Jee main 2014 qpaper key solutions

Sri Chaitanya IIT Academy., India
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JEE-MAIN

2014 JEE-MAIN Q. PAPER WITH SOLUTIONS
#CORPORATE OFFICE: PLOT NO: 304,KASETTY HEIGHTS, AYYAPPA SOCIETY, MADHAPUR – HYD
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2
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3
MATHEMATICS
1. The image of the line
x 1 y 3 z 4
3 1 5
  
 

in the plane 2x – y + z + 3 = 0 is the line:
1)
x 3 y 5 z 2
3 1 5
  
 

2)
x 3 y 5 z 2
3 1 5
  
 
 
3)
x 3 y 5 z 2
3 1 5
  
 

4)
x 3 y 5 z 2
3 1 5
  
 
 
Key: 1
Sol :
x 1 y 3 z 4
is parallel to 2x y z 3 0
3 1 5
  
     

Image
h 1 k 3 l 4
2
2 1 1
  
   

   h, k, l 3, 5, 2  
Required equation
x 3 y 5 z 2
3 1 5
  
 

2. If the coefficients of 3 4
x and x in the expansion of   
182
1 ax bx 1 2x   in powers
of ‘x’ are both zero, then (a, b) is equal to:
1)
251
16,
3
    
2)
251
14,
3
    
3)
272
14,
3
    
4)
272
16,
3
    

4
Key : 4
Sol : Equating coefficients of 3 4
x and x is zero
32 17
17a b 0
3
 
  
1615 – 32 a + 3b = 0
Solving (a, b) =
272
16,
3
    
3. If a R and the equation      
2 2
3 x x 2 x x a 0      (where [x] denotes the
greatest integer x ) has no integral solution, then all possible values of a lie in
the interval :
1)    1, 0 0, 1 
2) (1, 2)
3) (-2, -1)
4)    , 2 2,   
Key : 1
Sol : 2 2
3f 2f a 0   where   x x
2
1 1 3a
f 1
3
 
  1 a 1; a 0   
   a 1, 0 0, 1   
4. If
2
a b b c c a a b c             
       
then  is equal to:
1) 2
2) 3
3) 0
4) 1

5
Key : 4
Sol :
2
a b b c c a a b c            
       
5. The variance of first 50 even natural numbers is :
1)
833
4
2) 833
3) 437
4)
437
4
Key : 2
Sol : Variance of first ‘n’ even natural numbers is equal to
2
n 1
3

2
50 1
833
3

 
6. A bird is sitting on the top of a vertical pole 20 m high and its elevation from a
point ‘O’ on the ground is 0
45 . It flies off horizontally straight away from the
point ‘O’. After one second, the elevation of the bird from ‘O’ is reduced to 0
30 .
Then the speed (in m/s) of the bird is:
1)  40 2 1
2)  40 3 2
3) 20 2
4)  20 3 1

6
Key : 4
Sol :
P Q
A B C
450
300
0
A 45 AB PB 20 mts and CQ 20    
 0 20
Tan30 BC 20 3 1 PQ
20 BC
    

 distance
speed 20 3 1 mts
time
  
7. The integral 2
0
x x
1 4sin 4sin dx
2 2

  equals:
1) 4
2)
2
4 4 3
3

 
3) 4 3 4
4) 4 3 4
3

 
Key : 4
Sol :
3
0 0
3
x x x
1 2sin dx 1 2sin dx 2sin 1 dx
2 2 2

 

                  
3
0
3
x x
x 4cos 4cos x
2 2



   
      
      
4 3 4
3

  

7
8. The statement  ~ p ~ q is :
1) equivalent to p q
2) equivalent to ~ p q
3) a tautology
4) a fallacy
Key : 1
Sol : The statement  ~ p ~ q is
P q ~q p ~ q  ~ p ~ q p q
T T F F T T
T F T T F F
F T F T F F
F F T F T T
 ~ p ~ q  is equivalent to p q
9. If A is an 3 3 non-singular matrix such that 1
AA' A'A and B A A', then BB'
 
equals :
1) I + B
2) I
3) 1
B
4)  1
B
Key : 2
Sol :   
11 1 1 1
BB' A A A A 

 
11 1 1
A AA A I. I I

  

8
10. The integral
1
x
x
1
1 x e dx
x
       is equal to :
1)  
1
x
x
x 1 e c

 
2)
1
x
x
xe c


3)  
1
x
x
x 1 e c

 
4)
1
x
x
xe c

 
Key : 2
Sol :
1
x
x
1
1 x e dx
x
      
1
x
x
2
1
1 x 1 e dx
x
         

1 1
x x
x x
2
1
e dx x 1 e dx
x
        
By parts =
1
x
x
xe c


11. If ‘z’ is a complex number such that | z | 2, then the minimum value of
1
z
2
 :
1) is equal to
5
2
2) lies in the interval (1, 2)
3) is strictly greater than
5
3
4) is strictly greater than
3
2
but less than
5
2

9
Key: 2
Sol : z rcis then | z | r 2   
2 2
2 21 1
z rcos r sin
2 2
        
2 1
r rcos
4
   
17 9 25
2cos which lies in ,
4 4 4
             
1 3 5
z ,
2 2 2
 
  
  
Minimum value
3
2
lies in (1, 2)
12. If ‘g’ is the inverse of a function ‘f’ and    5
1
f ' x , then g' x
1 x


is equal to:
1) 5
1 x
2) 4
5x
3)
  
