JEE MAIN 2016
Test Paper Code – G
Questions with Solutions
Mathematics
1. A value of  for which
 1 
(1) sin 1 

 3
2  3i sin 
is purely imaginary is
1  2isin 
(2)

3
(3)

6
 3
(4) sin 1 

 4 
Ans (1)
2  3isin  1  2isin 
Z

1  2i sin  1  2isin 
Re(Z) = 0
2  6 sin2  = 0
1
1
1
sin 2    sin  
   sin 1
3
3
3
2. The system of linear equations
x + y  z = 0
x  y  z = 0
x + y  z = 0
has a non-trivial solution for
(1) exactly three values of 
(3) exactly one value of 
Ans (1)
For non-trivial solution | A | = 0
1  1
 1 1  0
1 1 
(2) infinitely many values of 
(4) exactly two values of 
1( + 1)   ( 2 + 1) 1 ( + 1) = 0
 = 0 or  =  1
 = 0, 1, 1
3. A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x
units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is
minimum, then:
(1) 2x = r
(2) 2x = ( + 4)r
(3) (4  )x = r
(4) x = 2r
Ans (4)
4x + 2r = 2
2x + r = 1
1  2x
r

To minimize area A
A = x2 + r2
2
 1  2x 
= x  

  
1
A  x 2  (1  2x) 2

dA
1
2 4x 

 2x  2(1  2x)(2)  2  x   
dx

 

2
For minima
dA
0
dx
2 4x
x 
0
 
 4 2
x 1   
  
x
2
4
1  2x
r


1
r
4
 x = 2r
1
4

4 

(4  )
4. A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on
the path, he observes that the angle of elevation of the top of the pillar is 30. After walking for 10
minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of
the pillar is 60. Then the time taken (in minutes) by him, from B to reach the pillar, is
(1) 5
(2) 6
(3) 10
(4) 20
Ans (1)
Let CD be height of tower
In ADC
h
h
tan 30 
tan 60 
xy
y
1
h

3y  h
… (2)
3 xy
x  y  3h
D
… (1)
From eq(1) and (2)
x + y = 3y
x  2y
… (3)
x = distance AB
x = speed  time
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h
A
60
30
B
x
y
2
C
x = s  10 m
… (1)
y=st
… (4)
from eq (3) and (4)
x=2st
s  10 = 2  s  t
t = 5 min
5. Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up
four, E2 is the event that die B shows up two and E3 is the event that the sum of numbers on both dice is
odd, then which of the following statements is NOT true?
(1) E1, E2 and E3 are independent
(2) E1 and E2 are independent
(3) E2 and E3 are independent
(4) E1 and E3 are independent
Ans (1)
P(E1).P(E2) P(E3)  P(E1  E2  E3)
6. If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true?
(1) 3a2  23a + 44 = 0
(2) 3a2  26a + 55 = 0
(3) 3a2  32a + 84 = 0
(4) 3a2  34a + 91 = 0
Ans (3)
Standard Deviation of 2, 3, a and 11 is 3.5
xi = 2, 3, a, 11  xi = 16 + a
x i2  4,9,a 2 ,121  x i2  134  a 2

3.5 
x i2  x i 


n  n 
2
134  a 2  16  a 


4
 h 
2
134  a 2 (16  a) 2

4
16
2
536  4a  256  a 2  32a
12.25 
16
2
196 = 280 + 3a  32a
3a2  32a + 84 = 0
(3.5) 2 
7. For x R, f(x) = | log 2  sin x| and g(x) = f(f(x)), then
(1) g is differentiable at x = 0 and g (0) =  sin (log 2)
(2) g is not differentiable at x = 0
(3) g (0) = cos (log 2)
(4) g (0) =  cos (log 2)
Ans (3)
g(x) = f[| log 2  sin x| ]
= | log 2  sin (log 2  sin x) |
g (x) =  cos (log 2 – sin x) [ cos x]
= cos x cos (log 2  sin x)
g(0) = cos (log 2)
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8. The distance of the point (1, 5, 9) from the plane x  y + z = 5 measured along the line x = y = z is
20
10
(1)
(2) 3 10
(3) 10 3
(4)
3
3
Ans (3)
x 0 y0 z0


