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AIEEE 2011
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3. The test is of 3 hours duration.
4. The Test Booklet consists of 90 questions. The maximum marks are 360.
5. There are three part in the question paper. The distribution of marks subjectwise in each part is as under for each correct
response.
Part A : MATHEMATICS (120 marks) - Question No. 1 to 30 consist Four (4) marks each for each correct
response.
Part B : CHEMISTRY (120 marks) - Question No. 31 to 60 consist Four (4) marks each for each correct response.
Part C : PHYSICS (120 marks) - Question No. 61 to 90 consist Four (4) marks each for each correct response..
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AIEEE 2011
[1]
Part - A Mathematics
1.
Consider 5 independent Bernulli's trials each with
probability of sucess p. If the probability of at least
one failure is greater than or equal to
lies in the interval
FG OP
H Q
F 3 11 O
(3) G 4 , 12 P
H Q
11
(1) 12 , 1
2.
3.
6.
31
, then p
32
F d y I F dy I
(1) – G dx J GH dx JK
H K
F d y I F dy I
(3) – G
H dx JK GH dx JK
F
GG
H
x2
2
FG 1 , 3 OP
H 2 4Q
F 1O
(4) G 0, 2 P
H Q
(2)
1  cos{2( x  2)}
x2
I
JJ
K
2
(3) equals 2
4.
7.
F d yI
(2) G
H dx JK
F d y I F dy I
(4) G dx J GH dx JK
H K
1
2
2
3
2
2
2
2
The number of values of k for which the linear
equations
4 x  ky  2 z  0
kx  4 y  z  0
2x  2 y  z  0
possess a non-zero solutoin is
(1) zero
(2) 3
(3) 2
(4) 1
8.
Statement -1
The point A (1, 0, 7) is the mirror image of the
(2) does not exist
point B(1, 6, 3) in the line
(4) equals  2
x y 1 z  2


1
2
3
Statement -2
Let R be the set of real numbers
x y 1 z  2


bisects the line
1
2
3
segment joining A (1, 0, 7) and B (1, 6, 3)
Statement -1
The line
A  {( x, y)  R  R : y  x is an integer} is an
equivalence relation on R.
(1) Statement -1 is false, Statement -2 is true
(2) Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for statement-1
Statement -2
B  {( x, y )  R  R : x  y for some rational
number  } is an equivalence relation on R.
(1) Statement -1 is false, Statement -2 is true
(2) Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for statement-1
(3) Statement-1 is true, Statement -2 is true, Statement-2 is not correct explanation for Statement-1
(4) Statement -1 is true, Statement-2 is false
z 2  z    0 has two distinct roots on the line
Consider the following statements
P : Suman is brilliant
Q : Suman is rich
R : Suman is honest
The negation of the statment "Suman is brilliant
and dishonest if and only if Suman is rich" can be
expressed as
(1) ~ ( P ~ R)  Q
Re z  1 , then it is necessary that
(1)   (1, )
(2)  (0, 1)
(3) ~ (Q  ( P  ~ R))
(3) Statement-1 is true, Statement -2 is true, Statement-2 is not correct explanation for Statement-1
(4) Statement -1 is true, Statement-2 is false
5.
1
2
1
(1) equals
3
2
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
lim
d2x
equal
dy 2
Let  ,  be real and z be complex number. If
(3)   ( 1, 0)
AIEEE 2011
(4)   1
9.
.
(2) ~ P  (Q ~ R)
(4) ~ Q ~ P  R
[2]
10.
The lines L1: y  x  0 and L2 : 2 x  y  0
value of  is
intersect the line L3 : y  2  0 at P and Q
respectivley. The bisector of the acute angle
(1) 5 3
(2) 2 3
(3) 3 2
(4) 2 5
between L1 and L2 intersects L3 at R.
Statement -1
16.
The ratio PR : RQ equals 2 2 : 5 .
f ( x) 
In any triangle, bisector an angle divides the triangle
into two similar triangles.
(1) Statement -1 is false, Statement -2 is true
(2) Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for statement-1
(3) Statement-1 is true, Statement -2 is true, Statement-2 is not correct explanation for Statement-1
(4) Statement -1 is true, Statement-2 is false
17.
The domain of the function
1
f ( x) 
x x
(1) ( , )
is
(2) ( , )  {0}
(3) (0, )
(4) ( , 0)
If the mean deviation about the median of the numbers a, 2a, . . ., 50 a is 50, then a equals
( 3, 1) and has eccentricity
(1) 5
(3) 3
2
5 is
(1) 5x 2  3 y 2  32  0
(2) 3x 2  5 y 2  32  0
(3) 5x  3 y  48  0
(4) 3x  5 y  15  0
2
2
2
19.
(1)
3
13
 A
4
16
(2)
(3)
13
 A1
16
(4) 1  A  2
The value of
z
(1) log 2
(3)

log 2
8
F
GH
j
e
j e j e
j
j
(2) 5
(4) 5
The value of p and q for which the function
2
3
2
is continuous for all x in R, are,
y 1 z  3

and
2

1
e
R| sin( p  1) x  sin x
,x0
x
||
f ( x)  S
q
,x0
|| x  x  x , x  0
|T x

log 2
2
the plane x  2 y  3z  4 is cos
AIEEE 2011
20.
(2)  log 2
If the angle between the line x 


1
3i  k and b  1 2i  3 j  6k , then
If a 
10
7




 
the value of 2a  b  a  b  a  2 b is
(1) 3
(3) 3
3
 A1
4
8 log(1  x )
dx is
1 x2
(4)
(2) 2
(4) 4
e
If A  sin 2 x  cos4 x then for all real x.
0
15.
and local minimum at 2
and 2
and 2
and local maximum at 2
Equations of the ellipse whose axes are the axes of
coordinates and which passes through the point
1
14.
t sin t dt
Then f has
(1) local maximum at 
(2) lcoal maximum at 
(3) lcoal minimum at 
(4) local minimum at 
A man saves Rs 200 in each of the first three months
of his services. 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) 21 months
(2) 18 months
(3) 19 months
(4) 20 months
2
13.
z
0
18.
12.
IJ
K
x
Statement -2
11.
FG
H
5
For x  0, 2 , define
I
JK
5
14 . Then
(1) p 
1
3
,q 
2
2
(2) p 
1
3
,q  
2
2
(3) p 
5
3
,q 
2
2
3
1
(4) p   , q 
2
2
[3]
21.
The number of ways of distributing 10 identical
balls in 4 distinct boxes such that no box is empty
The two circles x 2  y 2  ax and
x 2  y 2  c 2 (c  0) touch each other if
22.
(1) a  2c
(2) 2 a  c
(3) a  c
(4) a  2c
is 9 C3 .
Statement -2
The number of ways of choosing any 3 places
from 9 different places is 9 C3
(1) Statement -1 is false, Statement -2 is true
(2) Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for statement-1
(3) Statement-1 is true, Statement -2 is true, Statement-2 is not correct explanation for Statement-1
(4) Statement -1 is true, Statement-2 is false
Let be the purchase value of an equipment is I and
V ( t ) be the value after it has been used for t years.
The value V ( t ) depreciates at a rate given by differdV (t )
  k (T  t ) , where k  0 is
dt
a constant and T is the total life in years of the equipment. Then the scrap value V(T) of the equipment is
ential euqation
2
(2) T 
(1) e  kT
(3) I 
23.
24.
kT 2
2
(4) I 
1
k
curve x  y 2 is
4
k (T  t ) 2
2
If C and D are two events such that C  D and
P ( D )  0 , then the correct statement among the
following is
P ( D)
(1) P(C| D)  P(C )
(2) P(C| D)  P(C)
(3) P(C| D)  P(C)
(4) P(C| D)  P(C)
(1)
(3)
28.
3
4
8
3 2
8
(4)
3 2
The area of the region enclosed by the curves
b g
(1) 5/2 square units
(2) 1/2 square units
(3) 1 square units
(4) 3/2 square units
29.
A (BA) and (AB) A are symmetric matrices
If
bg
b g
dy
 y  3  0 and y 0  2, the y ln 2 is equal
dx
to:
(1) –2
(3) 5
(2) 7
(4) 13
(1) Statement -1 is false, Statement -2 is true
(2) Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for statement-1



The vectors a and b are no perpendicular and c and

   
d are two vectors satisfying : b  c  b  d and

 
a  d  0 . Then the vector d is equal to:
(3) Statement-1 is true, Statement -2 is true, Statement-2 is not correct explanation for Statement-1
(4) Statement -1 is true, Statement-2 is false
 

a c 
c

(1)
  b
a b

 b  c 
b

(2)
  c
a b
Statement -2
AB is symmetric matrix if matrix multiplication of
A with B commutative.
If  (  1) is a cube root of unity, and
(1   ) 7  A  B . Then (A, B) equals
(1) (–1, 1)
(2) (0, 1)
(3) (1, 1)
(4) (1, 0)
26
(2)
3
y  x, x  e, y  1 / x and the positive x-axis is
Let A and B be two symmetric matrices of order 3
Statement -1
25.
The shortest distance between line y  x  1 and
27
30.
FG IJ
H K
 
