground. When a light beam is directed horizontally, the Huygen’s
principle leads us to conclude that as it travels, the light beam:
(1) becomes narrower
(2) goes horizontally without any deflection
(3) bends downwards
(4) bends upwards
(4)
As R.I increases ray bend towards normal, hence it will bend upward.
Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of
25 cm, the minimum separation between two objects that human eye can resolve at 500
nm wavelength is:
(1) 1 m
(2) 30 m
(3) 100 m
(4) 300 m
(2)
1.22 y

d
D
1.22  500  109
y

2
0.50 10
25  102
y  30.50 106 m
 30  m
16
27.
Key:
Sol.
28.
Key:
29.
Key:
Sol.
30.
As an electron makes a transition from an excited state to the ground state of a hydrogenlike atom/ion:
(1) its kinetic energy increases but potential energy and total energy decrease
(2) kinetic energy, potential energy and total energy decrease
(3) kinetic energy decreases, potential energy increases but total energy remains same
(4) kinetic energy and total energy decrease but potential energy increase
(1)
kZe2
KE 
2r
kZe 2
PE 
r
kZe 2
TE 
2r
Match List-I (Fundamental Experiment) with List-II (its conclusion) and select the
correct option from the choices given below the list:
List-I
List-II
(A) Franck-Herts Experiment.
(i) Particle nature
(B) Photo-electric experiment
(ii) Discrete energy levels of atom
(C) Davision-Germer Experiment.
(iii) Wave nature of electron
(iv) Structure of atom
(1) (A) – (i)
(B) – (iv)
(C)-(iii)
(2) (A) – (ii) (B) – (iv) (C) – (iii)
(3) (A) – (ii)
(B) – (i)
(C) – (iii)
(4) (A) – (iv) (B) – (iii) (C) – (ii)
(3)
A signal of 5kHz frequency is amplitude modulated on a carrier wave of frequency
2 MHz. The frequency of the resultant signal is/are:
(1) 2 KHz only
(2) 2005 kHz, and 1995 kHz
(3) 2005 kHz, 2000 kHz and 1995 kHz
(4) 2000 kHz and 1995 kHz
(3)
Resultant frequency of signal if given by  fC  f M  ,  f R  ,  f C  f M 
Then possible frequencies are 2005 kHz , 2000 kGz ,1995 kHz
An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is
charged to Q0 and then connected the L and R as shown below:
17
R
L
C
2
If a student plots graphs of the square of maximum charge  QMax
 on the capacitor with
time (t) for two different values L1 and L2(L1 > L2) of L then which of the following
represents this graph correctly? (plots are schematic and not drawn to scale)
2
2
QMax
QMax
L1
(1)
L2
(2)
L2
L1
t
t
2
Max
Q
Q
L1
(3)
Q0 (For both L 1 and L 2)
(4)
L2
t
Key.
Sol:
2
Max
t
(1)
q
dI
 IR  L
0
C
dt
q  dq 
 dI 
  R  L 
C  dt 
 dt 
Comparing with equation of damped oscillation
R
Fdamp  bv    
L
R
Hence, b 
L
1

b
L
As A  A0 e
Q  Q0 e
2
2
0
 b 

t
 2m 
 b 

t
 2m 
Q Q e
b
  t
m
Topic & Level
18
Q.No.
1.
2.
3.
4.
5.
6.
7
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
S.No.
1
2
3
4
5
6
Topic
Kinematics
Experimental Physics
Friction
Collision
Centre of Mass
Rotational Motion
Gravitation
Elasticity
Heat & Thermodynamics
Heat & Thermodynamics
Heat & Thermodynamics
Simple Harmonic Motion
Waves
Electrostatics
Electrostatics
Electrostatics
Current Electricity
Current Electricity
Magnetism
Magnetism
Magnetism
E.M.I.
Electro magnetic waves
Optics
Optics
Optics
Modern Physics
Modern Physics
Modern Physics
E.M.I.
Major Topic
Mechanics
Heat & Thermodynamics
S.H.M. & Waves
Electricity & Magnetism
E.M.Waves & Optics
Modern Physics
Total
Level
Medium
Easy
Medium
Easy
Medium
Difficult
Difficult
Difficult
Difficult
Medium
Difficult
Easy
Easy
Medium
Difficult
Difficult
Easy
Easy
Easy
Medium
Easy
Easy
Easy
Medium
Easy
Difficult
Easy
Easy
Easy
Difficult
No. of Question
8
3
2
9
4
4
30
19
CHEMISTRY
31.
Key.
Sol.
The molecular formula of a commercial resin used for exchanging ions in water softening
is C8H7SO3Na (Mol.wt.206). What would be the maximum uptake of Ca2+ ions by the
resin when expressed in mole per gram resin?
1
1
2
1
(1)
(2)
(3)
(4)
103
206
309
412
(4)
Na+ ion of resin will be exchanged with Ca2+ ion
2C8 H 7SO3 Na  Ca 2 
  C8 H 7SO3 2 Ca  2Na 
So maximum uptake of Ca2+ ion by the resin 
1
1