5
1
1 g x
4)   
5
1 g x
Key : 4
Sol : If ‘g’ is the inverse of a function ‘f’, then   x f g x     f ' g x g' x 1 
 
 
  
51
g' x 1 g x
f ' g x
   
 
 

10
13. If  
   
     
     
n n
3 1 f 1 1 f 2
, 0, and f n and 1 f 1 1 f 2 1 f 3
1 f 2 1 f 3 1 f 4
 
        
  
=
     
2 2 2
K 1 1 , then K   is equal to :
1) 
2)
1

3) 1
4) -1
Key : 3
Sol : Given determinant = 2
2 2 2
1 1 1 1 1 1
1 . 1
1 1
   
   
K 1 
14. Let    k k
k
1
f x sin x cos x
k
  where x R and k 1  . Then    4 6f x f x equals:
1)
1
6
2)
1
3
3)
1
4
4)
1
12
Key : 4
Sol :    4 6f x f x = 2 2 2 21 1
[1 2sin xcos x] 1 3sin xcos x
4 6
     

11
15. Let and  be the roots of equation 2
px qx r 0, p 0    . If p, q, r are in A.P
and
1 1
4, 
 
then the value of | |  is :
1)
61
9
2)
2 17
9
3)
34
9
4)
2 13
9
Key : 4
Sol :
q
p
  
r
.
p
  
2q p q 
By given condition
1 1
4 
 
q
4
r
 
 
2
4      
16. Let A and B be two events such that
     
1 1 1
P A B , P A B and P A , where A
6 4 4
     stands for the complement of the
event A. Then the events A and B are :
1) mutually exclusive and independent
2) equally likely but not independent
3) independent but not equally likely
4) independent and equally likely

12
Key : 3
Sol :  
1 5
P A B 1
6 6
   
   
1 5
P A P B
4 6
  
 
3 1 5
P B
4 4 6
  
 
5 1 5 3 1
P B
6 2 6 3

   
     
3 1 1
P A .P B P A B
4 3 4
    
A, B are independent but not equally likely     P A P B
17. If ‘f’ and ‘g’ are differentiable functions in [0, 1] satisfying f(0) = 2 = g(1), g(0) = 0
and f(1) = 6, then for some c ]0, 1[ :
1)    2f ' c g' c
2)    2f ' c 3g' c
3)    f ' c g' c
4)    f ' c 2g' c
Key : 4
Sol :
 
 
   
   
 
f ' c f 1 f 0
for c 0, 1
g' c g 1 g 0

 

18. Let the population of rabbits surviving at a time ‘t’ be governed by the
differentiable equation
 
 
dp t 1
p t 200
dt 2
  . If p(0) = 100, then p(t) equals :
1) t/2
400 300 e
2) t/2
300 200 e

3) t/2
600 500 e
4) t/2
400 300 e


13
Key : 1
Sol :  
dp 1
p t 200
dt 2
 
This is linear d.e in ‘p’
1 1
t t
2 2
p.e 400e c
 
  
By given condition p(0) = 100, c = -300
1
t
2
p 400 300e  
19. Let ‘C’ be the circle with centre at (1, 1) and radius = 1. If ‘T’ is the circle
centred at (0, y), passing through origin and touching the circle ‘C’ externally,
then the radius of ‘T’ is equal to :
1)
3
2
2)
3
2
3)
1
2
4)
1
4
Key : 4
Sol :
   1 2c 1, 1 c 0, y 
1 2r 1, r y 
1 2 1 2c c r r 
   
2 2
1 y 1 1 y   
1= 4y

14
20. The area of the region described by   2 2 2
A x, y :x y 1 and y 1 x     is
1)
4
2 3


2)
4
2 3


3)
2
2 3


4)
2
2 3


Key : 1
Sol.
Required area =
1
0
4
2 1 x dx
2 2 3
 
   
21. Let a, b, c and d be non-zero numbers. If the point of intersection of the line
4ax + 2ay + c = 0 and 5bx + 2by + d = 0 lies in the fourth quadrant and is
equidistant from the two axes then :
1) 2bc – 3ad = 0
2) 2bc + 3ad = 0
3) 3bc – 2ad = 0
4) 3bc + 2ad = 0
Key : 3
Sol. Point of intersection
2ad 2bc 5bc 4ad
,
2ab 2ab
       
Equidistance from the both axis in
fourth quadrant 3bc 2ad 0 

15
22. Let PS be the median of the triangle with vertices P(2, 2), Q(6, -1) and R(7, 3).
The equation of the line passing through (1, -1) and parallel to PS is
1) 4x 7y 11 0  
2) 2x 9y 7 0  
3) 4x 7y 3 0  
4) 2x 9y 11 0  
Key : 2
Sol : Mid point
13
,1
2
    
Slope PS
2
9
 
23.
 2
2x 0
sin cos x
lim
x

is equal to :
1)
2

2) 1
3) 
4) 
Key : 4
Sol :
 2
2x 0
sin sin x
lim
x

 2 2
2 2x 0
sin sin x sin x
lim 1 1
sin x x
 
    

24. If  n
X 4 3n 1: n N    and   Y 9 n 1 :n N ,   where N is the set of natural
numbers, then X Y is equal to:
1) N
2) Y – X
3) X
4) Y

