SOLUTION : AIEEE 2011 (Booklet Code : Q)

PART A (PHYSICS)
1. The transverse displacement y(x,t) of a wave on a string is given by

2 2
2 ( )
( , )
ax bt ab xt
y x t e
+ +
=
This represents a :
(1) wave moving in x direction with speed
b
a


(2) standing wave of frequency b
(3) standing wave of frequency
1
b

(4) wave moving in

+ x direction with speed
a
b

Ans. (1)
Sol.
2
( )
( , )
a x bt
y x t e
+
= . This is a travelling wave in
x direction with speed of
b
a

2. A screw gauge gives the following reading when
used to measure the diameter of a wire.
Main Scale reading : 0 mm.
Circular Scale reading : 52 division
Given that 1mm on main scale corresponds to 100
division of the circular scale.
The diameter of wire from the above data is :
(1) 0.052 cm (2) 0.026 cm
(3) 0.005 cm (4) 0.52 cm
Ans. (1)
Sol. Least count of screw gauge =
1
0 01
100
. mm mm =
Diameter = Division on circular scale x least
count + main scale reading.

1
52 0
100
= + = 0.52 mm
Diameter = 0.052 cm
3. A mass of m hangs with the help of a string
wrapped around a pulley on a frictionless bearing.
The pulley has mass m and radius R. Assuming
pulley to be a perfect uniform circular disc, the
acceleration of the mass m, if the string does not
slip on the pulley, is:
(1) g (2)
2
3
g
(3)
3
g
(4)
3
2
g
Ans. (2)
Sol.
2
2
3
2
mg mg g
a
I m
m m
R
= = =
+ +

4. Work done in increasing the size of a soap bubble
from a radius of 3cm to 5cm is nearly (Surface
tension of soap solution = 0.03 Nm
1
) :
(1) 0.2t mJ (2) 2t mJ
(3) 0.4t mJ (4) 4t mJ
Ans. (3)
Sol. W T A = A =
2 2 4
0 03 2 4 5 3 10 . ( ) t


=
6
24 16 10 ( ) t

=
3
0 384 10 . t

Joule =
0 4 . mJ t
5. A thin horizontal circular disc is rotating about a
vertical axis passing through its centre. An insect
is at rest at a point near the rim of the disc. The
insect now moves along a diameter of the disc to
reach its other end. During the journey of the
insect, the angular speed of the disc:
(1) Continuously decreases
(2) Continuously increases
(3) First increases and then decreases
(4) Remains unchanged
Ans. (3)
Sol. L Ie = = Constant
If I first decreases, then increases.
Therefore e first increases and then decreases
6. Two particle are executing simple harmonic
motion of the same amplitude A and frequency
e along the x-axis. Their mean position is
separated by distance X
0
(X
0
> A). If the maximum
separation between them is (X
0
+ A), the phase
difference between their motion is:
(1)
3
t
(2)
4
t

(3)
6
t
(4)
2
t

Ans. (1)
Sol.
1 1
sin( ) x A t e | = +

2 2
sin( ) x A t e | = +

1 2 1 2
1 2
2
2 2
sin sin x x A t
| | | |
e
( + ( (
= +
( ( (



1 2
2
2
sin A A
| | | |
=
|
\ .


1 2
2 6
| | t
=

1
3
t
| =
7. Two bodies of masses m and 4 m are placed at a
distance r. The gravitational potential at a point
on the line joining them where the gravitational
field is zero is:
(1)
4Gm
r
(2)
6Gm
r

(3)
9Gm
r
(4) zero
Ans. (3)
Sol.
2 2
4 ( )
( )
Gm G m
x r x
=

1 2
x r x
=

, 2 r x x =
3x r = ,
3
r
x =
m 4m
r/3 2r/3
SOLUTION : AIEEE 2011 (Booklet Code : Q)

=
4
3 2 3
( )
/ /
Gm G m
r r

=
3 6 9 Gm Gm Gm
r r r
=
8. Two identical charged spheres suspended from a
common point by two massless strings of length l
are initially a distance d(d < < l) apart because of
their mutual repulsion. The charge begins to leak
from both the spheres at a constant rate. As a
result the charges approach each other with a
velocity v. Then as a function of distance x
between them,
(1) v x
1
(2) v x
1/2

(3) v x

(4) v x
1/2
Ans. (4)
Sol.


2
2
sin
kq
d
u = , cos mg u =

2
2
tan .
k q
mg x
u = ,
2
2
2
.
x k q
mg x
=
l


3 2
2k
x q
mg
=
l
, q
2
x
3
, q x
3/2


dq
dt

1 2
3
2
/
dx
x
dt
(dq/dt is constant)
C x
1/2
V, V x
1/2

9. A boat is moving due east in a region where the
earths magnetic field is
5
10 0 5

. NA
-1
m
1
due
north and horizontal. The boat carries a vertical
aerial 2m long. If the speed of the boat is 1.50 ms

1
, the magnitude of the induced emf in the wire of
aerial is:
(1) 0.75 mV (2) 0.50 mV
(3) 0.15 mV (4) 1 mV
Ans. (3)
Sol. Bvl c =
=
5
5 0 10 2 1 50 . .

=
5
15 10

= 0.15 mV

10. An object, moving with a speed of 6.25 m/s, is
decelerated at a rate given by:
2 5 .
dv
v
dt
=
Where v is the instantaneous speed. The time
taken by the object, to come to rest, would be:
(1) 2 s (2) 4 s
(3) 8 s (d) 1 s
Ans. (1)
Sol.
0
6 25 0
2 5
.
.
t
dv
dt
v
=
} }
,
0
6 25
2 2 5
.
. v t =

0
6 25
2 2 5
.
. v t = , 2 6 25 2 5 . . . t =
t = 2 sec.
11. A fully charged capacitor C with initial charge q
0

is connected to a coil of self inductance L at t = 0.
The time at which the energy is stored equally
between the electric and the magnetic field is:
(1)
4
LC
t
(2) 2 LC t
(3) LC (d) LC t
Ans. (1)
Sol. Since energy is halved when
0
2
q
q =

4
t
t
e =
1
4
t
L
t
=

4
t LC
t
=
12. Let the x z plane be the boundary between two
transparent media. Medium 1 in 0 z > has a
refractive index of 2 and medium 2 with z < 0
has a refractive index of 3 . A ray of light in
medium 1 given by the vector
6 3 8 3 10

A i j k = +
ur
is incident on the plane
of separation. The angle of refraction in medium 2
is:
(1) 45
0
(2) 60
0

(3) 75
0
(4) 30
0
Ans. (1)
Sol.
1 1 2
sin sin u u =

1
2 2
10 10 10
20
400
6 3 8 3 100
cos
( ) ( )
u = = =
+ +


0
1 1
1
60
2
cos , u u = =

0
2
2 60 3 sin sinu = ,
2
3
2 3
2
sinu =

2
1
2
sinu = ,
0
2
45 u =
13. A current I flows in an infinitely long wire with
cross section in the form of semicircular ring of
radius R. The magnitude of the magnetic
induction along its axis is :
(1)
0
2
2
I
R

t
(2)
0
2
I
R

t

(3)
0
4
I
R

t
(4)
0
2
I
R

t

Ans. (4)
Sol.
SOLUTION : AIEEE 2011 (Booklet Code : Q)


0
2
4
, ,
I I
dB I Rd
R R

u
t t
| |
= = =
|
\ .