molg 1 of resin.
206  2 412
o
32.
Sodium metal crystallizes in a body centered cubic lattice with a unit cell edge of 4.29 A .
The radius of sodium atom is approximately :
o
Key.
Sol.
(1) 1.86 A
(1)
For B.c.c. unit cell
4r  3a
o
(2) 3.22 A
o
(3) 5.72 A
o
(4) 0.93 A
o
3  4.29
 1.86 A
4
Which of the following is the energy of a possible excited state of hydrogen?
(1) 13.6 eV
(2) 6.8 eV
(3) 3.4 eV
(4)  6.8 eV
(3)
13.6 12
Energy of second shell of hydrogen  
 3.4eV / atom
22
 r
33.
Key.
Sol.
34.
Key.
Sol.
35.
The intermolecular interaction that is dependent on the inverse cube of distance between
the molecules is
(1) ion – ion interaction
(2) ion – dipole interaction
(3) London force
(4) hydrogen bond
(4)
Dipole –dipole interaction energy between stationary polar molecules (as in solids) is
1
proportional to 3 .
r
Hydrogen bond is a special case of dipole-dipole interactions.
The following reaction is performed at 298 K.
 2NO2  g 
2NO  g   O2  g  
The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the
standard free energy of formation of NO2(g) at 298 K?  K p  1.6  1012 
(1) R  298  ln 1.6 1012 
(2) 86600  R  298  ln 1.6  1012 
20
(3) 86600 
Key.
Sol.
36.
Key.
Sol.
37.
Key.
Sol.
ln 1.6  1012 
R  298 
(4) 0.5  2  86600  R  298  ln 1.6 1012  
(4)
G R  RT ln K p
G R  2   f G  NO 2    f G  NO  
 G  NO 
G f  NO 2   G R  f
2
R  298  ln1.6  1012
3
= 86.6 10 
2
The vapour pressure of acetone at 20oC is 185 torr. When 1.2 g of a non-volatile
substance was dissolved in 100 g of acetone at 20oC, its vapour pressure was 183 torr.
The molar mass (g mol-1) of the substance is
(1) 32
(2) 64
(3) 128
(4) 488
(2)
Po  P5
 X solute
P0
1.2
185  183
M

o
1.2
100
P

M 58
2
1.2 58
1.2
100 





 Assume

185 M 100
M
58 

1.2 
 M
 64.38
2 100
 B  C is 2494.2 J.
The standard Gibbs energy change at 300 K for the reaction 2A 
1
1
At a given time, the composition of the reaction mixture is  A   ,  B  2 and  C   .
2
2
The reaction proceeds in the : [ R = 8.314 J/K/mol, e = 2.718]
(1) forward direction because Q > KC
(2) reverse direction because Q > Kc
(3) forward direction because Q < Kc
(4) reverse direction because Q < Kc
(2)
G o   RT ln K c
2494.2  8.314  300ln Kc
ln Kc  1
1
1

