Answer solution with analysis aieee 2011 aakash

AIEEE 2011
Answers by
(Division of Aakash Educational Services Ltd.)

CODE CODE CODE
Q.N. P Q R S Q.N. P Q R S Q.N. P Q R S
01 2 1 4 1,3 31 2 2 1 3 61 4 2 2 4
02 3 1 4 1 32 3 4 1 1 62 1 1 3 1
03 1 2 2 3 33 4 2 4 2 63 4 4 3 2
04 4 3 4 1 34 3 4 4 4 64 2 3 1 4
05 3 3 1 3 35 2 1 4 2 65 3 2 1 3
06 2 1 1 3 36 3 3 2 4 66 4 2 2 4
07 1 3 3 3 37 1 4 3 4 67 3 4 3 2
08 3 4 3 1 38 1 2 2 1 68 3 4 4 4
09 4 3 3 2 39 2 4 1 3 69 3 4 4 3
10 1 1 4 1 40 2 3 1 2 70 1 4 1 4
11 1 1 1 3 41 2 2 3 2 71 3 2 1 1
12 3 1 2 2 42 2 3 2 1 72 2 4 4 2
13 4 4 2 1 43 2 2 4 4 73 4 4 2 2
14 1 3 2 2 44 1 4 4 4 74 2 2 3 4
15 3 3 2 3 45 4 4 2 3 75 2 2 4 1
16 4 3 1 3 46 2 4 2,4 3 76 1 1 1 4
17 3 1 4 3 47 4 1 2 1 77 1 3 3 4
18 3 1 4 1 48 4 3 4 1 78 1 2 3 2
19 3 1 2 2 49 2 1 1 3 79 3 4 1 4
20 1 2 4 4 50 2 1 4 4 80 1 3 1 1
21 4 4 3 2 51 3 1 2 2 81 2 3 1 3
22 2 1 3 1 52 4 1 1 1 82 3 3 3 4
23 3 2 3 4 53 2 1 1 4 83 3 1 3 2
24 4 1 3 4 54 4 2 2 1 84 2 2 3 4
25 4 1 3 1 55 2 1 4 4 85 2 2,4 1 4
26 1 3 3 1 56 3 1 2 1 86 2 3 3 2
27 2 4 3 2 57 2 1 4 4 87 4 3 4 3
28 4 1 4 2 58 4 2 3 3 88 2 3 3 2
29 1 2 2 2 59 2 2 2 4 89 1 4 3 2
30 1,3 3 1 3 60 4 3 1 3 90 2 2 3 4

Though every care has been taken to provide the answers correctly but the
Institute shall not be responsible for typographical error, if any.

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DATE : 01/05/2011
Code - R

Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075
Ph.: 011-47623456 Fax : 011-47623472

Time : 3 hrs.
Solutions Max. Marks: 360

for
AIEEE 2011
(Mathematics, Chemistry & Physics)

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AIEEE-2011 (Code-R)

PART–A : MATHEMATICS

Directions : Questions number 1-3 are based on the following
 1  cos{2(x  2)} 
paragraph. 3. lim  
x 2  x2 
1. Consider 5 independent Bernoulli’s trials each with  
probability of success p. If the probability of at least 1
(1) Equals (2) Does not exist
31 2
one failure is greater than or equal to , then p
32 (3) Equals 2 (4) Equals  2
lies in the interval
Ans. (2)
 11  1 3
(1)  , 1 (2)  ,
 12  2 4
  1  cos2( x  2) 
Sol. lim 
x2  x2


 3 11   1  
(3)  ,  (4)  0, 
 4 12   2
2|sin( x  2)|
Ans. (4)  lim
x2 x2
Sol. Probability of atleast one failure
which doesn't exist as L.H.L.   2 whereas
31
 1  no failure  R.H.L.  2
32
4. Let R be the set of real numbers.
31 Statement-1 :
 1  p5 
32
R )
A = {(x, y)  R × td.: y – x is an integer} is an
L
es
equivalence relation on R.
rvic
1
 p5 
Se
Statement-2 :
n al
32
B tio{(x, y)  R × R : x = y for some rational
u ca
=
 p
1
E d number } is an equivalence relation on R.
2
ka sh (1) Statement-1 is false, Statement-2 is true
f Aa
Also p  0
io no (2) Statement-1 is true, Statement-2 is true;

(D i vi s Statement-2 is a correct explanation for
Statement-1
 1
Hence p  0, 
 2 (3) Statement-1 is true, Statement-2 is true;
Statement-2 is not a correct explanation for
2. The coefficient of x 7 in the expansion of Statement-1
(1 – x – x2 + x3)6 is
(4) Statement-1 is true, Statement-2 is false
(1) 132 (2) 144 Ans. (4)
(3) –132 (4) –144 Sol. Statement-1 is true
Ans. (4) We observe that
Reflexivity
Sol. We have (1  x  x 2  x 3 )6  (1  x )6 (1  x 2 )6 xR x as x  x  0 is an integer, x  A
Symmetric
coefficient of x7 in
Let ( x , y ) A

1  x  x 
6
2
 x3  6C1 . 6C3  6C3 . 6C2  6C5 . 6C1  y – x is an integer
 x – y is also an integer
 6  20  20  15  6  6 Transitivity
 144 Let ( x , y ) A and ( y , z ) A

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(2)

AIEEE-2011 (Code-R)
 y – x is an integer and z – y is an integer 7. The number of values of k for which the linear
equations
 y – x + z – y is also an integer
4x + ky + 2z = 0
 z – x is an integer kx + 4y + z = 0
2x + 2y + z = 0
 (x , z ) A
possess a non-zero solution is
Because of the above properties A is an equivalence (1) Zero (2) 3
relation over R (3) 2 (4) 1
Statement 2 is false as 'B' is not symmetric on  Ans. (3)
We observe that Sol. For non-trivial solution of given system of linear
equations
0Bx as 0  0.xx  but (x ,0) B
4 k 2
5. Let ,  be real and z be a complex number. If z2 + k 4 1 0
z +  = 0 has two distinct roots on the line
2 2 1
Re z = 1, then it is necessary that
(1)   (1, ) (2)   (0, 1)  8  k (2  k )  2(2 k  8)  0

(3)   (–1, 0) (4) || = 1  k 2  6k  8  0
Ans. (1)  k 2  6k  8  0
Sol. Let the roots of the given equation be 1 + ip and  k  2,4
1 – ip, where p Clearly there exists two values of k.
8. Statement-1 :
   product of roots
The point A(1, 0, 7) is the mirror image of the point
x .) y  1 z  2
 (1  ip)(1  ip)  1  p2  1, p  B(1, 6, 3) in the line : Ltd  .