 F b c I 
(3) c  G   J c
H a b K
F I
GH JK

 F b  c I 
(4) b  G   J c
H a b K
Statement -1
AIEEE 2011
[4]
Part -B Chemistry
31.
32.
33.
In context of the lanthanoids, which of the following
statements is not correct?
(1) Availability of 4f electrons results in the formation of compounds in +4 state for all the members
of the series.
(2) There is a gradual decrease in the radii of the members with increasing atomic number in the series.
(3) All the members exhibit +3 oxidation state.
(4) Because of similar properties the separation of
lanthanoids is not easy.
37.
and NH 4 are respectively:
38.
(2) A 2 B
(3) AB2
(4) A 2 B3
The magnetic moment (spin only) of NiCl 4
. BM
(1) 141
. BM
(2) 182
(3) 5.46 BM
(4) 2.82 BM
is
Which of the following facts about the complex
b
Cr NH 3
g
6
(1) The complex gives precipitate with silver nitrate
solution.
35.
(1) 64 times
(3) 24 times
36.
(2) 10 times
(4) 32 times
'a' and 'b' are van der Waals' constants for gases. Chlorine is more easily liquefied than ethane because
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 –0.6 oC
The outer electron configuration of Gd (Atomic No :
64) is
(1) 4 f 7 5d 1 6s2
(2) 4 f 3 5d 5 6s2
(3) 4 f 8 5d 0 6s2
(4) 4 f 4 5d 4 6s2
The structure of IF7 is
Ozonolysis of an organic compound gives
formaldehyde as one of the products. This confirms
the presence of:
(1) an acetylenic triple bond
(2) two ethylenic double bonds
(3) a vinyl group
(4) an isopropyl group
42.
The degree of dissociation () of a weak electrolyte,
A x By is related to van't Hoff factor (i) by the expression:
(1)  
x  y 1
i 1
(2)  
(3)  
i 1
x  y 1
(4)  
(1) a for Cl 2  a for C 2 H 6 but b for Cl 2  b for C 2 H 6
(2) a and b for Cl 2  a and b for C 2 H 6
(2) 804.32 g
(4) 400.00 g
41.
The rate of a chemical reaction doubles for every
10 C rise of temperature. If the temperature is raised
by 50C the rate of the reaction increases by about:
(4) sp, sp 3 , sp 2
(1) pentagonal bipyramid
(2) square pyramid
(3) trigonal bipyramid
(4) octahedral
Cl 3 is wrong?
(2) The complex involves d 2 sp 3 hybridisation and is
octahedral in shape.
(3) The complex is paramagnetic
(4) The complex is an outer orbital complex.
(3) sp 2 , sp, sp 3
(1) 304.60 g
(3) 204.30 g
40.
34.
(2) sp, sp 2 , sp 3
mass of ethylene glycol  62 g mol 1 )
39.
2
(1) sp 2 , sp 3 , sp
will be: ( K f for water  186
. K kg mol 1 and molar
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 form one of the
face centred points, the formula of the compound is
(1) A 2 B5
The hybridisation of orbitals of N atom in NO 3 , NO 2
i 1
bx  y  1g
x  y 1
i 1
(3) a and b for Cl 2  a and b for C 2 H 6
(4) a for Cl 2  a for C 2 H 6 but b for Cl 2  b for C2 H 6
AIEEE 2011
[5]
43.
44.
45.
46.
47.
48.
A gas absorbs a photon of 355 nm and emists at two
wavelengths. If one of the emissions is at 680 nm, the
other is at:
(1) 518 nm
(2) 1035 nm
(3) 325 nm
(4) 743 nm
50.
Identify the compound that exhibits tautomerism.
(1) Phenol
(2) 2-Butene
(3) Lactic acid
(4) 2-Pentanone
51.
The entropy change involved in the isothermal
reversible expansion of 2 moles of an ideal gas from a
volume of 10 dm3 to a volume of 100 dm3 at 27C is
(1) 42.3 J mol 1 K 1
(2) 38.3 J mol 1 K 1
(3) 358
. J mol 1 K 1
(4) 32.3 J mol 1 K 1
Silver Mirror test is given by which one of the following compounds?
(1) Benzophenone
(2) Acetaldehyde
(3) Acetone
(4) Formaldehyde
Trichloroacetaldehyde was subjected to Cannizszaro's
reaction by using NaOH . The mixture of the products
contains sodium trichloroacetate and another
compound. The other compound is:
(1) Chloroform
(2) 2, 2, 2-Trichloroethanol
(3) Trichloromethanol (4) 2, 2, 2-Trichloropropanol
b g
(2) pb H g  1 atm and H
(3) pb H g  1 atm and H
(4) pb H g  2 atm and H

 2.0 M

 10
. M
2
2
2
49.
(1) MgCl 2
(2) FeCl 2
(3) SnCl 2
(4) AlCl 3
Boron cannot form which one of the following anions?
(1) BO 2
(2) BF63
(3) BH 4
(4) B OH
b g

4
52.
Sodium ethoxide has reacted with ethanoyl chloride.
The compound that is produced in the above reaction
is
(1) Ethyl ethanoate
(2) Diethyl ether
(3) 2-Butanone
(4) Ethyl chloride
53.
Which of the following reagents may be used to distinguish between phenol and benzoic acid
54.
(1) Neutral FeCl 3
(2) Aqueous NaOH
(3) Tollen's reagent
(4) Molisch reagent
A vessel at 1000 K 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 0.8 atm the value of K is
(1) 0.18
(2) 1.8
(3) 3
(4) 0.3
The reduction potential of hydrogen half cell will be
negative if:

(1) p H 2  2 atm and H  2.0 M
Among the following the maximum covalent character is shown by the compound:
55.
The strongest acid amongst the following compounds
is
(1) ClCH 2 CH 2 CH 2 COOH
(2) CH 3COOH
(3) HCOOH

b g
 10
. M
Phenol is heated with a solution of mixture of KBr and
KBrO 3 . The major product obtained in the above
reaction is:
(1) 2, 4, 6-Tribromophenol
(2) 2-Bromophenol
(3) 3-Bromophenol
(4) 4-Bromophenol
(4) CH 3CH 2 CH Cl CO 2 H
56.
Which one of the following orders presents the correct
sequence of the increasing basic nature of the given
oxides?
(1) K 2 O  Na 2 O  Al 2 O 3  MgO
(2) Al 2 O 3  MgO  Na 2 O  K 2 O
(3) MgO  K 2 O  Al 2 O 3  Na 2 O
(4) MgO  K 2 O  MgO  Al 2 O 3
AIEEE 2011
[6]
57.
A 5.2 molal aqueous solution of methyl alcohol,
59.
CH 3OH is supplied. What is the mole fraction of
methyl alcohol in the solution?
(1) 0.050
(2) 1.100
(3) 0.190
(4) 0.086
58.
Which of the following statement is wrong?
(1) N 2 O 4 has two resonance structures.
(2) The stability of hydrides increases form NH 3 to
BiH 3 in group 15 of the periodic table.
(3) Nitrogen cannot form d  p bond.
The presence or absence of hydroxy group on which
carbon atom of sugar differentiates RNA and DNA?
(1) 4th
(2) 1st
(3) 2nd
(4) 3rd
(4) Single N– N bond is weaker than the single
P  P bond.
60.
Which of the following statements regarding sulphur
is incorrect?
(1) The oxidation state of sulphur is never less than +4
in its compounds.
(2) S2 molecular is paramagnetic
(3) The vapour at 200 C consists mostly of S8 rings.
(4) At 600 C the gas mainly consists of S2
AIEEE 2011
[7]
72. A resistor 'R'and 2 F capacitor in series are connected
through a switch to 200 V direct supply. Across the
capacitor is a neon bulb that lights up at 120 V. Calculate
the value of R to make the bulb light up 5 s after the
b
switch has been closed. log10 2.5  0.4
g
(1) 3.3  107 
(2) 13
.  104 
(3) 17
.  105 
(4) 2.7  106 
73. A current I flows in an infinitely long wire with cross
section in the form of a semicircular ring of radius R. The
magnitude of the magnetic induction along its axis is:
(1)
0I
4R
(2)
0I
2R
(3)
0I
2 2 R
(4)
0I
2R
74. An object, moving with a speed of 6.25 m / s, is decelerated
at a rate given by
dv
 2.5 v
dt
where v is the instantaneous speed. The time taken by
the object, to come to rest, would be:
(1) 8 s
(2) 1 s
(3) 2 s
(4) 4 s
75. Direction:
The question has paragraph fllowed 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 plen-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 surffers a phase change of .
Statement - 2:
The centre of the interference pattern is dark.
(1) Statement - 1 is flase, Statement - 2 is true.
(2) Statement - 1 is true, Statement - 2 is flase.
(3) Statement - 1 is true, Statement - 2 is true and
Statement - 2 is the correct explanation of Statement - 1
(4) Statement - 1 is true, Statement - 2 is true and
Statement - 2 is not the correct explantion of
Statement - 1.
76. Two bodies of masses m and 4m are placed at a distance
r. The gravitational potential at a point on the line joining
them where the gravitational field is zero is:
9Gm
(1) 
r
(3) 
4Gm
r
AIEEE 2011
(2) zero
(4) 
6Gm
r
77. 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 signal 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 flase, Statement - 2 is true.
(2) Statement - 1 is true, Statement - 2 is flase.
(3) Statement - 1 is true, Statement - 2 is true and
Statement - 2 is the correct explanation of Statement - 1
(4) Statement - 1 is true, Statement - 2 is true and
Statement - 2 is not the correct explantion of
Statement - 1.
78. A fully charged capacitor C with initial charge q0 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)
LC
(2)  LC