2
2
/
/
cos B dB
t
t
u

=
}


2
0
2
2
/
/
cos d
t
t

u u
t

=
}
,
0 0
2
I
R

t t
= =
14. A thermally insulated vessel contains and ideal
gas of molecular mass M and ratio of specific
heats . It is moving with speed v and is suddenly
brought to rest. Assuming no heat is lost to the
surroundings, its temperature increased by:
(1)
2
1
2

( )
Mv
R
(2)
2
2
Mv
R

(3)
2
1
2
( )
Mv
R
(4)
2
1
2 1
( )
( )
Mv
R

+

Ans. (3)
Sol.
2
1
2
.
V
Mv C T = A ,
2
1
2 1
.
R
Mv T

= A

2 2
1 1
2 2
( ) ( ) mv Mv
T
R R

A = =
15. A mass M, attached to a horizontal spring,
executes S.H.M. with amplitude A
1
. When the
mass M passes through its mean position then a
smaller mass m is placed over it and both of them
move together with amplitude A
2
. The ratio of
1
2
A
A
| |
|
\ .
is:
(1)
M m
M
+
(2)
1
2
M
M m
| |
|
\ . +

(3)
1
2
M m
M
+ | |
|
\ .
(4)
M
M m +

Ans. (3)
Sol. C.O.L.M.
1 1 max max
( ) ,
new
MV m M V V Ae = + =

max
( )
new
MV
V
m M
=
+

Now,
2 2
.
new
V A e =

1
2
.
( ) ( )
M A K K
A
m M M m M
=
+ +


1 2
1
2 1
2
/
,
( )
A M m M
A A
m M A M
| |
= =
|
\ . +


16. Water is flowing continuously from a tap having
an internal diameter
3
8 10

m. The water velocity

below the
tap is close to:
(1)
3
7 5 10 . m

(2)
3
9 6 10 . m

(3)
3
3 6 10 . m

(3)
3
5 0 10 . m

Ans. (3)
Sol.
1 1 2 2
Av A v = ,
2 2
2 1
2 v v gh = + ,

2 2
2
0 4 2 10 0 2 ( . ) . v = +

2
0 4 4 ( . ) = + ,
2
2
4 16 . v = ,
2
4 16 . v =

2
2 04 . / v m s = - (i)
Also, A
1
V
1
= A
2
V
2


2 2
1 1 2 2
d V d V =

1
2 1
2
V
d d
V
= ,
3
0 4
8 10
2 04
.
.

=

3
3 54 10 . m

=
17. This equation has Statement-1 and Statement 2.
Of the four choices given after the statements,
choose the one that best describes the two
statements.
Statement-1 :
Sky wave signals are used for long distance radio
communication. These signals are in general, less
stable than ground wave signals.
Statement-2 :
The state of ionosphere varies from hour to hour,
day to day and season to season.
(1) Statement-1 is true, Statement2 is true,
Statement-2 is the correct explanation of
statement -1.
(2) Statement-1 is true, Statement -2 is true,
Statement-2 is not the correct explanation of
Statement -1
(3) Statement -1 is false, Statement -2 is true.
(4) Statement-1 is true, Statement -2 is false
Ans. (1)
Sol. Due to variation in composition of ionosphere,
the signals are unstable.
18. Three perfect gases at absolute temperatures T
1
,
T
2
and T
3
are mixed. The masses of molecules are
m
1
, m
2
and m
3
and the number of molecules are
n
1
, n
2
and n
3
respectively. Assuming no loss of
energy, the final temperature of the mixture is:
(1)
1 1 2 2 3 3
1 2 3
nT n T n T
n n n
+ +
+ +
(2)

2 2 2 2 2 2
1 1 2 2 3 3
1 1 2 2 3 3
n T n T n T
nT n T n T
+ +
+ +

SOLUTION : AIEEE 2011 (Booklet Code : Q)

(3)
2 2 2
1 1 2 2 3 3
1 1 2 2 3 3
nT n T n T
nT n T n T
+ +
+ +
(4)
( )
1 2 3
3
T T T + +

Ans. (1)
Sol. Final Temperature

3 1 2
1 2 3
3 1 2
n n n
T T T
N N N
T
n n n
N N N
+ +
=
+ +
=
1 1 2 2 3 3
1 1 2 2 3 3
+ +
+ +
nT n T n T
n T n T n T

19. A pulley of radius 2m is rotated about its axis by a
force F = (20t 5t
2
) Newton (where t is measured
in seconds) applied tangentially. If the moment of
inertia of the pulley about its axis of rotation of
the pulley about the axis of rotation is 10kgm
2
, the
number of rotations made by the pulley before its
direction of motion if reversed, is :
(1) more than 3 but less than 6
(2) more than 6 but less than 9
(3) more than 9 (4) less than 3
Ans. (1)
Sol. To reverse the direction 0 d t u =
}
(work done is
zero)

2 2
20 5 2 40 10 ( ) t t t t t = =

2
2
40 10
4
10
t t
t t
I
t
o

= = =

3
2
0
2
3
t
t
dt t e o = =
}
, e is zero at

3
2
2 0
3
t
t = , t = 6 sec.
dt u e =
}

3
6
2
0
2
3
( )
t
t dt =
}
6
3 4
0
2
36
3 12
t t (
= =
(

rad.
5 73 6
2
. n
u
t
= = <

20. A resistor R and 2F capacitor in series is
connected through a switch to 200V direct supply.
Across the capacitor is a neon bulb that lights up
at 120V. Calculate the value of R to make that
bulb light up 5s after the switch has been closed.
(log
10
2.5 = 0.4)
(1)
5
10 7 1 . (2)

6
10 7 2 .
(3)
7
10 3 3 . (4)

4
10 3 1 .
Ans. (2)
Sol. Charge on capacitor q = q
0
(1-e
-t/RC
)

q
V
c
= ,
0
1
/
( )
t RC
q
V e
c

= , 200 1
/
( )
t RC
V e

=
120 200 1
/
( )
t RC
e

= , 0 4
/
.
t RC
e

= ,
0 4 ( . )
t
In
RC

=

10
2 303 0 4
4
. .
t
In
RC
| |
= =
|
\ .