e
2.718
B
C
  
K c  e 1 =
Qc 
 A2
21
1
2 4

2
1
 2
 
 Reverse direction as Qc > Kc
Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper
deposited at the cathode is : (at.mass of Cu = 63.5 amu)
(1) 0 g
(2) 63.5 g
(3) 2 g
(4) 127 g
(2)
1 Faraday of electricity deposits 1 equivalent Cu
2 Faraday of electricity deposits 2 equivalents of Cu
1 mol of Cu = 63.5 g
Higher order (>3) reactions are rare due to
(1) low probability of simultaneous collision of all the reacting species
(2) increase in entropy and activation energy as more molecules are involved
(3) shifting of equilibrium towards reactants due to elastic collisions
(4) loss of active species on collision
(1)
Higher order (>3) reactions are rare due to low probability of simultaneous collision of all
the reacting species.
3 g of activated charcoal was added to 50 ml. of acetic acid solution (0.06 N) in a flask.
After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The
amount of acetic acid adsorbed (per gram of charcoal) is
(1) 18 mg
(2) 36 mg
(3) 42 mg
(4) 54 mg
(1)
No. of millimols of acetic acid before adsorption = 50  0.06  3
No. of millimols of acetic acid after adsorption = 50  0.042  2.1
No. of millimols of acetic acid adsorbed = 3  2.1  0.9
0.9  60
Weight of acetic acid adsorbed 
g  54 mg
103
54
Amount of acetic acid adsorbed per gm of charcoal =
mg = 18 mg
3
2
38.
Key.
Sol.
39.
Key.
Sol.
40.
Key.
Sol.
41.
Key:
Sol:
42.
The ionic radii (in Å) of N3, O2 and F are respectively
(1) 1.36, 1.40 and 1.71
(2) 1.36, 1.71 and 1.40
(3) 1.71, 1.40 and 1.36
(4) 1.71, 1.36 and 1.40
(3)
N 3  O 2  F
As the charge increase on anion its size increases due to decrease in Zeff.
In the context of the Hall – Heroult process for the extraction of Al, which of the
following statements is false?
(1) CO and CO2 are produced in this process
(2) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings
conductivity
22
Key:
Sol:
43.
Key:
Sol:
44.
Key:
Sol:
45.
Key:
Sol:
46.
Key:
Sol:
47.
Key:
Sol:
48.
(3) Al3+ is reduced at the cathode to form Al
(4) Na3AlF6 serves as the electrolyte
(4)
CaF2 or Na3AlF6 reduce melting point and increase the conductivity.
From the following statements regarding H2O2, choose the incorrect statement:
(1) It can act only as an oxidizing agent
(2) It decomposes an exposure to light
(3) It has to be stored in plastic or wax lined glass bottle in dark
(4) It has to be kept away from dust
(1)
It act as both OA/RA due to intermediate OS of oxygen.
Which one of the following alkaline earth metal sulphates has its hydration enthalpy
greater than its lattice enthalpy?
(1) CaSO4
(2) BeSO4
(3) BaSO4
(4) SrSO4
(2)
BeSO4 soluble in water, the hydration energy greater than its lattice energy.
Which among the following is the most reactive?
(1) Cl2
(2) Br2
(3) I2
(4) ICl
(4)
Inter Halogen compounds are more reactive due to improper overlapping than halogen.
Match the catalysts to the correct process:
Catalyst
Process
(A) TiCl3
(i) Wacker process
(B) PdCl2
(ii) Ziegler – Natta Polymerizatio n
