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PART A : PHYSICS
1. (2) Velocity of rain drop with respect tocar is
RC R CV V V=
which should be perpendicular to thewind screen. As a result the velocity
components of RV and cV along the
wind screen must cancel each other.That is,
( ) = R CV cos V cos 90
6cos 2 sin =
( ) = = 1tan 3. and hence tan 3
2. (2) ( )max
K.E = ( )21
m u cos2
at the
higher point
= 2x1
mu2
2max xk u
Given( )
( ) =
max1
max2
K.E 4
K.E 1
1
2
x
x
u 2
u 1=
Similarly maximum height
2yH u
= =
22 2yuu sin
H2g 2g
Given 1
2
H 4
H 1
=
=1
2
y
y
u 2
u 1
= =
2x y2u u2u sin cos
Rg g
= = =
1 1
2
x y1
2 x2 y
u uR 2 2 4.
R u u 1 1 1
BRILLIANTS
FULL SYLLABUS TEST 5
FOR OUR STUDENTS
TOWARDS
JOINT ENTRANCE EXAMINATION, 2013
JEE 2013
B.MAT 5 (MAIN) SOLNS
PHYSICS MATHEMATICS CHEMISTRY
SOLUTIONS
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3. (1) Second overtone of the closed pipeis 3rd harmonic. It has a frequencyof 5v /4 .
First overtone of the open pipe isthe second harmonic and it has a
frequency ( )= 2 v / 2
5v 2v100
4 2 =
(i.e.) =
v100
4or = v 400
Fundamental frequency of open
pipe is
=
1v200 s2 .
4. (1) Clearly ADBC is a rhombus. Leteach side have a length a whosecharge +Q is at A only, then
D0
QV
4 a=
and
20
QE
4 a=
along
AD. When the charge 2Q and Qare placed at the corners of B andC the new value of the potential is
D0
1 Q 2Q QV 0
4 a a a
= + =
The value of field will be
D 2 20 0
Q 120 2QE 2 Cos
24 a 4 a
=
20
QE
4 a
= =
5. (2) Let the particle leave the sphere atan angle with the vertical.
Energy conservation
( )21
mv mgR 1 cos2
= = (1)
= 2mv
mgcosR
(2)
solving (1) and (2),2
cos3
=
At this point the tangential acceler-ation is
= =T5
a g sin g3
6. (4) Momentum at the highest point
= =mv
mv cos452
Maximum height
=2 2v sin 45
h2g
2v / 4g=
Angular momentum
L momentum height= =
= =2 3
mv v mv4g2 4 2g
2vh or v 4gh
4g= =
( )=
3
24ghmL
g4 2
= 3m 2gh
7. (1) Time period T = 2h
i.e., = T 2 60 60 s
radius 8000 km. = 68 10 m
= =
62 r 22 8 10v 2
T 7 2 60 60
= 3 17 10 ms
According to Bohrs postulate
=
nhmvr
2
= 2 mvrnh
3 6
34
2 3.14 10 7 10 8 10
6.6 10
=
455.3 10=
This is the quantum number of orbitof the satellite.
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8. (2) Each photon has an energy
( ) ( )= = 34 14E h 6.63 10 Js 6.0 10 Hz 193.98 10 J=
If N is the number of photonsemitted by the source per second ,the power P transmitted in thebeam equals to N times energy perphoton E so that P = NE
= =
3
19
P 2 10 WN
E 3.98 10 J
155 10= photons per second
9. (3) Einsteins photo electron ic equation is
max 0 0hc
K h= =
Maximum kinetic energy 0k eV=
Work function 0 0hc
eV =
Given 102271A 2271 10 m = =
and 0V 1.3V=
work function
34 8
0 10
6.63 10 3 10
2271 10
=
191.6 10 1.3
19 198.76 10 2.08 10 =
= =
19
19
6.68 104.2 eV
1.6 10
10. (4) For a loop magnetic induction at
the center will be = 0 2B
4 R
When the loop subtends an angle
at the centre then
= 0B
4 R
In the problem3
2
= , given
3 / 2
=
3 32
2 2
4 3
2 2
=
0 0B4 2 R 8R
= =
11. (4) Current =V
Zwhere
( )= + 22
R L CV V V V and
( )= + 22
L CZ R X X
At resonance = =L CX X , Z R
and = =L C RV V , V V
= = =
R
3
V 100
0.1AR 1 10
Voltage across the inductance LV is
= = = =
L C 6
I 0.1V V I C
C 200 2 10
= 250 V
12. (2) r = 1m, B = 0.01T, and = =1 1
t sV 100
Induced e.m.f =d
edt
=
( )= = = =
22 0.01 1 1BA B re V
t t 1/100
Induced electric field
eE
2 r 2 1
= = = =
= 10.5 Vm
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13. (1)
Resolve E along CO and E along BO
into two r components.
