2. 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)
Important Instructions :
1. Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball Point Pen.
Use of pencil is strictly prohibited.
2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet,
take out the Answer Sheet and fill in the particulars carefully.
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 parts in the question paper A, B, C consisting of Mathematics, Chemistry and
Physics having 30 questions in each part of equal weightage. Each question is allotted 4 (four)
marks for each correct response..
6. Candidates will be awarded marks as stated above in Instructions No. 5 for correct response of each
question. ¼ (one-fourth) marks will be deducted for indicating incorrect response of each question.
No deduction from the total score will be made if no response is indicated for an item in the
answer sheet.
7. There is only one correct response for each question. Filling up more than one response in each
question will be treated as wrong response and marks for wrong response will be deducted accordingly
as per instruction 6 above.
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in the Room/Hall. However the candidates are allowed to take away this Test Booklet with them.
10. The CODE for this Booklet is R. Make sure that the CODE printed on Side-2 of the Answer Sheet is
the same as that on this booklet. In case of discrepancy, the candidate should immediately report
the matter to the Invigilator for replacement of both the Test Booklet and the Answer Sheet
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3. 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 x2
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
x2 x2
3 11 1
(3) , (4) 0,
4 12 2
2|sin( x 2)|
Ans. (4) lim
x2 x2
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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4. 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.xx 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;
Statement-2 is not a correct explanation for
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 z2
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.
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5. 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
(1) Statement-1 is true, Statement-2 is false
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
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6. 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 = 2log2
13. If A = sin2x + cos4x, then for all real x I = log2
3 13 3
(1) A (2) A1 y –1 z – 3
4 16 4 15. If the angle between the line x and
2
13
(3) A1 (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 A1
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 log1 tan d
0
=
2
3
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7. 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 , x0
neighbourhood of and – ve to +ve in the sh x
neighbourhood of 2 a ka f ( x ) q , x0
Hence f(x) has local maximum at x = ion local
and
of A
xx x
2
i vi s , x0
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 p2 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
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8. AIEEE-2011 (Code-R)
Ans. (3) Sol. We have
y-axis CD
CD=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;
Statement-2 is a correct explanation for
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
Statement-2 is not a correct explanation for
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 )
Statement-2 is not a correct explanation for
(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
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9. 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 y3
(3) (4) dx
8 3 2
Ans. (3) 1
dy dx
Sol. The equation of the tangent to x = y2 having slope y3
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
ab (2) b c
ab
(3) 1 square units (4) square units
2
ac bc
Ans. (4)
(3) c b
ab (4) b c
ab
Ans. (1)
1 y=x
y=x Sol. We have
x=e
(1,1) bc 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 bc
1
(a b)
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10. 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.
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11. 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 y1 (Di 1
i Sol. S nR ln
V2
(1) (2) x y 1 V1
i1
100
i1 x y1 S 2.303 2 8.314 log
(3) (4) 10
x y1 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 y1 Organic Compound
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12. 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)
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