AIEEE-2012 Solved papers by Triumph Academy

AIEEE 2012
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AIEEE 2012
[ 1 ]
Part A
1.
Truth table for system of four NAND gates as shown in
figure is
5.
A coil is suspended in a uniform magnetic field, with the
plane of the coil parallel to the magnetic lines of force.
When a current is passed through the coil it starts oscillating, it is very difficult to stop. But if an aluminium plate
is placed near to the coil, it stops. This is due to
(1) electromagnetic induction in the aluminium plate
giving rise to electromagnetic damping
(2) development of air current when the plate is placed
(1)
(3) induction of electrical charge on the plate
(4) shielding of magnetic lines of force as aluminium is a
paramagnetic material.
(2)
6.
A spectrometer gives the following reading when used
to measure the angle of a prisim
Main scale reading : 58.5 degree
Veriner scale reading: 09 divisions
(3)
2.
(4)
Given that 1 division on main scale corresponds to 0.5
degree. Total divisions on the vernier scale is 30 and
match with 29 divisions of the main scale. The angle of
the prism from the above data:
A thin liquid film formed between a U-shaped wire and a
light slider supports a weight of 1.5 × 10–2 N (See figure).
The length of the slider is 30 cm and its weight negligible.
The surface tension of the liquid film is
7.
(1) 59 degree
(2) 58.59 degree
(3) 58.77 degree
(4) 58.65 degree
In Young's double slit experiment, one of the slit is wider
than other, so that the amplitude of the light from one slit
is double of that from other slit. If Imbe the maximum intensity, the resultant intensity I when they interfere at
phase difference  is given by
3.
(1) 0.025 Nm–1
(2) 0.0125 Nm–1
(3) 0.1 Nm–1
(4) 0.05 Nm–1
Helium gas goes through a cycle ABCDA (consisting of
two isochoric and two isobaric lines) as shown in figure. Efficiency of this cycle is nearly.
(Assume the gas to be close to ideal gas)
8.
(1)
Im

(1  8 cos2 )
9
2
(2)
Im
(4  5 cos  )
9
(3)
Im

(1  2 cos2 )
3
2
(4)
Im

(1  4 cos2 )
5
2
Proton, Deuteron and alpha particle of the same kinetic
energy are moving in circular trajectories in a constant
magnetic field. The radii of proton, deuteron and alpha
particle are respectively rp , rd and r . Which one of the
following relations is correct?
4.
(2) r  rp  rd
(3) r  rp  rd
(4) r  rd  rp
Hydrogen atom is excited from ground state to another
state with principal quantum number equal to 4. Then the
number of spectral lines in the emission spectra will be
An object 2.4m in front of a lens forms a sharp image on a
film 12 cm behind the lens. A glass plate 1 cm thick, of
refractive index 1.50 is interposed between lens and film
with its plane faces parallel to film. At what distance (from
lens) should object be shifted to be in sharp focus on
film?
(1) 6
(2) 2
(1) 5.6 m
(2) 7.2 m
(3) 3
(4) 5
(3) 2.4 m
(4) 3.2 m
(1) 12.5%
(2) 15.4 %
(3) 9.1 %
(4) 10.5 %
AIEEE 2012
9.
(1) r  rd  rp
[ 2 ]
10. A liquid in a beaker has temperature  (t) at time t and  0
is temperature of surroundings, then according to Newton's law of cooling the correct graph between
b
13. The figures shows an experimental plot for discharge of a
capacitor in an R-C circuit. The time constant  of this
circuit lies betwen:
g
log e    0 and t is
(1)
(3)
(2)
(4)
11. This question has statement 1 and Statement 2. Of the
four choices given after the Statements choose the one
that best describes the two Statements
It two springs S1 and S2 of force constant k1 and k 2 ,
respectively, are streched by the same force, it is found
that more work is done on spring S1 than on spring S2 .
Statement 1: If stretched by the same amount, work done
on S1 , will be more than that on S2 .
Satement 2: k1  k 2
(1) Statement–1 is True,Statement–2 is True; Statement–2
is NOT a correct explanation for Statement–1
(2) Statement–1 is False, Statement–2 is True
(3) Statement–1 is True, Statement–2 is False
(4) Statement–1 is True,Statement–2 is True; Statement–2
is a correct explanation for Statement–1
12. This question 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: Davisson - Germer experiment established
the wave nature of electrons
(1) 100 sec and 150 sec
(2) 150 sec and 200 sec
(3) 0 and 50 sec
(4) 50 sec and 100 sec
14. Resistance of a given wire is obtained by measuring the
current flowing in it and the voltage difference applied
across it. If the percentage errors in the measurement of
the current and the voltage difference are 3% each, then
error in the value of resistance of the wire is
(1) 3%
(2) 6%
(3) zero
(4) 1%
15. A Carnot engine, whose efficiency is 40%, takes in heat
from a source maintained at a temperature of 500 K. It is
desired to have an engine of efficiency 60%. Then, the
intake temperature for the same exhaust (sink) temperature must be
(1) 600 K
(2) efficiency of Carnot engine cannot be made larger
than 50%
(3) 1200 K
(4) 750 K
16. A charge Q is uniformly distributed over the surface of
non-conducting disc of radius R. The disc rotates about
an axis perpendicular to its plane and passing through its
centre with an angular velocity  . As a result of this
rotation a magnetic field of induction B is obtained at the
centre of the disc. If we keep both the amount of charge
placed on the disc and its angular velocity to be constant
and vary the radius of the disc then the
variation of
the magnetic induction at the centre of the disc will be
represented by the figure.
(1)
(2)
(3)
(4)
Satement 2: If electrons have wave nature, they can interfere and show diffraction.
(1) Statement–1 is True,Statement–2 is True; Statement–2
is NOT a correct explanation for Statement–1
(2) Statement–1 is False, Statement–2 is True
(3) Statement–1 is True, Statement–2 is False
(4) Statement–1 is True,Statement–2 is True; Statement–2
is a correct explanation for Statement–1
AIEEE 2012
[ 3 ]
17. A radar has a power of 1 kW and is operating at a frequency of 10GHz. It is located on a mountain top of height
500 m. The maximum distance upto which it can detect
object located on the surface of the earth (Radius of earth
= 6.4×106 m) is
(1) 64 km
(2) 80 km
(3) 16 km
(4) 40 km
18. A partricle of mass m is at rest at the origin at time t  0 .
It is subjected to a force F (t )  Fo e bt in the x-direction.
Its speed v (t ) is depicted by which of the following
curves?
(1)
(2)
(3)
19. A cylindrical tube, open at both ends, has a fundamental
frequency, f, in air. The tube is dipped vertically in water
so that half of it is in water. The fundamental frequency
of the air-column is now.
(1) 2 f
(2) f
(3) f
(4) 3 f
4
20. The mass of a spaceship is 1000 kg. It is to be launched
from the earth's surface out into free space. The value of
' g ' and 'R' (radius of earth) are 10 m/s2 and 6400 km respectively. The required energy for this work will be
(1) 6.4 ×1010 Joules
(2) 6.4 ×1011 Joules
(3) 6.4 ×108 Joules
(4) 6.4 ×109 Joules
21. A boy can throw a stone up to a maximum height of 10 m.
The maximum horizontal distance that the boy can throw
the same stone up to will be
(1) 20 m
(3) 10 m
(4) 10 2 m
2
b
(2)
0.693
b
(3) b
(4)
1
b
AIEEE 2012
  