In this problem the sine componentswill cancel each other. Only thecosine components add up alongOA to give 2E cos 60.
the resultant field along
= AO 2E 2E cos60
2E E E= =
The resultant field will be E along
AO
14. (3) There are 50 divisions on the scale.
Full scale current for 50 divisions
= =
1 505 mA
10
= =
+
gg
505mA
I s G
+ =
g
s G
s
4 205 mA 30 mA
4
+ = =
15. (3) At time t = 2 sec the particle crossesthe mean position. At time t = 4 sec
its velocity is 14 ms .
For S.H.M = y a sin t
=
2y a sin t
T
=
1
2y a sin 2
16
= =
1a
y a sin4 2
After 4 sec or after 2 sec from mean
position velocity 14ms=
Velocity = 2 21a y
(i.e.) 2 22
4 a a / 216
=
= =
a 32 2or a
8 2
16. (1) Efficiency 2
1
T1
T =
Initially1
50 273 171
100 T
+=
(i.e.) 11
290 1or T 580K
T 2= =
Finally+
= = 1 1
60 273 17 290 21 or
100 T T 5
=1T 725 K
Change in source temperature
( )= =725 580 K 145 K
17. (4) Energy of n photonnhc
E=
Momentum gained by the body
= = =
E nhc nhP
c c
18. (2) Tension = 64N
length 10 cm 0.1m= =
mass per unit length
3210m 10 kg/m
0.1
= =
Frequency of fundamental mode
= =
1 Tn
2 m
= =
21 64
n 400Hz2 0.1 10
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The source is moving away from theobserver. The apparent frequency
= +
vn n
v v
n = 400 =v 300 m / s a = ?
n n 1 =
n n 1 400 1 399= = =
= +
300399 400
300 v
= =300
v 0.7518 m/ sec399
= 0.75 m/sec
19. (1) = = 1 0 1 2 1 2 1h E h h E E
2 0 2 2 12 1
hc hch E E E = =
2 12 1
1 1hc E E
=
1 2 2 11 2
hc E E
=
( )
( )
= =
2 1 1 2
11 2
E Eh 800 A
c
88 10 m=
= = 82 700A 7 10 m
191E 1.8eV 1.8 1.6 10 J
= =
192E 4.eV 4 1.6 10 J
= =
= 8c 3 10 m/sec
( )
( )
=
19 19
8 8
8 8 8
4 1.6 10 1.8 1.6 10
8 10 7 10h
3 10 8 10 7 10
= 346.57 10 J.s
20. (1) According to Newtons law
+ =
1 2 1 20k
t 2 where 0
is the surrounding temperature.
60 40 60 40k 10
7 2
+ =
20k 40
7= (1)
Let be the temperature after
next 7 minutes
+ =
40 40k 10
7 2 ...(2)
Dividing (1) by (2)
=
= +
20 40 2
40 20
i.e. + = 20 160 4
= = = =140
5 160 20 140 or 28 C5
21. (2) Mass of the cavity
2
1R
m
3
=
Mass of the remaining portion
22
2R
m R3
=
=
28 R
9
Let the position of the centre ofmass of the remaining portion be
. From the centre of the disc,
( )
+ =+
2
1 2
2Rm m 3
0m m
12 1
2
m2R 2Rm m i.e.