(C) CuCl2
(iii) Contact process
(D) V2O5
(iv) Deacon’s process
(1) (A) – (iii), (B) – (ii), (C) – (iv), (D) – (i) (2) (A) – (ii), (B) – (i), (C) – (iv), (D) – (iii)
(3) (A) – (ii), (B) – (iii), (C) – (iv), (D) – (i) (4) (A) – (iii), (B) – (i), (C) – (ii), (D) – (iv)
(2)
TiCl3 + (C2H5)3Al Ziegler Natta Catalyst
PdCl2 used in Wacker process.
CuCl 2
HCl + O2 
 H2O + Cl2 Decan process
V2 O5
SO2 + O2 
 SO3 Contact process.
Which one has the highest boiling point?
(1) He
(2) Ne
(3) Kr
(4) Xe
(4)
Boiling point of inert gas increases down the group due to Vander Waal force increases
with molar mass.
The number of geometric isomers that can exist for square planar
[Pt(Cl)(py)(NH3)(NH2OH)]+ is (py – pyridine)
23
Key:
Sol:
(1) 1
(2) 3
(2)
planar [Pt(Cl)(py)(NH3)(NH2OH)]
Cl
NH3
Cl
Pt
Key:
Sol:
Py
(4) 6
Cl
Pt
HONH2
49.
(3) 4
Py
HONH2
Py
Pt
NH3
NH3
NH2OH
The color of KMnO4 is due to
(1) M  L charge transfer transition
(2) d – d transition
(3) L  M charge transfer transition
(4)  - * transition
(3)
MnO 4
O
O
Mn
O
O
Color is due to charge transfer from oxide ion to empty ‘d’ orbital of Mn.
50.
Key:
Sol:
51.
Key
Sol.
52.
Assertion :
Nitrogen and Oxygen are the main components in the atmosphere but
these do not react to form oxides of nitrogen.
Reason:
The reaction between nitrogen and oxygen requires high temperature.
(1) Both assertion and reason are correct, and the reason is the correct explanation for the
assertion
(2) Both assertion and reason are correct, but the reason is not the correct explanation for
the assertion.
(3) The assertion is incorrect, but the reason is correct.
(4) Both the assertion and reason are incorrect
(1)
The bond enthalpy of N2 is much higher hence the process is endothermic.
N2 + O2  2NO - Heat
In Carius method of estimation of halogen, 250 mg of an organic compound gave 141 mg
of AgBr. The percentage of bromine in the compound is: (at. Mass Ag = 108, Br = 80)
(1) 24
(2) 36
(3) 48
(4) 60
(1)
Mass of AgBr precipitated 0.141 g
80  0.141
% of Br =
 100  24
188  0.250
Which of the following compounds will exhibit geometrical isomerism?
24
Key
Sol.
(1) 1 – Phenyl – 2 – butene
(3) 2 – Phenyl – 1 – butene
(1)
CH3
PhCH2
H
PhCH2
C=C
H
(2) 3 – Phenyl – 1 – butene
(4) 1, 1, - Diphenyl – 1 – propene
C=C
H
H
CH3
Geometrical Isomers
53.
Which compound would give 5 – keto – 2 – methyl hexanal upon ozonolysis?
CH3
CH3
CH3
CH3
CH3
H3C
(1)
(2)
(3)
CH3
CH3
Key
(4)
(2)
CH3
O
ozonolysis
H3 C – C– CH2CH2 – CH –CHO
Sol.
CH3
CH3
5 - keto - 2 methyl - hexanal
54.
Key
Sol.
55.
The synthesis of alkyl fluorides is best accomplished by :
(1) Free radical fluorination
(2) Sandmeyer’s reaction
(3) Finkelstein reaction
(4) Swarts reaction
(4)
Swart
H3C  Br  AgF 
 CH 3  F  AgBr
reaction
In the following sequence of reaction :
KMnO4
SOCl2
H 2 / Pd
Toluene 
A 
 B 
C
BaSO 4
Key
The product C is
(1) C6H5COOH
(4)
(2) C6H5CH3
(3) C6H5CH2OH
(4) C6H5CHO
O
CH3
COOH
KMnO4
C – Cl
SOCl2
H2/Pd
Sol.
BaSO4
(A)
56.
CHO
(B)
In the reaction
25
NH2
NaNO2 / HCl
Cu / KCN