 (1, ) Statement-2 : rvi ce s 1 2 3
e
lS
2
d x a ti
The line :on a x  y  1  z  2 bisects the line segment
du c
6. equals 1 2 3
dy 2 E
ka sh joining A(1, 0, 7) and B(1, 6, 3).
Aa
3 1
 d 2 y   dy   d2y  (1) Statement-1 is false, Statement-2 is true
f
no
(1)   2    (2)  2 
i si o
 dx   dx   dx  (2) Statement-1 is true, Statement-2 is true;
v
(D i
Statement-2 is a correct explanation for
1 3 2
 d 2 y   dy   d 2 y   dy  Statement-1
(3)   2    (4)  2   
 dx   dx   dx   dx  (3) Statement-1 is true, Statement-2 is true;
Ans. (1) Statement-1
Sol. We have (4) Statement-1 is true, Statement-2 is false
Ans. (3)
d 2 x d  dx  d  1  Sol. The mid point of A(1, 0, 7) and B(1,6,3) is (1, 3,5)
  
dy 2 dy  dy  dy  dy 
 
 dx  x y 1 z2
which lies on the line  
1 2 3

d  1  dx 1 d2 y 1 Also the line passing through the points A and B is
 .  . 2. perpendicular to the given line, hence B is the
dx  dy  dy 2
 dy  dx  dy 
   dx  mirror image of A, consequently the statement-1 is
 dx   dx   
  true.
Statement-2 is also true but it is not a correct
3
 d2 y  explanation of statement-1 as there are infinitely
1 d2 y
 dy 
 .     2 many lines passing through the midpoint of the line
3 2
 dx   dx 
 dy  dx   segment and one of the lines is perpendicular
 dx  bisector.
 
(3)

AIEEE-2011 (Code-R)
9. Consider the following statements 11. A man saves Rs. 200 in each of the first three
months of his service. In each of the subsequent
P : Suman is brilliant months his saving increases by Rs. 40 more than
the saving of immediately previous month. His total
Q : Suman is rich saving from the start of service will be Rs. 11040
after
R : Suman is honest
(1) 21 months (2) 18 months
The negation of the statement "Suman is brilliant (3) 19 months (4) 20 months
and dishonest if and only if Suman is rich" can be
Ans. (1)
expressed as
Sol. Total savings = 200 + 200 + 200 + 240 + 280 + ... to
(1) ~ (P  ~ R)  Q (2) ~ P  (Q  ~ R) n months = 11040
(3) ~ (Q  (P  ~ R)) (4) ~ Q  ~ P  R
Ans. (3)  400 
n –2
2

400   n – 3 . 40  11040 
Sol. Suman is brilliant and dishonest is P  ~ R
 (n – 2) ( 140 + 20n) = 10640
Suman is brilliant and dishonest iff suman is rich is
 20n2 + 100n – 280 = 10640
Q  (P  ~ R)
 n2 + 5n – 546 = 0
Negative of statement is expressed as
 (n – 21) (n + 26) = 0
~ (Q  (P  ~ R))
 n = 21 as n  – 26
10. The lines L1 : y – x = 0 and L2 : 2x + y = 0 intersect
12. Equation of the ellipse whose axes are the axes of
the line L3 : y + 2 = 0 at P and Q respectively. The
coordinates and which passes through the point
bisector of the acute angle between L 1 and L 2
intersects L3 at R. 2
.)
(– 3, 1) and has eccentricity
td 5
is
Statements 1 : The ratio PR : RQ equals 2 2 : 5 . L
Statement 2 : In any traingle, bisector of an angle i
(1) 5 x 2  erv2 – 32  0
3y
ce s
divides the triangle into two similar triangles.
n al S
(2) io 2
cat 3 x  5y – 32  0
2
u
(2) Statement-1 is true, Statement-2 is true; h Ed
Statement-2 is the correct explanation aka
of
s (3) 5 x 2  3 y 2 – 48  0
Statement-1 of A
i on (4) 3 x 2  5 y 2 – 15  0
(3) i vi s
Statement-1 is true, Statement-2 is true;
(D Ans. (2)
Statement-2 is the not the correct explanation of
Statement-1
x2 y2
(4) Statement-1 is false, Statement-2 is true Sol. Let the equation of the ellipse be 2
 =1
a b2
Ans. (4)
9 1
Sol. The figure is self explanatory Which will pass through (– 3, 1) if 2
 2 1
y=x a b
L2 Y
L1 b2 2
and eccentricity = e = 1– 2

a 5
X
O y + 2x = 0
2 b2
(–2, –2) (1, – 2)   1– 2
L3 5 a
P R Q y+2=0

b2 3
 2

PR OP 2 2 a 5
 
RQ OQ 5
3 2
Statement-2 is false  b2 = a
5

(4)

AIEEE-2011 (Code-R)

9 1  /4
Thus  2 1  1 – tan  
a 2
b = 8  log 1  1  tan   d
0
 
9 5
 1
a 2
3a 2  /4
 2 
 27 + 5 = 3a2 = 32 = 8  log  1  tan   d
0
 

32 3 32 32
 a  , b   
2 2

3 5 3 5 = 8  log2 . –I
4
Required equation of the ellipse is 3x2 + 5y2 = 32
 2I = 2log2
13. If A = sin2x + cos4x, then for all real x  I = log2
3 13 3
(1) A (2)  A1 y –1 z – 3
4 16 4 15. If the angle between the line x   and
2 
13
(3)  A1 (4) 1  A  2  5 
16 the plane x + 2y + 3z = 4 is cos–1  14  , then 
 
Ans. (2)
equals
Sol. We have,
A = sin2x + cos4x = 1 – cos2x + cos4x 5 2
(1) (2)
3 3
2
 1 1
= 1   cos x –
2
 2  –4
 3 2
(3) (4)
2
L t d. ) 5
es
rvic
2
3  1 3 Ans. (2)
=   cos x –  
2
4  2 4
l Se
Sol. The direction ratios of the given line
a
x ca
i n
tyo– 1 z – 3
3 u
Clearly  A1
h E d1  2   are

k as
4

fA a 1, 2, 
8log 1  x  no
1

 io Hence direction cosines of the line are
14. The value of dx is
vi s
0
1  x2
(D i 1
,
2
,


(1) log 2 (2) log2 5 2
5  2
5  2

  Also the direction cosines of the normal to the plane
(3) log2 (4) log2
8 2 1 2 3
are , , .
Ans. (2) 14 14 14
Sol. We have, Angle between the line and the plane is ,
1  4  3
8 log 1  x 
1
then cos(90° –) = = sin
I 
0
1 x 2
dx
 5  2  14

Put x = tan 3 5  3


 14   5 14
2

log 1  tan  
4
 I=  8. sec 
2
sec 2  d   9  2  45  9 2  30  25
0
 30 = 20
 /4

= 8  log1  tan  d 
0
=
2
3
(5)

AIEEE-2011 (Code-R)