1
1
1
x 1 y  5 z  9
PQ :



1
1
1
Q = ( + 1,   5,  + 9)
+1+5++9=5
 =  10
Q(9, 15, 1)
Let P(1, 5, 9)
PQ  102  102  102  300  3 102  10 3
9. The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its
conjugate axis is equal to half of the distance between its foci, is
4
4
2
(1) 3
(2)
(3)
(4)
3
3
3
Ans (4)
2b 2
 8  b 2  4a
a
1
2b   2ae
2
4b 2  a 2e2
b 2 e2

a2 4
Also e  1 
 e2  1 
b2
a2
e2
2
e
4
3
10. Let P be the point on the parabola, y2 = 8x which is at a minimum distance from the centre C of the
circle, x2 + (y + 6)2 = 1. Then the equation of the circle, passing through C and having its centre at P is
(1) x2 + y2  4x + 9y + 18 = 0
(2) x2 + y2  4x + 8y + 12 = 0
x
(3) x2 + y2  x + 4y  12 = 0
(4) x 2  y 2   2y  24  0
4
Ans (2)
Let S  PC  4t 4  (4t  6) 2
S2 = 4t4 +16t2 + 48t + 36
d 2
(S )  16t 3  32t  48
dt
d 2
(S )  0  16t 3  32t  48  0
dt
 t3 + 2t + 3 = 0  t = 1
 P(2, 4), C = (0, 6)
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P(2t2, 4t)
C(0 , 6)
4
PC  4  4  8
 Equation of the circle is (x  2)2 + (y + 4)2 = 8
x2 + y2  4x + 8y + 20 = 8
x2 + y2  4x + 8y + 12 = 0
5a b 
T
11. If A  
 and A adj A = A A , then 5a + b is equal to
3
2


(1) 13
(2) 1
(3) 5
Ans (3)
2 b
AdjA  

 3 5a 
(4) 4
Given
 A. Adj A = A. AT
5a b   2 b  5a b   5a 3 
 3 2   3 5a    3 2   b 2 


 


2
2
0   25a  b 15a  2b 
10a  3b


 0
10a  3b   15a  2b
13 

Comparing elements 5a + b = 5
 1  sin x
12. Consider f (x)  tan 1 
 1  sin x
point
 
(1)  ,0 
4 


 
also passes through the
 , x   0,  . A normal to y = f(x) at x 
 2
6

(2) (0, 0)
 2 
(3)  0, 
 3 
Ans (3)
f (x)  tan 1
1  sin x
1  sin x
 x
 x
 x  x
 tan 1 cot 2     tan 1 cot     tan 1 tan     
4 2
4 2
4 2 4 2
1
f '(x)  = m1
2
m2 = slope of normal = 2

3
1  sin

6  tan 1 2  tan 1 3  
x1  , y1  tan 1

1
6
3
1  sin
6
2
 
(x1, y1) is  , 
6 3



Equation of normal is y   2  x  
3
6

 2 
 0,  satisfies the equation
 3 
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 
(4)  ,0 
6 
13. Two sides of a rhombus are along the lines, x  y + 1 = 0 and 7x  y  5 = 0. If its diagonals intersect at
(1, 2), then which one of the following is a vertex of this rhombus?
 10 7 
1 8
(1)   ,  
(2) (3, 9)
(3) (3, 8)
(4)  ,  
 3 3
3 3
Ans (4)
Let AD = x  y + 1 = 0
CD = 7x  y  5 = 0
Solving these equations
x = 1, y = 2
 D(1, 2)
OB is mid point of BD
 1  3 2  y 
(1, 2)  
,