16
Key : 4
Sol : X and Y are multiplies of 9 and X Y
X Y Y  
25. The locus of the foot of perpendicular drawn from the centre of the ellipse
2 2
x 3y 6  on any tangent to it is :
1)  
22 2 2 2
x y 6x 2y  
2)  
22 2 2 2
x y 6x 2y  
3)  
22 2 2 2
x y 6x 2y  
4)  
22 2 2 2
x y 6x 2y  
Key : 3
Sol. Given that ellipse
2 2
x y
1
6 2
 
C
P
A
B
Slope CP = 1
1
y
x
Slope of AB = 1
1
x
y
 = m
Equation of the tangent is 2
y mx 6m 2  
By above equations we get  
22 2 2 2
x y 6x 2y  

17
26. Three positive numbers from an increasing G.P. If the middle term in this G.P is
doubled, the new numbers are in A.P. Then the common ratio of the G.P is :
1) 2 3
2) 3 2
3) 2 3
4) 2 3
Key : 4
Sol : 2
a,ar,ar G.P .
2
a ar
2ar
2


2
4ar a ar 
2
r 4r 1 0  
4 16 4
r
2
 

r 2 3 
r 2 3  
27. If              
9 1 8 2 7 9 9
10 2 11 10 3 11 10 ..... 10 11 k 10 ,     then ‘k’ is equal to
1)
121
10
2)
441
100
3) 100
4) 110

18
Key : 3
Sol :            
9 1 8 2 7 9
S 10 2 11 10 3 11 10 .....10 11     ……… (1)
         
2 3 6 108 711
S 11.10 2 11 10 3 11 10 .....9 11 11
10
     ……… (2)
(2) ... (1)
  29 8 7 9 101
S 10 11.10 11 10 .... 11 11
10
      
 9
S 100 10 
K 100 
28. The angle between the lines whose direction cosines satisfy the equations
2 2 2
m n 0 and m nl l     is :
1)
3

2)
4

3)
6

4)
2

Key : 1
Sol :
2 2 2
l m n 0  
 l m n  
 
2 2 2
m n m n 0   
2mn = 0
m = 0 (or) n = 0
If m = 0
l = -(0 + n)
l = -n
l : m : n = -n : 0 : n = -1 : 0 : 1
similarly if n = 0 then
l : m : n = -1, : 1 : 0
1 0 0
cos
2 2
 
 
3

 

19
29. The slope of the line touching both the parabolas 2 2
y 4x and x 32y  is :
1)
1
2
2)
3
2
3)
1
8
4)
2
3
Key : 1
Sol : Common tangent of  
2/32 2 1/3 1/3
y 4ax and x 4by is a x b y ab 0    
Here 4a = 4 and 4b = -32
Slope of the common tangent = 1/2
30. If x = -1 and x = 2 are extreme points of   2
f x log | x | x x    then:
1)
1
6,
2
   
2)
1
6,
2
   
3)
1
2,
2
   
4)
1
2,
2
   
Key : 3
Sol :    f ' 1 0 and f ' 2 0  
Solving above equations we get
1
2,
2
   

20
PHYSICS
31. When a rubber-band is stretched by a distance x , it exerts a restoring force of
magnitude 2
F ax bx  where a and b are constants. The work done in stretching
the unstretched rubber-band by L is:
1)
2 3
1
2 2 3
aL bL 
 
 
2) 2 3
aL bL
3)  2 31
2
aL bL
4)
2 3
2 3
aL bL

Key: 4
Sol:  
2 3
2
0
2 3
L
aL bL
w Fdx ax bx dx     
32. The coercivity of a small magnet where the ferromagnet gets demagnetized is
3 1
3 10 Am
 . The current required to be passed in a solenoid of length 10 cm and
number of turns 100, so that the magnet gets demagnetized when inside the
solenoid, is:
1) 6A
2) 30 mA
3) 60 mA
4) 3 A
Key: 4
Sol: 3100
I=3 10
0.1
nI H   
3I A

21
33. In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80
Wand 1heater of 1 kW. The voltage of the electric mains is 220 V. The minimum
capacity of the main fuse of the building will be:
1) 14A
2) 8A
3) 10A
4) 12A
Key: 4
Sol:  
1
40 15 100 5 80 5 1000
220
P IV I        
2500
11.4 12A
220
A  
34. An open glass tube is immersed in mercury in such a way that a length of 8 cm
extends above the mercury level. The open end of the tube is then closed and
sealed and the tube is raised vertically up by additional 46cm. What will be
length of the air column above mercury in the tube now?
(Atmospheric pressure = 76 cm of Hg)
1) 6 cm
2) 16 cm
3) 22 cm
4) 38 cm
Key: 2
Sol:  76 8 22x x  
By trail and error method 16x 

22
35. A bob of mass m attached to an inextensible string of length l is suspended from
a vertical support. The bob rotates in a horizontal circle with an angular speed 
rad/s about the vertical. About the point of suspension:
1) angular momentum changes both in direction and magnitude
2) angular momentum is conversed
3) angular momentum changes in magnitude but not in direction
4) angular momentum changes in direction but not in magnitude
Key: 4
Sol: Angular Momentum Changes in directions
36. The current voltage relation of diode is given by  1000 /
1 ,V T
I e mA  where the
applied voltage V is in volts and the temperature T is in degree Kelvin. If a
student makes an error measuring 0.01V while measuring the current of 5mA at
300 K, what will be the error in the value of current in mA?
1) 0.05 Ma
2) 0.2 mA
3) 0.02 mA
4) 0.5 mA
Key: 2
Sol: I+1 = 1000 /V T
e
 