 
(2) X || B and k || B  E
  
 
(3) X || E and k || E  B
  
 
(4) X || B and k || E  B
24. This question has statement 1 and Statement 2. Of the
four choices given after the Statements, choose the one
that best describes the two Statements
An insulating solid sphere of radius R has a uniformly
positive charge density  . As a result of this uniform
charge distribution there is a finite value of electric potential at the centre of the sphere, and also at a point out
side the sphere. The electric potential at infinity is zero.
Statement 1: When a charge 'q' is taken from the centre
of the surface of the sphere, its potential energy charges
Satement 2: The electric field at a distance r ( r  R ) from
the centre of the sphere is
r
3 0
(1) Statement–1 is True,Statement–2 is True; Statement–2
is NOT a correct explanation for Statement–1
(2) Statement–1 is False, Statement–2 is True
(3) Statement–1 is True, Statement–2 is False
(4) Statement–1 is True,Statement–2 is True; Statement–2
is a correct explanation for Statement–1
25. A diatomic molecule is made of two masses m1 and m2
which are separated by a distance r. If we calcualte its
rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by:
(n is an integer)
(1)
(m1  m2 )n 2 h 2
2m1m2 r
2
(2)
(m1  m2 ) 2 n 2 h 2
2m12 m2 2 r 2
(2) 20 2 m
22. If a simple pendulum has significant amplitude (up to a
factor of 1/e of original) only in the period between t = 0
s to t   s , then  may be called the average life of the
pendulum. When the spherical bob of the pendulum
suffers a retardation (due to viscous drag) proportional
to its velocity, with 'b' as the constant of proportionality,
the average life time of the pendulum is (assuming damping
is small) in seconds
(1)
  
 
(1) X || E and k || B  E
q
by 3 
0
(4)
2
23. An electromagnetic wave in vacuum has the electric and


magnetic field E and B which are always perpendicular
to each other. The direction of polarization is given by


X and that of wave propagation by k . Then
n2h2
(3)
2(m1  m2 )r
2n 2 h 2
2
(4)
(m1  m2 )r 2
26. Two cars of masses m1 and m2 are moving in circles of
radii r1 and r2 , respectively. Their speeds are such that
they make complete circles in the same time t. The ratio of
their centripetal acceleration is:
(1) 1 : 1
(2) m1 r1 : m2 r2
(3) m1 : m2
(4) r1 : r2
[ 4 ]
27. In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as a function of distance from the centre. The graph which would correspond
to the above will be
(1)
(2)
(3)
(4)
28. Assume that a neutron breaks into a proton and an electron. The energy released during this process is
(Mass of neutron) = 1.6725 × 10–27 kg
Mass of Proton = 1.6725 × 10–27 kg
Mass of electron = 9 × 10–31 kg)
(1) 5.4 MeV
(2) 0.73 MeV
(3) 7.10 MeV
(4) 6.30 MeV
AIEEE 2012
29. A wooden wheel of radius R is made of two semicircular
parts (see figure). The two parts are held together by ring
made of a metal strip of cross sectional area S and length
L. L is slightly less than 2R . To fit the ring on the wheel,
it is heated so that its temperature rises by T and its
just steps over the wheel. As it cools down to surrounding temperature, it presses the semicircular parts together.
If the coefficient of linear expansion of the metal is  ,
and its Young's modulus is Y, the force that one part of
the wheel applies on the other part is
(1) 2SYT
(2) 2 SYT
(3) SYT
(4)  SYT
30. Two electric bulbs marked 25 W-220V and 100W-220V
are connected in series to a 440 V supply. Which of the
bulbs will fuse?
(1) neither
(2) both
(3) 100 W
(4) 25 W
[ 5 ]
Part -B Mathematics
31.
36.
Statement 1: The sum of the series
b
g b
g b
g
b
g
1  1  2  4  4  6  9  9  12  16 .... 361  380  400
parabola y2  16 3x and the ellipse 2 x 2  y 2  4 is
is 8000
y  2x  2 3
n
Statement 2:
 ek  bk  1g j  n , for any natural
3
3
3
Statement 2: If the line y  mx 
k 1
number n.
(1) Statement-1 is True, Statement-2 is False.
(2) Statement-1 is False, Statement-2 is True.
(3) Statement-1 is True, Statement-2 is true, Statement2 is a correct explanation for Statement-1.
(4) Statement-1 is True, Statement-2 is true, Statement2 is not a correct explanation for Statement-1.
32.
bx  1g
2
g
(1) Statement-1 is True, Statement-2 is False.
(2) Statement-1 is False, Statement-2 is True.
(3) Statement-1 is True, Statement-2 is true, Statement2 is a correct explanation for Statement-1.
(4) Statement-1 is True, Statement-2 is true, Statement2 is not a correct explanation for Statement-1.
37.
the centre of the ellipse is at the origin and its axes are
the coordinate axes, then the equation of the ellipse is:
2
2
(1) x  4y  16
(2) 4 x  y  4
(3) x 2  4 y 2  8
(4) 4 x 2  y 2  8
2
38.
point 2, 3 is:
(1)
2
5
(2)
3
8
(3)
1
5
(4)
1
4
bg
z
x
(1) 5 / 3
(2) 10 / 3
(3) 3 / 5
(4) 6 / 5
b
g
If g x  cos 4 t dt then g x   equals:
0
bg bg
(3) gb x g  g b  g
(1) g x . g 
bg
bg
(4) gb xg  g b g
(2)
gx
g 
Let P and Q be 3×3 matrices with P  Q . If
(1) –1
(2) – 2
Assuming the balls to be identical except for difference
in colours, the number of ways in which one or more
balls can be selected from 10 white, 9 green and 7 black
balls is
(1) 879
(2) 880
(3) 1
(4) 0
(3) 629
P 3  Q 3 and P 2 Q  Q2 P , then determinant of
eP
35.
q
mum is 3, given that their maximum is 6 is:
the x-axis at the point 1, 0 and passes through the
34.
l
ment form 1, 2, 3,...,8 . The probability that their mini-
The length of the diameter of the circle which touches
b g
Three numbers are choosen at random without replace-
2
b g
g
ellipse 2 x 2  y 2  4 , then m satisfies m4  2 m2  24
 y 2  1 as its semiminor axis and a diameter
b
b
4 3
, m  0 is a
m
common tangent to the parabola y2  16 3x and the
An ellipse is drawn by taking a diameter of the circle
of the circle x 2  y  2 2  4 as its semi-major axis. If
33.
Statement 1: An equation of a common tangent to the
2
39.
j
 Q 2 is equal to:
If n is a positive integer, then
e
j e
3 1
2n