   25  26
16. For x   0,  , define Sol. From the given data, median = a = 25.5a
 2 2
Required mean deviation about median
x
f  x   t sint dt 2 | 0.5  1.5  2.5  ...  24.5 |
= | a |  50
0 50
Then f has  |a| = 4
(1) Local maximum at  and local 2   1
1 ˆ ˆ ˆ
(2) Local maximum at  and 2 19. If a  (3i  k ) and b  (2i  3 ˆ  6 k ) , then the
ˆ j
10 7
(3) Local minimum at  and 2      
value of (2 a  b )  [( a  b )  ( a  2 b)] is
(4) Local minimum at  and local maximum at 2
(1) 3 (2) –5
Ans. (1)
(3) –3 (4) 5
Sol. We have,
Ans. (2)
x Sol. We have
f(x) =  t sin t dt  

 
 
 
2a – b .  a  b  a  2b    
0
 
   
 f (x) = x sin x  
= 2a – b . b – 2a 
 
= –  2a – b 
For maximum or minimum value of f(x), f (x) = 0 2

 x = n, n Z
 2  2  
We observe that = –  4 | a |  | b | – 4a. b 
 .) 
– ve = – [4 + 1] = – 5 s Lt d
–ve +ve e
rvic
0  2 5 Se
20. The values of p and q for which the function
al
 n
u c a ti o  sin( p  1)x  sin x

Ed
f (x) changes its sign from +ve to – ve in the , x0
neighbourhood of  and – ve to +ve in the sh  x

neighbourhood of 2 a ka f ( x )  q , x0

Hence f(x) has local maximum at x = ion local
and
of A 
 xx  x
2

i vi s  , x0
minima at x = 2  x 3/2
(D is continuous for all x in R, are
1 1 3 1 3
17. The domain of the function f ( x )  is (1) p  ,q (2) p  ,q
| x|  x 2 2 2 2
5 1 3 1
(1) (–, ) – {0} (2) (–, ) (3) p  ,q (4) p   , q 
2 2 2 2
(3) (0, ) (4) (–, 0) Ans. (4)
Ans. (4) Sol. The given function f is continuous at x = 0 if
Sol. The given function f is well defined only
lim f (0  h )  f (0)  lim f (0  h )
h 0 h 0
when |x| – x > 0
1
 x<0  p2 q 
2
Required domain is (– , 0) 3 1
 p ,q
18. If the mean deviation about the median of the 2 2
numbers a, 2a, ......, 50a is 50, then |a| equals
21. The two circles x2 + y2 = ax and x2 + y2 = c2 (c > 0)
(1) 5 (2) 2 touch each other, if
(3) 3 (4) 4 (1) |a| = 2c (2) 2|a| = c
Ans. (4) (3) |a| = c (4) a = 2c

(6)

AIEEE-2011 (Code-R)
Ans. (3) Sol. We have
y-axis CD
 CD=C
 C  P(C  D) P(C )
 P     P(C )
D P(D) P(D)

(c, 0) as 0  P(D)  1
(0, 0) x-axis
a/2 24. Let A and B be two symmetric matrices of order 3.
Sol.
Statement-1 : A(BA) and (AB)A are symmetric
matrices.
Statement-2 : AB is symmetric matrix if matrix
multiplication of A with B is commutative.
The figure is self explanatory (1) Statement-1 is false, Statement-2 is true
Clearly c = |a| (2) Statement-1 is true, Statement-2 is true;
22. Let I be the purchase value of an equipment and Statement-1
V(t) be the value after it has been used for t years.
(3) Statement-1 is true, Statement-2 is true;
The value V(t) depreciates at a rate given by
dV ( t ) Statement-1
differential equation   k(T  t ) , where k > 0
dt (4) Statement-1 is true, Statement-2 is false
is a constant and T is the total life in years of the
Ans. (3)
equipment. Then the scrap value V(T) of the
equipment is Sol. Clearly both statements are true but statement-2 is
not a correct explanation )of statement-1.
.td
(1) e–kT (2) T 2 
I
25. If ( 1) is a cubece s Lof unity, and (1 + )7 = A + B.
k Then (A, B)Ser
vi root
equals
al
(1) (–1,n1)
kT 2 k(T  t )2
ti o (2) (0, 1)
uca(1, 1)
(3) I  (4) I 
Ed
2 2 (3) (4) (1, 0)
Ans. (3)
k a sh (3)
Ans.
a
of A Sol. (1 + )7 = A + B
i on
V (T ) T

vi s
 (–2)7 = A + B
 dV (t )   k(T  t )dt
(D i
Sol.
I t 0  –2 = A + B
T  1 +  = A + B
 (T  t )2 
 V (T )  I  k  2   A = 1, B = 1

 0
  (A, B) = (1, 1)
T 2  26. Statement-1 : The number of ways of distributing
 V (T )  I   k   10 identical balls in 4 distinct boxes such that no
 2 
box is empty is 9C3.
kT 2 Statement-2 : The number of ways of choosing any
 V (T )  I 
2 3 places from 9 different places is 9C3.

23. If C and D are two events such that C  D and (1) Statement-1 is false, Statement-2 is true
P(D)  0, then the correct statement among the (2) Statement-1 is true, Statement-2 is true;
following is Statement-2 is the correct explanation for
Statement-1
P( D)
(1) P(C|D)  (2) P(C|D) = P(C) (3) Statement-1 is true, Statement-2 is true;
P(C )
(3) P(C|D)  P(C) (4) P(C|D) < P(C) Statement-1
Ans. (3) (4) Statement-1 is true, Statement-2 is false

(7)

AIEEE-2011 (Code-R)
Ans. (3)
1 e
=  [ln x ]1
Sol. Number of ways to distribute 10 identical balls in 2
four distinct boxes such that no box remains empty
= 10–1C4–1 = 9C3 1 3
=  1  0  sq. units
2 2
Number of ways to select 3 different places from 9
places = 9C3
dy
29. If  y  3  0 and y(0) = 2, then y(ln 2) is equal
Clearly statement-2 is not a correct explanation of dx
statement-1 to
27. The shortest distance between line y – x = 1 and (1) –2 (2) 7
curve x = y2 is
(3) 5 (4) 13
4 3 Ans. (2)
(1) (2)
3 4 Sol. We have

8 dy
3 2 y3
(3) (4) dx
8 3 2
Ans. (3) 1
 dy  dx
Sol. The equation of the tangent to x = y2 having slope y3
1
1 is y  x   ln |(y + 3)| = x + k, where k is a constant of
4
integration
1  (y + 3) = c ex
1
Hence shortest distance = 4 Initially when x = 0, y.) 2
=
2 L td
 c=5
rvi ce s
3 3 2 Finally l Serequired solution is y + 3 = 5ex
the
=
4 2

8
units
na
 tio (ln2) = 5e ln2 – 3 = 10 – 3 = 7
a y
uc
d
28. The area of the region enclosed by the curves y = x,
s hE  
a ka
30. The vectors a and b are not perpendicular and
1
x = e, y  and the positive x-axis is
o fA      
c and d are two vectors satisfying b × c  b × d
i on
x
v is   
(1)
5
square units (2)
1
(D i
square units
and a  d  0 . Then the vector d is equal to
   
2 2   a c     b c  
3 (1) c      b
 ab  (2) b      c
 ab 
(3) 1 square units (4) square units    
2    
  ac     bc 
Ans. (4)
(3) c      b
 ab  (4) b      c
 ab 
   
Ans. (1)
1 y=x
y=x Sol. We have
x=e    
(1,1) bc  b d
     
 a  (b  c )  a  (b  d )
Sol.
         