1000V
ln I 1
T
 
1000
V
I+1 T
I
 
31000
I= 6 10 0.01
3

   
= 0.2 mA

23
37. From a tower of height H, a particle is thrown vertically upwards with a speed u.
The time taken by the particle to hit the ground, is n times that taken by it to
reach the highest point of its path. The relation between H, u and n is:
1)   2
2gH n u 
2) 2 2
2gH n u
3)  
2 2
2gH n u 
4)  2
2 2gH nu n 
Key: 4
Sol:
H
u
Let 1t is the time taken to reach max height 1
u
t
g

Let 2t is the time taken to reach ground from tower equation
2
2 2
1
2
H ut gt  
2
2
2 4 8
2
u u gH
t
g
 

2
2
2u u gH
g g

 
Given that
2 1t nt
2
2
2u u gH nu
g g g

  ,  
2 2
2 2
2
1
u gH u
n
g g

 
 
22
2 1 1gH u n   
 
2 2
2 2gH u n n   
 2
2 2gH u n n 

24
38. A thin convex lens made from crown glass
3
2

 
 
 
has focal length f . When it is
measured in two different liquids having refractive indices
4
3
and
5
3
, it has the
focal lengths 1f and 2f respectively. The correct relation between the focal lengths
is:
1) 1f and 2f both become negative
2) 1 2f f f 
3) 1f f and 2f becomes negative
4) 2f f and 1f becomes negative
Key: 3
Sol: From  
1 2
1
f R
 
     
1 1
f R
  3 2/ 
f R
For medium 4 3/ 
3
1 2 12 1
4 3 4f / R R
 
 
   
 
 
1 4f R
for medium
2
1 3 2 2
5 3 1
5 3
/
M /
f / R
 
    
 
1
5R
 
2 5f R 

25
39. A parallel plate capacitor is made of two circular plates separated by a distance
of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the
electric field in the dielectric is 4
3 10 /V m , the charge density of the positive plate
will be close to:
1) 4 2
6 10 /C m
2) 7 2
6 10 /C m

3) 7 2
3 10 /C m

4) 4 2
3 10 /C m
Key: 2
Sol: 4
0
3 10
K

 

12 4
2.2 8.8 10 3 10 
    
7
6 10 
 
40. In the circuit shown here, the point ‘C’ is kept connected to point ‘A’ till the
current flowing through the circuit becomes constant. Afterward, suddenly, point
‘C’ is disconnected from point ‘A’ and connected to point ‘B’at time t=0. Ratio of
the voltage across resistance and the inductor at t = L/R will be equal to:
L
R
B
A C
1)
1 e
e

2)
1
e
e
3) 1
4) 1
Key: 4
Sol: 0R LV V  always for 0t 

26
41. Two beams, A and B, of plane polarized light with mutually perpendicular planes
of polarization are seen through a polaroid. From the position when the beam A
has maximum intensity (and beam B has zero intensity), a rotation of polaroid
through 30makes the two beams appear equally bright. If the initial intensities
of two beams are AI and BI respectively, then A
B
I
I
equals:
1)
1
3
2) 3
3)
3
2
4) 1
Key: 1
Sol: 1 1 2 2
30 60A B A BI I I cos I cos  
2
2
cos 60 1
cos 30 3
A
B
I
I
  
42. There is a circular tube in a vertical plane. Two liquids which do not mix and of
densities 1d and 2d are filled in the tube. Each liquid subtends 90 angle at centre.
Radius joining their interface makes an angle  with vertical. Ratio 1
2
d
d
is:

1d
2d
1)
1 sin
1 cos




2)
1 sin
1 sin




3)
1 cos
1 cos




4)
1 tan
1 tan





27
Key: 4
Sol: A BP P
1 1 2 2h d g h d g 
   1 2cos sin cos sinR d R d      
1
2
cos sin 1 tan
cos sin 1 tan
d
d
  
  
 
 
 
A
B
R
2h
2d
1d
1h
R 
43. The pressure that has to be applied to the ends of a steel wire of length 10 cm to
keep its length constant when its temperature is raised by 100 C is: (For steel
Young’s modulus is 11 2
2 10 N m
 and coefficient of thermal expansion is 5 1
1.1 10 K 
 )
1) 6
2.2 10 Pa
2) 8
2.2 10 Pa
3) 9
2.2 10 Pa
4) 7
2.2 10 Pa
Key: 2
Sol: Pressure 11 5 8
α 2 10 1.1 10 100 2.2 10y  
        

28
44. A block of mass m is placed on a surface with a vertical cross section given by
3
6
x
y  . If the coefficient of friction is 0.5, the maximum height above the ground
at which the block can be placed without slipping is:
1)
1
2
m
2)
1
6
m
3)
2
3
m
4)
1
3
m
Key: 2
Sol: At the instant of just sliding

H=?
tan 
0.5
dy
dx

2
1
2
x
x   
1
' ' 1
6
H y at x m  

29
45. Three rods of Copper, Brass and Steel are welded together to form a Y – shaped
structure. Area of cross – section of each rod 2
4cm . End of copper rod is
maintained at 100 C where as ends of brass and steel are kept at 0 C . Lengths of
the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are
thermally insulated from surroundings except at ends. Thermal conductivites of
copper, brass and steel are 0.92,0.26 and 0.12 CGS units respectively. Rate of heat
flow through copper rod is:
1) 6.0 cal/s
2) 1.2 cal/s
3) 2.4 cal/s
4) 4.8 cal/s
Key: 4
Sol: Let T is the junction temperature then
cu Brass steelH H H 
   100cu B S
cu B S
K A T K A T K AT
l l l