 2
2 
D
x y+1=0
A
 x = 3, y = 6
 B = (3, 6)
Let equation of BC be x  y + k = 0
But this is passing through B
 3 + 6 + k = 0
k = 3
 Equation of BC is x  y  3 = 0
Solve BC, CD
7x  y  5 = 0
xy3=0
() (+) (+)
6x = 2
1
x
3
8
y
3
 1 8 
 one of the vertex is  , 
3 3 
7x – y  5 = 0
(1, 2) O
C
B
14. If a curve y = f(x) passes through the point (1, 1) and satisfies the differential equation,
 1
y(1 + xy) dx = x dy, then f    is equal to
 2
4
2
4
2
(1)
(2) 
(3) 
(4)
5
5
5
5
Ans (1)
y(1  xy)dx  xdy
dy y

  y2
dx x
dy y
  y2
dx x
1 dy 1 1

1
y 2 dx x y
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1
1 dy dt
 t,  2

y
y dx dx
dt 1
  t 1
dx x
dt 1
 t  1
dx x
1
dx
IF is e  x  eln x  x
 tx   (1)xdx
x2
C
2
x
x2
  C
y
2
1
1    C
2
1
 C
2
xt  
 Solution is
 y
x x2 1

 or 2x = y (x2 + 1)
y 2 2
2x
1 x2
1
2  
1
 
24
f 
 2  1 1
5
4
Alternatively
ydx  xdy + xy2 dx = 0
ydx  xdy
 xdx  0
y2
d x
 xdx  0
dx  y 
x x2

C
y 2
Passing through (1, 1)
1
1
1   C  C  
2
2
2
x x
1
x
1 x2
 
   
y 2
2
y
2 2
x
y
 f (x)
1 x2
 
2 2
1
1


1


2  24
f   
 2 1 1 5 5
2 8
8
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15. If all the words (with or without meaning) having five letters, formed using the letters of the word
SMALL and arranged as in a dictionary; then the position of the word SMALL is
(1) 58th
(2) 46th
(3) 59th
(4) 52nd
Ans (1)
ALLMS
4!
A
 12
2!
L
4!  24
M
S A
S L
4!
 12
2!
3!
3
2!
3!  6
S M A L L 1 1
= 58
16. If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P, then the common ratio of this G.P is
7
8
4
(1)
(2)
(3)
(4) 1
4
5
3
Ans (3)
T2 = a + d
T5 = a + 4d
T9 = a + 8d
(a + 4d)2 = (a + d) (a + 8d)  T2 ,T5 and T9 are in GP
 a2 + 16d2 + 8ad = a2 + 9ad + 8d2
 8d2  ad = 0
 d = 0 or a = 8d
 T2 = 9d, T5 = 12d, T9 = 16d
12d 16d 4
 r


9d 12d 3
n
 2 4 
17. If the number of terms in the expansion of 1   2  , x  0 is 28, then the sum of the coefficients of
 x x 
all the terms in this expansion is
(1) 729
(2) 64
Ans
Number of terms = 2n + 1
Given number of terms = 28
 2n + 1 = 28 which is not possible
(3) 2187
2
2
(4) 243
2
2
16
 3  2  1
 4
18. If the sum of the first ten terms of the series 1    2    3   42   4   ..., is
m , then m
5
 5  5  5
 5
is equal to
(1) 99
Ans (3)
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(2) 102
(3) 101
8
(4) 100
2
2
2
 8   12   16 
 25 
2
      4  
 4 
5  5   5 
(4  2) 2 (4  3) 2 (4  4) 2


 .....
52
52
52
42  2 2
2  3  42  .....112 
2
5
42  2
1  22  32  ...  112  12 
52
42 11(11  1)(22  1)
1
52
6
42
11(23)  2  1
52
16
16
(505)  m
25
5
 m = 101

19. If the line,
2

x 3 y 2 z 4


lies in the plane, lx+ my  z = 9, then l2 + m2 is equal to
2
1
3
(2) 26
(3) 18
(4) 5
(1) 2
Ans (1)
x 3 y 2 z 4