6
5
2 10 2 303 0 4 . .
R

=

, =
6
2 7 10 . O

21. A Carnot engine operating between temperatures
T
1
and T
2
has efficiency
6
1
. When T
2
is lowered
by 62K, its efficiency increases to
3
1
. Then T
1
and
T
2
are, respectively:
(1) 372 K and 330 K (2) 330 K and 268
K
(3) 310 K and 248 K (4) 372 K and 310
K
Ans. (4)
Sol.
2
1
1
1
6
T
T
q = =
2
1 5
1
1 6 6
T
T
= =

2
1
62 1
1
3
( ) T
T

=
2
1
62 2
3
T
T

=

2
5 62 2
3
( ) T
T

= 5T
2
310 = 4T
2

T
2
= 310 and
1
6 310
5
T

=
T
1
= 372K.
22. If a wire is stretched to make it 0.1% longer, its
resistance will:
(1) increase by 0.2% (2) decrease by
0.2%
(3) decrease by 0.05% (4) increase by
0.05%
Ans. (1)
Sol. R l
2


2 R l
R l
A A
=
23. Direction:
The question has a paragraph followed by two
statements, Statement -1 and Statement -2. Of the
given four alternatives after the statements,
choose the one that describes the statements

A thin air film is formed by putting the convex
surface of a plane-convex lens over a plane glass
plate. With monochromatic light, this film gives
an interference pattern due to light reflected from
the top (convex) surface and the bottom (glass
plate) surface of the film.
Statement 1:
When light reflects from the air-glass plate
interface, the reflected wave suffers a phase
change of t .
Statement 2:
The centre of the interference pattern is dark.
SOLUTION : AIEEE 2011 (Booklet Code : Q)

(1) Statement-1 is true, Statement 2 is true,
Stateent-2 is the correct explanation of
statement -1.
(2) Statement-1 is true, Statement -2 is true,
Statement-2 is not the correct explanation of
Statement -1
(3) Statement -1 is false, Statement -2 is true.
(4) Statement-1 is true, Statement -2 is false.
Ans. (2)
Sol. The centre of interference pattern is dark,
showing that the phase difference between two
interfering waves is t . But this does not imply
statement-1.

24. A car is fitted with a convex side view mirror of
focal length 20cm. A second car 2.8m behind the
first car is overtaking the first car at a relative
speed of 15 m/s. The speed of the image of the
second car as seen in the mirror of the first one is
(1)
1
15
/ m s (2) 10 m/s
(3) 15 m/s (4)
1
10
/ m s
Ans. (1)
Sol. From mirror formula :

1 1 1
v u f
+ = ,
2 2
1 1
0
dv du
dt dt v u

=

2
2
2
dv v du f du
dt dt f u dt u
| |
= =
|
\ .

=
2
20
20 280
du
dt
| |

|
\ . +
=
2
1 1
15 15
/
du
m s
dt
| |
=
|
\ .

25. Energy required for the electron excitation in Li
++

from the first to the third Bohr orbit is :
(1) 36.3 eV (2) 108.8 eV
(3) 122.4 eV (4)` 12.1 eV
Ans. (2)
Sol. Every required

2
2 2
1 1
13 6 3
3 1
. ( ) E
(
=
(

=
8
13 6 9
9
. = 108.8
eV
26. The electrostatic potential inside a charged
spherical ball is given by b ar + =
2
where r is
the distance from the centre; a, b are constant.
Then the charge density inside the ball is:
(1)
0
6a r c (2)
0
24 a t c
(3)
0
6ac (4)
0
24 a r t c
Ans. (3)
Sol. 2
d
E ar
dt
|
= = ,
0
.
q
E dS
c
=
}
uuv

3
0
8 q a r c t = ,
0
3
6
4
3
,
q
a
r
c
t
= =
27. A water fountain on the ground sprinkles water all
around it. If the speed of water coming out of the
fountain is v, the total area around the fountain
that gets wet is:
(1)
4
2
v
g
t (2)
4
2
2
v
g
t

(3)
2
2
v
g
t (4)
2
v
g
t
Ans. (1)
Sol.
2 2
2
max
sin
v v
R
g g
u = =
area =
2
R t
4
2
v
g
t =
28. 100g of water is heated from 30
0
C to 50
0
C.
Ignoring the slight expansion of the water, the
change in its internal energy is (specific heat of
water is 4184J / K g / K):
(1) 8.4 kJ (2) 84 kJ
(3) 2.1 kJ (4) 4.2 kJ
Ans. (1)
Sol. , , Q M S T A = A
=
3 3
100 10 4 184 20 8 1 10 . .

=
84 0 , Q kJ W A = A =
8 4 . Q kJ A =
29. The half life of a radioactive substance is 20
minutes. The approximate time interval (t
2
t
1
)
between the time t
2
when
3
2
of its has decayed
and time t
1
when
3
1
of it has decayed is:
(1) 14 min (2) 20 min
(3) 28 min (4) 7 min
Ans. (2)
Sol.
1
0 0
2
3
t
N N e

= ,
2
0 0
1
3
t
N N e

= ,
2 1
2
( ) t t
e

=

2 1
2 ( ) t t n l
2 1
2
20 ( )
n
t t

=
l
min.
30. This equation has Statement-1 and Statement2.
Of the four choices given after the statements,
chose the one that best describes the two
statements.
Statement 1:
A metallic surface is irradiated by a
monochromatic light of frequency v > v
0
(the
threshold frequency). The maximum kinetic
energy and the stopping potential are K
max
and V
0

respectively. If the frequency incident on the
surface is doubled, both the K
max
and V
0
are also
doubled.
Statement 2 :
The maximum kinetic energy and the stopping
potential of photoelectrons emitted from a surface
are linearly depended on the frequency of incident
light.
SOLUTION : AIEEE 2011 (Booklet Code : Q)

(1) Statement-1 is true, Statement 2 is true,
Stateent-2 is the correct explanation of
statement -1.
(2) Statement-1 is true, Statement -2 is true,
Statement-2 is not the correct explanation of
Statement -1
(3) Statement -1 is false, Statement -2 is true.
(4) Statement-1 is true, Statement -2 is false.
Ans. (3)
Sol.
max
k hf | = , 2
max
'
k hf | =
2
max
'
max
k k | = + 2
max
'
max
k k >

PART- B (MATHEMATICS)
31. The lines L
1
: y x = 0 and L
2
: 2x + y = 0
intersect the line L
3
: y + 2 = 0 at P and Q
respectively. The bisector of the acute angle
between L
1
and L
2
intersects L
3
at R.
Statement 1 : The ratio PR : RQ equals 2 2 : 5
Statement 2 : In any triangle, bisector of an angle
divides the triangle into two similar triangles.
(1) Statement-1 is true, Statement-2 is true;
Statement-2 is not a correct explanation for
Statement -1.
(2) Statement-1 is true, Statement-2 is false.
(3) Statement-1 is false, Statement-2 is true.
(4) Statement-1 is true, Statement-2 is true;
Statement-2 is a correct explanation for
Statement-1.
Ans. (2)
Sol. The figure is self explanatory

2 2
5
PR OP
RQ OQ
= =
Statement-2 is false
32. If
2 4
sin cos A x x = + , then for all real x :
(1)
13
1
16
A s s (2) 1 2 A s s
(3)
3 13
4 16
A s s (4)
3
1
4
A s s
Ans. (4)
Sol. A = sin
2
x + cos
4
x = sin
2
x + (1- sin
2
x)
2

= sin
4
x - sin
2
x + 1 =
2
2
1 3
sin
2 4
x
| |
+
|
\ .