 D 
 E  N2

0 5 o C
CH3
The product E is
COOH
(1)
(2)
H3 C
CH3
CH3
CN
CH3
(3)
(4)
CH3
Key
(3)
NH2
Sol.
CH3
57.
Key
Sol.
58.
Key
Sol.
59.
Key
Sol.
60.
N2 Cl
CN
NaNO2/HCl
CuCN/ K CN
0-5°C

CH3
(D)
N2 +
CH3
(E)
Which polymer is used in the manufacture of paints and lacquers?
(1) Bakelite
(2) Glyptal
(3) Polypropene
(4) Poly vinyl chloride
(2)
Glyptal is used as paints and lacquers
Which of the vitamins given below is water soluble
(1) Vitamin C
(2) Vitamin D
(3) Vitamin E
(4) Vitamin K
(1)
Vitamin A, D, E and K are fat Soluble. Vitamin C is water soluble.
Which of the following compounds is not an antacid?
(1) Aluminium hydroxide
(2) Cimetidine
(3) Phenelzine
(4) Ranitidine
(3)
Phenelzine is antidepressant, so not an antacid. All others are antacid
Which of the following compounds is not colored yellow?
26
Key
Sol.
(1) Zn2[Fe(CN)6]
(3) (NH4)3[As(Mo3O10)4]
(1)
Zn2[Fe(CN)6] is bluish white.
(2) K3[Co(NO2)6]
(4) BaCrO4
Q. NO.
TOPIC
31
Physical Chemistry
32
Physical Chemistry
33
Physical Chemistry
34
Physical Chemistry
35
Physical Chemistry
36
Physical Chemistry
37
Physical Chemistry
38
Physical Chemistry
39
Physical Chemistry
40
Physical Chemistry
41
Inorganic Chemistry – Periodic table
42
Inorganic Chemistry – Metallurgy of Al
43
Inorganic Chemistry – Hydrogen and its compounds
44
Inorganic Chemistry (Alkaline earth Metals)
45
Inorganic Chemistry – Halogens
46
Inorganic Chemistry – Transition elements
47
Inorganic Chemistry – 18th group
48
Inorganic Chemistry – Coordination compounds
49
Inorganic Chemistry – d-block elements
50
Inorganic Chemistry – p block elements
51
Organic Chemistry
52
Organic Chemistry
53
Organic Chemistry
54
Organic Chemistry
55
Organic Chemistry
56
Organic Chemistry
57
Organic Chemistry
58
Organic Chemistry
59
Organic Chemistry
60
Organic Chemistry
Organic Chemistry – 9 questions
Physical chemistry – 10 questions
Inorganic chemistry – 11 questions
LEVEL
Simple
Medium
Simple
Simple
Medium
Medium
Difficult
Simple
Simple
Medium
Simple
Medium
Simple
Medium
Simple
Medium
Simple
Medium
Medium
Simple
Medium
Simple
Simple
Simple
Simple
Medium
Simple
Simple
Medium
Medium
27
MATHEMATICS
61.
Key.
Sol.
62.
Let A and B be two sets containing four and two elements respectively. Then the number
of subsets of the set A  B , each having at least three elements is :
(1) 219
(2) 256
(3) 275
(4) 510
(1)
No. of elements in A  B will be 4  2  8
Now no. of subsets of A  B having at least 3 elements is
8
C3  8C4  8C5  ....  8C8  28  1  8  28
 256  37  219
A complex number z is said to be unimodular if z  1 . Suppose z1 and z2 are complex
numbers such that
Key.
Sol.
z1  2z 2
is unimodular and z2 is not unimodular. Then the point of z1
2  z1 z2
lies on a
(1) straight line parallel to x-axis
(3) circle of radius 2
(3)
z1  2z 2
1
2  z1 z2
2
 z1  2z 2  2  z1 z2
(2) straight line parallel to y-axis
(4) circle of radius 2
2
  z1  2z 2  z1  2z2    2  z1 z2  2  z1z 2 
 z1 z1  2z1 z2  2z1z2  4z2 z2 = 4  2z1z 2  2z1 z2  z1 z1z2 z2
2
2
2
 z1  4 z 2  4  z1 z 2
2
2
2
2
2
 z1 z 2  z1  4 z 2  4  0

z
1
2
4
 z
 z1  2 or
2
2

1  0
z2  1
But z 2  1 z1  2
 z1 lie on a circle of radius 2
28
63.
Key.
Sol.
Let  and  be the roots of equation x 2  6x  2  0 . If a n   n   n , for n  1 , then
a  2a 8
the value of 10
is equal to
2a 9
(1) 6
(2) –6
(3) 3
(4) –3
(3)
a n   n   n   n 1   n 1   n 1   n 1   n 1   n 1
   n 1   n 1     n 1   n 1     n  2   n  2 
      n 1   n 1     n  2   n  2 
a n  6a n 1  2a n 2 
a n  2a n  2
3
2a n 1
Put n = 10
a  2a 8
 10
3
2a 9
64.
Key.
Sol.
1 2 2 
If A   2 1 2  is a matrix satisfying the equation AA T  9I , where I is 3  3 identity
 a 2 b 
matrix, then the ordered pair (a,b) is equal to
(1) (2, –1)
(2) (–2, 1)
(3) (2, 1)
(4) (–2, –1)
(4)
1 2 2 
A   2 1 2 
 a 2 b 
1 2 a 
 A   2 1 2 
 2 2 b 
 1 2 2  1 2 a 
T
A.A   2 1 2  2 1 2 
 a 2 b   2 2 b 
9
0
a  4  2b 



0
9
2a  2  2b   9I
 a  4  2b 2a  2  2b a 2  4  b2 
 a  4  2b  0 … (1) and 2a  2  2b  0 … (2)
Solving (1) and (2) a  2, b  1
The set of all values of  for which the system of linear equations
2x1  2x 2  x 3   x1
T
65.
2x1  3x 2  2x3   x 2
29
Key.
Sol.
x1  2x 2   x3
has a non-trivial solution,
(1) is an empty set
(2) is a singleton
(3) contains two elements
(4) contains more than two elements
(3)
Rewrite the equation
 2    x1  2x 2  x 3  0
2x1     3 x 2  2x 3  0
x1  2x 2   x3  0
For non trivial solution
2
2
1
2
    3 2  0
1
2

    1   2  2  3  0
66.
Key.
Sol.
67.
    1   3   1  0    1,  3
The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6 7
and 8, without repetition, is :
(1) 216
(2) 192
(3) 120
(4) 72
(2)
Case (i) number of ‘4’ digit numbers will be 3  4  3  2  72
Case (ii) number of ‘5’ digit numbers will be 5!  120
So total number  120  72  192