2
1
3
x = 2 + 3, y =   2, z = 3 4
This fixed point satisfies the plane lx + my  z = 9
l (2  3)  m(  2)  (3  4)  9
2l  3l  m  2m  3  4  9  0
(3l  2m  5)  (2l  m  3)  0
The line passes through the fixed point
Hence 3l  2m  5 = 0
2l  m – 3 = 0
Solving eqns (1) and (2)
l = 1, m = 1
 l2 + m2 = 2
… (1)
… (2)
20. The Boolean Expression (p  ~ q)  q  (~ p  q) is equivalent to
(1) p  ~ q
Ans (4)
(p  ~ q)  q  (~ p  q)
(2) ~ p  q
(3) p  q
 (p  ~q)  (~p  q)  q
 [p  (~p  q)]  [~q  (~p  q)]  q
 (p  ~p)  (p  q)  [(~q  ~p)  (~q  q)]
pq
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(4) p  q
21. The integral
2x12  5x 9
 (x 5  x 3  1)3 dx is equal to
(1)
 x10
C
2(x 5  x 3  1)2
(2)
x 5
C
(x 5  x 3  1) 2
(3)
x10
C
2(x 5  x 3  1)2
(4)
x5
C
2(x 5  x 3  1)2
Where C is an arbitrary constant
Ans (3)
Dividing numerator and denominator by x15
2x 3  5x 6
 (1  x 2  x 5 )3 dx
Put 1 + x2 + x5 = t
(2x3 + (5) x6) dx = dt
dt t 2
1
1
x10



C


C
or
C
 t 3 2
2 (1  x 2  x 5 )
2(x 5  x 3  1)2
22. If one of the diameters of the circle, given by the equation, x2 + y2  4x + 6y  12 = 0 is a chord of a
circle S, whose centre is at (3, 2), then the radius of S is
(1) 10
(2) 5 2
(3) 5 3
(4) 5
Ans (3)
x2 + y2  4x + 6y  12 = 0
r  4  9  12  5
S
A (3, 2)
C
Centre (2, 3) AB  25  25  5 2
2
2
B(2, 3)
2
AC = AB + BC = 25 + 50
AC  5 3
1/n
 (n  1)(n  2)...3n 
23. lim 

n  
n 2n

(1) 3 log3  2
is equal to
(2)
18
e4
(3)
27
e2
(4)
Ans (3)
1
 (n  1)(n  2)...(n  2n)  n
lim 

n 
n  n.... n


1
  1  2   2n   n
lim  1  1   ... 1 

n   
n  n  
n 
ln y 

1  1
 2
 2n  
ln 1    ln 1    ...  ln 1 


n  n
 n

n  
1 2n 
r
ln  1  

n r 1  n 
2
2
  ln (1  x)dx   (1  x)ln (1  x)  (1  x) 0  (3ln 3  3)  (1)  3ln 3  2
0
y  23ln 32 
JEE-MAIN 2016
e3ln 3 27
 2
e2
e
10
9
e2
24. The centres of those circles which touch the circle x2 + y2  8x  8y  4 = 0, externally and also touch
the x-axis, lie on
(1) a parabola
(2) a circle
(3) an ellipse which is not a circle
(4) a hyperbola
Ans (1)
y
Centre of the given circle is (4, 4)
2
2
2
(x  4) + (y  4) = (6 + y)
x2  8x + 16 + y2 + 16  8y = 36 + y2 + 12y
(4, 4)
(x, y)
x2  8x  8y  12y  4 = 0
6
2
y
x  8x  20y  4 = 0
x
O
r1  16  16  4  6

 

  

3  
25. Let a, b and c be three unit vectors such that a   b  c  
b  c . If b is not parallel to c , then the
2


angle between a and b is
5
3

2
(1)
(2)
(3)
(4)
6
4
2
3
Ans (1)
     