3
1
4
A s s
33. The coefficient of x
7
in the expansion of (1-x-
x
2
+x
3
)
6
is:
(1) -132 (2) -144 (3) 132 (4) 144
Ans. (2 )
Sol. Coeff. of x
7
in
( )
6
2 3
1 x x x +
Coeff. of x
7
in ( ) ( )
6
6
2
1 1 x x
=
6 6 6 6 6 6
1 3 3 2 5 1
. . . C C C C C C + = -144
34.
( ) { }
2
1 cos 2 2
lim
2
x
x
x

| |

|
|
\ .
:
(1) equals 2 (2) equals - 2
(3) equals
1
2
(4) does not exist
Ans. (4)
Sol.

( )
2
1 cos 2 2
lim
2
x
x
x

| |

|
|
\ .
=
( )
2
2 sin 2
lim
2
x
x
x

| |
|

\ .

LHL = 2 ; RHL = 2 ; LHL = RHL
Limit doesnt Exist.
35. Statement-1 : The number of ways of distributing
10 identical balls in 4 distinct boxes such that no
box is empty is
9
3
C .
Statement 2 : The number of ways of choosing
any 3 places from 9 different places is
9
3
C .
(1) Statement -1 is true, Statement -2 is true;
Statement-2 is not a correct explanation for
Statement-1.
(2) Statement-1 is true, Statement-2 is false.
(3) Statement-1 is false, Statement-2 is true.
(4) Statement-1 is true, Statement-2 is true;
Statement-2 is a correct explanation for
Statement-1.
Ans. (1)
Sol. Number of ways to distribute 10 identical balls in
four distinct boxes such that no box remains
empty =
101
C
41
=
9
C
3

Number of ways to select 3 different places from
9 places =
9
C
3
Clearly statement-2 is not a correct explanation of
statement-1
36.
2
2
d x
equals
dy
:
(1)
1
3
2
2
d y dy
dx dx

| |
| |

|
|
\ .
\ .
(2)
2
2
2
d y dy
dx dx

| |
| |
|
|
\ .
\ .

(3)
3
2
2
d y dy
dx dx

| |
| |

|
|
\ .
\ .
(4)
1
2
2
d y
dx

| |
|
\ .

Ans. (1)
Sol.
2
2
1 1 d x d dx d d dx
dy dy
dy dy dy dy dx dy
dx dx
| | | |
| | | |
= = =
| | |
\ .
| |
\ . \ .


3
2 2 2
2 3 2 2 2
1 1 1
. . . .
d y d y dy d y
dy dx dx dx dx dy dy
dx
dx dx

| | | |
= =
| |
\ . | | \ .
| | | |
|
| |
\ .
\ . \ .

SOLUTION : AIEEE 2011 (Booklet Code : Q)

37. If 3 0
dy
y
dx
= + > and ( ) 0 2 y = , then ( ) ln 2 y =
(1) 5 (2) 13 (3) -2 (4) 7
Ans. ( 4 )

( ) 3 log 3
3
dy dy
y dx y x c
dx y
= + = + = +
+
3
x
y ce + =
0, 2 5 x y c = = = ( )
log 2
log2 5 3 7 y e = =
38. Let R be the set of real numbers.
Statement-1 :
( ) { } A , x : integer x y R R y x is an = e is an
equivalence relation on R.
Statement-2 :
( ) { B , x : x y R R x y for for o = e = some rational
number } o is an equivalence relation on R.
(1) Statement-1 is true, Statement-2 is true;
Statement-2 is not a correct explanation for
Statement-1.
(2) Statement-1 is true, Statement-2 is false.
(3) Statement-1 is false, Statement-2 is true.
(4) Statement-1 is true, Statement-2 is true;
Statement-2 is a correct explanation for
Statement-1.
Ans. (1)
Sol. Statement 1 is true but Statement-2 is false
because it is not symmetric.
39. The value of
( )
1
2
0
8log 1
1
x
dx
x
+
+
}
is :
(1) log 2
8
t
(2) log 2
2
t

(3) log2 (4) log 2 t
Ans. (4)
Sol.
( )
1
2
0
8log 1
1
x
I dx
x
+
=
+
}
, Put tan x u =

( )
/ 4
2
2
0
log 1 tan
8 sec
sec
I d
t
u
u u
u
+
=
}
( )
/ 4
0
8 log 1 tan d
t
u u = +
}


/ 4
0
1 tan
8 log 1
1 tan
d
t
u
u
u
| |
= +
|
\ . +
}
=
/ 4
0
2
8 log
1 tan
d
t
u
u
| |
=
|
\ . +
}

( ) 8 log2 .
4
I
t
= 2 2 log2 log2 I I t t = =
40. Let , o | be real and z be a complex number. If
2
0 z z o | + + = has two distinct roots on the line
Re z = 1, then it is necessary that :
(1) ( ) 1, 0 | e (2) 1 | =
(3) ( ) 1, | e (4) ( ) 0,1 | e
Ans. ( 3)
Sol. Let roots are p iq + and p iq ; where , p q R e .
As root lie on line Re(z) = 1 1 p = . Now product
of roots =
2 2 2
1 p q q | + = = + ( 1 > ) ( ) 1, | e
41. Consider 5 independent Bernoullis trials each
with probability of success p. If the probability of
at least one failure is greater than or equal to
31
32
,
then p lies in the interval :
(1)
3 11
,
4 12
| (

(
\

(b)
1
0,
2
(
(


(3)
11
,1
12
| (

(
\

(4)
1 3
,
2 4
| (

(
\


Ans. ( 2)
Sol. Probability of atleast one failure
= 1 prob of no failure.