The sum of coefficients of integral powers of x in the binomial expansion of 1  2 x
50

is
(1)
Key.
Sol.
(1)
Let
1 50
 3  1
2
(2)
1 50
3 
2
(3)
1 50
 3  1
2
(4)
1 50
 2  1
2
x y
50
1  2y 
 a 0  a1 y  a 2 y2  a 3 y3  a 4 y 4  ....
Put y = 1
1  a 0  a1  a 2  a 3  a 4  ....
Put y = –1
350  a 0  a1  a 2  a 3  a 4  ....
… (1)
… (2)
Adding (1) and (2)
30
68.
Key.
Sol.
350  1
 a 0  a 2  a 4  ....
2
If m is the A.M. of two distinct real numbers  and n (  , n > 1) and G1,G2 and G3 are
three geometric means between  and n, then G14  2G 24  G 34 equals.
(1) 4  2 mm
(2) 4m 2 n
(2)
n
 m     n   2m
2
,G1 ,G 2 ,G 3 , n are in G.P
Let common difference be r
(3) 4  mn2
(4) 4  2m2 n2
1/4
n
n  .r  r   

4
1/ 4
n
G1  .    3/ 4 .n1/ 4

1/ 2 1/ 2
G2   n
G 3  1/ 4 .n 3/ 4
Then G14  2G 24  G 34 
3n  2 2 n 2  n 3
 n   2  2 n  n 2 
2
 n    n   4m 2 n
69.
Key.
13 13  23 13  23  33
The sum of first 9 terms of the series 

 .... is
1 1 3
1 3  5
(1) 71
(2) 96
(3) 142
(4) 192
(2)
n
Sol.
Tn 
3

n 2  n  1
4.n 2
2
n2
2
 n  1

4
n 2  1  2n
Tn 
4
1
  n 2  2n  1
4
1
Sn   Tn    n 2  2 n  n 
4
n  n  1

1  n  n  1 2n  1
 
2
 n
4
6
2

31
S9 
1  9  10  19
9  10 
2
 9

4
6
2

1
15 19  9 10  9  96
4
1  cos 2x  3  cos x  is equal to
lim
x 0
x tan 4x

70.
(1) 4
Key.
(3) 2
(4)
1
2
(3)
lim
Sol.
(2) 3
1  cos 2x  3  cos x 
x 0
x tan 4x
2
2 sin x  3  cos x 
x 0
 tan 4x 
x2 
 .4
 4x 
2
 4  2
4
 Lt
71.
Key:
Sol:
k x  1, 0  x  3
If the function : g  x   
, is differentiable, then the value of k + m is
mx  2, 3  x  5
16
10
(1) 2
(2)
(3)
(4) 4
5
3
(1)
 g(x) is continuous at x = 3
 k  3  1 = 3m + 2
2k = 3m + 2
(1)
for g(x) to be differentiable,
g '  3   g '  3 
k
 m  k = 4m
4
From (1) and (2), 8m = 3m + 2
 m = 2/5
Option (1) is correct.

72.
Key:
Sol:
(2)
and
k = 4  (2/5) = 8/5
k + m = 10/5 = 2
The normal to the curve, x2 + 2xy – 3y2 = 0, at (1, 1) :
(1) does not meet the curve again
(2) meets the curve again in the second quadrant
(3) meets the curve again in the third quadrant
(4) meets the curve again in the fourth quadrant
(4)
dy
dy
2x + 2y + 2x
 6y
0
dx
dx
32
 dy 
  
=1
 dx 1,1
Hence, equation of normal, y – 1 = 1(x – 1)
y – 1 = x + 1
y+x=2
(1)
x2 + 2x(2 – x) – 3(2 – x)2 = 0
x2 + 4x – 2x2 – 3(4 + x2 – 4x) = 0
4x2 + 16x – 12 = 0
 x = 1, 3 and y = 1, 1
 meets the curved again in fourth quadrant.
 option (4) is correct.
73.
Key:
Sol:
74.
Let f(x) be a polynomial of degree four having extreme values at x = 1 and x = 2. If
 f x 
lim 1  2   3, then f(2) is equal to
x 0
x 

(1) 8
(2) 4
(3)
Let f(x) = ax4 + bx3 + cx2
 f(1) = 0  4a + 3b + 2c = 0
f(2) = 0
 32a + 12b + 4c = 0
 f x
 lim 1  2  = 3
x 0
x 