3  
(a  c)b  (a  b)c 
(b  c)
2
 
 
3
3
a c 
ab  
2
2


3
cos 1 
 1 = 30 (1 is angle between a and c )
2


3
cos 2  
(2 angle between a and b )
2
5
2 = 180  30 = 150 =
6
26. Let p  lim (1  tan 2 x )1/2x then log p is equal to
x 0
1
4
Ans (4)
(1)
(2) 2
(3) 1
1
p  lim (1  tan 2 x )1/2x  lim (sec 2 x ) 2x
x 0
x 0
p  lim  sec 2 x 
1
2x
x 0
1
ln sec 2 x
x 0 2x
1
1
 2sec x  sec x  tan x 
2
2 x (L – Hospital rule)
 lim sec x
x 0
2
1
ln p 
2
ln p  lim
JEE-MAIN 2016
11
(4)
1
2
27. If 0  x < 2, then the number of real values of x, which satisfy the equation
cos x + cos 2x + cos 3x + cos 4x = 0, is
(1) 9
(2) 3
(3) 5
(4) 7
Ans (4)
cos x  cos 2x  cos3x  cos 4x  0
2 cos 2x cos x  2cos3x cos x  0
2cos x(cos 2x  cos3x)  0
 5x 
x
cos x = 0 or 2cos   cos    0
 2 
2
 5x 
x
 cos x = 0 or cos    0 or cos    0
 2 
2

5x

x

 x  (2n  1) or
 (2n  1) or  (2n  1)
2
2
2
2
2


 x  (2n  1) or x  (2n  1) or x  (2n  1)
2
5
 3  3 7  9
 x  , , , , , ,
2 2 5 5
5 5
28. The sum of all real values of x satisfying the equation (x 2  5x  5) x
2
 4x  60
(3) 4
(1) 5
(2) 3
Ans (3)
x2  5x + 5 > 0 x2  5x + 5  1

5 5 
x   ,
   4,  

2 
 1 is
(4) 6
x2 + 4x  60 = 0
x2  10x + 6x  60 = 0
x = 6, x = 10
Sum of real values of x = 10 + 6 = 4
29. The area (in sq. units) of the region {(x, y) : y2  2x and x2+ y2  4x, x  0, y  0} is
 2 2
4
8
4 2
(1) 
(2)  
(3)  
(4)  
2
3
3
3
3
Ans (4)
y2 = 2x
… (1)
x2 + y2 = 4x
… (2)
Centre (2, 0) and radius = 2
y
Solving eq(1) and (2) x = 0 and x = 2
2
Required area = A   4x  x 2  2 x dx
0
2
A   22  (x  2) 2  2x1/2dx
(0, 0)
0
x2 2
22
x 3/2
 x2
A
2  (x  2) 2  sin 1 
  2 3
2
2
 2 
2
JEE-MAIN 2016
2
0
12
(2, 0)
x
     4

A  [0]   2       2  0 

2

3




 
4 2
3
1
30. If f (x)  2f    3x, x  0 and S = {x  R : f(x) = f(x)}; then S:
x
(1) contains more than two elements
(2) is an empty set
(3) contains exactly one element
(4) contains exactly two elements
Ans (4)
1
Given, f (x)  2f    3x
x
6
1
Consider 2f    4f (x) 
x
x
… (1)
… (2)
Solving eqn(1) and (2)
2
f (x)  x 
x
 f(x) = f(x)
2
2
 x   x 
x
x
4
2x 
x
2
2x  4
x2 = 4
x 2
JEE-MAIN 2016
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Physics
31. A combination of capacitors is set up as shown in the figure. The magnitude of
the electric field, due to a point charge Q (having a charge equal to the sum of
the charges on the 4 F and 9 F capacitors), at a point distance 30 m from it,
would equal:
(1) 480 N/C
(2) 240 N/C
(3) 360 N/C
(4) 420 N/C
Ans (4)
q = CV
q1
3 F
q2 = 2 × 8 × 10–6
V2
4 F
q2 = 16 C
V1
9 F
V1 + V2 = 8 V
q2
q1 q1
 8
2 F
4 12
4q1
8
12
8V
q1 = 24 C.
q
24
V1  1 
 6V , V2 = 2V
4
4
V2 = 2V
The charge on 9F capacitor is q1  9F  2  18 C
Given: Q  q1  q1
= 24 + 18
= 42 C
Electric field at a distance d from a point charge Q is
1 Q
E
40 d 2
9  109  42  106
30  30
42  103