  ,  
0 96 0 26 0 12
100
46 13 12
. . .
T T T  
200 5T
40T C 
Heat flow per sec through copper
 100KA T
l


 0 96 4 60
46
. 
 4 8. cal/sec
0 C0 C
100 C
Brass
Copper
Steell=13 cm
k=0.26
l=12 cm
k=0.12
l=46
k=0.96
T

30
46. A mass ‘m’ is supported by a massless string wound around a uniform hollow
cylinder of mass m and radius R. If the string does not slip on the cylinder, with
what acceleration will the mass fall on release?
m
R
m
1) g
2)
2
3
g
3)
2
g
4)
5
6
g
Key: 3
Sol: mg T ma 
2 2
.
a
TR mR mR T ma
R
   
2ma mg  or / 2a g

31
47. Match List-I (Electromagnetic wave type) with List – II (Its
association/application) and select the correct option from the choices given
below the lists:
List-I List-II
(a) Infrared waves (i) To treat muscular strain
(b) Radio waves (ii) For broadcasting
(c) X-rays (iii) To detect fracture of bones
(d) Ultra violet rays (iv) Absorbed by the ozone layer
of the atmosphere
(a) (b) (c) (d)
1) (i) (ii) (iii) (iv)
2) (iv) (iii) (ii) (i)
3) (i) (ii) (iv) (iii)
4) (iii) (ii) (i) (iv)
Key: 1
Sol: Conceptual
48. The radiation corresponding to 3 2 transition of hydrogen atom falls on a
metal surface to produce photoelectrons. These electrons are made to enter a
magnetic field of 4
3 10 T
 . If the radius of largest circular path followed by these
electrons is 10.0 mm, the work function of the metal is close to:
1) 1.6 eV
2) 1.8 eV
3) 1.1 eV
4) 0.8 eV

32
Key: 3
Sol: In magnetic field ,
 2 KE m
r
Bq

 
28 19 42 2 2
31 19
9 10 1.6 10 10 1
2 2 9 10 1.6 10
B q r
KE
m
  
 
   
  
  
0.8ev
1 1
13.6 1.89
4 9
E eV
 
   
 
0 1.89 0.8 1.1W E KE eV    
49. During the propagation of electromagnetic waves in a medium:
1) Both electric and magnetic energy densities are zero
2) Electric energy density is double of the magnetic energy density
3) Electric energy density is half of the magnetic energy density
4) Electric energy density is equal to the magnetic energy density
Key: 4
Sol: Conceptual
50. A green light is incident from the water the air – water interface at the critical
angle   . Select the correct statement.
1) The entire spectrum of visible light will come out of the water at various angles
to the normal
2) The entire spectrum of visible light will come out of the water at an angle of
90 to the normal
3) The spectrum of visible light whose frequency is less than that of green light
will come out to the air medium
4) The spectrum of visible light whose frequency is more than that of green light
will come out to the air medium

33
Key: 3
Sol: With increase in frequency, critical angle c decreases i.e., present angle of
incidence in water will be more than respective critical angles of high frequencies
radiation. Hence high frequency radiations will get TIR and low frequencies
radiations will come out in to air
51. Four particles, each of mass M and equidistant from each other, move along a
circle of radius R under the action of their mutual gravitational attraction. The
speed of each particle is:
1)  1
1 2 2
2
GM
R

2)
GM
R
3) 2 2
GM
R
4)  1 2 2
GM
R

Key: 1
Sol:
V
R
R 2
2
2
2 2
mv 1 1 1
2
2 42
Gm
R R R
 
    
 
 2 1
v 2 2 1
4
Gm
R
 
 1
v 2 2 1
2
Gm
R
 

34
52. A particle moves with simple harmonic motion in a straight line. In first s ,
after starting from rest it travels a distance a, and in next s it travels 2a, in
same direction, then:
1) time period of oscillations is 6
2) amplitude of motion is 3a
3) time period of oscillations is 8
4) amplitude of motion is 4a
Key: 1
Sol:
a a a
 0 2
6
T
 
53. A conductor lies along the z – axis at 1.5 1.5z m   and carries a fixed current of
10.0 A in ˆza direction (see figure). For a field 4 0.2
ˆ3.0 10 x
yB e a T 
 

, find the power
required to move the conductor at constant speed to 2.0x m , 0y m in 3
5 10 s
 .
Assume parallel motion along the x  axis.
x
yB
I
2.0
1.5
1.5
1) 29.7 W
2) 1.57 W
3) 2.97 W
4) 14.85 W

35
Key: 3
Sol: 4 0.2
Instant 3
2
3 10 3
5 10
x
P B V e 

    