(3) 0
 c=2
By using (1), (2) and (3),
a = 1/2, b = 2 and c = 2.
x4
Hence, f(x) =
 2x 3  2x 2
2
 f(2) = 0
(3)
x
The integral
dx
2
x
4
3/4
 1
(4) 4
(1)
(2)
equals
1/4
 x4 1 
(1)  4 
 x 
1/4
(2)  x 4  1  C
C
1/4
1/ 4
(3)   x  1
4
Key:
(4)
Sol:
x
dx
2
x
2
 1
3/ 4
 x4 1 
(4)   4 
 x 
C

C
dx
1 

x 5 1  4 
 x 
3/ 4
33
1
=t ,
x4
Put, 1 +
4x 5 dx  dt ,
1/4
75.
Key:
1/4
  x 4  1
1 

C
I =  1  4  
x
 x 
 (4) is correct.
4
log x 2
The integral 
dx is equal to
2
2
2 log x  log  36  12x  x 
(1) 2
(3)
(2) 4
log  x 2 
4
Sol:
I
2
4
I
2
(3) 1
(1)
log x 2  log  36  12x  x 2 

log  6  x 

2
2
(4) 6

(2)

log  6  x   log x 2
4
Adding (1) and (2), 2I =  dx
2
2I = 2  I = 1
 Option (3) is correct
76.
Key:
Sol:
The area (in sq. units) of the region described by
{(x, y): y2  2x and y  4x – 1} is
7
5
15
(1)
(2)
(3)
32
64
64
(4)
y2 = 2x, y = 4x – 1
2y2 = y + 1
2y2 – y – 1 = 0
y = 1, y = 1/2
x = ½, x = 1/8
(4)
9
32
1
1
 y  1 y2 
 y 2 y y3 
 dy      =
Area =  
4
2
 8 4 6 1/2
1/2 
9
32
77.
Let y(x) be the solution of the differential equation  x log x 
Key:
then y(e) is equal to
(1) e
(3)
(2) 0
(3) 2
dy
 y  2x log x,  x  1 .
dx
(4) 2e
34
Sol:
dy
1

y2
dx x log x
1
I. F. = e
 x log x dx
e
log  log x 
 log x
Solving, y log x   2.log x.dx  2  log x  x  x   C
78.
Key:
y. log x = 2(log x – 1)x + C, since C = 2  y (e) = 2
The number of points, having both co-ordinates as integers, that lie in the interior of the
triangle with vertices (0, 0), (0, 41) and (41, 0) is
(1) 901
(2) 861
(3) 820
(4) 780
(4)
Number of points = 39 + 38 +
(0, 41)
37 + ….+ 1
(1, 40)
39  40
=
 780
2
(1, 39)
(1, 1)
(0, 0)
79.
Key:
Sol:
80.
Key:
Sol:
(1, 0)
(40, 1)
(40, 0)
(41, 0)
Locus of the image of the point (2, 3) in the line (2x – 3y + 4) + k(x – 2y + 3) = 0, k  R ,
is a
(1) straight line parallel to x-axis
(2) straight line parallel to y-axis
(3) circle of radius 2
(4) circle of radius 3
(3)
The point of concurrency of the given family of lines is (1, 2)
So, therefore the locus of the image of the point (2, 3) in the line
(2x – 3y + 4) + k(x – 2y + 3) = 0 will be a circle; whose centre is (1, 2) and radius is
distance between (1, 2) and (2, 3).
Hence the equation of the required locus will be (x – 1)2 + (y – 2)2 = 2.
The number of common tangents to the circles x2 + y2 – 4x – 6y – 12 = 0 and
x2 + y2 + 6x + 18y + 26 = 0, is
(1) 1
(2) 2
(3) 3
(4) 4
(3)
Given circles are
x2 + y2 – 4x – 6y – 12 = 0
(1)
2
2
and x + y + 6x + 18y + 26 = 0
(2)
C1 = (2, 3); r1 = 4  9  12  5
9  81  26  64  8
C1C2 = 25  144  13
C1C2 = r1 + r2
Circles touch externally each other, hence number of common tangent are three.
C2 = (3, 9); r2 =
35
81.
Key
Sol.
The area (in sq. units ) of the quadrilateral formed by the tangents at the end points of the
x 2 y2
latera recta to the ellipse