5 5
31 1 1
1
32 32 2
p p p > s s
As probability > 0; Hence
1
0,
2
p
(
e
(


42. A man saves Rs. 200 in each of the first three
months of his service. In each of the subsequent
months his saving increases by Rs. 40 more than
the saving of immediately previous month. His
total saving from the start of service will be Rs.
11040 after :
(1) 19 months (2) 20 months
(3) 21 months (4) 18 months
Ans. (3)
Sol. Total saving = 200 + 200 + 200 + 240 + 280 +
to n months = 11040
( ) ( )
2
400 400 3 .40 11040
2
n
n

+ + =

2
5 546 0 21 26 n n n as n + = = =
43. The domain of the function ( )
1
f x
x x
=

is :
(1) ( ) 0, (2) ( ) , 0
(3) ( ) { } , 0 (4) ( ) ,
Ans. (2)
Sol. For ( ) f x to be defined 0 0 x x x > <
44. If the angle between the line
1 3
2
y z
x

= = and
the plane 2 3 4 x y z + + = is
1
5
cos
14

| |
|
\ .
, then
equals :
(1)
3
2
(b)
2
5
(3)
5
3
(4)
2
3

Ans. (4 )
Sol. The angle is
1
3
sin
14

( )( )
( ) ( )
2
2
2
1 4 3 3
14 3 5 9 x 14 5
14
1 4 1 4 9

+ +
= + = +
+ + + +

2 2
2
9 30 25 9 45
3
+ + = + =
SOLUTION : AIEEE 2011 (Booklet Code : Q)

45. If
( )
1

3
10
a i k = +
r
and
( )
1

2 3 6
7
b i j k = +
r
, then
the value of
( ) ( ) ( )
2 x x 2 a b a b a b
(
- +

r r r
r r r
is :
(1) -3 (2) 5 (3) 3 (4) -5
Ans. ( 4)
Sol.
( ) ( ) ( ) ( ) ( ) ( )
2
2 . x x 2 2 . 2 2 a b a b a b a b b a a b
(
+ = =

r r r r r r
r r r r r r
= 5
46. Equation of the ellipse whose axes are the axes of
coordinates and which passes through the point (-
3, 1) and has eccentricity
2
5
is :
(1)
2 2
5 3 48 0 x y + = (2)
2 2
3 5 15 0 x y + =
(3)
2 2
5 3 32 0 x y + = (4)
2 2
3 5 32 0 x y + =
Ans. (1,4)
Sol.
2 2
2 2 2 2
9 1
1 1..................(1)
x y
a b a b
+ = + =
Case-1 (when a > b)
( ) ( )
2 2 2 2 2 2 2
1 1 2/ 5 5 3 ................( ) b a e b a b a ii = = =

From (i) and (ii) :
2 2
32 32
&
5 3
b a = =
Hence : equation of ellipse is
2 2
3 5 32 0 x y + =
Case 2 : (when b > a)
Equation of ellipse is
2 2
5 3 48 0 x y + =
47. Let I be the purchase value of an equipment and
V(t) be the value after it has been used for t years.
The value V(t) depreciates at a rate given by
differential equation
( )
( )
dV t
k T t
dt
= , where k
> 0 is a constant and T is the total life in years of
the equipment. Then the scrap value V(T) of the
equipment is :
(1)
2
I
2
kT
(2)
( )
2
I
2
k T t

(3)
kT
e

(4)
2
I
T
k

Ans. (1 )
Sol. ( )
( )
( ) ( )
( )
2
0
0
2
T
V T T
I t
T t
dV t k T t dt V T I k
=
(

= = (
(

} }

( ) ( )
2 2
2 2
T kT
V T I k V T I
(
= =
(


48. The vectors a and b
r
r
are not perpendicular and
c and d
r
r
are two vectors satisfying :
x x b c b d =
r r r
r
and 0 a d =
r
r
g . Then the vector d
r

is equal to :
(1)
a c
c b
a b
| |

|
\ .
r r
r
g r
r
r
g
(2)
b c
b c
a b
| |

|
\ .
r
r
r
g r
r
r
g

(3)
a c
c b
a b
| |
+
|
\ .
r r
r
g r
r
r
g
(4)
b c
b c
a b
| |
+
|
\ .
r
r
r
g r
r
r
g

Ans. (1)
Sol.
( ) ( )
x x x x x x b c b d a b c a b d = =
r r r r r r
r r r r

( ) ( ) ( ) ( )
. . . . a c b a b c a d b a b d =
r r r r r r
r r r r r r
( ) ( ) ( )
( )
( )
.
. . .
.
a c
a b d a c b a b c d b c
a b
= + = +
r r
r r r r r r
r r r r r r
r
r

49. The two circles
2 2
x y ax + = and
( )
2 2 2
0 x y c c + = > touch each other if :
(1) a c = (2) 2 a c =
(3) 2 a c = (4) 2 a c =
Ans. (1)
Sol. ( )
2 2
...... 1 x y ax + =

1 1
, 0
2 2
a a
centre c and radius r
| |
=
|
\ .

( )
2 2 2
............. 2 x y c + =
( )
2 2
0, 0 centre c and radius r c =
Now
1 2 1 2
c c r r a c = =
50. If C and D are two events such that C D c and
( ) 0 P D = , then the correct statement among the
following is :
(1) ( ) ( ) | P C D P C > (2) ( ) ( ) | P C D P C <
(3) ( )
( )
( )
|
P D
P C D
P C
= (4) ( ) ( ) | P C D P C =
Ans. (1)
Sol. We have C D c C D C =

( ) ( )
( )
( ) ( )
C P C D P C
P P C
D P D P D
| |
= = >
|
\ .

As 0 1 ( ) P D < s
51. The number of values of k for which the linear
equations : 4 2 0 x ky z + + = , 4 0 kx y z + + = ,
2 2 0 x y z + + = . Possess a non-zero solution is :
(1) 2 (2) 1 (3) zero (4) 3
Ans. (1)
Sol. For non-trivial solution of given system of linear
equations

4 2
4 1 0
2 2 1
k
k = 8 2 2 2 8 0 ( ) ( ) k k k + + =

2
6 8 0 k k + =
2
6 8 0 k k + = 2 4 , k =
52. Consider the following statements :
P : Suman is brilliant; Q : Suman is rich
R : Suman is honest
The negation of the statement Suman is brilliant and
dishonest if and only if Suman is rich can be
expressed as
(1) ( ) ( ) Q P R . : : (2) Q P R . : :
SOLUTION : AIEEE 2011 (Booklet Code : Q)

(3) ( ) P R Q . : : (4) ( ) P Q R . : :
Ans. (1)
Sol. Suman is brilliant and dishonest is P. ~ R
Suman is brilliant and dishonest if Suman is rich
is ( ~ ) Q P R .
Negative of statement is expressed as
~ ( ( ~ )) Q P R .
53. The shortest distance between line 1 y x = and
curve
2
x y = is :
(1)
3 2
8
(2)
8
3 2

(3)
4
3
(4)
3
4

Ans. (1)
Sol. The equation of the tangent to x = y
2
having
slope 1 is
1
4
y x = +
Hence shortest distance =
1
1
4
2

=
3 3 2
8
4 2
=
units.