 ( a  c )b  ( a  b )c  ( a  d )b  ( a  b )d
      
 ( a  b )d  ( a  c )b  ( a  b)c
Required area
e  
1 1
=  1  1   dx  (a  c )  
2 x  d     bc
1
(a  b)
(8)

AIEEE-2011 (Code-R)

PART–B : CHEMISTRY
31. In context of the lanthanoids, which of the following Sol. Cr +3 in octahedral geometry always form inner
statements is not correct? orbital complex.
(1) Availability of 4f electrons results in the 35. The rate of a chemical reaction doubles for every
formation of compounds in +4 state for all the 10°C rise of temperature. If the temperature is raised
members of the series by 50°C, the rate of the reaction increases by about
(2) There is a gradual decrease in the radii of the (1) 64 times (2) 10 times
members with increasing atomic number in the (3) 24 times (4) 32 times
series
Ans. (4)
(3) All the members exhibit +3 oxidation state
(4) Because of similar properties the separation of rate at (t + 10)°C
Sol. Temperature coefficient () =
lanthanoids is not easy rate at t°C
Ans. (1)  = 2 here, so increase in rate of reaction = ()n

Sol. Lanthanoid generally show the oxidation state of +3. When n is number of times by which temperature is
raised by 10°C.
32. In a face centred cubic lattice, atom A occupies the
corner positions and atom B occupies the face 36. 'a' and 'b' are van der Waals' constants for gases.
centre positions. If one atom of B is missing from one Chlorine is more easily liquefied than ethane
of the face centred points, the formula of the because
compound is (1) a for Cl2 > a for C2H6 but b for Cl2 < b for C2H6
(1) A2B5 (2) A2B (2) a and b for Cl2 > a and b for C2H6
(3) AB2 (4) A2B3 (3) a and b for Cl2 < a and b for C2H6
Ans. (1) (4) a for Cl2 < a for C2tH.6) but b for Cl2 > b for C2H6
d L
es
Sol. ZA 
8 Ans. (2)
e rvic
8 Sol. Both a and S for Cl is more than C2H6.
al b
5 ti o n
u ca
ZB  37. The hybridisation of orbitals of N atom in
2
EdNO3 – , NO2 and NH4 are respectively
So formula of compound is AB5/2
ka sh
i.e., A2B5
o f Aa (1) sp2, sp3,sp (2) sp, sp2,sp3

33. The magnetic moment (spin only) of [NiCls]onis
(3) sp2, sp,sp3 (4) sp, sp3,sp2
4i
2–

(1) 1.41 BM (2) 1.82 BM(D
i vi Ans. (3)
Sol. NO3– – sp2 ; NO2+ – sp ; NH4+ – sp3
(3) 5.46 BM (4) 2.82 BM
38. Ethylene glycol is used as an antifreeze in a cold
Ans. (4) climate. Mass of ethylene glycol which should be
Sol. Hybridisation of Ni is sp3 added to 4 kg of water to prevent it from freezing at
–6°C will be: (Kf for water = 1.86 K kgmol–1 and
Unpaired e– in 3d8 is 2
molar mass of ethylene glycol = 62gmol–1)
So   n(n  2) BM (1) 304.60 g (2) 804.32 g
= 2  4  8  2.82 BM (3) 204.30 g (4) 400.00 g

34. Which of the following facts about the complex Ans. (2)
[Cr(NH3)6]Cl3 is wrong? Sol. Tf = Kf.m for ethylenglycol in aq. solution
(1) The complex gives white precipitate with silver w 1000
nitrate solution Tf  K f 
mol. wt. wt. of solvent
(2) The complex involves d 2 sp3 hybridisation and
is octahedral in shape 1.86  w  1000
6
(3) The complex is paramagnetic 62  4000

(4) The complex is an outer orbital complex  w = 800 g

Ans. (4) So, weight of solute should be more than 800 g.

(9)

AIEEE-2011 (Code-R)
39. The outer electron configuration of Gd (Atomic No.: 43. A gas absorbs a photon of 355 nm and emits at two
64) is wavelengths. If one of the emissions is at 680 nm,
(1) 4f7 5d1 6s2 (2) 4f3 5d5 6s2 the other is at :
(1) 518 nm (2) 1035 nm
(3) 4f8 5d0 6s2 (4) 4f4 5d4 6s2
(3) 325 nm (4) 743 nm
Ans. (1)
Ans. (4)
Sol. Gd = 4f 75d 16s 2
40. The structure of IF7 is  1 1
Sol.  
(1) Pentagonal bipyramid  1  2

(2) Square pyramid  1 1
 
355 680  2
(3) Trigonal bipyramid
(4) Octahedral 1 680  355

Ans. (1)  2 355  680

Sol. Hybridisation of iodine is sp 3d 3  2 = 743 nm
So, structure is pentagonal bipyramid. 44. Identify the compound that exhibits tautomerism.
41. Ozonolysis of an organic compound gives (1) Phenol (2) 2–Butene
formaldehyde as one of the products. This confirms (3) Lactic acid (4) 2–Pentanone
the presence of :
Ans. (4)
(1) An acetylenic triple bond
O
(2) Two ethylenic double bonds
(3) A vinyl group Sol.
(4) An isopropyl group 2-Pentanone
t d. )
has -hydrogen & hence it will exhibit tautomerism.
sL
Ans. (3)
ce
45. The entropy ichange involved in the isothermal
 Ozonolysis
lS e rv
reversible expansion of 2 moles of an ideal gas from
on a
Sol. 
C = CH2 HCHO
a volume of 10 dm3 at 27°C is to a volume of 100 dm3

a ti
Vinylic group
E duc 42.3 J mol—1 K—1
(1) (2) 38.3 J mol—1 K—1
42. The degree of dissociation () of a weak electrolyte, as
h
A xB y is related to van't Hoff factor (i) by A ak (3) 35.8 J mol—1 K—1
f the Ans. (2)
(4) 32.3 J mol—1 K—1

expression: io no
vi s
x y1 (Di 1
i Sol. S  nR ln
V2
(1)   (2)    x  y  1  V1
i1
100
i1 x y1 S  2.303  2  8.314 log
(3)   (4)   10
x y1 i 1
S = 38.3 J/mole/K
Ans. (2)
46. Silver Mirror test is given by which one of the
Sol. Van’t Hoff factor (i) following compounds?
Observed colligative property
(1) Benzophenone (2) Acetaldehyde
= Normal colligative property
(3) Acetone (4) Formaldehyde
A x B y  xA  y  yB  x