j
3 1
2n
(1) a rational number other than positive integers
(2) an irrational number
(3) an odd positive integer
is
40.
(4) 630
If 100 times the 100th term of an AP with non zero common difference equals the 50 times its 50th term, then
the 150th term of this AP is
(1) zero
(2) – 150
(3) 150 times its 50th term
(4) 150
(4) an even positive integer
AIEEE 2012
[6]
41.
y
The area bounded between the parabolas x 2  and
4
47.
other, then the angle between a and b is:
x2  9 y , and the straight line y  2 is
(1) 10 2
(3)
42.
10 2
3
(2) 20 2
(4)
20 2
3
An equation of the plane parallel to the plane
x  2 y  2 z  5  0 and at a unit distance from the origin is
(1) x  2 y  2 z  5  0
(2) x  2 y  2 z  3  0
(3) x  2 y  2 z  1  0
(4) x  2 y  2 z  1  0
Let a and b be two unit vectors. If the vectors


c  a  2 b and d  5a  4 b are perpendicular to each
48.
(1)

4
(2)

6
(3)

2
(4)

3
A line with negative slope is drawn through the point
(1, 2) to meet the coordinate axes at P and Q such that
it forms a triangle OPQ, where O is the origin. If the
area of the triangle OPQ is least then the slope of the
line PQ is:
1
2
(2) –
(3) – 4
(4) –2
(1) –
43.
44.
The equation esin x  e  sin x  4  0 has
(1) exactly four real roots.
(2) infinite number of real roots.
(3) no real roots
(4) exactly one real root.
49.
(2) I will become a teacher and I will not open a school
50.
(4) Neither I will become a teacher nor I will open a
school.
45.
The population p( t ) at time t of a certain mouse species satisfies the differential equation
46.
(2) 2 n 18
(3) n 9
(4) 1/2 n 18
51.
z
5 tan x
dx  x  a ln sin x  2 cos x  k
tan x  2
then a is equal to:
(1) 2
(2) –1
(4) 1
(2) 52
(3) 35
(4) 25


Let ABCD be a parallelogram such tht AB  q ,



AD  p and  BAD be an acute angle. If r is the
vector that coincides with the altitude directed form

the vertexd B to the side AD, then r is given by:
b g
b g
 

 bp  qg 
r


q

(3)
bp  p g p
If the integral
(3) –2
(1) 53
 
 

 3 pq 
  3 pq 
(1) r  3q    p (2) r  3q    p
q q
q q
dp( t )
 0.5 p( t )  450 . If p(0)  850 , then the time at
dt
which the population becomes zero is:
(1) n 18
q
Let X  1, 2, 3, 4, 5 . The number of different ordered
pairs (Y, Z) that can be formed such that
Y  X, Z  X and Y  Z is empty, is
The negation of the statement
"If I become a teacher, then I will open a school", is :
(1) I will not become a teacher or I will open a school
(3) Either I will not become a teacher or I will not open
a school.
l
1
4
52.
b g
b g
 
  bp  q g 
r

q

(4)
bp  p g p
If the line 2x  y  k passes through the point which
divides the line segment joining the points (1, 1) and (2,
4) in the ratio 3 : 2, then k equals:
(1) 11 / 5
(2) 29 / 5
(3) 5
(4) 6
In a PQR, if 3 sin P  4 cos Q  6 and
4 sin Q  3cos P  1, then the angle R is equal to:
AIEEE 2012
(1)
3
4
(2)
5
6
(3)

6
(4)