LC
(4) 2 LC
4
79. This equation has Statement - 1 and Statement - 2. Of the
four choices given after the statements, choose the one
that best describe the two statements.
Statement - 1:
A metallic surface is irradiated by a monochromatic light
of frequency v  v0 (the threshold frequency). The
maximum kinetic energy and the stopping potential are
K max and V0 respectively. If the frequency incident on
(3)
the surface is doubled, both the K max and V0 are also
doubled.
Statement - 2:
The maximum kinetic energy and the stopping potential
of photoelectrons emitted from a surface are linearly
dependent on the frequency of incident light.
(1) Statement - 1 is flase, Statement - 2 is true.
(2) Statement - 1 is true, Statement - 2 is flase.
(3) Statement - 1 is true, Statement - 2 is true and
Statement - 2 is the correct explanation of Statement - 1
(4) Statement - 1 is true, Statement - 2 is true and
Statement - 2 is not the correct explantion of
Statement - 1.
80. Water is flowing continuously from a tap having an
internal diameter 8  10 3 m . The water velocity as it
leaves the tap is 0.4 m s 1 . The diameter of the water
stream at a distance 2  10 1 m below the tap is close to:
(1) 3.6  10 3 m
(2) 5.0  10 3 m
(3) 7.5  10 3 m
(4) 9.6  10 3 m
[9]
81. A mass M, attached to a horizontal spring, executes SHM
with amplitude A1 . 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 A2 . The ratio
FG A IJ
H A K is:
1
of
2
F M  mIJ
(1) G
H M K
1
2
(2)
M
M m
F M IJ
(4) G
H M  mK
M m
(3)
M
1
2
82. Two particles are executing simple harmonic motion of
the same amplitude A and frequency  along the
x-axis. Their mean position is separated by distance
b
g
86. A screw gauge gives the following reading when used to
measure the diameter of a wire.
Main scale reading:
0mm.
Circular scale reading:
52 dividisons
Given that 1 mm on main scale corresponds to 100
divisions of the circular scale.
The diameter of wire from the above data is
(1) 0.005 cm
(2) 0.52 cm
(3) 0.052 cm
(4) 0.026 cm
87. A mass 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
ciruclar disc, the acceleration of the mass m, if the string
does not slip on the pulley, is
(1)
g
3
(3) g
(2)
3
g
2
(4)
2
g
3
X 0 X 0  A . If the maximum separation between them
b
g
is X 0  A , the phase difference between their motion
is
b g
88. The transverse displacement y x, t of a wave on a string
is given by
b g
e
 ax 2 bt 2  2 abxt
j

(1)
6

(2)
2
This represents a:

(3)
3

(4)
4
(1) standing wave of frequency
y x, t  e
1
83. If a wire is stretched to make it 0.1% longer, its resistance
will
(1) decrease by 0.05%
(2) increase by 0.05%
(3) increase by 0.2%
(4) decrease by 0.2%
84. 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) 
(3) 
v2
g
(2) 
2
v4
(4)
g2
v2
g
 v4
2 g2
85. A thermally insulated vessel contains an 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 increases by:
(1)
b  1g Mv K
(3)
b  1g Mv K
2
2R
2
2R
AIEEE 2011
b g
b g
 1
2
(2) 2   1 R Mv K
(4)
Mv 2
K
2R
b
(2) wave moving in +x direction with speed
a
b
(3) wave moving in –x direction with speed
b
a
(4) standing wave of frequency
b
89. A car is fitted with a convex side-view mirror of focal
length 20 cm. A second car 2.8 m 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) 15 m / s
(3)
(2)
1
m/s
15
1
m/ s
10
(4) 10 m / s
90. Let the x - z plane be the boundary between two
transparent media. Medium 1 in z  0 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 A  6 3i  8 3 j  10k is incident on the plane
of separation. The angle of refraction in medium 2 is:
(1) 75o
(2) 30o
(3) 45o
(4) 60o
[ 10 ]
Triumph Academy
AIEEE-2011 Main Test Paper
Solutions
AIEEE 2011MAIN TEST PAPER SOLUTIONS
Mathematics (Solution)
1. [4]
[For   1 , roots are equal]
P (at least one failure) =1 - P (all success)
1 1 p 
5
5

0  p  1

1
0  p5 
2
31
32
31
1

32 32
FG IJ 5
H K
L 1O
p  M 0, P
N 2Q

F
H
2
I6
K
e
C0  C1 x  C2 x ... C6 x
6
e
2
6
je
j
6
j .e
6
e
C0  C1 x  C2 x ... C6 x
6
2
j e
6
je
4
  C0   C5  C2  C3   C3  C1
6
6
6
6
6
6
6
12
j
j
4
1  cos 2 ( x  2 )

x2
 lim
x 2
k
2
k
4 1 0
2
2 1
k 2  6k  8  0
( k  2) ( k  4)  0  k  2, 4
x2
Number of solution = 2
 lim
x 2 
 2 sin( x  2 )
x2
8. [3] mid pt of AB  (1, 3, 5) lies on the line
dr of AB  ( 0, 6,  4)
dr of given line  (1, 2 , 3)
2 sin( x  2 )
x 2
d 2 y dx

dx 2 dy

 2
RHL = lim


x2
x 2
FG dy IJ
H dx K
2
2 sin( x  2)
2 sin( x  2 )
LHL = lim
1
I
JK
7. [3] For non-zero
 144
x 2
d dx
d
1

dy dy
dy dy / dx
FG IJ
H K
 36  ( 300)  120
3. [2] lim

d2y
2
d 2x
  dx 3
2
dy
dy
dx
= Coeff of x 7 in the expansion of
6
2

2. [4] Coeff of x 7 in the expansion of (1  x ) (1  x )
6
F I F
GH JK GH
dy
d F 1 I dx

G J
dx H dy / dx K dy
d2x
6. [1]
x2
 lim
x 2 
2 sin( x  2 )
 2
x2
lines are perpendicular 
0 (1)  6( 2)  ( 4 ) ( 3)  0


AB is perpendicular to the given line
B is mirror image of A
Statement (1) is true .
Statmeent (2) is also true But not sufficient.
LHL  RHL
The limit does not exist
4. [4] Statement-1 ( x , y )  A  ( x  x ) integer
9. [3] Given Statement is equivalent to
x  x  0  x  A is reflexive
( P  ~ R)  Q
( y  x ) integer  ( x  y ) integer
b
g
( y  x )  ( z  y )  ( z  x ) integer

Hence x Az  A in transitive

g b
Negation is ~ ( P  ~ R )  Q  ~ (Q  ( P  ~ R)
A is symmetric

( y  x ) integer and ( z  y ) integer
L1
L2
A is equivalance
Statement-2 ( x , y )  B  x  y , for  , some rational
( 0, 1)  B as 0  0.1, 0 is a rational no.
but (1, 0)  B as 1   . 0 there is no rational  existing.
Hence B is not symmetric
O
10. [4]
Q
P
(–2, –2)
R
(1, –2)
L3
5. [1] The coefficients of quadratic equation
z 2  z    0
are real. Therefore complex roots exists in conjugate pair. Let
z1  1  iy , z2  1  iy
z1  z2      2
  z1z2  1  y 2  (1,  )
PR OP
2 2