1 
x y
Ans. (2, 4 )
Total moles = 1 –  + x + y Sol. Both formaldehyde and acetaldehyde will give this
test
= 1 + (x + y – 1)
[Ag(NH ) ]
1  (x  y  1) HCHO  Ag   Organic compound
3 2

i Silver mirror
1
[Ag (NH3 )2 ]
i 1 CH 3 – CHO  Ag  

 
Silver mirror
x y1 Organic Compound

(10)

AIEEE-2011 (Code-R)
47. Trichloroacetaldehyde was subject to Cannizzaro's
Sol. KBr  KBrO 3  Br2  .....

reaction by using NaOH. The mixture of the
products contains sodium trichloroacetate and Br
another compound. The other compound is : OH OH
H2O
(1) Chloroform + Br2
Br Br
(2) 2, 2, 2-Trichloroethanol
(3) Trichloromethanol 2, 4, 6-Tribomophenol
50. Among the following the maximum covalent
(4) 2, 2, 2-Trichloropropanol
character is shown by the compound
Ans. (2) (1) MgCl2 (2) FeCl2
(3) SnCl2 (4) AlCl3
Cl Cl Ans. (4)
NaOH Sol. According to Fajans rule, cation with greater charge
Sol. Cl–C–CHO Cl–C–COONa +
Cannizaros reaction and smaller size favours covalency.
Cl Cl 51. Boron cannot form which one of the following
anions?
Cl
(1) BO  (2) BF3
2 6
Cl–C–CH2–OH
2 1 (3) BH (4) B(OH)
4 4
Cl Ans. (2)
2, 2, 2-trichloroethanol
Sol. Due to absence of low lying vacant d orbital in B.
sp3d2 hybridization is not possible hence BF63– will
48. The reduction potential of hydrogen half-cell will be not formed.
negative if : 52. Sodium ethoxide has reacted with ethanoyl chloride.
The compound that is) produced in the above
(1) p(H2) = 2 atm and [H+] = 2.0 M .
reaction is Lt d
(2) p(H2) = 1 atm and [H+] = 2.0 M es
e rvic
(1) Ethyl ethanoate (2) Diethyl ether

al S
(3) p(H2) = 1 atm and [H+] = 1.0 M (3) 2–Butanone (4) Ethyl chloride
Ans. (1) tion
u ca
(4) p(H2) = 2 atm and [H+] = 1.0 M
d
Ans. (4)
s hE
ka
Aa
O
1 f
no
 
Sol. H  e  H 2

–

io Sol. CH3—CH2ONa + Cl—C—CH3
i vi s
2
Sodium ethoxide
Apply Nernst equation (D
1

0.059 PH 2 O
E  0 log 2
1 [H ]
CH3 —CH2—O—C—CH3

0.059 2 1/2 Cl
E log –
1 1 –Cl

Therefore E is negative. O
49. Phenol is heated with a solution of mixture of KBr CH3—CH2—O—C—CH3
and KBrO3 . The major product obtained in the Ethylacetate
above reaction is :
(1) 2, 4, 6-Tribromophenol
53. Which of the following reagents may be used to
(2) 2-Bromophenol distinguish between phenol and benzoic acid?
(3) 3-Bromophenol
(1) Neutral FeCl (2) Aqueous NaOH
3
(4) 4-Bromophenol
(3) Tollen's reagent (4) Molisch reagent
Ans. (1)

(11)

AIEEE-2011 (Code-R)
Ans. (1) Al2O3 < MgO < Na2O

Sol. Neural FeCl3 gives violet colored complex Increasing basic strength
OH
3– and while descending in the group basic character


of corresponding oxides increases.
+FeCl3 3H
+
 O Fe + 3HCl

6

 6 Na2O < K2O

Increasing basic strength
Violet coloured
complex  Correct order is
54. A vessel at 1000 K contains CO2 with a pressure of
Al2O3 < MgO < Na2O < K2O
0.5 atm. Some of the CO2 is converted into CO on
the addition of graphite. If the total pressure at 57. A 5.2 molal aqueous solution of methyl alcohol,
equilibrium is 0.8 atm, the value of K is CH3OH is supplied. What is the mole fraction of
(1) 0.18 atm (2) 1.8 atm methyl alcohol in the solution?
(3) 3 atm (4) 0.3 atm (1) 0.050 (2) 0.100
Ans. (2) (3) 0.190 (4) 0.086
Sol. CO2(g) + C(s) 2CO Ans. (4)
No. of moles of compound
at 1000 K P = 0.5 atm 0 Sol. x 
Total no. of mole of solution
0.5 – p 2p
For 1 kg solvent
According to given condition 0.5 + p = 0.8 5.2 5.2
x   0.086
p = 0.3 55.5  5.2 60.7
(0.6)2 0.36 58. The presence or absence of hydroxy group on which
So, K p    1.8 atm carbon atom of sugar differentiates RNA and DNA?
(0.2) 0.2
55. The strongest acid amongst the following (1) 4th
td .) (2) 1st
compounds is : sL
(3) 2nd
i ce (4) 3 rd
(1) ClCH CH CH COOH
2 2 2 Ans. (3) e rv
lS
(2) CH COOH
a ti
Sol. Fact. on a
du c
3
(3) HCOOH
E
h 59. Which of the following statement is wrong?
(4) CH CH CH(Cl)CO H ka s
Aa
3 2 2 (1) N2O4 has two resonance structures
f
Ans. (4)
io no (2) The stability of hydrides increases from NH3 to
O (D i vi s BiH3 in group 15 of the periodic table
(3) Nitrogen cannot form d-p bond
Sol. CH3—CH2—CH—C—OH (4) Single N - N bond is weaker than the single
P - P bond
Cl
Ans. (2)
Presence of electron withdrawing group nearest to
the carboxylic group increase the acidic strength to Sol. Stability of hydrides from NH3 to BiH3 decreases
maximum extent due to decreasing bond strength.
56. Which one of the following orders presents the 60. Which of the following statements regarding
correct sequence of the increasing basic nature of sulphur is incorrect?
the given oxides? (1) The oxidation state of sulphur is never less than
(1) K O  Na O  Al O  MgO +4 in its compounds
2 2 2 3
(2) S2 molecule is paramagnetic
(2) Al O  MgO  Na O  K O
2 3 2 2
(3) The vapour at 200°C consists mostly of S8 rings
(3) MgO  K O  Al O  Na O
2 2 3 2
(4) At 600°C the gas mainly consists of S2 molecules
(4) Na O  K O  MgO  Al O
2 2 2 3 Ans. (1)
Ans. (2 )
Sol. The oxidation state of S may be less than +4.
Sol. While moving from left to right in periodic table
basic character of oxide of elements will decrease. i.e., in H2S it is –2.
(12)