54. If the mean deviation about the median of the
numbers a, 2a, .,50a is 50, then a equals :
(1) 3 (2) 4 (3) 5 (4) 2
Ans. (2)
Sol. Median = 25.5 a; Mean deviation about median
= 50
1
25.5
24.5 23.5 ..... 0.5 0.5 ... 24.5 2500
50
x a
a a a a a

+ + + + + + =

( )
25
3 5 ..... 49 2500 50 2500 4
2
a a a a a a + + + + = = =

55. Statement 1 : The point A(1, 0, 7) is the mirror
image of the point B(1, 6, 3) in the line
1 2
1 2 3
x y z
= = .
Statement 2 : The line :
1 2
1 2 3
x y z
= =
bisects the line segment joining ( ) A 1, 0, 7 and
( ) B 1, 6, 3 .
(1) Statement 1 is true, Statement-2 is true;
Statement-2 is not a correct explanation for
Statement-1.
(2) Statement-1 is true, Statement-2 is false.
(3) Statement-1 is false, Statement-2 is true.
(4) Statement-1 is true, Statement-2 is true;
Statement-2 is a correct explanation for
Statement-1.
Ans. (1)
Sol. The mid point of A(1,0,7) and B(1,6,3) is (1,3,5)
which lies on the line
1 2
1 2 3
x y z
= =
Also the line passing through the points A and B
is perpendicular to the given line, hence B is the
mirror image of A, consequently the statement -1
is true.
Statement -2 is also true but it is not a correct
explanation of statement -1 as there are infinitely
many lines passing through the midpoint of the
line segment and one of the lines is perpendicular
bisector.
56. Let A and B be two symmetric matrices of order
3.
Statement-1 : A(BA) and (AB)A are symmetric
matrices.
Statement-2 : AB is symmetric matrix if matrix
multiplication of A with B is commutative.
(1) Statement-1 is true, Statement-2 is true;
Statement-2 is not a correct explanation for
Statement-1.
(2) Statement-1 is true, Statement-2 is false.
(3) Statement-1 is false; Statement-2 is true.
(4) Statement-1 is true, Statement-2 is true;
Statement-2 is a correct explanation for
Statement-1.
Ans. (1)
Sol. if AB = BA (AB)
T
= A
T
B
T
AB is symmetric
Statement-2 is true (ABA)
T
= A
T
B
T
A
T

Take A = I and B = some non
symmetric
ABA always A(BA) and (AB)A are
symmetric
Statement-1 is true nut does not
depend
on Statement-2
57. If ( ) 1 e = is a cube root of unity, and
( )
7
1 A B e e + = + . Then (A, B) equals :
(1) (1, 1) (2) (1, 0)
(3) (-1, 1) (4) (0, 1)
Ans. (1)
Sol.
7
1 ( ) A B e e + = +
2 7
( ) A B e e = +

2
A B e e = + 1 A B e e + = +
1 1 , A B = = 1 1 ( , ) ( , ) A B =
58. The values of p and q for which the function
( )
( )
2
3/ 2
sin 1 sin
, 0
, 0
, 0
p x x
x
x
f x q x
x x x
x
x

+ +
<

= =

>

Is continuous for all x is R, are :
(1)
5 1
,
2 2
p q = = (2)
3 1
,
2 2
p q = =
(3)
1 3
,
2 2
p q = = (4)
1 3
,
2 2
p q = =
SOLUTION : AIEEE 2011 (Booklet Code : Q)

Ans. (2)
Sol. The given function f is continuous at x = 0 if

0 0
0 0 0

= = + lim ( ) ( ) lim ( )
h h
f h f f h

1
2
2
p q + = =
3 1
2 2
, p q = =

59. The area of the region enclosed by the curves
1
, , y x x e y
x
= = = and the positive x axis is :
(1) 1 square units (2)
3
2
square units
(3)
5
2
square units (4)
1
2
square units
Ans. (2)
Sol. Required area = OAB + ACDB

1
1 1
1
2
e
dx
x
= +
}

1
1
2
( )
e
nx = + l
3
2
= square
units.
60. For
5
0,
2
x
t | |
e
|
\ .
, define ( )
0
sin dt
x
f x t t =
}
. Then
( ) f x has :
(1) local minimum at t and 2t .
(2) local minimum at t and local maximum at
2t .
(3) local maximum at t and local minimum at
2t .
(4) local maximum at t and 2t .
Ans. (3)
Sol.
0
( ) sin
x
f x t tdt =
}
'( ) sin f x x x =
local maximum at t and local minimum at 2.
54. If the mean deviation about the median of the
numbers a, 2a, .,50a is 50, then a equals :
(1) 3 (2) 4 (3) 5 (4) 2
Ans. (2)
Sol. A = Rs.200 d = Rs.40
Savings in first two monts = Rs.400
Remained savings = 200 + 240 + 280 + .. upto n
terms

| | 400 1 40 11040 400
2
( )
n
n = + =
200n + 20n
2
20n 10640 = 0
20n
2
+ 180n 10640 = 0
n
2
+ 9n 532 = 0, (n+28 )(n-19) =0
n = 19
55. Statement 1 : The point A(1, 0, 7) is the mirror
image of the point B(1, 6, 3) in the line
1 2
1 2 3
x y z
= = .
Statement 2 : The line :
1 2
1 2 3
x y z
= =
bisects the line segment joining ( ) A 1, 0, 7 and
( ) B 1, 6, 3 .
(5) Statement 1 is true, Statement-2 is true;
Statement-2 is not a correct explanation for
Statement-1.
(6) Statement-1 is true, Statement-2 is false.
(7) Statement-1 is false, Statement-2 is true.
(8) Statement-1 is true, Statement-2 is true;
Statement-2 is a correct explanation for
Statement-1.
Ans. (1)
Sol. The mid point of A(1,0,7) and B(1,6,3) is (1,3,5)
which lies on the line
1 2
1 2 3
x y z
= =
Also the line passing through the points A and B
is perpendicular to the given line, hence B is the
mirror image of A, consequently the statement -1
is true.
Statement -2 is also true but it is not a correct
explanation of statement -1 as there are infinitely
many lines passing through the midpoint of the
line segment and one of the lines is perpendicular
bisector.
56. Let A and B be two symmetric matrices of order
3.
Statement-1 : A(BA) and (AB)A are symmetric
matrices.
Statement-2 : AB is symmetric matrix if matrix
multiplication of A with B is commutative.
(1) Statement-1 is true, Statement-2 is true;
Statement-2 is not a correct explanation for
Statement-1.
(2) Statement-1 is true, Statement-2 is false.
(3) Statement-1 is false; Statement-2 is true.
(4) Statement-1 is true, Statement-2 is true;
Statement-2 is a correct explanation for
Statement-1.
Ans. ( 1 )
Sol. if AB = BA (AB)
T
= A
T
B
T
AB is
symmetric
Statement-2 is true (ABA)
T
= A
T
B
T
A
T

Take A = I and B = some non
symmetric ABA always
A(BA) and (AB)A are symmetric
Statement-1 is true nut does not
depend on Statement-2
57. If ( ) 1 e = is a cube root of unity, and
( )
7
1 A B e e + = + . Then (A, B) equals :
(1) (1, 1) (2) (1, 0)
(3) (-1, 1) (4) (0, 1)
Ans. (1)
Sol.
7
1 ( ) A B e e + = +
2 7
( ) A B e e = +