3 m 3
= =
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2
2
R
2R R3
3 128 R
9
= =
Moment of inertia of the system
about 2O is given by
( ) = +
2
2 22 2
o
R RR R
2 12
( )2 2 2 2R/3 R R 3R2 3 3 4
+
4 1 1 1 1R2 144 162 16
= +
40.44 R=
22. (3) Let be the length of the block
immersed in liquid when the block is
floating.
= mg A g
If the block is given a vertical
displacement M then the effective
restoring force
( )( )F A y g mg)= +
( )A y g A g A g = + = y
F y . As this F is directed towards its
equilibrium position of block, if the
block is left free it will executesimple harmonic motion. Hence
inertia factor = mass of block = m
Spring factor = A g
i.e. Time period
= =
m 1T 2 i.e., T
A g A
23. (1) Applying conditions of rotationalequilibrium
O
1/4mg3/4mg
mgx
+ = 3 1mgx mg mg4 4 2
3 mg mgmg x
4 2 4=
3 mgmg x
4 4=
3x
4 4=
or x3
=
24. (2) Voltage drop across. 150
resistor is given by
( )5 0.5 V 4.5V =
=V
iR
4.5i A 30 mA
150= =
25. (4) Applying snells law at air waterinterface.
Air
Glass slab =g 1.5
Water flim
60
=w4
3
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= 4sin60 sinr3
= =
3 3 3 3sin r
4 2 8
=
1 3 3r sin8
From the figure it is clear that theangle of incidence on the glass slab
is
= =
1 3 3i r sin8
26. (1) If 0 is the intensity of unpolarised
light, then intensity of lighttransmitted through first polariser
0
2
=
Intensity of light transmitted throughthe second polarizer is
(by Malus, Law)
= = =
22 0 0I 1cos 45
2 42
27. (2) ( )1y 0.5 sin 3 t 2x=
( )2y 0.5 sin 3 t 2x=
According to the principle of superposition, the resultant displacementis
= + =1 2y y y
( ) ( ) + + 0.05 sin 3 t 2x sin 3 t 2x
= y 0.05 2 sin 3 t cos 2x
( )= = y 0.1cos 2x sin 3 t A sin 3 t
where A = 0.1 cos 2x = amplitude ofthe resultant standing wave
At x = 0.5 m A = 0.1 cos 2x = 0.1cos 2 0.5
A 0.1cos1= radian
=
1800.1cos
0.1cos 57.3=
= = =A 0.1 0.54 m 0.054 m 5.4 cm
28. (3) In circuit A diodes are forwardbiased, current flowing in the circuitis
8V 8V4A
24 4
4 4
= = =
+
( ) 4 and 4 are parallel
4
4
4
4
In the circuit upper diode is forwardbiased and lower diode is reversebiased. Current flowing into circuit
= = =
8vB I 2A
4
29. (1) The magnitude of electric field at adistance r from a thin infinitely longstraight wire of uniform linear
charge density is
=0
1 2.1E .
4 r where r is the r
distance of the point from the wire.
= 11
cm3
= = 2r 18cm 18 10 m
=
9
2
19 10 2
3E
18 10
=
91
2
9 10 2NC
3 18 10
= 11 11
10 NC3
= 11 10.33 10 NC
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30. (3) As = L
T 2g
squaring on both sides
=2
2 4 LTg
=
2
2
4 Lor g
T
T and Tn n
= =
The percentage of error in g is
g L 2 T100 100g L T
= +
The errors in both L and T are leastcount.