AIEEE-2011 (Code-R)

PART–C : PHYSICS

61. A carnot engine operation between temperatrues T1 63. Three perfect gases at absolute temperatures T1, T2
1 and T3 are mixed. The masses of molecules are m1,
and T 2 has efficiency . When T2 is lowered by m2 and m3 and the number of molecules are n1, n2
6
1 and n3 respectively. Assuming no loss of energy, the
62 K, its efficiency increases to . Then T1 and T2 final temperature of the mixture is
3
are respectively n1 T12  n2 T22  n3 T32
2 2 2
T1  T2  T3
(1) 310 K and 248 K (2) 372 K and 310 K (1) n T  n T  n T (2)
1 1 2 2 3 3 3
(3) 372 K and 330 K (4) 330 K and 268 K n1T1  n2T2  n3T3 n1T12  n2T22  n3T32
Ans. (2) (3) n1  n2  n3 (4) n1T1  n2T2  n3T3
1 T T2 5 Ans. (3)
Sol.  1 2  
6 T1 T1 6
n1 n n
T1  2 T2  3 T3
T  62 2
Also, 2  Sol. U = N N N
T1 3 n1 n2 n3
 
T2 5 N N N

T1 6
n1T1  n2 T2  n3T3
=
T2  62 4 n1  n2  n3
 
T2 5 64. A boat is moving due east in a region where the
or T2 = 62 × 5 earth's magnetic field is 5.0 × 10–5 NA–1 due north
)
and horizontal. The boat.carries a vertical aerial 2 m
or T2 = 310 K
s Lt d
long. If the speed of the boat is 1.50 ms –1 , the
 T1 = 372 K magnitude of rvice
the induced emf in the wire of aerial
62. A pulley of radius 2 m is rotated about its axis by is
na l Se
a force F = (20t – 5t2) newton (where t is measured (1) atio mV
0.15 (2) 1 mV
uc
in seconds) applied tangentially. If the moment of
Ed 0.75 mV
(3) (4) 0.50 mV
inertia of the pulley about its axis of rotation is
as h
10 kg m2 , the number of rotations made by theak Ans. (1)
pulley before its direction of motion if reversed of
A
n is Sol.  = Bvl
v i si o
(D i
(1) More than 9 = 5.0 × 10–5 × 2 × 1.50
(2) Less than 3 = 15 × 10–5
(3) More than 3 but less than 6 = 0.15 mV
(4) More than 6 but less than 9 65. A thin horizontal circular disc is rotating about a
Ans. (3) vertical axis passing through its centre. An insect is
at rest at a point near the rim of the disc. The insect
Sol. T = F × r = 40t – 10t2
now moves along a diameter of the disc to reach its
 = 4t – t2 other end. During the journey of the insect the
angular speed of the disc
t3
2t 2  0 (1) First increase and then decrease
3
 t=6s (2) Remains unchanged
(3) Continuously decreases
2t 3 t 4
Also,  =  (4) Continuously increases
3 12
Ans. (1)
2666
=  108 Sol. L = I = Constant
3
 I first decreases, then increases.
n  5.73
2   first increases and then decreases

(13)

AIEEE-2011 (Code-R)
66. Two identical charged spheres suspended from a Ans. (4)
common point by two massless strings of length l
are initially a distance d(d << l) apart because of N
Sol. At t  t2 , N 2  (Left)
their mutual repulsion. The charge begins to leak 3
from both the spheres at a constant rate. As a result 2N
the charges approach each other with a velocity v. t  t1 , N 1 
3
Then as a function of distance x between them
 t2 – t1 = half life = 20 min
1
 69. Energy required for the electron excitation in Li++
(1) v  x (2) v  x 2

from the first to the third Bohr orbit is
1
(1) 122.4 eV (2) 12.1 eV
(3) v  x–1 (4) v  x 2
(3) 36.3 eV (4) 108.8 eV
Ans. (2)
Ans. (4)
F
Sol. tan   1 1
mg Sol. E = 13.6 (3)2  2  2 
l  3 1 
8
x kq 2 T = 13.6  9 
or  9
2l mgx 2
F = 108.8 eV
x3
kq 2 x
 70. The electrostatic potential inside a charged spherical
2l mg mg
ball is given by  = ar2 + b where r is the distance
from the centre ; a, b are constants. Then the charge
dx dq
3x 2 2 kq density inside the ball is
dt  dt
(1) –6 a0 (2) –24  a0r
2l mg
Also, q  x3/2
(3) –6 a0r
td .) (4) –24  a0
sL
vice
Ans. (1)
dx x 3/2
 2 , i.e., v  x 1/2 Sol.  = ar2 +Ser
b

dt x al
d i on
at  2 ar  E  2 ar
67. 100 g of water is heated from 30°C to 50°C. Ignoring
E ducdr
the slight expansion of the water, the change in sh
its internal enetgy is (specific heat of water ak
a E  4r 2 
q
f A is 0
4184 J/kg/K) no io
vi s  q = –80ar3
(D i
(1) 2.1 kJ (2) 4.2 kJ
dq
(3) 8.4 kJ (4) 84 kJ  =
dV
Ans. (3)
240 r 2 dr
Sol. Q = U + W =
4r 2 dr
100 = –60a
As W = 0  U = Q = 4184   20
1000
71. Work done in increasing the size of a soap bubble
= 8368 from a radius of 3 cm to 5 cm is nearly (Surface
= 8.368 kJ tension of soap solution = 0.03 Nm–1)
 8.4 kJ (1) 0.4 m J (2) 4 m J
68. The half life of a radioactive substance is 20 minutes. (3) 0.2 m J (4) 2 m J
The approximate time interval (t2 – t1) between the Ans. (1)
2
time t2 when of its has decayed and time t1 when Sol. W = SE
3
1 = 2 × T × 4(52 – 32) × 10–4
of its had decayed is = 2 × 0.03 × 4(16) × 10–4
3
(1) 28 min (2) 7 min = 3.84 × 10–4
(3) 14 min (4) 20 min = 0.4 mJ

(14)