1 0.212
10
5
x
P e 
 
2.97range
pdx
P W
dx
 


54. The forward biased diode connection is:
1)
2V 2V
2)
2V2V
3)
3V3V
4)
2V 4V
Key: 2
Sol: Conceptual
55. Hydrogen 1
1 H , Deuterium 2
1 H , singly ionized Helium  4
2 He

and doubly
ionized lithium  6
3 Li

all have one electron around the nucleus. Consider an
electron transition from 2n  to 1n  . If the wavelengths of emitted radiation are
1 2 3, ,   and 4 respectively then approximately which one of the following is
correct?
1) 1 2 3 42 3 4     
2) 1 2 3 44 2 2     
3) 1 2 3 42 2     
4) 1 2 3 44 9     

36
Key: 4
Sol: 2
2
1 1 1
1
4
Rz
z


 
   
 
1 2  as 1z  for both 1
1H and 2
1 H
2
31
1 32
3 1
4 4
z
z

 

   
2
1 4
1 42
4 1
9 9
z
z

 

   
 1 2 3 44 9     
56. On heating water, bubbles being formed at the bottom of the vessel detach and
rise. Take the bubbles to be spheres of radius R and making a circular contact of
radius r with the bottom of the vessel. If ,r R and the surface tension of water
is T, value of r just before bubbles detach is: (density of water is w )
R
2r
1) 2 3 wg
R
T

2) 2
3
wg
R
T

3) 2
6
wg
R
T

4) 2 wg
R
T


37
Key: 3(This is an ambigious Question, so expected key is 3)
Sol: sinTdl   buoyancy force on bubble
R

r r
 2
r
T r
R
  34
3
wR g 
4
2 2
3
wR g
r
T


2 2
3
wg
r R
T

 
57. A pipe of length 85 cm is closed from one end. Find the number of possible
natural oscillations of air column in the pipe whose frequencies lie below 1250
Hz. The velocity of sound in air is 340 m/s.
1) 4
2) 12
3) 8
4) 6
Key: 4
Sol: From 2
340
100
4 4 85 10
V
n HZ
l 
  
 
Possible frequencies in closed pipe is odd harmonics only So
100 300 500 700 900HZ, HZ, HZ, HZ, HZ and 1100 1300 1500HZ, HZ, HZ..... etc below
1250HZ , 6 natural oscillations are possible

38
58. Assume that an electric field
2ˆ30E x i

exists in space. Then the potential
difference A OV V
, where OV
is the potential at the origin and AV
the potential at
2x m is:
1) 80 J
2) 120 J
3) – 120 J
4)– 80 J
Key: 4
Sol:
0
2
2
0
0 0
30 80
AV A
A
V
dv Edx V V x dx J         
59. A student measured the length of a rod and wrote it as 3.50 cm. Which
instrument did he use to measure it?
1) A screw guage having 50 divisions in the circular scale and pitch as 1mm
2) A meter scale
3) A vernier calipers where the 10 divisions in vernier scale matches with 9
division in main scale has 10 divisions in 1cm
4) A screw guage having 100 divisions in the circular scale and pitch as 1mm

39
Key: 3
Sol: Least count of vernier
1
10
mm
60. One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in
figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K,
800 K and 600 K respectively. Choose the correct statement:
P
V
A
400K
C
B 800 K
600 K
1) The change in internal energy in the process BC is 500R
2) The change in internal energy in whole cyclic process is 250 R
3) The change in internal energy in the process CA is 700 R
4) The change in internal energy in the process AB is 350R
Key: 1
Sol: 1) du for isochoric process VnC dT  
3
1 400 600
2
R
R  
2) for Process BC
   1 2 200
500
1 1 7 / 5
nR T T R
du R
r
 
   
 
3) for whole process 0du 

40
CHEMSITRY
61. Which one is classified as a condensation polymer?
1) Acrylonitrile
2) Dacron
3) Neoprene
4) Teflon
Key: 2
Sol: Dacron is the condensation polymer
n
Neoprene
Add polymers
Teflon



62. Which one of the following properties is not shown by NO?
1) It’s bond order is 2.5
2) It is diamagnetic in gaseous state
3) It is a neutral oxide
4) It combines with oxygen to form nitrogen dioxide
Key: 2
Sol: NO is paramagnetic

41
63. Sodium phenoxide when heated with 2CO under pressure at 0
125 C yields a product
which on acetylation produces C.
2CO
0
125
5Atm

2
H
AC O
B C


ONa
The major product C would be:
1) 3OCOCH
COOH
2) 3OCOCH
COOH
3) OH
3COCH
3COCH
4) OH
3COOCH
Key: 2

42
Sol:
ONa
0125 C
5atm2
CO 
OH
COOH H
AC O2


COOH
3
OCOCH
(B) (C)
Aspirin
64. Given below are the half-cell reactions:
2 0
2 ; 1.18Mn e Mn E V 
   
 3 2 0
2 ; 1.51Mn e Mn E V  
   
The 0
E for 2 3
3 2Mn Mn Mn 
  will be:
1) - 0.33 V; the reaction will occur
2) – 2.69 V; the reaction will not occur
3) – 2.69 V; the reaction will occur
4) – 0.33 V; the reaction will not occur
Key: 2
Sol: 0
1.18 1.51E   
= –2.69 V

43
65. For complete combustion of ethanol.        2 5 2 2 23 2 3C H OH l O g CO g H O l   , the
amount of heat produced as measured in bomb calorimeter, is 1364.47 1
kJmol
at
0
25 C . Assuming ideally the Enthalpy of combustion, cH , for the reaction will be:
 1
8.314R kJmol