g 0.1 1100 2 100
g 20.0 90
= +
( )0.05 0.2 100= +
= = =0.205 100 20.5 21%
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y
A
B
AO ( )2, 0( )2, 0
A
B CD
E
F
3 6
y
y
36
31. (4)( )
( )2x 2
f xLt 3
x 2=
( )0
f 2 00
=
form
Also( )
( )x 2
f xLt
2 x 2
is
0
0 form
( )f 2 0 =
Now( )
( )x 2
f xLt 3 f 2 6
2
= =
32. (2) ( )4 2
2f x 1
x x 2x 4=
+
( ) ( )=
+ +2 22
21
x 1 x 1 2
which is
1but 0<
The range is [0, 1)
33. (2) Nowx x
b c ,2 2
x xb b
2 2 +
andx x
b a
2 2
are in HP
(given)
Let
x x xa A , b B and c C
2 2 2 = = =
B C, 2B, B A are in HP
( ) ( )2 B C B A2B
B C B A
=
+
2B AC =
x x xa , b and c
2 2 2
are in G.P
34. (3)
The third vertex lies on the
perpendicular bisector of the line
joining ( )A 2, 0 and ( )A 2, 0 i.e., B
lies on y axis. (Above x axis). The le is equilateral.
From A BO, =BO
tan602
BO 2 3 = . B is 0, 2 3 Since thele is equilateral. orthocenter is the
centroid which is2 3
0,3
which
lies on ( )3 x y 2 3+ =
35. (1)
Let AE = y then AF = y Evidently
BF = BD = 3 = inradius = B 90
( ) ( ) ( )2 2 2
6 3 y 3 y 6 y 9+ + + = + =
Area of the triangle 1 9 122
=
= 54
36. (2) Let 1 2 3E ,E ,E denote the events that
the bag contains 2 white, 3 whiteand 4 white balls respectively. Allthe three are equally likely. Hence
( ) ( ) ( )1 2 31
P E P E P E3
= = = . Let A denote
the event that the 2 balls drawn are
= =
22
412
C 1APE 6C
= = = =
3 42 2
4 42 32 2
C C1A AP ; P 1E E2C C
By Bayes theorem
= = + +
3
11
3E 3PA 1 1 1 1 1 5
13 6 3 2 3
PART B: MATHEMATICS
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37. (4) Now1 2 1 2
1 1 1 1
a a d a a =
2 3 2 3
1 1 1 1
a a d a a
=
100 101 100 101
1 1 1 1
a a d a a
=
Adding + +1 2 2 3 100 101
1 1 1...
a a a a a a
1 1011 1 1d a a
=
1 101
1 100d 1
d a a 14= =
1 101a a 1400 =
38. (1) The equation of the ellipse is
2 2x y1
12 4+ = ( )P is 2 3 cos , 2sin
3 3M is
2 2
Let ( ), be the mid point of PM.
Then
32 2 3 cos ,
2 =
32 2 sin
2 =
Eliminating
2 23 3
4 41
3 1
+ =
Locus of ( ), is
2 23 3
x y4 4
13 1
+ =
which is an ellipse with centre3 3
,4 4
39. (3) ( )323y x 2= +
Differentiating w. r. to x
( )2dy
6y 3 x 2dx
= +
Sub normal( )
2x 2dy
ydx 2
+= =
Sub tangent( )
2
2
y 2y
dy x 2dx
= =+
( )
2
x 23= +
( )2
Sub tangent 8
Sub normal 9 =
40. (4) The slope of the normal to y = f(x) at
x = 0 is 5
slope of the tangent ( )1
f 05
= =
Now the limit is
( ) ( ) ( )2 2 2x 02x
Lt2x f x 48x f 4x 80x f 8x +
1
1 1 124 40
5 5 5
=
+
1 1
1 315
5
= =
41. (2)
2px qx r 0+ + = has exactly two
roots say and . suppose p 0,>
then the graph is as above. Then
the curve 2y px qx r= + + has a
minimum below and . Now,
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P
Q RQ / 2
1P
P / 2 P / 2
( ) ( )2px 10p q x 5q r+ + + +
( ) ( )f x 5 f x= +
Let ( ) ( ) ( )g x f x 5f x= +
( ) ( )g 0 5f = +
( )5f=
( ) ( )g 5f =
Now ( ) ( )g and g are of
opposite signs. The same result
holds good if p 0< ( )g x 0 = has atleast one positive
root.