4
[7]
53.
F1
A  G2
GH 3
Let
I
0J . If
J
1K
0 0
1
2
matrices such that Au1
57.
u1 and u 2 are column
F 1I
 G 0J and Au
GH 0JK
2
F 0I
 G 1J , then
GH 0JK
(2) either on the real axis or on a circle passing through
the origin.
(3) on a circle with centre at the origin.
(4) either on the real axis or on a circle not passing
through the origin.
u1  u 2 is equal to
54.
F 1I
G J
(1) G 1J
H 1K
F 1I
G 1J
(2) G J
H 0K
F 1I
G 1J
(3) G J
H 1K
F 1I
G 1J
(4) G J
H 0K
f : R  R is
If
bg
f x  x cos
a
function
z2
is real, then the point represented
z 1
by the complex number z lies:
(1) on the imaginary axis
If z  1 and
58.
bg
Consider the function, f x  x  2  x  5, x  R
bg
Statement 1: f ' 4  0
Statement 2: f is continuous in [2, 5], differentiable in
(2, 5) and f (2)  f (5)
defined
(1) Statement-1 is True, Statement-2 is False.
(2) Statement-1 is False, Statement-2 is True.
(3) Statement-1 is True, Statement-2 is true, Statement2 is a correct explanation for Statement-1.
(4) Statement-1 is True, Statement-2 is true, Statement2 is not a correct explanation for Statement-1.
by
FG 2x  1IJ  , where [x] denotes the greatH 2 K
est integer function, then f is
(1) continuous only at x  0
(2) continuous for every real x
59.
arithmetic mean and  2 be their variance.
(3) discontinuous only at x = 0
(4) discontinuous only at non-zero integral values of x
55.
56.
Statement 1: Variance of 2x1, 2x2 ,....2x n is 4 2
A spherical balloon is filled with 4500  cubic meters of
helium gas. If a leak in the balloon causes the gas to
escape at the rate of 72  cubic meters per minute, then
the rate (in meters per minute) at which the radius of
the balloon decreases 49 minutes after the leakage
began is
(1) 9 / 2
(2) 9 / 7
(3) 7 / 9
(4) 2 / 9
Statement 2: Arithmetic mean of 2 x1 , 2 x 2 ,....2 x n is
4x
(1) Statement-1 is True, Statement-2 is False.
(2) Statement-1 is False, Statement-2 is True.
(3) Statement-1 is True, Statement-2 is true, Statement2 is a correct explanation for Statement-1.
(4) Statement-1 is True, Statement-2 is true, Statement2 is not a correct explanation for Statement-1.
Let a, b  R be such that the function f given by
bg
f x  ln x  bx 2  ax , x  0 has extreme values at
x  1 and x =2.
Statement 1: f has local maximum at x  1 and at
x2 .
Let x1 , x 2 , ..... x n be n observation, and let x be their
60.
x3 yk z
x 1 y 1 z 1
and




1
2
1
2
3
4
intersect, then k is equal to :
(1) 0
(2) –1
If the lines
(3)
2
9
(4)
9
2
1
1
and b 
.
2
4
Statement-1 is True, Statement-2 is False.
Statement-1 is False, Statement-2 is True.
Statement-1 is True, Statement-2 is true, Statement2 is a correct explanation for Statement-1.
Statement-1 is True, Statement-2 is true, Statement2 is not a correct explanation for Statement-1.
Statement 2: a 
(1)
(2)
(3)
(4)
AIEEE 2012
[8]
Part - C Chemistry
61.
62.
Which of the following compounds can be detected
by Molisch's test?
(1) Primary alcohols
(2) Nitro compounds
(3) Sugars
(4) Amines
69.
(1)
The increasing order of the ionic radii of the gtiven
isolectronic speices is
2

(3) S , Cl , Ca
2
,K
70.

(4) Ca 2  , K  , Cl  , S2 
63.
64.
65.
66.
Which one of the following statements is correct?
(1) All amino acids except glutamic acid are optically
active.
(2) All amino acids except lysine are optically active.
(3) All amino acids are optically active.
(4) All amino acids except glycine are optically active
2 – Hexyne gives trans – 2 – Hexene on treatement with
(1) LiAlH 4
(2) Pt / H 2
(3) Li / NH 3
(4) Pd / BaSO 4
71.
72.
(2) LiAlH 4
(3) HNO 3
(4) AlCl 3
67.
68.
(1) 1  10 7
(2) 3  10 1
(3) 1  10 3
(4) 1  10 5
The incorrect expression among the following is
G system
S total
o
/ RT
 T
w reversible   nRT ln
(4) ln K 
Vf
Vi
H o  TSo
RT
73.
Iodoform can be prepared from all except:
(1) Isobutyl alcohol
(2) Ethyl methyl ketone
(3) Isopropyl alcohol
(4) 3 – Methyl – 2 – butanone
74.
In the given transformation, which of the following is
the most appropriate reagent?
(2) X  Ni, Y  Fe
(3) X  Ni, Y  Zn (4) X  Fe, Y  Zn,
Lithium forms body centred cubic structure. The length
of the side of its unit cell is 351 pm. Atomic radius for
the lithium will be
(1) 152 pm
(2) 75 pm
(3) 300 pm
(4) 240 pm
The pH of a 0.1 molar solution of the acid HA is 3. The
value of the ionization constant, Ka of this acid is
(3) In isothermal process,
X  Y 2   X 2   Y will be spontaneous when
(1) X  Zn, Y  Ni
x
 p1
m
x
 p1/ n
m
(4) All the above are correct for different ranges of
pressure.
The density of a solution prepared by dissolving 120 g
of urea (mol. mass = 60 u) in 1000 g of water is 1.15 g
/ mL. The molarity of this solution is
(1) 2.30 M
(2) 0.50 M
(3) 1.78 M
(4) 1.02 M
(2)
The standard reduction potentials for Zn 2 / Zn,
Ni 2 / Ni, and Fe 2 / Fe are –0.76, –0.236 and
–0.44
V
respectively.
The
reaction
(2)
(1) K  e G
The speices which can best serve as an initiator for the
cationic polymerization is
(1) BuLi
x
 p0
m
(3)