1) 1
1350.50kJmol

2) 1
1366.95kJmol

3) 1
1361.95kJmol

4) 1
1460.50kJmol

Key: 2
Sol: H U nRT   
  3
1364 1 8.314 298 10
    
66. For the estimation of nitrogen, 1.4 g of an organic compound was digested by
Kjeldahl method and the evolved ammonia was absorbed in 60 mL of
10
M
sulphuric acid. The unreacted acid required 20 mL of
10
M
sodium hydroxide for
complete neutralization. The percentage of nitrogen in the compound is:
1) 5%
2) 6%
3) 10%
4) 3%

44
Key: 3
Sol: Meq of 2 4
1
H SO 60 2 12
10
   
Meq of base used for back titration
1
20 2
10
  
= Meq of acid left
Meq of acid reacted with liberated
Ammonia = 12-2 = 10 meq
= meq of 3
NH
Millimoles of 3
NH liberated = 10 [ n-factor = 1]
Moles of 3
3
NH 10 10
 
Moles “N” 3
10 10
 
Wt of “N” 3
10 10 14
  
% of “N” in the organic compound
3
10 10 14
100 10%
1.4

 
  
67. The major organic compound formed by the reaction of 1, 1, 1 – trichloroethane
with silver powder is:
1) 2 – Butene
2) Acetylene
3) Ethene
4) 2 – Butyne
Key: 4

45
Sol:
3
2CH C Cl 
Cl
Cl
3 3
6Ag CH C C CH    
2-Butyne
+ 6 AgCl
68. The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 :
4. The ratio of number of their molecule is:
1) 3 : 16
2) 1 : 4
3) 7 : 32
4) 1 : 8
Key: 3
Sol: Ratio of the molecules =
1 4
:
32 28
= 7: 32
69. The metal that cannot be obtained by electrolysis of an aqueous solution of its salt
is:
1) Cr
2) Ag
3) Ca
4) Cu
Key: 3
Sol: Ca cannot be obtained by the electrolysis of aqueous solution of its salt.

46
70. The equivalent conductance of NaCl at concentration C and at infinite dilution
are C and , respectively. The correct relationship between C and is given as
(where the constant B is positive)
1)  C B C  
2)  C B C  
3)  C B C  
4)  C B C  
Key: 4
Sol:  C B C  
71. The correct set of four quantum numbers for the valence electrons of rubidium
atom (Z=37) is:
1)
1
5,0,1,
2

2)
1
5,0,0,
2

3)
1
5,1,0,
2

4)
1
5,1,1,
2

Key: 2
Sol: 1
5s
72. Consider separate solutions of 0.500 M  2 5 ,C H OH aq    3 4 2
0.100 ,M Mg PO aq
 0.250M KBr aq and  3 40.125M Na PO aq at 0
25 C . Which statement is true about these
solutions, assuming all salts to be strong electrolytes?
1)  2 50.500M C H OH aq has the highest osmotic pressure
2) They all have the same osmotic pressure
3)    3 4 2
0.100M Mg PO aq has the highest osmotic pressure
4)  3 40.125M Na PO aq has the highest osmotic pressure

47
Key: 2
Sol: iCST 
73. The most suitable reagent for the conversion of 2R CH OH R CHO    is:
1) PCC (Pyridinium Chlorochromate)
2) 4KMnO
3) 2 2 7K Cr O
4) 3CrO
Key: 1
Sol: Most suitable reagent for the conversion of 0
1 alcohol to aldehyde is “PCC”
74. CsCl crystallises in body centred cubic lattice. If ‘a’ is its edge length then which
of the following expressions is correct?
1) 3Cs Cl
r r a  
2) 3Cs Cl
r r a  
3)
3
2Cs Cl
a
r r  
4)
3
2Cs Cl
r r a  
Key: 4
Sol: c a3a 2(r r ) 
c a
3a
r r
2
 

48
75. In which of the following reactions 2 2H O acts as reducing agent?
(a) 2 2 22 2 2H O H e H O 
  
(b) 2 2 22 2H O e O H 
  
(c) 2 2 2 2H O e OH 
 
(d) 2 2 2 22 2 2H O OH e O H O 
   
1) (b), (d)
2) (a), (b)
3) (c), (d)
4) (a), (c)
Key: 1
Sol: While acting as reducing agent it gives up electrons and release oxygen gas.
76. For which of the following molecule significant 0  ?
a)
Cl
Cl b)
CN
CN c)
OH
OH d)
SH
SH
1) (c) and (d)
2) Only (a)
3) (a) and (b)
4) Only (c)
Key: 1

49
Sol:
Cl
Cl
0 
CN
CN
0 ;
0  0 ;
O H
O
H
S H
S
H
But dipole moment of
"SH"
SH is less than dipolemoment of
OH
OH , but is significant
77. On heating an aliphatic primary amine with chloroform and ethanolic potassium
hydroxide, the organic compound formed is:
1) an alkyl isocyanide
2) an alkanol
3) an alkanediol
4) an alkyl cyanide
Key: 1
Sol: 2 3
R NH CHCl KOH R NC     alkylisocyanide