2
A B e e = + 1 A B e e + = +
1 1 , A B = = 1 1 ( , ) ( , ) A B =
58. The values of p and q for which the function
SOLUTION : AIEEE 2011 (Booklet Code : Q)

( )
( )
2
3/ 2
sin 1 sin
, 0
, 0
, 0
p x x
x
x
f x q x
x x x
x
x

+ +
<

= =

>

Is continuous for all x is R, are :
(1)
5 1
,
2 2
p q = = (2)
3 1
,
2 2
p q = =
(3)
1 3
,
2 2
p q = = (4)
1 3
,
2 2
p q = =
Ans. (2)
Sol. The given function f is continuous at x = 0 if

0 0
0 0 0 lim ( ) ( )lim ( )
h h
f h f f h

= +

1
2
2
p q + = =
3 1
2 2
, p q = =
59. The area of the region enclosed by the curves
1
, , y x x e y
x
= = = and the positive x axis is :
(1) 1 square units (2)
3
2
square units
(3)
5
2
square units (4)
1
2
square units
Ans. ( 2)
Sol. Required area = OAB + ACDB

1
1 1
1
2
e
dx
x
= +
}

1
1
2
( )
e
nx = + l
2
2
= square
unit.
60. For
5
0,
2
x
t | |
e
|
\ .
, define ( )
0
sin dt
x
f x t t =
}
. Then
f has :
(1) local minimum at t and 2t .
(2) local minimum at t and local maximum at
2t .
(3) local maximum at t and local minimum at
2t .
(4) local maximum at t and 2t .
Ans. (3)
Sol.
0
( ) sin
x
f x t tdt =
}
; '( ) sin f x x x =
local maximum at t and local minimum at 2.

PART- C (CHEMISTRY)
61. Among the following the maximum covalent
character is shown by the compound :
(1) SnCl
2
(2) AlCl
3
(3) MgCl
2
(4) FeCl
2

Ans. ( 2 )
Sol. According to Fajans rule, cation with greater
charge and smaller size favours covalency.
62. The presence or absence of hydroxyl group on
which carbon atom of sugar differentiates RNA and
DNA ?
(1) 2
nd
(2) 3
rd
(3) 4
th
(4) 1
st
Ans. ( 1 )
Sol. RNA contains D ribose while DNA
contains D2deoxyribose.
63. Trichloroacetaldehyde was subjected to
Cannizzaros reaction by using NaOH. The mixture
of the products contains sodium trichloroacetate and
another compound. The other compound is :
(1) Trichloromethanol (2) 2,2,2-
Trichloropropanol
(3) Chloroform (4) 2,2,2-
Trichloroethanol
Ans. (4)
Sol. In connizzaros reaction CCl
3
CHO undergoes
disproportionation reaction as :

64. Sodium ethoxide has reacted with ethanoyl chloride.
The compound that is produced in the above
reaction is :
(1) 2-Butanone (2) Ethyl chloride
(2) Ethyl ethanoate (4) Diethyl ether
Ans. ( 3 )
Sol.

65. The reduction potential of hydrogen half-cell will be
negative if :
(1) p(H
2
) = 1 atm and [H
+
] = 1.0 M
(2) p(H
2
) = 2 atm and [H
+
] = 1.0 M
(3) p(H
2
) = 2 atm and [H
+
] = 2.0 M
(4) p(H
2
) = 1 atm and [H
+
] = 2.0 M
Ans. ( 2 )
Sol.
SOLUTION : AIEEE 2011 (Booklet Code : Q)

Therefore E is (-
)ve
66. The strongest acid amongst the following
compounds is :
(1) HCOOH (2)
CH
3
CH
2
CH(Cl)CO
2
H
(3) ClCH
2
CH
2
CH
2
COOH (4) CH
3
COOH
Ans. ( 2)
Sol.

Presence of electron withdrawing group nearest to the
carboxylic group increase the acidic strength to maximum
extent
67. The degree of dissociation ( ) o of a weak
electrolyte, A B
x y
is related to Vant Hoff factor (i)
by the expression:
(1)
1
1
i
x y
o

=
+ +
(2)
1
1
x y
i
o
+
=

(3)
1
1
x y
i
o
+ +
=

(4)
( )
1
1
i
x y
o

=
+

Ans. (4)
Sol. Vant Hoff factor (i)
Observed colligative property
Normal colligative property
=

1
y x
x y
x
y
A B xA yB
o
o
o
+

+
Total moles after dissociation = 1 x y o o o + +

( ) 1 1
1
x y
i
o + +
=
1
1
i
x y
o

=
+

68. a and b are van der Waals constants for gases.
Chlorine is more easily liquefied than ethane
because
(1) a and b for Cl
2
< a and b for C
2
H
6

(2) a for Cl
2
< a and C
2
H
6
but b for Cl
2
> b for C
2
H
6

(3) a for Cl
2
> a and C
2
H
6
but b for Cl
2
< b for
C
2
H
6

(4) a and b for Cl
2
> a and b for C
2
H
6

Ans. (3)
Sol. a is a measure of attraction between the molecules
and b is the size of the molecules.
69. A vessel at 1000K contains CO
2
with a pressure of
0.5 atm. Some of the CO
2
is converted into CO on
the addition of graphite. If the total pressure at
equilibrium is 0.8atm, the value of K is :
(1) 3 atm (2) 0.3 atm
(3) 0.18 atm (4) 1.8 atm
Ans. (4)
Sol. ( ) ( ) ( )
2
0.5
2
atm
CO g C s CO g +
Total pressure = 0.5-P+2P=0.8; P = 0.3

( )
( )
( )
( )
2
2 2
2
2 0.6
1.8
0.5 0.5 0.3
CO
p
CO
P P
K
P P
= = =


70. Boron cannot form which one of the following
anions ?
(1)
4
BH

(2) ( )
4
B OH


(3)
2
BO

(4)
3
6
BF


Ans. (4)
Sol. Due to lack of vacant d orbital in B. sp
3
d
2
hybridization
is not possible hence BF6
3-
will not form.
71. Which of the following facts about the complex
[Cr(NH
3
)
6
]Cl
3
is wrong ?
(1) The complex is paramagnetic
(2) The complex is an outer orbital complex
(3) The complex gives white precipitate with silver
nitrate solution
(4) The complex involves d
2
sp
3
hybridisation and is
octahedral in shape.
Ans. (2)
Sol. Cr
+3
in octahedral geometry always form inner orbital
complex
72. Ethylene glycol is used as an antifreeze in a cold
climate. Mass of ethylene glycol which should be
added to 4 kg of water to prevent it from freezing at
-6
0
C will be : (K
f
for water = 1.86 K kg mol
-1
, and
molar mass of ethylene glycol = 62g mol
-1
).
(1) 204.30 g (2) 400.00g
(3) 304.60g (4) 804.32g
Ans. (4 )
Sol.