 1 , is:
9
5
27
27
(1)
(2)18
(3)
(4)27
4
2
(4)
2x y
Eq. of tangent at (2, 5/3) is
 1
9 3
(0,3)
(2, 5/3)
(9/2, 0)
1 9
Area  4    3  27 sq.units
2 2
82.
Key
Sol.
Let O be the vertex and Q be any point on the parabola, x 2  8y . If the point P divides
the line segment OQ internally in the ratio 1:3, then the locus of P is:
(1) x 2  y
(2) y2  x
(3) y2  2x
(4) x 2  2y
(4)
Let P   h, k 
.t  3  0
h
t = 4h …(1)
4
t2
.  3  0
k 8
t2 = 32k …(2)
4
Eliminating t, we get
x 2  2y
83.
Key
2
x =8y
1:3
P
Q(t, t2 /8)
O
The distance of the point (1,0,2) from the point of intersection of the line
x  2 y 1 z  2


and the plane x-y+z=16, is:
3
4
12
(1) 2 14
(2)8
(3) 3 21
(4) 13
(4)
36
Sol.
Let point of intersection   3t  2, 4t  1,12t  2
  3t  2    4t  1  12t  2   16
t=1
Point of intersection (5,3,14)
distance 

84.
Key
Sol.
85.
Sol.
2
2
  3  0   14  2 

 

86.
2
=13
The equation of the plane containing the line 2x-5y+z=3; x+y+4z=5, and parallel to the
plane, x+3y+6z=1, is :
(1) 2x  6y  12z  13
(2) x  3y  6z  7
(3) x  3y  6z  7
(4) 2x  6y  12z  13
(3)
Required plane is  2x  5y  z  3    x  y  4z  5  0
Or
x  2     y    5  z  4  1   3  5   0 ……..(i)
x  3y  6z  1  0 ………..(ii)
As (i)and (ii) are parallel
2     5 4  1



1
3
6
11

2

Required plane is
x+3y+6z=7
 

Let a , b and c be three non-zero vectors such that no two of them are collinear and


   1  
a  b  c  b c a. If  is the angle between vectors b and c , then a value of sin  is:
3
2 2
 2
2 3
2
(1)
(2)
(3)
(4)
3
3
3
3
(1)
   
1  
c.b a   c.a  b   b c a
3




1 
 c.b   b c
and c.b  0
3
1

cos   
3
2 2
sin  
3
If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of
the boxes contains exactly 3 balls is :

Key
 5  1

37
11
(1)
Key
Sol.
87.
Key
10
55  2 
 
3 3
2
(2) 55  
3
12
1
(3) 220  
 3
11
1
(4) 22  
3
(1)
Required probability
12C3  29

312
11
12  11 10  29 55  2 

  
3  3
6  312
The mean of the data set comprising of 16 observations is 16.if one of the observation
valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data,
then the mean of the resultant data, is:
(1) 16.8
(2)16.0
(3)15.8
(4)14.0
(4)
16
 xi
Sol.
i 1
16
 16
16
 xi  256

i 1
16
252
=14
18
i 1
If the angles of elevation of the top of a tower from three collinear points A,B and C, on
a line leading to the foot of the tower, are 30 0 , 450 and 60 0 respectively, then the ratio is,
AB:BC, is :
(1) 3 :1
(2) 3 : 2
(3) 1: 3
(4) 2 : 3
(1)
x
tan 60o   x  y 3
…(1)
y
x
tan 45o 
 x  BC  y
…(2)
BC  y
x
tan 30o 
 x 3  AC  y
y  AC
 x i  16  3  4  5  252
88.
Key
Sol.
3x  y  AB  BC

….(3)
y 3  BC  y
BC  y



3 1
…(4)

3 y 3  y  AB  BC
38
N
x
30°
A
45°
AB  BC  2y

60°
y
C
B
M
…(5)

AB  2y  y 3  y  2y  3y  y
AB 2  3  1
3 1


 2 3  2  3  3  3 1  3
BC
3 1
3 1
89.
Key
Sol.
1
 2x 
Let tan 1 y  tan 1 x  tan 1 
,where x 
.Then a value of y is :
2 
3
1 x 
3x  x 3
3x  x 3
3x  x 3
3x  x 3
(1)
(2)
(3)
(4)
1  3x 2
1  3x 2
1  3x 2
1  3x 2
(1)
Let x  tan  , where    / 6,  / 6 
tan 1 y    2  3
3x  x 3


, where   3 
2
1  3x
2
2
The negation of ~s  (~r  s) is equivalent to
(1) s  ~r
(2)s  (r  ~s)
(4)
y  tan 3 
90.
Key
Sol.
r
T
T
F
F
(3)s  (r  ~s)
(4) s  r
s
~s
~r
~r  s
~s   ~r  s 
~ [~ s   ~ r  s 
sr
T
F
T
F
F
T
F
T
F
F
T
T
F
F
T
F
F
T
T
T
T
F
F
F
T
`F
F
F
39