2
2

(1) K , S , Ca , Cl
(2) Cl  , Ca 2  , K  , S2 
According to Freundlich adsorption isotherm, which
of the following is correct?
The electrons identified by quantum numbers n and 1:
(a) n  4, l  1
(b) n  4, l  0
(c) n  3, l  2
(d) n  3, l  1
can be placed in order of increasing energy as
(1)
(2)
(3)
(4)
bag  bcg  bbg  bdg
bcg  bdg  bbg  bag
bdg  bbg  bcg  bag
b b g  b d g  ba g  bc g
AIEEE 2012

(1) NaBH 4
(2) NH 2 NH 2 , O H
(3) Zn  Hg / HCl
(4) Na , Liq. NH 3
[9]
75.
Very pure hydrogen (99.9%) can be made by which of
the following processes?
(1) Reaction of salt like hydrides with water
(2) Reaction of methane with steam
(3) Mixing natural hydrocarbons of high molecular weight
(4) Electrolysis of water
76.
Which among the following will be named as
dibromidobis (ethylene diamine) chromium (III)
bromide?
b g
Cr beng Br
(1) Cr en Br2 Br
(3)
77.
78.
79.
80.
81.
2
2
Br
Which branched chain isomer of the hydrocarbon with
molecular mass 72u gives only one isomer of mono
substituted alkyl halide?
(1) Neohexane
(2) Tertiary butyl chloride
(3) Neopentane
(4) Isohexane
84.
The equilibrium constant
bg
bg
What is DDT among the following:
(1) Non - biodegradable pollutant
(2) Greenhouse gas
(3) A fertilizer
(4) Biodegradable pollutant
K f for water is 1.86 K kg mol–1. If your automobile
radiator holds 1.0 kg of water, how many grams of
b
g
(3) 2.5  10
85.
bg b g
Tibsg  2I bgg
523 K
bg
1700 K
Ti s  2I 2 g   TiI 4 g  
2
(1) Van Arkel
(3) Cupellation
AIEEE 2012
(2) Zone refining
(4) Poling
at temperature T is
bg
(2) 0.02
2
(4) 4  10 4
bg
For a first order reaction, A  products, the
(1) 173
.  10 4 M / min
(2) 173
.  10 5 M / min
(3) 3.47  10 4 M / min
(4) 3.47  10 5 M / min
86.
Aspirin is known as
(1) Methyl salicylic acid
(2) Acetyl salicylic acid
(3) Phenyl salicylate
(4) Acetyl salicylate
87.
The molecule having smallest bond angle is
88.
(1) PCl 3
(2) NCl 3
(3) AsCl 3
(4) SbCl 3
The compressibility factor for a real gas at high pressure is
(1) 1
pb
RT
(3) 1
89.
freezing point of the solution lowered to –2.8 oC?
(1) 27 g
(2) 72 g
(3) 93 g
(4) 39 g
Which method of purification is represented by the
following equation:
for the reaction
concentration of A changes from 0.1 M to 0.025 M in
40 minutes. The rate of reaction when the concentration
of A is 0.01 M, is
ethylene glycol C 2 H 6O 2 must you add to get the
82.
bg
is
(1) 50.0
How many chiral compound are possible on
monochlorination of 2 - methyl butane?
(1) 6
(2) 8
(3) 2
(4) 4
Iron exhibits +2 and +3 oxidation states. Which of the
following statements about iron is incorrect?
(1) Ferrous compound are more easily hydrolysed
than the corresponding ferric compounds.
(2) Ferrous oxide is more basic in nature than the ferric
oxide.
(3) Ferrous compounds are relatively more ionic than
the corresponding ferric compounds.
(4) Ferrous compounds are less volatile than the
corresponding ferric compounds.
c
NO g  21 N 2 g  21 O 2 g at the same temperature
4
Ortho-Nitrophenol is less soluble in water than p- and
m-Nitrophenols because
(1) Melting point of o -Nitrophenol is lower than those
of m - and p - isomers.
(2) o - Nitrophenol is more volatile in steam than those
of m - and p - isomers.
(3) o - Nitrophenol shows Intramolecular H - bonding
(4) o - Nitrophenol shows Intermolecular H - bonding
bg
bK g
4  10 4 . The value of K c for the reaction,
3
3
bg
N 2 g  O 2 g  2 NO g
b g Br
Cr ben gBr
(2) Cr en
(4)
83.
90.
(2) 1 RT
pb
pb
(4) 1
RT
Which of the following on thermal-decomposition
yields a basic as well as an acidic oxide?
(1) NH 4 NO 3
(2) NaNO 3
(3) KClO 3
(4) CaCO 3
In which of the following pairs the two speices are not
isostructural?
(1) AlF6 3 and SF6
(2) CO 32 and NO 3 
(3) PCl 4  and SiCl 4
(4) PF5 and BrF5
[ 10 ]
AIEEE 2012
Part -A Physics (Solutions)
1.
[2] The given circuit represents XOR gate.

Ans.
Ans.
2.
[1] For equilibrium
2SL  W
S
2SL
b g
i 5b2 p2 gV
d
0
 5 p0V0
0
Ans. r  rp  rd
9.
[1] Due to the introduction of sheet, the point of con
vergence shifts away from its original position by
FG
H
13
p0V0
2
1
cm before the film i.e. at a distance of
3
1
35
v   12cm - cm = cm . If x be the distance of
3
3
shifted object then
1
1
1


35  x
f
3
b g
g
N N 1
6
2
b
5.
[1] Electromangetic induction in the aluminium plate
giving rise to electromagnetic damping
6.
[4] LC 
S 0.5

degree
N 30
10.
   0e  kt
b
j
2
 9 I0
g
log e    0   kt
So the graph must be a straight line with negative
slope.
[1] Let the intensities be I0 and 4I0.
I0  4 I0
. . . (ii)
[2] According to Newton's law of cooling
b g
e
g
From above equations, x = 5.6 m
Ans. 5.6 m
Reading  58.5  9 LC = 58.65 degree
Ans. 58.65 degree
Im 
. . . (i)
Previously 1  1  1
12 240
f
Ans. 6
7.
IJ
K
form image
W
pV
2