AIEEE-2011 (Code-R)
72. A resistor 'R' and 2F capacitor in series is connected 74. An object, moving with a speed of 6.25 m/s, is
through a switch to 200 V direct supply. Across the decelerated at a rate given by
capacitor is a neon bulb that lights up at 120 V.
Calculate the value of R to make the bulb light up dv
 2.5 v
5 s after the switch has been closed. (log102.5 = 0.4) dt
(1) 3.3 × 107  (2) 1.3 × 104  where v is the instantaneous speed. The time taken
(3) 1.7 × 105  (4) 2.7 × 106  by the object, to come to rest, would be
Ans. (4) (1) 8 s (2) 1 s
Sol. Charge on capacitor q = q0(1 – e–t/RC) (3) 2 s (4) 4 s
q0 Ans. (3)
or V  (1  e t /RC )
C
0 t
V = 200(1 – e–t/RC) dv
Sol.  v
 — 2.5 dt
120 = 200(1 – e–t/RC) v 0

e–t/RC = 0.4
2 v  2.5 t
t
 ln (0.4)
RC
2 6.25  2.5t
t  10  t=2s
 ln    2.303  0.4
RC  4 
75. Direction : The question has a paragraph followed by
5 two statements, Statement-1 and Statement-2. Of the
R = 6 given four alternatives after the statements, choose the
2  10  2.303  0.4
.)
one that describes the statements.
= 2.7 × 106  Lt d
A thin air film cesformed by putting the covex
73. A current I flows in an infinitely long wire with surface of a erv
i is
plane-convex lens over a plane glass
lS
ona pattern due to light reflected from the
corss section in the form of a semi-circular ring of plate. With monochromatic light, this film gives an
raidus R. The magnitude of the magnetic induction a ti
interference
along its axis is
E duc (coNvex) surface and the bottom (glass plate)
top
ka sh surface of the film.
(1)
0 I
(2)
0 I
o f Aa Statement-1 : When light reflects from the air-glass
4 R 2 R n
v i si o plate interface, the reflected wave
0 I 0 I (D i suffers a phase change of .
(3) (4)
22 R 2 R Statement-2 : The centre of the interference pattern
Ans. (2) is dark.

dl (1) Statement-1 is false, Statement-2 is true

Sol. (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
B   dB sin 
Statement-1
0 di
B sin  (4) Statement-1 is true, Statement-2 is true and
2 R
Statement-2 is not the correct explanation of
Statement-1
0  i  
2 R  R  
   R sin  d
0
Ans. (4)
Sol. The centre of interference pattern is dark, showing
0 i
 that the phase difference between two interfering
2 R waves is  . But this does not imply statement-1.

(15)

AIEEE-2011 (Code-R)
76. Two bodies of masses m and 4 m are placed at a Ans. (3)
distance r. The gravitational potential at a point on
Sol. q  q0 cos t
the line joining them where the gravitational field is
zero is q0
energy is halved when q 
9Gm 2
(1)  (2) Zero
r 
 t 
4Gm 6Gm 4
(3)  (4) 
r r 1 
 t
Ans. (1) Lc 4
Sol. Field is zero at O. 
 t  Lc
r/3 2r/3 4
79. This question has Statement-1 and Statement-2. Of
m o 4m the four choices given after the statements, choose
the one that best describes the two statements.
GM 4GM Statement-1 : A metallic surface is irradiated by a
V – – monochromatic light of frequency
r /3 2r /3
v > v0 (the threshold frequency). The
9GM maximum kinetic energy and the
–
r stopping potential are Kmax and V0
77. This question has Statement-1 and Statement-2. Of respectively. If the frequency incident
the four choices given after the statements, choose on the surface is doubled, both the
the one that best describes the two statements. Kmax and V0 are also doubled.

Statement-1 : Sky wave signals are used for long Statement-2 : The maximum kinetic energy and the
.)
Ltd from a surface are linearly
distance radio communication. These stopping potential of photoelectrons
signals are in general, less stable es
emitted
than ground wave signals. rvicdependent on the frequency of
Se incident light.
Statement-2 : The state of ionosphere varies from n al
ti o
hour to hour, day to day and season
d uca Statement-1 is false, Statement-2 is true
(1)
E
to season.
sh
a ka
of A
(1) Statement-1 is false, Statement-2 is true
n (3) Statement-1 is true, Statement-2 is true and
i si o
Div
(
(3) Statement-1 is true, Statement-2 is true and Statement-1
Statement-2 is the correct explanation of (4) Statement-1 is true, Statement-2 is true and
Statement-1 Statement-2 is not the correct explanation of
(4) Statement-1 is true, Statement-2 is true and Statement-1
Statement-2 is not the correct explanation of Ans. (1)
Statement-1
Sol. kmax = hf – 
Ans. (3)
k'max = 2hf – 
Sol. Because of variation in composition of ionosphere,
the signals are unstable. k'max = 2kmax + 

78. A fully charged capacitor C with initial charge q0 is k'max > 2kmax
connected to a coil of self inductance L at t = 0. The 80. Water is flowing continuously from a tap having an
time at which the energy is stored equally between internal diameter 8 × 10–3 m. The water velocity as
the electric and the magnetic fields is it leaves the tap is 0.4 ms–1. The diameter of the
water stream at a distance 2 × 10–1 m below the tap
(1) LC (2)  LC is close to
 (1) 3.6 × 10–3 m (2) 5.0 × 10–3 m
(3) LC (4) 2  LC
4 (3) 7.5 × 10–3 m (4) 9.6 × 10–3 m

(16)

AIEEE-2011 (Code-R)
Ans. (1) 82. Two particles are executing simple harmonic motion
Sol. a1v1 = a2v2 of the same amplitude A and frequency  along the
x-axis. Their mean position is separated by distance
a1
v22 = v12 + 2gh v1 X0(X0 > A). If the maximum separation between
v22 = (0.4)2 + 2 × 10 × 0.2 them is (X0 + A), the phase difference between their
motion is
= (0.4)2 + 4
a2  
v22 = 4.16 v2 (1) (2)
6 2
v2  4.16  
(3) (4)
3 4
v2 = 2.04 m/s
Ans. (3)
Also, a1V1 = a2V2 Sol. x1 = A sin t
d V1 = d V2
1
2
2
2
x2 = x0 + A sin(t + )
V1 Separation = x0 + A sin (t + ) – A sin t
d2 = d1 V
2

= x0  2 A sin   cos t
3 0.4 2
= 8  10
2.04 
Maximum separation = x0  2 A sin  x0  A
= 3.54 × 10 m –3 2
81. A mass M, attached to a horizontal spring, executes 
 
S.H.M. with amplitude A1. When the mass M passes 3
through its mean position then a smaller mass m is
83. If a wire is stretched to make it 0.1% longer, its
placed over it and both of them move together with
resistance will )
A  td. (2)
amplitude A2. The ratio of  1  is sL
(1) Decrease by 0.05% Increase by 0.05%
 A2  i ce
(3) Increase erv0.2%
by (4) Decrease by 0.2%
 M  m
(1) 
1/2
M
(2) M  m Ans. (3) iona
lS

t
u a
 M 
1/2 Sol.E dR c l2
as h
Mm  M 
(3) (4)  
 M  m a k R 2 l
of A
M 
i on
R l
Ans. (1)
is
Sol. Dv
At mean position, initial velocity = A( i 1 1
84. A water fountain on the ground sprinkles water all
around it. If the speed of water coming out of the
MA1 1 fountain is v, the total area around the fountain
New velocity =
Mm that gets wet is
MA1 1
 A2 2  v2 v2
M m (1)  (2) 
g2 g
A2  M  1
 