42. (1) Equation to the perpendicular
bisector of the chord joining (2, 4)
and (6, 4) is x = 4. Solving with
y x 6= + , we get centre of the
circle is (4, 10).
Slope of one diagonal6
32
= =
Slope of another diagonal
6 32
= =
Angle between the diagonals
1 6tan1 9
=
1 3tan4
=
43. (2) The equation is
( )109 109 1094x 1 4 x =
( ) ( )1 2
108 107C C109 4x 109 4x ... 1 0 + =
sum of the roots 2
1
C
C
109 1
109 4=
27
2=
44. (4)( )
( ) ( )
14 2 2
2 2 4
0
1 x x xI dx
1 x 1 x x
+ +=
+ +
( )
11
2
2 23
00
dx 1 3xdx
31 x 1 x
= ++ +
( )1
1 1 3
0
1tan x tan x
3
= +
4 12 3
= + =
45. (2)
Join 1QP
1PPQP Q2
= +
In 1PQP
1P
PP 2 4sin Q2
= +
Q R8 sin Q
2 2 2
= +
Q R8cos
2 2
=
1P Q R P
PP cos 8cos cos2 2 2
=
Q R Q R8cos sin
2 2
+ =
1P
PP cos 4 sinQ sinR2
= +
similarly 1Q
QQ cos 4 sinR sinP2
= +
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7
37
and 1RRR cos 4 sinP sinQ2
= +
1P
PP cos 8 sinP sinQ sinR2
= + +
1P
PP cos2 8
sinP
=
46. (4) ( )a b a b a b a b =
( ) ( ){ }a b b a b a b=
( ) ( ) ( )2
a a b b a b=
( )2 2 2
a b a b a b a b + =
47. (1) Total number of functions from A to
B 53= . Let us find the number of
nononto functions from A to B.
(1) 3 functions 1 2 3f , f , f in which all
the elements go to 1, 2, 3
(2) Select any two elements of B in3
2C i.e., 3 ways. Then number of
nononto function ( )3 52C 2 2=
Required number of onto functions
( ){ }5 523 3 3C 2 2 150= + =
48. (2) Minimum values of a, b, c are
respectively equal to 1
we get 3 a b c 7 + +
This a b c 3, a b c 2, ... + + = + + =
a b c 7+ + =
Total number of positive integral
solutions
2 3 4 5 62 2 2 2 2C C C C C= + + + +
3 3 4 5 63 2 2 2 2C C C C C= + + + +
4 4 5 63 2 2 2C C C C= + + +
( )n 1n nr r 1 rFormula C C C
+
+ =
5 5 63 2 2C C C= + +
6 6 73 2 3C C C 35= + = =
49. (2) Even power of a function is always
positive. Hence we can ignore
even powers
Inequality is ( )( )
3
53x 7 0x 7
By wavy curve method
7x , 7
3
Integral values are 3, 4, 5, 6Their sum = 18
50. (3) Since is a 6throot of unity
2 3 4 51 0+ + + + + =
( )2 3 41 1 + + + = +
2 3 41 1+ + + = +
11= +
Now cos i sin3 3
= +
21 2cos i2sin cos6 6 6
+ = +
2cos cos i sin6 6 6
= +
31 2cos 2 3
6 2
+ = = =
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51. (3) 2tanx tan x 1 =
2tanx sec x =
The given expression is
( )5
4 2sec x sec x 32 +
( )5
2 2tan x sec x 32= +
1 32 31= + =
52. (4)2 2 2x y a =
dy dy x2x 2y 0dx dx y
= =
The point P is nearer to the line y =
3x + 5 means, the tangent at P is
parallel to the above line.
x3 x 3y 0
y = = which is the
locus required.