 0 0 
Qabs 13 p V
13
0 0
2
  15.4%
Ans. 15.4%
b
IJ b g FG
K
H
1
1
1
 1 cm 1 
 cm
n
3/ 2
3
So now to form the image on the film, the lens should
t  t 1 
[1] Number of spectral lines 
2mK
qB
r  rp  rd

3
p0V0
2
Qabs  QBC  QAB 
IJ
K
IJ
K
m
q
W
QBC  n C p T 
4.
mv

qB
r
[2] Work done = Area under the p-V graph
W  p0V0
QAB 
FG
H
Im

1  8 cos2
9
2
[3] r 
W 15
.  10 2

 0.025 N / m
2L
2 0.3
Ans. 0.025 N / m
3.
8.
FG
H
Im

1  8 cos2
9
2
Ans.
I
4
I1  I 0  m and I 2  4 I 0  I m
9
9
Resultant intensity at phase difference 
I  I1  I 2  2 I1 I 2 cos 

Im 4 Im
4 2

2
I m cos
9
9
81

Im
5  4 cos
9
b
g
11. [2] W 

F2
1
 W
2k
k
W1  W2  k1  k 2 (For same force)
So Statement 2 is true.
and W 
1 2
kx
2
[ 1 ]
AIEEE 2012
W1  W2 when stretched by same amount
so statement 1 is false.
Ans. statement 1 is false, Statement 2 is true

18. [4]
dv F0  bt

e
dt
m
z
t
12. [4] Davision Germer experiment establised the wave nature of electrons. Thus electron can interfere and show
diffraction.
v
or
0
e
F0 bt
F
e dt  0 1  e bt
m
mb
j
Ans.
So both Statement are correct.
Ans. Statement 1 is true, Statement 2 is true, Statement
2 is the correct explanation for Statement 1
13. [1] The time constant  is time when potantial drop to
b
0.37 time of its peak value i.e. 0.37  25  9.25 V
g
19. [2] For open pipe, f 
graph shows that  lies between 100 – 150 s
Ans. 100 to 150 seconds
F R IJ
Thus GH
RK
V
I
max
F V IJ  I  3%  3%
G
HVK I
The maximum error in R is 6%
Ans. 6%
v
v

 f
L
2L
4
2
FG IJ
H K
Ans. f
20. [1] E 
b g
1 2 1
mve  m 2 gR  mgR  6.4  1010 J
2
2
Ans. 6.4  1010 J
TC
TC
15. [4] For Carnot engine   1  T  0.4  1  500
H
TC  300K
300
For desired engine 0.6  1 
(Sink is same)
TH
or
TH  750 K
Ans. 750 K
21. [1] As hmax 
u2
u2
,
and Rmax 
2g
g
so
Rmax  2hmax  20 m
Ans. 20 m
22. [1] The equation of motion is
dv
k
  x  bv
dt
m
and amplitude is given by
16. [2] Consider a diffrential element in the
form of a ring of radius r [Fig.] and width
dr, the current is
di 
A  A0e
FG Q IJ b2r gdr
H R K

bt
2
As per question at t   , A 
2
and corresponding magnetic field at centre,
dB 
So
z
b g
b
g
z
R
 0Q
 Q 1
dr  0

2
R
R
R
0
Ans.
17. [2] The maximum distance of coverage d  2 Rh
 2  6.4  10  500 m  80 m
Ans. 80 km
so solving we get  
 0 di

Q
 0  2 2rdr 
2r
2 r R
B  dB 
L
length
2
For closed pipe having
f
14. [2] Resistance R 
v
2L
Ans.
A0
e
2
b
2
b

23. [3] The direction of polarization X is defined to be the
 

direction in which E is vibrating i.e. X || E .
 

Wave propagation k is along E  B
 
  
Ans. X || E and k || E  B
24. [2]
b
g b g
U  q V r  R  V r  0
L kQ  3 kQ OP   kqQ   1
 qM
N R 2 R Q 2 R 4
q

0
FG 4 R IJ
H3 K
3
2R
[ 2 ]
AIEEE 2012

27. [4] For uniformly charged sphere
E r
rR
1

rR
r
R q
6 0
2
Hence statement-1 is wrong
E
Statement-2 for r  R,
r
3 0
Ans.
is correct.
Ans. Statement 1 is false Statment 2 is true
FG nh IJ
H 2 K
2
L
25. [1] K 

2I
2
F m m IJ r
2G
Hm m K
1 2
1

b
n 2 h 2 m1  m2
28. [2] As per the given data,the reaction is endoergic
g
so it should be bonus
Ans. (Bonus)
8  2 m1m2 r 2
2
2
29. [1] The tension produced due to thermal stress is
It matches none of the given options hence bonus
must be awarded.
bm  m gn h
2 2
1
Ans.
2
2m1m2 r 2
Ans.
26. [4] Both cars have same angular speed so from a   2 r ,
a1 r1

a2 r2
Ans. r1 : r2
F  SYT
Here the force is exerted on one part due to other at
both ends and hence the force that one part of the
wheel applies on the other part is 2SYT .
2SYT
V2
, 25 W bulb has greater resistance so in
P
series, voltage across it will be more. Therefore, 25 W
bulb will fuse
30. [4] From R 
Ans. 25 W
Part -B Mathematics (Solutions)
31. [3]
b
g b
gb
g
1  1  2  4  4  6  9  9  12  16
b
g
... 361  380  400
e
j e
j  e2  2.3  3 j
...e19  19.20  20 j
  eb k  1g  b k  1gk  k j
 0  0.1  1  1  1.2  2
2
2
2
2
2
2
2
2
34. [4]