A1  M  m  2 v4  v4
(3)  (4)
g2 2 g2
k
1  Ans. (3)
M
2
k Sol. Area  Rmax
2 
M m
v2 sin 90 v2
A2 M Rmax  
  g g
A1 m M

A1 m M v 4
  Area 
or g2
A2 M

(17)

AIEEE-2011 (Code-R)
85. A thermally insulated vessel contains an ideal gas 88. The transverse displacement y(x, t) of a wave on a
of molecular mass M and ratio of specific heats . It string is given by
is moving with speed v and is suddenly brought to
rest. Assuming no heat is lost to the surroundings, y(x , t )  e

– ax 2  bt 2  2 ab xt 
its temperature increases by
This represents a
(  – 1) (  – 1) 2
(1) Mv 2 K (2) 2(   1)R Mv K 1
2R (1) Standing wave of frequency
b
(  – 1) Mv 2
(3) Mv 2 K (4) K a
2 R 2R (2) Wave moving in +x direction with speed
b
Ans. (1)
b
1 2 n  RT (3) Wave moving in –x direction with speed
Sol. mv  a
2 (   1)
(4) Standing wave of frequency b
MV 2 (   1)
 T  Ans. (3)
2R
86. A screw gauge gives the following reading when ( a x  b t )2
Sol. y(x , t )  e
used to measure the diameter of a wire.
This is a wave travelling in –x direction with speed
Main scale reading : 0 mm
Circular scale reading : 52 divisions b

Given that 1 mm on main scale corresponds to 100 a
divisions of the circular scale. 89. A car is fitted with d.)convex side-view mirror of
a
Lt
The diameter of wire from the above data is es
focal length 20 cm. A second car 2.5 m behind the
rvic
first car is overtaking the first car at a relative speed
(1) 0.005 cm (2) 0.52 cm of 15 m/s. e
l S The speed of the image of the second car
(3) 0.052 cm (4) 0.026 cm i on a
as tseen in the mirror of the first one is
d u ca
hE
Ans. (3) 1
m/s
Diameter = main scale reading + l.c. × circular scale as
(1) 15 m/s (2)
Sol. k 10
f Aa
no
reading 1
i si o
(3) m/s (4) 10 m/s
v 15
 0  52 
1
100
mm (D i Ans. (3)
Sol. For mirror
 0.52mm  0.052cm
87. A mass m hangs with the help of a string wrapped 1 1 1
 
around a pulley on a frictionless bearing. The pulley v u f
has mass m and radius R. Assuming pulley to be a
perfect uniform circular disc, the acceleration of the 1 dv 1 du
mass m, if the string does not slip on the pulley, is  0
v 2 dt u2 dt
g 3
(1) (2) g 2
3 2 dv  v2 du  f  du
 2   
dt u dt  f  u  dt
2
(3) g (4) g
3 2
 20  du
Ans. (4)   
 20  280  dt
mg mg 2g
Sol. a    2
I m 3  1  du 1
m 2 m    m/s
R 2  15  dt 15

(18)

AIEEE-2011 (Code-R)
90. Let the x-z plane be the boundary between two Sol. The data is inconsistant
transparent media. Medium 1 in z  0 has a Taking boundary as x-y plane instead of x-z plane,
refractive index of 2 and medium 2 with z < 0 the angle of incidence is given as
has a refractive index of 3 . A ray of light in  i j 
( k ).(6 3   8 3   10 k )
 cos  
ˆ
medium 1 given by the vector A  6 3 i  8 3 ˆ – 10 k
ˆ j 108  192  100
is incident on the plane of separation. The angle of 1
    60
refraction in medium 2 is 2
(1) 75° (2) 30° Now, 2 sin 60  3 sin r

(3) 45° (4) 60° 1
 sin r 
2
Ans. (3)
 r  45



td .)
sL
i ce
lS e rv
n a
u c a ti o
d
s hE
ka
f Aa
io no
(D i vi s

(19)

(Division of Aakash Educational Services Lt d. )

ANALYSIS OF PHYSICS PORTION OF AIEEE 2011
XII XI XII XI XII XII XI XI
Heat & Modern Unit and
Electricity Magnetism Mechanics Optics Waves Total
Thermodynamics Physics Measurements

Easy 0 3 1 7 3 0 2 1 17

Medium 1 1 2 2 0 2 0 1 9

Tough 2 0 0 1 0 1 0 0 4

Total 3 4 3 10 3 3 2 2 30

XI syllabus 18 XII syllabus 12

Distribution of Level of Questions in Physics Electricity Topic wise distribution in Physics

Heat &
Thermodynamics 7% 7%
13% Magnetism 10% 10%
13%
Mechanics
30% 57% 33%
Modern Physics
10%
10%
Optics

Unit and
Measurements
Waves
Easy Medium Tough

Percentage Portion asked from Syllabus of Class XI & XII

40%
60%

XI syllabus XII syllabus


ANALYSIS OF CHEMISTRY PORTION OF AIEEE 2011

Organic Chemistry Inorganic Chemistry Physical Chemistry Total

Easy 0 3 3 6
Medium 8 6 8 22
Tough 1 0 1 2
Total 9 9 12 30


Distribution of Level of Question in Chemistry Topic wise distribution in Chemistry

40% 30%
7% 20%

73% 30%

Easy Medium Tough Organic Chemistry Inorganic Chemistry Physical Chemistry


30%

70%

XI syllabus
XII syllabus


ANALYSIS OF MATHEMATICS PORTION OF AIEEE 2011
XII XI XII XI XI XI XI XII XII
Trigonom Algebra Coordinate
Calculus Algebra (XI) Probability Statistics 3-D (XII) Vectors Total
etry (XII) Geometry

Easy 5 1 1 4 2 0 1 2 0 16

Medium 5 0 1 1 1 2 0 0 2 12

Tough 1 0 0 1 0 0 0 0 0 2

Total 11 1 2 6 3 2 1 2 2 30


Distribution of Level of Questions in Calculus Topic wise distribution in Mathematics
Mathematics
Trigonometry

7% 7%
Algebra (XII)
7% 3%
7% 36%
Algebra (XI)

40% 53% Coordinate Geometry 20%
Probability 10%

Statistics 7% 3%

3-D (XII)

Easy Medium Tough Vectors


43%
57%

XI syllabus XII syllabus

Answer solution with analysis aieee 2011 aakash

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