102
E = 420 NC–1
32. An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the
observer the tree appears:
(1) 20 times nearer.
(2) 10 times taller.
(3) 10 times nearer.
(4) 20 times taller.

Ans (1)
Magnification due to a telescope

m

i.e., the visual angle of the final image is 20 times more than that due to the object with respect to a bare
eye. So the tree appears 20 times nearer.
JEE-MAIN 2016
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33. Hysteresis loops for two magnetic materials A and B are given below.
These materials are used to make magnets for electric generators, transformer core and electromagnet
core. Then it is proper to use:
(1) B for electromagnets and transformers.
(2) A for electric generators and transformers.
(3) A for electromagnets and B for electric generators.
(4) A for transformers and B for electric generators.
Ans (1)
Loop (B) has a smaller hysteresis area. i.e., hysteresis loss is less. Such materials are used as cores for
electromagnets and transformers.
34. Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the
samples have equal number of nuclei. After 80 minutes, the ratio of decayed numbers of A and B nuclei
will be:
(1) 5 : 4
(2) 1 : 16
(3) 4 : 1
(4) 1 : 4
Ans (1)
For radioactive element A,
Half life TA = 20 minutes
Number of half lives in 80 minutes = 4.
Number of nuclei remaining undecayed after 4 – half lives is
15N 0
N
N
1

N A  40  0 . The number decayed is x   1   N 0 
2
16
16
 16 
For radioactive element B,
Half life, TB = 40 minutes
Number of half lives in 80 minutes = 2.
Number of nuclei remaining undecayed after 2 half lives is
N
N
N B  20  0 . The number decayed is
2
4
1
3


y  1   N 0  N 0
4
 4
x 15N 0
4
5
Now, 


y
16
3N 0 4
35. A particle of mass m is moving along the side of square of side ‘a’, with a uniform speed v in the x-y
plane as shown in the figure:

Which of the following statements is false for the angular momentum L about the origin?
 mv
(1) L 
Rkˆ when the particle is moving from D to A.
2

mv ˆ
(2) L  
Rk when the particle is moving from A to B.
2

R

(3) L  mv 
 a  kˆ when the particle is moving from C to D.
 2


R

(4) L  mv 
 a  kˆ when the particle is moving from B to C.
 2

Ans (1), (3)
JEE-MAIN 2016
15
For (3) component of position vector perpendicular to velocity is
R
y
a
2

 R

Hence L  mv 
 a  kˆ
 2

For (1) component of position vector perpendicular to velocity x 
R
2
 mvR
Hence L 
kˆ
2
 
36. Choose the correct statement:
(1) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in
proportion to the frequency of the audio signal.
(2) In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in
proportion to the amplitude of the audio signal.
(3) In amplitude modulation the frequency of the high frequency carrier wave is made to vary in
proportion to the amplitude of the audio signal.
(4) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in
proportion to the amplitude of the audio signal.
Ans (2)
Conceptual
37. In an experiment for determination a refractive index of glass of a prism by i – , plot, it was found that a
ray incident at angle 35, suffers a deviation of 40 and that it emerges at angle 79. In that case which of
the following is closest to the maximum possible value of the refractive index?
(1) 1.8
(2) 1.5
(3) 1.6
(4) 1.7
Ans (2)
i1 = 35
d = 40
i2 = 79
d = i1 + i2 – A  A = 74
Limiting angle of prism is A = 2C  C = 37
In the given situation, i1 = 35
A
 A < 2C or C 
i.e.,
C  37
2
3
sin C  sin 37  sin C 
5
5
1