20
2
2

K 1
20
b gj
 b20g  b1  1g  b20g

e
k3  k  1
35. [2]
3
K1
3
32. [1]
3
33. [2]
 8000
x2 y2
2
2
 2  1  4 x  y  16
2
2
4
Let the radius of the circle  r

b2  1g  b3  r g

r
2
2
 r2
10
5
 2r 
3
3
2
2
2
2
2
2
2n
2n
2 n1
2 n 3
2n
1
3
2 n 2
2n
2n
1
2 n 4
2n
1
2n
3
1
 2 3  an integer
= an irrational number
x2 y2
2
2

 1  x  4 y  16
42 2 2
or
b g b g
P b P  Qg  Q b P  Qg  O ( Null matrix)
eP  Q jbP  Qg  O
 PQ
det e P  Q j  0
e 3  1j  e 3  1j
L
O
 2M C e 3j
 C e 3j
... C e 3 jP
N
Q
L
O
 2 3 M C e 3j
 C e 3j
... C P
N
Q
P 2 P  Q  Q2 Q  P
2n
3
Length of semiminor axis =2
Length of semimajor axis =4
Required ellipse is
P 3  P 2 Q  Q3  Q 2 P
36. [3]
Any tangent to parabola is
4 3
. . . (1)
m
Any tangent to ellipse is
y  mx 
y  mx  2m2  4 . . . (2)
(1) and (2) represent same line
4 3
  2m 2  4
m

4 3
  2m 2  4
m
[ 3 ]
AIEEE 2012
or

m4  2m2  24  0
m  2
(m2  6) (m2  4)  0
or
43. [3]
e j  4e
L1 O
 M , eP =
Ne Q
Let e sin x
So equation of tangent is y  2 x  2 3
2
sin x
 1  esin x  2  5
But e sin x
37. [3]
Let E1 : minimum of two number choosen is 3
E 2 : maximum of two number choosen is 6
so
P
FH E E IK  P( E  E )  n( E  E )  1
P( E )
n( E )
5
1
1
2
2
1
2
2

44. [2]
2
Both solutions are rejected.
p : I become a teacher
q : I will open a school
If I become a teacher, then I will open a school
b
z
 pq
x
38. [3,4] g ( x )  cos 4t dt
0
z
z
 x
and
g (  x ) 

0
z
b
b g
z
45. [2]
2  /2  x
= g ( ) 
cos 4t dt
b g
0 2  / 2
= g ( )  g ( x ) = 0  g( x)
z
z
1
dp ( t )  dt
0.5 p ( t )  450
p(t)
 450  t  k
2

2n

p(0)  850
39. [1]
Required ways = (11) (10) (8)  1  879

2 log 25  k
40. [1]
Let first term of A.P. is a and common difference
is d then, given 100 (a + 99d) = 50 (a + 49 d)

2 n

50a 
d (100) (149)
0
2
a  149 d  0 

Required area  2
z
0
F3
GH
I dy
J
2 K
2 y
x=4
x = 9y

42. [2]
z
y dy  5 
0
e j
2
 y 3/ 2
3
5 sin x
dx
sin x  2 cos x

A  2, B  1

I2
z
b
g b
d
sin x  2 cos x  B sin x  2 cos x
dx
z
cos x  2 sin x
dx  dx
sin x  2 cos x
 2 log sin x  2 cos x  x  k
47. [4]
2
0

 a2

c. d  0
ea  2b j  e5a  4b j  0
1
 a  b 
2
If  be the angle between a and b , then
10
20 2
2 2
3
3
Plane parallel to x  2 y  2 z  5  0
is x  2y  2z  k  0
According to sum,
z
Let 5 sin x  A
x
O
2
I
2
y=2
5
2
46. [1]
y
y
y
2
t  2n18
150th term is 0
2
41. [4]
p( t )
 450  2 log 25  t
2
when p( t )  0 , then 2n 450  2n25  t
 (50) a  149 d  0

g
= I will become a teacher and I will not open a school
cos 4t dt

0
g b
Now ~ p  q  ~ ~ p  q  p ~ q
x
cos 4t dt  cos 4t dt 
g
k
1 
3
cos   a  b =


1
2

=
3
k3
[ 4 ]
g
AIEEE 2012
48. [4]

b g
Any line through (1, 2) is y  2  m x  1 , m  0
 1
y-intercept
2m
Area of the triangle (A) 
LM
N
1
4
 4 m
2
m
FG
H
49. [3]
IJ b
K
Au 2
[Applying AM  GM ]
1
4
 m2  4
m
1
Let x1  X then
x1  P, x1  Q
or
x1 Q, x1  P
so required ways
F 0I
 G 1J
GH 0JK
. . . (2)
or
=3
2
2
I F xI F x
I F 1I
JJ GG yJJ  GG 2 x  y JJ  GG 1JJ
K H z K H 3x  2 y  zK H 0K
0
0
1

x  1, 2 x  y  1, 3x  2 y  z  0

x  1, y  1, z  1

F 1I
 G 1J
GH 1JK
x1  P, x1  Q
5
F 1I
)  G 1J
GH 0JK
F xI
Let u  u  G yJ
GH zJK
F1 0
A( u  u )  G 2 1
GH 3 2
A4
b m  0g
. . . (1)
Adding A ( u1  u 2
1
4
m 
2
m
m  2
5
is rejected.
6
F 1I
 G 0J
GH 0JK
g
1
2
1
2m
2
m
A is minimum when  m 

53. [1] Au1
OP
Q
LMb g
OP
b g PQ
MN
1
4
 2  2 b mg
2
b mg
2

2
m
x-intercept
R
54. [2]
u1  u2
FG 2 x  1IJ 
H 2 K
f ( x )  [ x ] cos
if x  Z then f ( x ) is a continuous function
if x,  Z, x  I then
50. [3]
FG 2 x  1IJ   0  f ( I
H 2 K
f ( I  )  lim [ x ]cos
x I 