53. (1) When < < <
( )f x 0 =
Hence
( ) ( ) ( )
2e
e 2e
e1
1
f x dx f x dx f x dx
= +
ee
3 32
11
x e 1x dx 0
3 3
= + = =
54. (2)
Let OA a OB b= =
OC a b c = + =
Let AD a=
OE OD BD = +
2a b d= + =
Since a c= quadrilateral OAEC is
a rhombus
Diagonals AC and OE are
perpendicular b d 0 =
55. (3) ( ) ( )(x
2
3
2
1f x 3t 2f t dt
x
=
Differentiating w.r. to x
( ) ( ){ 231
f x 3xx
2f x .1 0 =
( )( )x
24
2
33t 2f t dt
x
+
[Idifferentiation is by Lebnitz Rule
and II is ordinary differentiation]
( ) ( ){1
f 2 12 2f 2 08
= +
( )3 1
f 22 4
=
( )5 3
f 24 2
=
( )6
f 25
= slope of the normal to
( )5
y f x at x 2 is6
= =
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56. (4) Now ( )1 1 p 1 da f
= +
( )1 1
q 1 db f
= +
( )1 1
r 1 dc f
= +
where f is the I term of the HP
and d, C-D of the corresponding
AP
Det
1 1 1
a b c
abc p q r
1 1 1
=
( ) ( ) ( )1 1 1
p 1 d q 1 d r 1 df f f
abc p q r
1 1 1
+ + +
=
Now 2 1 3 1C C , C C gives
( ) ( ) ( )
( )
1p 1 d q p d r p d
fabc p q p r p
1 0 0
+
=
= abc. 0 = 0
57. (3)2 25x x 3 0 x 5x 2 0 +
Range of I term is 0,2
(closed
interval)
Range of II term is
0,
2 (open
interval)
Sum of the two terms of LHS can
never be equal to
Number of solutions is zero
58. (2) The required number of ways
= Coeff of 40t in
( ) ( )3 1
0 15 0 20t .. t t ... t+ +
= Coeff of 40t in
3 116 211 t 1 t
1 t 1 t
= Coeff of 40t in
( ) ( ) ( ) 416 32 48 211 3t 3t t 1 t 1 t +
= Coeff of 40t in ( )4
1 t
( )4 40 1 43 4340 40 3C C C
+ = = =
59. (4) The condition for the existence ofnon trivial solution is
a sin cos
1 cos sin 0
1 sin cos
=
a sin2 cos2 = +
2 sin 24
= +
Maximum of a is 2
Minimum of a is 2
60. (4) The equation can be written as
( ) 2 22 2xdx ydy xdy ydx
x y1 x y
+ =
+ +
( )2 2 2 2xdx ydy
x y 1 x y
+
+ +
( )2 2xdy ydx
x y
=
+
In LHSput 2 2u x y= +
( )2 2
1du 2xdx 2ydy
2 x y= +
+
LHS 1 1 2 22
du sin u sin x y1 u
= = +
RHS 12
2
yd
yxtan
xy1
x
=
+
Solution is
1 2 2 1 ysin x y tan kx
+ = +
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61. (4) 25.55 4.9 125.195 = In multiplica-
tion the result will be expressed
correct to same number of
significant digits as that of the
operand having the least significant
digits. Since 4 9 is the number with
least significant digit (two) the final
result is limited to two significant
digits. Thus the result is
2
1 3 10
( ) 2125.195 i s rounded to 1 3 10
62. (3) + +
2 2 2 2 42 mo1mo 1mo
SO C 2H O H SO 2HC
Therefore, 2 22 mol SO C on hydro-
lysis will give 2 mol 2 4H SO and 4 mol
HC
( )2 4 422H SO 2Mg OH 2MgSO+
24H O+
( ) 224HC 2Mg OH 2MgC+
24H O+
Thus for complete neutralization we
need + = 22 2 4 mols Mg(OH)
63. (4) orbital angular momentum is given
by ( )h
12
+ +
for a 3d orbital
2=
orbital angular momentum
( )h h
2 2 1 62 2
= + =
64. (3)
( )23in SOS :
As C, N and B have no d orbitals,
they do not form p d bonds.
65. (3)
3 3NH and NF :
Due to greater electronegativity of
F, the N F bond electrons are
more towards fluorine in N F
bonds. Hence there is less bond
pair- bond pair repulsion in 3NF
than in 3NH . Hence 3NF has lower
bond angle than 3NH
3 3PF andPH : Partial double bond
between P and F due to d p
back bonding in 3PF , so bond
angle is greater in 3PF than in 3PH .