3 sin C
5
n
3
5
or n  or n  1.6
3
Maximum value of RI of material of prism is less than 1.6.
JEE-MAIN 2016
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38. ‘n’ moles of an ideal gas undergoes of process A  B as shown in the figure.
The maximum temperature of the gas during the process will be:
9P V
9 P0 V0
(1) 0 0
(2)
nR
4nR
3 P0 V0
9 P0 V0
(3)
(4)
2nR
2nR
Ans (2)
PV = constant at end points A and B
 Temperature is maximum at the mid-point of the curve AB.
At the mid-point,
3P
3V
P 0, V 0
2
2
PV 9P0 V0
T 

nR
4nR
39. Two identical wires A and B, each of length ‘l’, carry the same current I. Wire A is bent into a circle of
radius R and wire B is bent to form a square of side ‘a’. If BA and BB are the values of magnetic field at
B
the centres of the circle and square respectively, then the ratio A is:
BB
(1)
2
8 2
Ans (1)
For circular loop:
l = 2R
l
R
2
I
BA  0
2R
0 I

 l 
2 
 2 
 I
 0
l
For square loop:
l = 4a
 I
BB  0 8 2
4 a
 I
 0 8 2
4 l
4
 0 4I
BB 
8 2
4 l
B
2
 A 
BB 8 2
JEE-MAIN 2016
(2)
2
8
(3)
A
B
R
a
a
17
2
16 2
(4)
2
16
40. A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to a circular scale
with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the
measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th
division coincides with the main scale line and that the zero of the main scale is barely visible. What is
the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the
main scale line?
(1) 0.50 mm
(2) 0.75 mm
(3) 0.80 mm
(4) 0.70 mm
Ans (3)
pitch
Least count of screw gauge =
No. of divisions on head scale
0.5 mm
1


 0.01 mm
50
100
Zero error = –5
Zero correction = + 5 × 0.01 mm = 0.05 mm
Given reading: PSR: 0.5 mm
HSR: 25
 Obtained Reading = [PSR + (HSR × LC)] mm
= 0.5 + (25 × 0.1) mm
= 0.75 mm
Applying zero correction;
Corrected reading = 0.75 + 0.05 = 0.8 mm
41. For a common emitter configuration, if  and  have their usual meanings, the incorrect relationship
between  and  is:
2
1 1


(1)  
(2)   1
(3)  
(4)  
2
1 
 
1 
1 
Ans (1), (3)
IE = IB + IC
IE I B

1
IC IC
1 1
  1  option (2) is a correct relation
 
1 1 



 option (4) is a correct relation


1   


2

.
1       2 
2
which is option (1), is an incorrect relation.
1  2


  is negative, since  > 1
1   
Thus,
 option (3) is an incorrect relation.
JEE-MAIN 2016
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42. The box of a pin hole camera, of length L, has a hole of radius a. It is assumed that when the hole is
illuminated by a parallel beam of light of wavelength  the spread of the spot (obtained on the opposite
wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would
then have its minimum size (say bmin) when:
 2 2 
2
2
(1) a  and b min  
(2)
a