PQ OQ 
5
Statement (1) is True
Statment (2) is False
11. [1] Let number of months = n
PCM
81-B/3, Lohagal Road, Ajmer 305001 Tel: 0145-2628805, 2628178
[1]
Triumph Academy
AIEEE-2011 Main Test Paper
Solutions
Total saving = 200 + 200 +
z
LM
MN
 /4
d200  240  280... to (n  2) termsi
2I 
8 log(1  tan  )  log
0
n2
2 ( 200)  ( n  3) 40
2
d
11040  400 

d
10640  ( n  2 ) 200  ( n  3) 20

i
i
8 log 2  d  8 log 2
0
FI
GH 4 JK
I   log 2
532  ( n  2) (10  n  3)

z
 /4

F 2 I OP  d
GH 1  tan  JK P
Q

 ( n  2) (n  7)

n 2  5n  546  0
dr's of line :

( n  21) (n  26)  0
dr's of normal plane (1, 2, 3)

n  21 ( n  0)
15. [2]
so,

Let angle between line and plane is
then
(1, 2,  )
1  1  2.2    3
cos( 90   ) 
  4 1  9  4 1
2
2
12. [2] e  1 

b
2
a
2
b2
2  5
14
 1  e  1
2
2 3

5 5
given cos 
a 2 : b2  5 : 3

Equation of ellipse

14   5
2
(1) passes through ( 3, 1)
( 3)
(1)
9 1 32
k

  
5
3
5 3 15
Equation of ellipse
2
5
3
 sin  
14
14
3  5
Hence
2
x2 y

 k . . . (1)
5
3

3  5
sin  
a2

9(   5)  ( 3  5)

30  20


2
3
14
2
2
2
3
z
x
2
x2 y
32


5
3
15
16. [1]
applying newton leibnitz formula
13. [2] A  sin 2 x  cos 4 x
f’
 cos x  cos x  1
4
2
FG
H
 cos2 x 
1
2
IJ
K
F
Now 0  G cos
H

2

I
z
1 x
Local minima
IJ
K
f  ( x )  ( x  sin x )  1  0
2

1

4
z
z
0

2
17. [4]
 dx
8 log(1  tan  )
1  tan 2 
1
Now x  x  0 . . . (1)
case 1  x  0
then
xx0
.sec  dx
2
00 x 
8 log(1  tan  ) d .
. . (1)
0
L F  IO
I  z 8 log M1  tan G   J P . d
MN H 4 K PQ
L (1  tan  ) OP  d
I  z 8 log M1 
N 1  tan  Q
F 2 I  d . . . (2)
I  z 8 log G
H 1  tan  JK
case 2 x  0 then
(1)   x  x  0

 /4
also
f ( x) 
x x
 /4
I
f  ( x )  x  sin x
Hence f ( x ) is maximum at x  
and local minima at x  2
dx  sec 2 
I
+
2
Local maxima
x  tan 
 /4
–
0
3
4
1
x
2
8 log(1  x )
0
+
f
3
 A1
4
1
14. [2]
2
t sin t dt
0
3x 2  5 y 2  32

f ( x) 
2 x  0
x  0  x R
Hence x  (  , 0)
0
 /4
18. [4]
xi  a , 2a . . . 25a, 26a, . . . 50a
0
Median =
 /4
0
e
25a  26a
 25.5 a
2
j
Now xi  x  24.5 a , 23.5 a . . . 0.5 a , 0.5 a . . .
using equation (1) and (2)
23.5 a , 24.5 a
PCM
81-B/3, Lohagal Road, Ajmer 305001 Tel: 0145-2628805, 2628178
[2]
Triumph Academy

FH
xi  x  2 0.5 a  a  15
. a ...24.5 a
LM
N
25
0.5 a  24.5 a
2
 2
AIEEE-2011 Main Test Paper
Solutions
IK
22. [3]
OP
Q
d i
z zd
V T
I
Mean deviation 
625 a
50

xi  x

n
F
t I
V dT i  I  G  kTt  k J
2K
H

 
 

( 2 a  b )  ( a  b )  ( a  2b )
23. [3]
 
  
  
 ( 2 a  b )  ( a  b )  a  2( a  b )  b
LM
N
 
     
     
 ( 2 a  b )  ( a  a ) b  ( a  b ) a  2 ( a  b ) b  (b  b ) a
{
  

 ( 2 a  b )  (b  2 a )
2 2
 4 a  b
}OPQ


24. [3]
  4  1  5
di
V T I
kT 2
2
F C I  Pd C  D i  P d C i
GH D JK Pd Di Pd Di
O  Pd Di  1
F CI
P G J  P dC i
H DK
P
d i d i
d ABAi  A B
20. [4]
x 0
 lim


x 3/ 2

1 x 1
 lim
x 0
x
(1  x )  1
1

x 0
x
1 x 1
25. [3]
 lim
sin( p  1) x  sin x
q  lim f ( x )  lim


x
x 0
x 0
x 0

d ABi
T
 B T A T  BA  AB
AB is symmetric
statement 2 is TRUE but NOT a correct explanationof statement 1.
b
A  B  1  
A  B  1  

A  1, B  1
26 [3]
LM sin( p  1) x  ( p  1)  sin x OP
x Q
N ( p  1) x
p q2 
A T  ABA
statement 1 is TRUE

g  e j
7
2
7
  14
Statement 1 number of way's of distributing n identical things into m
person when each receive at least one
 n 1Cm 1
hence total ways  101 C41  9 C3
 ( p  1)  1  p  2

T
A  B   2
1

2
 lim
T
Given AB  BA
x  x2  x
x 0 
( C  D )
A BA  AB A  ABA
T
q  lim f ( x )
0
T2
kT 2

2
2
2

T
2
  kT  k


 
a  1, b  1, a . b  0
19. [2]
i
 kT  kt dt
0
 50
a 4

T
dV 
 625 a

dV
  kT  kt
dt
Statement 2 is correct but not correct explanation of staement 1.
1
3
2 
2
2
27 [3]
e j on the parabola. Distance of the line
2
Let a point P t , t
y  x 1
from P

t2  t 1
t  t 1
2

2
2
FG t  1 IJ
H 2K

21. [3]
2


t2  t 1 0
3
4
2
2
2
Diameter of x  y  ax = Radius of
x 2  y 2  c2 
a c
3
 4
2

3
4 2
PCM
(when t 

1
)
2
3 2
8
81-B/3, Lohagal Road, Ajmer 305001 Tel: 0145-2628805, 2628178
[3]
Triumph Academy
   
30. [1] b  c  b  d
z z
1
e
28. [4] Area  x dx  1 dx
x
0
1
Fx I
 G J  d ln x i
H2K
2
1
e
1

1
3
1
2
2
0
29. [2]
dy
 y3
dx
d i
Now y d 0i  2


PCM
ln y  3  x  C
y  5e x  3
AIEEE-2011 Main Test Paper
Solutions

dy
 dx
y3

y  3  C e
 23 C

d i
   
b  c d 0
e
j


 
c  d  b

 

d  c  b
Now
 
a d  0

 
 
a  c   a b  0

 
a c
  
a b

e j
  F a  c I 
d  c  G   Jb
H a b K
x
 C 5
y ln 2  5e
ln 2
 3 7
81-B/3, Lohagal Road, Ajmer 305001 Tel: 0145-2628805, 2628178
[4]
[1]
AIEEE 2011
Mathematics
1.
Consider 5 independent Bernulli's trails each with
probability of sucess p. If the probability of at least
one failure is greater than or equal to
lies in the interval
F 11 O
(1) G 12 , 1P
H Q
F 3 11 O
(3) G 4 , 12 P
H Q
Ans:
Ans:
x2
F 1 3O
(2) G 2 , 4 P
H Q
F 1O
(4) G 0, 2 P
H Q
LHL =  lim
x2
RHL = lim
x2


4.
5
2.
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:
[4] Coeff of x 7 in the expansion of (1  x ) (1  x 2 )
e
j
Coeff of x 7 in the expansion of
6
0
6
je
j
 e C j  e  C j  C e C j  e C je  C j
C1x  6C2 x 2 ... 6C6 x 6 . 6 C0  6 C1x 2  6C2 x 4 ... 6C6 x12
6
6
0
6
6
5
2
6
3
6
3
 36  ( 300)  120
3.
F
limG
GH
x2
 144
1  cos{2( x  2)}
x2
2
(3) equals 2
1
Let R be the set of real numbers
Statement -1
A  {( x , y )  R  R : y  x is an integer} is an
equivalence relation on R.
Ans:
[4] Statement-1 ( x , y )  A  ( x  x ) integer
x  x  0  x  A is reflexive
( y  x ) integer  ( x  y ) integer
A is symmetric

( y  x ) integer and ( z  y ) integer

( y  x )  ( z  y )  ( z  x ) integer
Hence x Az  A in transitive
A is equivalance