50
78. In 2NS reactions, the correct order of reactivity for the following compounds:
 3 3 2 3 2
, ,CH Cl CH CH Cl CH CHCl and  3 3
CH CCl is
1)    3 3 2 3 32 3
CH CHCl CH CH Cl CH Cl CH CCl  
2)    3 3 3 2 32 3
CH Cl CH CHCl CH CH Cl CH CCl  
3)    3 3 2 3 32 3
CH Cl CH CH Cl CH CHCl CH CCl  
4)    3 2 3 3 32 3
CH CH Cl CH Cl CH CHCl CH CCl  
Key: 3
Sol:    3 3 2 3 32 3
CH Cl CH CH Cl CH CHCl CH CCl  
79. The octahedral complex of a metal ion 3
M 
with four monodentate ligands
1 2 3, ,L L L and 4L absorb wavelengths in the region of red, green, yellow and blue,
respectively. The increasing order of ligand strength of the four ligands is:
1) 1 2 4 3L L L L  
2) 4 3 2 1L L L L  
3) 1 3 2 4L L L L  
4) 3 2 4 1L L L L  
Key: 3
Sol: Order of wavelength of colours
Red >Yellow > Green > Blue

51
80. The equation which is balanced and represents the correct product(s) is:
1)  4 2 2 44
4CuSO KCN K Cu CN K SO    
2) 2 22 2Li O KCl LiCl K O  
3)   2
3 45
5 5CoCl NH H Co NH Cl

   
      
4)      
2 24
2 26
6excess NaOH
Mg H O EDTA Mg EDTA H O
 
        
Key: 3
Sol:   2
3 45
CoCl NH 5H Co 5NH Cl

   
      
81. In the reaction, 54 .
3 ,PClLiAlH Alc KOH
CH COOH A B C   the product C is:
1) Acetyl chloride
2) Acetaldehyde
3) Acetylene
4) Ethylene
Key: 4
Sol:
54
3 3 2 3 2 2 2
PClLiAlH Alc.KOH
CH C OH CH CH OH CH CH Cl CH CH         
O
82. The correct statement for the molecule, 3CsI , is:
1) it contains ,Cs I 
and lattice 2I molecule
2) it is a covalent molecule
3) it contains Cs
and 3I
ions
4) it contains 3
Cs 
and I 
ions

52
Key: 3
Sol: 3
CsI contains Cs
and 3
I
ions
83. For the reaction      2 2 3
1
2g g g
SO O SO  , if  
x
p cK K RT where the symbols have
usual meaning then the value of x is: (assuming ideality)
1) 1
2) 1
3)
1
2

4)
1
2
Key: 3
Sol: n
P CK K (RT)

3 1
n 1
2 2

  
84. For the non-stoichiometre reaction 2 ,A B C D   the following kinetic data
were obtained in three separate experiments, all at 298 K.
Initial
Concentration
(A)
Initial Concentration
(B)
Initial rate of formation
of C  mol L S 
0.1M
0.1 M
0.2 M
0.1 M
0.2 M
0.1 M
3
1.2 10

3
1.2 10

3
2.4 10

The rate law for the formation of ‘C’ is?
1)  
dc
k A
dt

2)   
dc
k A B
dt

3)    
2dc
k A B
dt

4)    
2dc
k A B
dt


53
Key: 1
Sol: m m
r K(A) (B)
85. Resistance of 0.2 M solution of an electrolyte is 50. The specific conductance of
the solution is 1
1.4 S m
. The resistance of 0.5 M solution of the same electrolyte is
280. The molar conductivity of 0.5 M solution of the electrolyte in 2 1
S m mol
is:
1) 2
5 10
2) 4
5 10

3) 3
5 10

4) 3
5 10
Key: 2
Sol:
l
R
a
 
1 l
50
1.4 a
 70
a
l

280 (70)  4 
K = 0.25
3
410 0.25
5 10
0.5



  
86. Among the following oxoacids, the correct decreasing order of acid strength is:
1) 2 4 3HClO HClO HClO HOCl  
2) 2 3 4HOCl HClO HClO HClO  
3) 4 2 3HClO HOCl HClO HClO  
4) 4 3 2HClO HClO HClO HOCl  
Key: 4
Sol: With increase in unprotonated oxygen atoms acidic strength increases

54
87. Which one of the following bases is not present in DNA?
1) Thymine
2) Quinoline
3) Adenine
4) Cytosine
Key: 2
Sol: Qunoline is the base which is not present in DNA
88. Considering the basic strength of amines in aqueous solution, which one has the
smallest bpK value?
1) 6 5 2C H NH
2)  3 2
CH NH
3) 3 2CH NH
4)  3 3
CH N
Key: 2
Sol:  3 2
CH NH Stronger base in aqueous medium than 3
Me N . So its pKb value is
smallest
89. If Z is a compressibility factor, vander Waals equation at low pressure can be
written as:
1) 1
Pb
Z
RT
 
2) 1
RT
Z
Pb
 
3) 1
a
Z
VRT
 
4) 1
Pb
Z
RT
 

55
Key: 3
Sol: 2
a
P V RT
V
 
  
 
a
PV RT
V


a
Z 1
RTV
 
90. Which series of reactions correctly represents chemical relations related to iron
and its compound?
1)
0 0
2 , ,600 ,700
3 4
O heat CO C CO C
Fe Fe O FeO Fe  
2)  2 4 2 4 2dil ,
4 2 4 3
H SO H SO O heat
Fe FeSO Fe SO Fe  
3) 2 2 4, dil
4
O heat H SO heat
Fe FeO FeSO Fe  
4) 2 , ,
3 2
Cl heat heat air Zn
Fe FeCl FeCl Fe  
Key: 1
Sol: O heat CO CO2
600 7003 4
Fe Fe O FeO Fe  

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