1000 1.86 x x 1000
6
. . . 62 x 4000
f f
w w
T K x
mol wt wt of solvent
A = =

800 w g = ; So, weight of solute should be more than
800g.
73. Which one of the following orders presents the
correct sequence of the increasing basic nature of the
given oxides ?
(1) MgO < K
2
O < Al
2
O
3
< Na
2
O
(2) Na
2
O < K
2
O < MgO < Al
2
O
3

(3) K
2
O < Na
2
O < Al
2
O
3
< MgO
(4) Al
2
O
3
<MgO < Na
2
O < K
2
O
Ans. (4)
Sol. As metallic character of element attached to oxygen atom
increases, the difference between the electronegativity
values of element and oxygen increases and thus basic
character of oxides increases and vice-versa. Hence the
increasing correct order of basic nature is Al2O3 < MgO
< Na2O < K2O.
74. The rate of a chemical reaction doubles for every
10
0
C rise of temperature. If the temperature is raised
by 50
0
C, the rate of the reaction increases by about :
(1) 24 times (2) 32 times
SOLUTION : AIEEE 2011 (Booklet Code : Q)

(3) 64 times (4) 10 times
Ans. (2)
Sol. ( ) ( ) 1
50
5
10
Rate at 50C
2 2 2 32
Rate at T C
T
T
times
A
= = = =
75. The magnetic moment (spin only) of
| |
2
4
NiCl

is :
(1) 5.46 BM (2) 2.82 BM
(3) 1.41 BM (4) 1.82 BM
Ans. (2)
Sol. Hybridisation of Ni is sp
3

Unpaired e

in 3d
8
is 2
So, ( ) 2 n n BM = + 2 x 4 8 2.82BM = = =
76. The hybridization of orbitals of N atom in
3 2 4
NO , NO NH and
+ +
are respectively :
(1)
2 3
, , sp sp sp (2)
3 2
, , sp sp sp
(3)
2 3
, , sp sp sp (4)
2 3
, , sp sp sp
Ans. (1)
Sol.
2 3
3 2 4
; ; NO sp NO sp NH sp
+ +

77. In context of the lanthanoids, which of the following
statements is not correct ?
(1) All the members exhibit +3 oxidation state
(2) Because of similar properties the separation of
lanthanoids is not easy.
(3) Availability of 4f electrons results in the
formation of compounds in +4 state for all the
members of the series.
(4) There is a gradual decrease in the radii of the
members with increasing atomic number in the
series
Ans. (3)
Sol. All the lanthanoids does not exhibit +4 oxidation
state.
78. A 5.2 molal aqueous solution of methyl alcohol,
CH
3
OH, is supplied. What is the mole fraction of
methyl alcohol in the solution ?
(1) 0.190 (2) 0.086 (3) 0.050 (4) 0.100
Ans. (2)
Sol. Mole fraction of methanol

5.2
0.086
1000
5.2
18
Mole of methanol
total moles
= = =
+

79. Which of the following statement is wrong ?
(1) Nitrogen can not form d p t t bond.
(2) Single N-N bond is weaker than the single P P
bond
(3) N
2
O
4
has two resonance structures
(4) The stability of hydrides increases from NH
3
to
BiH
3
in group 15 of the periodic table
Ans. (4)
Sol. Stability of hydrides from NH3 to BiH3 decreases due
to decreasing ionic character.
80. The outer electron configuration of Gd (Atomic no :
64) is:
(1) 4f
8
5d
0
6s
2
(2) 4f
4
5d
4
6s
2

(3) 4f
7
5d
1
6s
2
(4) 4f
3
5d
5
6s
2
Ans. (3)
Sol.
7 1 2
4 5 6 Gd f d s =
81. Which of the following statements regarding sulphur
is incorrect ?
(1) The vapour at 200
0
C consists mostly of S
8

rings.
(2) At 600
0
C the gas mainly consists of S
2

molecules.
(3) The oxidation state of sulphur is never less than
+4 in its compounds.
(4) S
2
molecule is paramagnetic
Ans. (3)
Sol. Sulphur exhibit + 2, + 4, + 6 oxidation states but + 4
and + 6 are more common.
82. The structure of IF
7
is :
(1) trigonal bipyramid (2) octahedral
(3) pentagonal bipyramid (4) square pyramid
Ans. (3 )
Sol. Hybridisation of iodine is sp
3
d
3
So, structure is
pentagonal bipyramid
83. Ozonolysis of an organic compound gives
formaldehyde as one of the products. This confirms
the presence of :
(1) a vinyl group (2) an isopropyl
group
(3) an acetylenic triple bond (4) two ethylenic double
bonds
Ans. ( 1)
Sol.

84. A gas absorbs a photon of 355nm and emits at two
wavelengths. If one of the emissions is at 680nm,
the other is at :
(1) 325 nm (2) 743 nm
(3) 518 nm (4) 1035 nm
Ans. (2 )
Sol.
1 2 2
1 1 1 1 1 1
355 680
= + = +

2
742.76nm =
85. Silver Mirror test is given by which one of the
following compounds ?
(1) Acetone (2) Formaldehyde
(3) Benzophenone (4) Acetaldehyde
Ans. (2 ,4)
Sol. :

86. Which of the following reagents may be used to
distinguish between phenol and benzoic acid ?
(1) Tollens reagent (2) Molisch reagent
(3) Neutral FeCl
3
(4) Aqueous NaOH
Ans. ( 3 )
Sol. Neural FeCl
3
gives violet coloured complex
SOLUTION : AIEEE 2011 (Booklet Code : Q)


87. Phenol is heated with a solution of mixture of KBr
and KBrO
3
. The major product obtained in the above
reaction is :
(1) 3-Bromophenol (2) 4-Bromophenol
(3) 2, 4,6- Tribromophenol (4) 2-Bromophenol
Ans. ( 3)
Sol.


88. In a face centred cubic lattice, atom A occupies the
corner positions and atom B occupies the face centre
positions. If one atom of B is missing from one of
the face centred points, the formula of the compound
is :
(1) AB
2
(2) A
2
B
3

(3) A
2
B
5
(4) A
2
B
Ans. (3 )
Sol.
8
8
A
Z = ;
5
2
B
Z = ,
So, formula of compound is AB
5/2
; i.e., A
2
B
5

89. The entropy charge involved in the isothermal
reversible expansion of 2 moles of an ideal gas from
a volume of 10dm
3
to a volume of 100 dm
3
at 27
0
C
is :
(1) 35.8 J mol
1
K
1
(2) 32.3J mol
1
K
1

(3) 42.3 J mol
1
K
1
(4) 38.3 J mol
1
K
1
Ans. (4 )
Sol.
2
1
ln
V
S nR
V
A = ;
100
2.303 x 2 x 8.314log
10
S A =
38.3 / / S J mole K A =
90 Identify the compound that exhibits tautomerism :
(1) Lactic acid (2) 2-Pentanone
(3) Phenol (4) 2-Butene
Ans. ( 2)
Sol.

O
2-Pentanone

has o -hydrogen & hence it will exhibit
tautomerism