Let position vector of M is  p

 


also AD  AB  ( p)  (p  q)  0
 
p q
 2
p


Hence BM 
2

sin R 

R
1
2
5

or
6
6
)
Also f ( I )  0
F p . q I  
GGH p JJK p  q
51. [4] Co-ordinate of point is b8 / 5, 14 / 5g
8
14
So 2FG IJ 
H 5K 5  k  k  6
52. [3] b3 sin P  4 cos Qg  b4 sin Q  3 cos Pg  6  1
 9  16  24 sinb P  Qg  37


55. [4]
2
2
2
2
f is continuous for all x  Z also
Vi  4500  m3 ,
Vf  4500   72   49  972  
972  3

4

rf3 

dV
 72 m3 / mn
dt
V

4 3
rf
3
rf3  93  rf  9
4 3
dV
dr
r 
 4 r 2
3
dt
dt
dr dV
1
1
2


 72  

dt
dt 4 r 2
4  9  9 9
[ 5 ]
AIEEE 2012
bg
R|7  2x
f b xg  S 3
|T2x  7
f x  ln x  bx2  ax
56. [3]
bg
f' x 

58. [4]
bg
1
 2 bx  a
x
 f  x   1  2 b
x2
 a  1  2b
b g
How f ' 1  0
bg
1
1
, b
2
4
1 1
f '' x   2   0 x  0
2
x
Solving a 
b g
2
bg
f" 1  0 and f" 2  0
g e
b
jb
2
x 2  y 2  2ixy x  1  iy
x  iy
z2
w


2
z 1 x 1 i y
x 1  y2
b g
Arithmetic mean of 2 x1 , 2x 2 , .....2 x n  2 x
g
60. [4]
ex  y jb yg  2xybx  1g  0
ye x  y  2 x  2xj  0
ye x  y  2 x j  0

2

2
2

2
2
2 k  1 1
2
3
4 0
1
w0
Im
5
Variance of 2 x1 , 2 x 2 , .....2 x n is 22   2  4 2
59. [1]
Let z  x  iy
57. [2]
x5
3
bg

2 x5
1
 4b  a  0
2

f' 2  0
x2
2
2
1
b g b gb g

2 5  k  1 2  1  0

k
9
2
2
Part -C Chemistry (Solutions)
61.
62.
[3] Molisch test used for identification of sugar.
Z
[4] In isoelectronic series as
increases size
e
decreases.
Z
e
Ca 2 
20
18
K
19
18
Cl 
17
18
S2 
16
18
68.
b g
is minimum should be filled first. If n   is same lower
69.
value of n, should be filled first.
[4] Freundlich adsorption isotherm apply for different
range of pressures
70.
[1] Volume of solution  1000 mL
115
.
Size Ca 2  , K  , Cl  , S2 
63.
64.
[4] Structure of glycine is H 2 N  CH 2  COOH
No chiral carbon,
H
C
71.
C
CH3
H
67.
no of moles
Volume of solution in L
M
120  115
.  1000
 2.30 M
60  1000
bg
 H   aq   A   aq 
[4] HA  aq  
0.1
0
0
0.1  x
[4] for cationic polymerization, the species which
[1] Zn  Ni
[1] for bcc
2

 Zn
2
r

 Ni
3a  4 r
1732
.
 351
 1519
.  152 pm
4
x
x
e j
3 2
helps in the formation of cation is Lewis acid i.e. AlCl 3 .
66.
M
/ NH 3
[3] H 3C  H 2 C  H 2 C  C  C  CH 3 Li


CH3CH2CH2
65.
b g
[3] According to afbau principle orbital whose n  
10
x2

01
.  x 0.1  10 3
~ 1  10 5
72.
[4] G o   RT ln K
H o  TSo
  ln K
RT
[ 6 ]
AIEEE 2012
84.
TS o  H o
hence ln K 
RT
73.
NO 
 21 N 2  21 O 2 K' c  ?
1
K' c 
[1] CH 3  CH  CH 2 (isobutyl alcohol) does not
|
|
CH 3 OH
contain CH 3  CH group.
|
OH
74.
85.
[3] A
0
Kc
1

4  10
A
A
76.
[3] Cr en 2 Br2 Br
k  0.035 min 1
77.
[3] ortho-Nitrophenol is less soluble in water because
it does not form. Intermomlecular hydrogen bonding
with H2O.
rate  k A
k
b g
H
 0.035  0.01
CH 2Cl
+
Cl
M
M
 35
.  10 4
min
min
O
O—C—CH3
H
86.
[2]
87.
[4] Due to less electronegativity of Sb lone pair-bond
pair repulsion dominates, consequently bond angle
decreases.
88.
[4]
H
Cl
80.
[1] The incorrect statement is that ferrous compounds
are more easily hydrolysed. In fact ferric compounds
are more easily hydrolysed due to more covalent
character because of greater charge to radius ratio
[1] DDT is non-biodegradable pollutant
81.
[3] Tf  K f m
2
2
I bV  nbg  nRT
JK
PV  RT  Pb
62  2.8
w0 
 93.33 g
186
.
PV RT Pb


RT RT RT
[1] It is Van Arkel process of purification
[3] Neopentane
CH 3
CH 3
|
|
Cl 2
CH 3  C  CH 3   CH 3 – C  CH 2 Cl
hv
|
|
CH 3
CH 3
F P  an
GH V
at high pressure
PV  Pb  RT
w0
2.8  186
. 
62  1
molecular mass 72
IJ
K
+
+
82.
83.
t
2.303
01
.
log
40
0.025
COOH
79.
100
 50.0
2
0
FG
H
75.
CC
[4] By electrolysis of water
[4]

 0.1
kt  2.303 log
CH 2  group without affecting –OH group and
78.
4
A t  0.025
[2] NH 2 NH 2 , OH converts  C  O group into
CH2Cl
H
K c  4  10 4
[1] N 2  O 2 
 2 NO
Z  1
Pb
RT

89.
[4] CaCO 3  CaO  CO 2
90.
[4] PF5  sp 3d hybridization (Trigonal bi pyramidal)
BrF5  sp 3d 2 hybridization (square pyramidal)
[ 7 ]