PART C : CHEMISTRY
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+
+
+ +
+ = = = = +
+ = + = + =
+ = =
3
2 0 0
3 2 0 0
Fe 3e Fe; E 0.036 V; G nFE 3Fx 0.036V 0.108 FV
Fe Fe 2e ; E 0.439V; G 2Fx 0.439 0.878 FV
Adding (1) and (2), we get
Fe e Fe ; G nFE 0.770 FV
3F
2F
0E = 0nFE
74. (3) For l mole,
H U P V = +
( ) ( )V 2 1 2 1C T T R T T=
( ) ( )V 2 1C R T T= +
Since 2 1T T T 0 , = = =H 0 f or the
process.
75. (4) In
= 2
OCH CH CH CHO,
the hydrogen is lost as H+ be-
cause the resulting anion is
stabilised by resonance and is most
likely to be formed. The reaction is
crossed aldol reaction as follows:
= 3O
OHCH CH CH CHO
= 2
OCH CH CH CHO
[
= CH CHO + OH
Product
76. (2)
heat2Ca +
77. (3)
78. (4) Neutral dichlorocarbene ( )2: CC is
the elctrophile formed in the
reaction.
+ + 3 2 3
OOHCC OH H O CC
+ 3 2Dichlorocarbene
O: CC : CC C
73. (2)
In the above reaction n = 1
Then = +0E 0.77 V
3CH CHO
2
2
CC
OH ; H O
2CCi i
2CHC
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79. (2)
3 3CH C CHC>
C
CC
C
(4) 3 3CH C CH F>
3CH C has greater dipole moment
than 3CH F due to larger C C
distance than C F distance though F
is more electronegative than C
80. (1)
The chain with greater number of
side chains is selected as principal
chain. Hence the name is 2, 3, 5-
trimethyl-4-propyl heptane
81. (4)
( ) 2, 3 dibromobutane
82. (1) (1) is most stable. The two C are
farthest from each other in (1). In (2)
and (3) the two C are close to
each other causing electronic
repulsion. In (4) there is instability
due to both electronic as well as
steric repulsion.
83. (4) is a stronger base than
because of the + and
hyper conjugation of 3CH group
is more strongly
deactivated by 2NO group com-
pared to CN. Hence the order is:
2
OH
H O
2H O
2H O
OH
2 4Br /CC
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84. (2)
85. (1) Molarity is the number of moles perlitre of solution.
1mL of 2 2'10 - volume' H O
gives 10ml 2O at STP 1L (1000 mL) of
10 volume. 2 2H O gives 10000 mL
2O at STP
22400 mL 2O at STP is given by
2240 mL or 2.24 L of 10 vol 2 2H O
solution.
2240 mL or 2.24 L of 2 2H O
(10 volume) solution contains
2 mol 2 2H O
Molatity of 10 volume
2 22 moL
H O 0.892 M2.24 litre
= =
86. (3) The smallest ion ( )F will have the hig-
hest value for hydration en-
ergy.(most negative).
87. (1) Due to gradual increase in size of
the 2M + cation both lattice energy
and hydration energy decrease
down the group. Decrease in lattice
energy causes increase in solubility
but decrease in hydration energy
tends to decrease it .But since
lattice energy has a dominating
role the overall effect is an increase
in solubility down the group.
88. (1) Oxidizing power in the decreasing
order is:
2 3 4HC O HC O HC O HC O> > > .
As the oxidation number of C
increases the C O bond
becomes stronger and the thermal
stabilities of oxychlorides increase.
Hence HC O is the strongest
oxidizer and 4HC O is the weakest.
89. (3)2
4
Mn C :
2Mn :+
Magnetic moment
( ) ( )= + = +n n 2 5 5 2 BM
= 5.9 BM
90. (4)2Mg + is not precipitated by 2H S in
ammoniaccal medium.
2 2
2 34 68 g
H O
=