Statement-2 ( x , y )  B  x  y , for  , some rational
( 0, 1)  B as 0  0.1, 0 is a rational no.
I
JJ
K
but (1, 0)  B as 1   . 0 there is no rational 
existing.
Hence B is not symmetric
1
(1) equals
2 sin( x  2)
2 sin( x  2)
 lim
 2
x2
x2
x 2 
Statement -2
B  {( x , y )  R  R : x  y for some rational
number  } is an equivalence relation on R.
(1) Statement -1 is false, Statement -2 is true
(2) Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for statement-1
(3) Statement-1 is true, Statement -2 is true, Statement-2 is not correct explanation for Statement-1
(4) Statement -1 is true, Statement-2 is false
6
Coeff of x 7 in the expansion of (1  x ) 6 (1  x 2 ) 6
eC
2 sin( x  2)
 2 sin( x  2)
 lim
x2
x2
x2 
LHL  RHL
The limit does not exist
31 1

32 32
FG 1 IJ
H 2K
L 1O
p  M0, P
N 2Q
2 sin( x  2)
x2
 2
31
1 1 p 
32
0  p5 
1  cos 2( x  2)
x2
 lim
5
0  p5  1 
x2
31
, then p
32
[4] P (at least one failure) =1 - P (all success)

[2] lim
(2) does not exist
(4) equals  2
[2]
5.
AIEEE 2011
Let  ,  be real and z be complex number. If
7.
z 2  z    0 has two distinct roots on the line
Re z  1 , then it is necessary that
(1)   (1, )
(2)  (0, 1)
Ans:
kx  4 y  z  0
2x  2 y  z  0
possess a non-zero solutoin is
(1) zero
(2) 3
(3) 2
(4) 1
(4)   1
(3)   ( 1, 0)
The number of values of k for which the linear
equations
4 x  ky  2 z  0
[1] The coefficients of quadratic equation
z 2  z    0
are real. Therefore complex roots exists in conjugate pair.
Let
Ans:
[3] For non-zero
4
k
2
z1  1  iy , z2  1  iy
k
4 1 0
z1  z2      2
2
2 1
  z1z2  1  y 2  (1,  )

k 2  6k  8  0
[For   1 , roots are equal]

( k  2) ( k  4)  0  k  2, 4

Number of solution = 2
2
6.
d x
equal
dy 2
8.
F d y I F dy I
(1) – G dx J GH dx JK
H K
F d y I F dy I
(3) – G
H dx JK GH dx JK
F d yI
(2) G
H dx JK
F d y I F dy I
(4) G dx J GH dx JK
H K
3
2
2
1
2
[1]
3
2
FG IJ
H K
FG
H
FG
H
IJ
K
d
d
dx
dy dy / dx dy

1
FG dy IJ
H dx K
2
d 2 y dx
 2 
dx dy
d2y
2
d 2x
  dx 3
2
dy
dy
dx
FG IJ
H K
x y 1 z  2


1
2
3
Statement -2
x y 1 z  2


bisects the line
1
2
3
segment joining A (1, 0, 7) and B (1, 6, 3)
(1) Statement -1 is false, Statement -2 is true
(2) Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for statement-1
(3) Statement-1 is true, Statement -2 is true, Statement-2 is not correct explanation for Statement-1
(4) Statement -1 is true, Statement-2 is false
2
The line
2
d 2 x d dx
d
d


2
dy
dy
dy
dy
/ dx
dy

point B(1, 6, 3) in the line
2
2
Ans:
1
2
Statement -1
The point A (1, 0, 7) is the mirror image of the
IJ
K
Ans:
[3] mid pt of AB  ( 2 , 3, 5) lies on the line
dr of AB  ( 0, 6,  4)
dr of given  (1, 2 , 3)
0 (1)  6( 2)  ( 4 ) ( 3)  0
AB
is perpendicular to the given line

B is mirror image of A

Statement (1) is true .
Statmeent (2) is also true But not sufficient.
[3]
9.
.
Consider the following statements
P : Suman is brilliant
Q : Suman is rich
R : Suman is honest
The negation of the statment "Suman is brilliant
and dishonest if and only if sumna is rich" can be
expressed as
(1) ~ ( P ~ R)  Q
(2) ~ P  (Q ~ R)
(3) ~ (Q  ( P  ~ R))
Ans:
AIEEE 2011
11.
A man saves Rs 200 in each of the first three months
of his services. 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 wil be Rs. 11040 after
(1) 21 months
(2) 18 months
(3) 19 months
(4) 20 months
Ans:
[1] Let number of months = n
(4) ~ Q ~ P  R
Total saving = 200 + 200 +

d200  240  280... to (n  2) termsi
n2
11040  400 
c2(200)  (n  3)40h
2
10640  (n  2) c200  (n  3)20h
The lines L1: y  x  0 and L2 : 2 x  y  0

532  ( n  2) (10  n  3)
intersect the line L3 : y  2  0 at P and Q
respectivley. The bisector of the acute angle

[3] Given Statement is equivalent to
( P  ~ R)  Q

Negation is ~ ( P  ~ R)  Q
10.

between L1 and L2 intersects L3 at R.
Statement -1
The ratio PR : RQ equals 2 2 : 5 .
Statement -2
In any triangle, bisector an angle divides the trianle
into two similar triangles.
(1) Statement -1 is false, Statement -2 is true
(2) Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for statement-1
(3) Statement-1 is true, Statement -2 is true, Statement-2 is not correct explanation for Statement-1
(4) Statement -1 is true, Statement-2 is false
Ans:
12.
 ( n  2 ) (n  7 )

n 2  5n  546  0
( n  21) (n  26)  0

n  21 (Q n  0)
Equations of the ellipse whose axes are the axes of
coordinates and which passes through the point
( 3, 1) and has eccentricity
Ans:
2
5 is
(1) 5x 2  3 y 2  32  0
(2) 3x 2  5 y 2  32  0
(3) 5x 2  3 y 2  48  0
(4) 3x 2  5 y 2  15  0
2
[2] e  1 
[4]
b2
a2

b2
2 3
 1  e2  1  
5 5
a2

a 2 : b2  5 : 3
Equation of ellipse
x2 y2

 k . . . (1)
5
3
(1) passes through ( 3, 1)

( 3) 2 (1) 2 9 1 32

  
5
3
5 3 15
Equation of ellipse
k
x 2 y 2 32


5
3
15
PR OP

PQ OQ

2 2
5
Statement (1) is ture
Statment (2) is False

3x 2  5 y 2  32
[4]
13.
If A  sin 2 x  cos4 x then for all real x.
(1)
(3)
Ans:
AIEEE 2011
3
13
 A
4
16
(2)
13
 A1
16
(4) 1  A  2
15.
3
 A1
4
If the angle between the line x 
y 1 z  3

and
2

1
the plane x  2 y  3z  4 is cos
[2] A  sin 2 x  cos4 x
(1) 5 3
(2) 2 3
(3) 3 2
(4) 2 5
F
GH
5
14
I
JK
 cos4 x  cos2 x  1
FG
H
 cos2 x 
1
2
IJ
K
2

Ans:
3
2
FG
H
1
2
Now 0  cos x  2
IJ
K
[2] Let angle between line and plane is  then
dr's of line : (1, 2,  )
2

dr's of normal plane (1, 2, 3)
1
4
3
 A 1
4

z
1
14.
The value of
0

log 2
8
(3)
z
1
Ans:
[2] I 
0
14
(4)

log 2
2
x  tan 
z
 /4

0
I
z
LM
MN
8 log 1  tan 
0
z
z
 /4

0
 /4

0
16.
8 log(1  tan  )
.sec 2  dx
1  tan 2 
 /4
also
z LMMN
8 log(1  tan  )  log
0
z
 /4

8 log 2  d  8 log 2
0
I   log 2
FG  IJ
H 4K
3
14

9( 2  5)  (3  5) 2

30  20


2
3
FG
H
IJ
K
5
For x  0, 2 , define
z
t sin t dt
0
Then f has
(1) local maximum at 
(2) lcoal maximum at 
(3) lcoal minimum at 
(4) local minimum at 
z
and local minimum at 2
and 2
and 2
and local maximum at 2
x
Ans:
[1] f ( x ) 
t sin t dt
0
using equaiton (1) and (2)
2I 

x
8 log 1 
 /4
14   5
2
f ( x) 
FG    IJ OP . d
H 4 K PQ
LM (1  tan ) OP  d
N 1  tan Q
F 2 IJ  d . . . (2)
8 log G
H 1  tan K
5
3
 sin  
14
14
3  5
Hence
8 log(1  x )
 dx
1  x2
dx  sec 2 
2  5
given cos 
(2)  log 2
(1) log 2
  4 1  9  4 1
2
3  5
sin  
8 log(1  x )
dx is
1 x2
1  1  2.2    3
cos( 90   ) 
so,
FG 2 IJ OP  d
H 1  tan K PQ
applying newton leibnitz formula
[5]
AIEEE 2011
 x  x  2e0.5 a  a  15. a ...24.5 a j
i
 2
f ( x)  ( x  sin x )  1  0
 625 a
f ( x )  x  sin x

Mean deviation 
Hence f ( x ) is maximum at x  
and local minima at x  2
17.
1
f ( x) 
x  x is
(3) (0, )
Ans:

The domain of the function
(1)
[4]
(4) ( , 0)
f ( x) 

(2) ( , )  {0}
19.
x x
case 1  x  0
xx0
Ans:

2 x  0
x  0  x R
Hence x  ( , 0)
(2) 2
(4) 4
xi  a , 2a . . . 25a, 26a, . . .
25a  26a
 25.5 a
2
i
Now xi  x  24.5 a , 23.5 a . .. 0.5 a , 0.5 a . . .
23.5 a , 24.5 a
 1
If a  1 3i  k and b  2i  3 j  6k , then
7
10

 
  
the value of 2a  b  a  b a  2b is
e
e
j
je
je
j
j
(2) 5
(4) 5


 
[2] a  1, b  1, a . b  0
o

 

 ( 2 a  b )  (b  2 a )
t
2
2
 4 a 4 b
bers a, 2a, . . ., 50 a is 50, then a equals
d
a 4
 
     
     
 (2a  b )  ( a  a ) b  (a  b ) a  2 (a  b ) b  (b  b ) a
If the mean deviation about the median of the num-
Median =
 50
 
  
  
 ( 2a  b )  ( a  b )  a  2 ( a  b )  b
x  x  0
[4]
50a
i
n

 
 

(2a  b )  (a  b )  (a  2b )
case 2 x  0
Ans:
50
 x x
(1) 3
(3) 3
00 x 
(1) 5
(3) 3
625 a
e
1
Now x  x  0
18.
25
0.5 a  24.5 a
2
  4  1  5
20.
The value of p and q for which the function
R| sin( p  1) x  sin x
,x0
x
||
f ( x)  S
q
,x0
|| x  x  x , x  0
|T x
2
3
2
is continuous for all x in R, are,
(1) p 
1
3
,q 
2
2
(2) p 
1
3
,q  
2
2
(3) p 
5
3
,q 
2
2
3
1
(4) p   , q 
2
2
[6]
Ans:
AIEEE 2011
Ans:
[4] q  lim f ( x )
x 0
x 0
 lim
x0
z zb
bg
V T
x  x2  x
x 3/ 2
 lim
dV
  kT  kt
dt
[3]
T
dV 
I
1 x 1
x
0
F
GH
bg
t2
 V T  I   kTt  k 2
(1  x )  1
1

x
x0
1 x 1
 lim
  kT 2  k
1
2

q  lim f ( x )  lim
x 0
x 0 
sin( p  1) x  sin x
x
L sin( p  1) x  ( p  1)  sin x OP
 lim M
x Q
N ( p  1) x

 ( p  1)  1  p  2
21.
1
3
pq2 2 
2
2
x  y  c ( c  0) touch each other if
2
2
(1) a  2c
(2) 2 a  c
(3) a  c
(4) a  2c
Ans:


[3]
(2) P(C| D)  P(C)
(3) P(C| D)  P(C)
(4) P(C| D)  P(C)
[3] P
kT 2
2
(4) I 
[3] A BA  AB A  ABA
b g b g
b ABAg

1
k
k (T  t ) 2
2
( C  D )
b g
F CI
P G J  PbC g
H DK
differential euqation
(3) I 
FG C IJ  PbC  Dg  PbCg
H D K P b Dg P b Dg
O  P D 1
The value V ( t ) depreciates at a rate given by
2
(2) T 
kT 2
2
Ans:
Let be the purchase value of an equaipment and
V ( t ) be the value after it has been used for t years.
(1) e  kT
T2
kT 2

2
2
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 commutative.
(1) Statement -1 is false, Statement -2 is true
(2) Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for statement-1
(3) Statement-1 is true, Statement -2 is true, Statement-2 is not correct explanation for Statement-1
(4) Statement -1 is true, Statement-2 is false
x 2  y 2  c2
dV (t )
  k (T  t ) , where
dt
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
0
24.
Diameter of x 2  y 2  ax = Radius of
22.
T
P ( D)
(1) P(C| D)  P(C )
The two circles x 2  y 2  ax and
2
Ans:
bg
V T I
I
JK
If C and D are two events such that C  D and
P ( D )  0 , then the correct statement among the
following is
23.
x0

g
 kT  kt dt
 AT B T AT  ABA
statement 1 is TRUE
Given AB  BA
b ABg

T
T
 B T AT  BA  AB
AB is symmetric
statement 2 is TRUE but NOT a correct explanationof
statement 1.
[7]
AIEEE 2011
3
4

2
If  (  1) is a cube root of unity, and
25.
(1   ) 7  A  B . Then (A, B) equals
(1) (–1, 1)
(2) (0, 1)
(3) (1, 1)
(4) (1, 0)
b
[3] A  B  1  
Ans:
g  e j
7
2

26
3
4 2
28.
The area of the region enclosed by trhe curves
b g
(1) 5/2 square units
(2) 1/2 square units
(3) 1 square units
(4) 3/21 square units
A  1, B  1
Statement -1
The number of ways of distributing 10 identical
balls in 4 distinct boxes such that no box is empty
is C3 .
Statement -2
The number of ways of choosing any 3 places
z z
1
from 9 different places is C3
(1) Statement -1 is false, Statement -2 is true
(2) Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for statement-1
(3) Statement-1 is true, Statement -2 is true, Statement-2 is not correct explanation for Statement-1
(4) Statement -1 is true, Statement-2 is false
0

29.
F x I  bln xg
GH 2 JK
2
1
e
1
Ans: [2]

Statement 2 is correct but not correct explanation of
staement 1.
curve x  y 2 is
4
(2)
3
3
4
dy
 y3
dx

(4)
3 2
e j
Ans: [3] Let a point P t 2 , t on the parabola. Distance of the
line y  x  1 from P
 23 C
b g



The vectors a and b are no perpendicular and c and

   
d are two vectors satisfying : b  c  b  d and

 
a  d  0 . Then the vector d is equal to:

 b  c 
(2) b    c
a b
F I
GH JK

 F b  c I 
(4) b  G   J c
H a b K
FG IJ
H K
 
 F b c I 
(3) c  G   J c
H a b K


t  t 1
2
FG t  1 IJ
H 2K


2
2

t  t 1
3
4
2
2
 t2  t  1  0
 C5
 y ln 2  5eln 2  3  7
y  5e x  3


   
b  c d 0

e
Ans: [1] b  c  b  d
2
dy
 dx
y3
 y  3  C  ex
 

a c 
(1) c    b
a b
8
3 2
8
30.
b g
(2) 7
(4) 13
b g
Now y b0g  2
hence total ways  10 1C4 1  9C3
1
3
1
2
2
bg
 ln y  3  x  C
The shortest distance between line y  x  1 and

dy
 y  3  0 and y 0  2, the y ln 2 is equal
dx
If
to:
(1) –2
(3) 5
 n 1Cm 1
(3)
1
1
dx
x
0
Ans: [3] Statement 1 number of way's of distributing identical
things into m person when each receive at least one
e
Ans: [4] Area  x dx 
9
(1)
3 2
8
y  x, x  e, y  1 / x and the positive x-axis is
9
27

7
  14   2  1  

1
)
2
(when t 


 
c  d  b
 
Now a  d  0

 
a c
  
a b



j
 

d  c  b
 
 
a c   a b  0
e j
  F a  c I 
d  c  G   Jb
H a b K
[1]
AIEEE 2011
PHYSICS
PART C — PHYSICS
61. A Carnot engine operating between temperatures T1 and
63. Three perfect gases at absolute temperatures T1 , T2 and
1
. When T2 is lowered by 62 K, its
6
T3 are mixed. The masses of molecules are m1 , m2 and
T2 has efficiency
1
efficiency increases to
. Then T1 and T2 are,
3
respectively:
(1) 310 K and 248 K
(2) 372 K and 310 K
(3) 372 K and 330 K
(4) 330 K and 268 K
Sol: [2] As on reducing T2 , efficiency is increasing means T1  T2 .
So, from   1 
1
and 1 
T2
,
T1
T2 1

T1 6
T2  62 1

T1
3
m3 and the number of molecules are n1 , n2 and n3
respectively. Assuming no loss of energy, the final
temperature of the mixtures is:
n12 T12  n22 T22  n32 T32
(1)
n1T1  n2 T2  n3T3
(2)
1
2
3
3
(3)
n1T1  n2 T2  n3T3
n1  n2  n3
(4)
n1T12  n2 T22  n3T32
n1T1  n2 T2  n3T3
Sol: [3] From conservation of internal energy
b
g
n1Cv T1  n2 Cv T2  n3Cv T3  n1  n2  n3 TCv
Solving we get T2  310 K and T1  372 K
62. A pulley of radius 2 m is rotated about its axis by a force
e
bT  T  T g
j

T
n1T1  n2T2  n3T3
n1  n2  n3
F  20t  5t 2 newton (where t is measured in seconds)
applied tangentially. If the moment of inertia of the pulley
about is axis of rotation is 10 kg m2 , the number of
rotations made by the pulley before its direction of motion
if reversed, is:
(1) more than 9
(2) less than 3
(3) more than 3 but less than 6
(4) more than 6 but less than 9
Sol: [3]  
z ze

or
jb g
e
20t  5t 2 2


I
10
t
d 
0

j
4t  t 2 dt

  2t 2 
t3
3
t3
0
3

t  6s
d
t3
 2t 2 
dt
3
z FGH
6
 
0
2t 2 
I
JK
t3
dt  36 rad
3
So number of rotations 
than 6 bat greater than 3.
the speed of the boat is 150
. ms1 , the magnitude of the
induced emf in the wire of aerial is:
(1) 0.15 mV
(2) 1 mV
(3) 0.75 mV
(4) 0.50 mV
  
Sol: [1] As B  v  
  Bv  5  105  15
.  2  0.15  10 3 V
0
  2t 2 
magnetic field is 5.0  10 5 N A 1 m 1 due north and
horizontal. The boat carries a vertical aerial 2 m long. If
so, motional emf is
The direction of motion reverses when  becomes zero.

64. A boat is moving due east in a region where the earth's
36 18

 5.73 which is lesser
2 
65. 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 dimater of the disc to reach its other end. During
the journey of the insect, the angular speed of the the
disc:
(1) first increases and then decreases
(2) remains unchanged
(3) continuously decreases
(4) continuously increases
Sol: [1] As during the said journey or insect, moment of inertia
first decreases then increases so from angular momentum
conservation, angular speed first increases then decreases.
[2]
AIEEE 2011
PHYSICS
66. Two identical charged spheres suspended from a common
point by two massless strings of length  are initially a
b
distance d d  
g
Sol: [4] As the amount present is reducing by half amount in the
given time interval so the time interval must be equal to a
half life which according to question is 20 min.
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
(2) v  x
(3) v  x 1
(4) v  x

1
2

1
2
69. Energy required for the electron excitation in Li  from
the first to the third Bohr orbit is:
(1) 122.4 eV
(2) 12.1 eV
(3) 36.3 eV
(4) 108.8 eV
Sol: [4] For hydrogen like ions, the excitation energy is given by
E  136
. Z2
or
70. The electrostatic potential inside a charged spherical ball
or
l
x
3/ 2
FG 2k IJ q
H mg K
l
T
x
kq 2

2 mgx 2
x3 
F
T
F
x
d
mg
2
is given by   ar 2  b where r is the distance from the
centre ; a, b, are constants. Then the charge density
inside the ball is
mg
(1) 6 a 0
(2) 24 a 0 r
(3) 6 a 0 r
(4) 24 a 0
dV
we have E  2ar
dr
As the electric field is directly proportional to r so the
charge density must be constant. For a uniform spherical
Sol: [1] From E  
2 k
q
mg

Differentiating w.r.t. time we have
1
3 2
x v
2

vx

charge distrubition from E 
2 k dq
 constant
mg dt
r
we get
3 0
r
 2ar
3 0
1
2

67. 100 g of water is heated fvrom 30 oC to 50 oC. Ignoring the
slight expansion of the water, the change in its internal
energy is (specific heat of water is 4184 J / kg / K):
(1) 2.1 kJ
(2) 4.2 kJ
(3) 84 kJ
(4) 8.4 kJ
Sol: [4] On ignoring expansion, from first law of thermodynamics,
we have
Q  U
or
2
2
E  108.8 eV
kq 2 / x 2
mg
x/2
For d  , tan  

so
2
1
For Z  3, and n1  1, n2  3 , on solving we get
Sol: [2] For the figure shown, in equilibrium.
tan  
F1 1I
GH n n JK
U  msT  01
.  20  4184  8.4 kJ
  6a 0
71. Work done in increasing the size of a soap bubble from a
radius of 3 cm to 5 cm is nearly (Surface tension of soap
solution  0.03Nm 1 ):
(1) 0.4 mJ
(2) 4 mJ
(3) 0.2 mJ
(4) 2 mJ
Sol: [1] Work done in increasing radius of a soap bubble from r1 to
r2 is
68. The half life of a radioactive substance is 20 minutes. The
b
g
approximate time interval t 2  t1 between the time t 2
2
1
of it has decayed and time t1 when of it had
3
3
decayed is
(1) 28 min
(2) 7 min
(3) 14 min
(4) 20 min
when
e
j
e
j
W  8S r2 2  r12  8    0.03 52  32  104  0.4  mJ
72. A resistor 'R'and 2 F capacitor in series is connected
through a switch to 200 V direct supply. Across the
capacitor is a neon bulb that lights up at 120 V. Calculate
the value of R to make the bulb light up 5 s after the
b
switch has been closed. log10 2.5  0.4
g
[3]
AIEEE 2011
PHYSICS
(1) 3.3  107 
(2) 13
.  104 
(3) 17
.  105 
(4) 2.7  106 
where v is the instantaneous speed. The time taken by
the object, to come to rest, would be:
(1) 8 s
(2) 1 s
(3) 2 s
(4) 4 s
Sol: [4] For a RC circuit during growth
F
GH
V  V0 1  e
so
or

t
RC
F
GH
120  200 1  e
I
JK

t
RC
I
JK


t
5
Cn
2
6.25
v
t
 2.5dt
0
t  2s
The question has paragraph fllowed by two statements,
Statement-1 and Statement-2. Of the given four
alternatives after the statements, choose the one that
describes the statements.
t

or
dv
75. Direction:
t
5
 n
RC
2
R
z z
0
dv
 2.5 v
Sol: [3] Given,
dt
C  2.3 log10
5
2
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.
5
 2.7  106 
6
2  10  2.3  0.4
73. A current I flows in an infinitely long
wire with cross section in the form
of a semicircular ring of radius R. The
magnitude of the magnetic
induction along its axis is:
(1)
0I
4R
(2)
0I
2R
(3)
0I
2 2 R
(4)
0I
2R

dB
d

 / 2 
R
Statement - 1:
0
Sol: [2] Consider an element in the form of a is a wire of width
Rd along the length of the cylinder. The current through
the element is
FG I IJ b Rd g  I d .
H R K

The magnetic field at O due to the element is
When light reflects from the air-glass plate interface, the
reflected wave surfers a phase change of .
Statement - 2:
The centre of the interference pattern is dark.
(1) Statement - 1 is false, Statement - 2 is true.
(2) Statement - 1 is true, Statement - 2 is false.
(3) Statement - 1 is true, Statement - 2 is true and
Statement - 2 is the correct explanation of Statement - 1
(4) Statement - 1 is true, Statement - 2 is true and
Statement - 2 is not the correct explanation of
Statement - 1.
FI I
 G d J
H K
Sol: [3] Statement 1 and 2 both are true and statement 2 is a correct
explanation of statement-1
2R
76. Two bodies of masses m and 4m are placed at a distance
r. The gravitational potential at a point on the line joining
them where the gravitational field is zero is:
0
From symmetry of figure, the net field is
z

B
0
0
FG I d IJ
H  K sin   I
2R
0
2
 R
74. An object, moving with a speed of 6.25 m / s, is decelerated
at a rate given by
dv
 2.5 v
dt
(1) 
9Gm
r
(2) zero
(3) 
4Gm
r
(4) 
Sol: [1] Refer figure shown
x
m
r–x
P
4m
6Gm
r
[4]
AIEEE 2011
PHYSICS
For gravitating field at P to be zero
b g
b g

Gm G 4m

2
x2
rx
77. This question has Statement - 1 and Statement - 2. Of the
four choices given after the statements, choose the one
that best decrribes the two statements.
q0
2
for an LC circuit, from q  q0 cost for q 
r
 x
3
So, gravitational potential at P will be
F Gm IJ  FG  Gx4m IJ  FG Gm IJ  FG G  4m IJ   9Gm
V  G 
H x K H r  x K H r / 3 K H 2r / 3 K r
q
cost 
q0
2
we get
1
2
or
t 
t

4
 

2C
 4
79. This question has Statement - 1 and Statement - 2. Of the
four choices given after the statements, choose the one
that best decrribes the two statements.
Statement - 1:
Sky wave signals are used for long distance radio
communication. These signals are in gneeral, less table
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 false, Statement - 2 is true.
(2) Statement - 1 is true, Statement - 2 is false.
(3) Statement - 1 is true, Statement - 2 is true and
Statement - 2 is the correct explanation of Statement - 1
(4) Statement - 1 is true, Statement - 2 is true and
Statement - 2 is not the correct explanation of
Statement - 1.
Sol: [3] Statement - 1 is true, Statement - 2 is true and
Statement - 2 is the correct explanation of Statement - 1
78. A fully charged capacitor C with iniital charge q0 is
connected to a coil of self inductance L at t  0. The time
at which the enrgy is stored equally between the electric
and the magnetic fields is:
(1)
(3)
(2)  LC
LC

4
(4) 2 LC
LC
Sol: [3] As per question, at the said instent U B  U e  U net / 2
So it at this instant if the charge on capacitor plates is q
then
so
F I
GH JK
q 2 1 q02
Ue 

2C 2 2C
Statement - 1:
A metallic surface is irradiated by a monochromatic light
of frequency v  v0 (the threshold frequency). The
maximum kinetic energy and the stopping potential are
K max and V0 respectively. If the frequency incident on
the surface is doubled, both the K max and V0 are also
doubled.
Statement - 2:
The maximum kinetic energy and the stopping potential
of photoelectrons emitted from a surface are linearly
dependent on the frequency of incident light.
(1) Statement - 1 is false, Statement - 2 is true.
(2) Statement - 1 is true, Statement - 2 is false.
(3) Statement - 1 is true, Statement - 2 is true and
Statement - 2 is the correct explanation of Statement - 1
(4) Statement - 1 is true, Statement - 2 is true and
Statement - 2 is not the correct explanation of
Statement - 1.
Sol: [1] Let on doubling v, new value of maximum KE be
  h 2v    2 Kmax  
Kmax
 , then Kmax
b g
As, Kmax  hv  
Thus on doubling v, maximum KE and hence stopping
potential become more than double so statement 1 is false.
80. Water is flowing continuously from a tap having an
internal diameter 8  10 3 m . The water velocity as it
leaves the tap is 0.4 ms1 . The dimaeter of the water
stream at a distance 2  10 1 m below the tap is close to:
(1) 3.6  10 3 m
(2) 5.0  10 3 m
(3) 7.5  10 3 m
(4) 9.6  10 3 m
[5]
AIEEE 2011
PHYSICS
b
Sol: [1] If v2 be the speed at depth h below the tap, then
v2 
v12
 2 gh
phase difference between their motion is
2
(1)

6
(2)

2
v1
 d1
v2
(3)

3
(4)

4
From equation of continuty
v1d12
or
d2 
 v2 d 2
v1

ev12  2 ghj
1/ 2
Sol: [3] The positions of two particles w.r.t. a common origion O
are
 d1
x1  A sin t
0.4
1/ 2
.  4g
b016
 x0
 3.6  10 m
81. A mass M, attached to a horizontal spring, executes SHM
with amplitude A1 . When the mass M passes through its
mean position then a smaller mass m is placed over it and
FG A IJ
H A K is:
2
M
M m
F M IJ
(4) G
H M  mK
M m
(3)
M
gb g
A F M  mIF  I
G
JG J
A H M KH  K
F k I
F M  m IJ GH M  m JK 
G
H M K F kI
GH M JK
b g

1
2
1

.
3
 2 2


A
A
V
Here volume of the wire remains same so for small changes
in length, fractional change in resistance is given by
1
2
g b
2
b g
 A 2 cosbt   / 2g sin  / 2
Sol: [3] Resistance of a wire R 
M  1 A1  M  m  2 A2
or
O’
sin  / 2 must be half. For that
R 2 

 0.2%
R

Sol: [1] From momentum conservation just before and after putting
the mass m, we have
b
x0
83. If a wire is stretched to make it 0.1% longer, its resistance
will
(1) decrease by 0.05%
(2) increase by 0.05%
(3) increase by 0.2%
(4) decrease by 0.2%
1
(2)
O
As the maximum separation is x0  A so the value of
both of them move together with amplitude A2 . The ratio
1
2
g
x  x2  x1  x0  A sin t    sin t
3
F M  mIJ
(1) G
H M K
b
P2
The separation between them is
 8  103
 0.2  8  103
of
P1
x2  x0  A sin t  
on substituting values of v1, h and d1, we obtain
d2 
g
the maximum separation between them is X 0  A , the
84. 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) 
M m
M
(3) 
82. Two particles are executing simple harmonic motion of the
same amplitude A and frequency  along the x-axis. Their
b
g
mean position is separated by distance X 0 X 0  A . If
v2
(2) 
g2
v4
(4)
g2
v2
g
 v4
2 g2
Sol: [3] Maximum area over which water sprinkels  Rmax 2
where Rmax is the maximum possible range, given by
Rmax 
v2
g
and thus maximum area is
v 4
g2
[6]
AIEEE 2011
PHYSICS
85. A thermally insulated vessel contains an 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 increases by:
(1)
b  1g Mv K
(3)
b  1g Mv K
b g
b g
 1
2
(2) 2   1 R Mv K
2
2R
2
(4)
2R
So
T 
b  1g Rv
F mR I   F mR I FG a IJ . . . (ii)
GH 2 JK GH 2 JK H R K
2
2
On Solving (i) and (ii) we get, a 
2g
3
b g
88. The transverse displacement y x, t of a wave on a string
is given by
Mv 2
K
2R
b g
y x, t  e
Sol: [1] As the vessel is suddenly brought to rest, the kinetic
associated with the motion of centre of mass (ordered
kinetic energy) will be abosrbed by the gas and is now
energy appear as the inerease in the disordered kinetic
energy of gas and results in increase in the temperature of
the gas, thus
1 2
mv  nCv T
2
TR 
FG IJ FG R IJ T
H K H   1K
e
 ax 2 bt 2  2 abxt
j
This represents a:
1
(1) standing wave of frequency
b
(2) wave moving in +x direction with speed
a
b
(3) wave moving in –x direction with speed
b
a
1 2
m
mv 
2
M

2
2M
(where m is the mass of gas enclosed in the container)
(4) standing wave of fcrequency
b
Sol: [3] The given expression can be written
86. A screw gauge gives the following reading when used to
measure the diameter of a wire
Main scale reading:
0mm.
Circular scale reading:
52 divisions
Given that 1 mm on main scale corresponds to 100
divisions of the circular scale.
The diameter of wire from the above data is
(1) 0.005 cm
(2) 0.52 cm
(3) 0.052 cm
(4) 0.026 cm
Sol: [3] As, least count of screw gauge is given by
pitch
1

 0.01 mm
Number of disions on circular scale 100
So the given reading is  52  LC  0.52 mm  0.052 cm
87. A mass 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
3
(3) g
(2)
b g
y x, t  e
Sol: [4] For motion of mass m mg  T  ma . . . (i)
For rotational motion of pulley (Disc)
e
a x  bt
j
2
which represents a wave travelling in –x direction with
speed
b
.
a
89. A car is fitted with a convex side-view mirror of focal
length 20 cm. A second car 2.8 m 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) 15 m / s
(3)
(2)
1
m/s
15
1
m/ s
10
(4) 10 m / s
Sol: [3] On differentiating the mirror equation w.r.t. time, we
obtain.

3
g
2
2
(4) g
3

or
1 dv 1 du

0
v 2 dt u 2 dt
FG IJ du
H K dt
F 20 IJ  15  1 m / s
G
H 280  20 K
15
dv v 2 du
f
 2

dt u dt
u f
2
2
[7]
AIEEE 2011
PHYSICS
90. Let the x - y plane be the boundary between two
transparent media. Medium 1 in z  0 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 A  6 3i  8 3 j  10k is incident on the plane
of separation. The angle of refraction in medium 2 is:
(1) 75o
(2) 30o
o
(3) 45
(4) 60o
Sol: [3] In our opinion this question can be solved only if we
consider the boundary between the two transparent media
to x - y plane rather than x-z plane given wrongly in
question.
If  1 be the angle of incident ray with the normal to z-axis
then using the direction cosine formula with z-axis we have
cos 1 

10
e6 3j  e8 3j  b10g
2
2
 1  60o
From Snell's law
n1 sin 1 : n2 sin  2 we have
e 2 j sin 60  e 3 j sin
o
sin  2 

2 

4
2
3

2
3
1

2
2
2

10 1

20 2