and b min  4L

L
L
 L 
 2 2 
(3) a  L and b min  

 L 
(4) a  L and b min  4L
Ans (4)
wall
C

A
A
a
2a O
bmin
O
a
B

B
D
L
Considering the geometrical spreading of light alone, OA= OA = a
2
Fresnel’s distance, L 
…(1)
2a
Considering the Fresnel’s class of diffraction,
Angle of diffraction is given by the relation, a = 
…(2)
(assuming first diffraction minimum at C)
a
From figure,  
…(3)
L
a
Eq. (2): a     a  L
L
From figure, b min  2a  4L
43. A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1m 1000 times.
Assume that the potential energy lost each time he lowers to mass is dissipated. How much fat will he
use up considering the work done only when the weight is lifted up? Fat supplies 3.8 × 107 J of energy
per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms–2:
(1) 12.89 × 10–3 kg
(2) 2.45 × 10–3 kg
(3) 6.45 × 10–3 kg
(4) 9.89 × 10–3 kg
Ans (1)
Given: conversion efficiency = 20%
Mechanical energy
… (1)
 20%
Energy from
Mechanical
Energy from fat
fat
M.E. = 1000 × mgh
 1000  10  9.8  1m
= 9.8 × 104
JEE-MAIN 2016
19
 Energy from fat = mass of fat burnt × 3.8 × 107
= m × 3.8 × 107
9.8  104
 m × 3.8 × 107 
 12.89  103 kg
0.2  3.8  107
44. Arrange the following electromagnetic radiations per quantum in the order of increasing energy:
A: Blue light
B. Yellow light
C: X-ray
D. Radiowave.
(1) B, A, D, C
(2) D, B, A, C
(3) A, B, D, C
(4) C, A, B, D
Ans (2)
hc
Energy of a radiation quantum. E 

1
E

 Radio  waves   Yellow   Blue   x  ray
Obviously,
E Radio  waves  E Yellow  E Blue  E x  ray
45. An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains
constant. If during this process the relation of pressure P and volume V is given by PVn = constant, then
n is given by (Here CP and CV are molar specific heat at constant pressure and constant volume,
respectively):
C  CV
C
C  CP
C C
(1) n 
(2) n  P
(3) n 
(4) n  P
C  CP
CV
C  CV
C  CV
Ans (3)
C = constant
R
C  CV 
1 n
R
C  CV 
1 n
R
1 n 
C  CV
R
1
n
C  CV
C  CV  R
n
C  CV
C  CP
n
C  CV
46. A satellite is revolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth R;
h << R). the minimum increase in its orbital velocity required, so that the satellite could escape from the
earth’s gravitational field, is close to: (Neglect the effect of atmosphere.)
(1)
gR


2 1
(2)
2gR
(3)
Ans (1)
v0  gR
ve  2gR
v 

JEE-MAIN 2016

2 1
gR
20
gR
(4)
gR / 2
47. A galvanometer having a coil resistance of 100  gives a full scale deflection, when a current of 1 mA is
passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving
a full scale deflection for a current of 10 A, is:
(1) 3 
(2) 0.01 
(3) 2 
(4) 0.1 
Ans (2)
100 
10  103  S  103 102
S
101
 102
0.999
G
1 mA
S
3
(10  10 mA)
100 
1 mA
G
48. Radiation of wavelength , is incident on a photocell. The fastest emitted electron has speed v. If the
3
wavelength is changed to
, the speed of the fastest emitted electron will be:
4
1
1
1
 3 2
 4 2
(1)  v  
(2)  v  
4
3
Ans (2)
1
hc
mv12 

2

1
9 c
mv 22 

2
3
1
41

mv 22   mv12     
2
32

4
4 2
2
v 22  v12    
3
3 m
m
4
2
1
 
v 22  v12    
3
m 3
4
2
 v12  
3
m
4
Since   0, v 22  v12
3
1
 4 2
(3)  v  
3
 4 2
(4)  v  
3
1
 4 2
v 2  v1  
3
49. If a, b, c, d are inputs to a gate and x is its output, then, as per the following time graph, the gate is:
(1) NAND
(2) NOT
(3) AND
(4) OR
Ans (4)
When any one of the inputs is high, the output (x) is high. Hence the gate is OR-gate.
JEE-MAIN 2016
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50. The region between two concentric spheres of radii ‘a’ and ‘b’, respectively (see
A
figure), has volume charge density   , where A is a constant and r is the
r
distance from the centre. At the centre of the spheres is a point charge Q. The value
of A such that the electric field in the region between the spheres will be constant,
is:
2Q
Q
Q
Q
(1)
(2)
(3)
(4)
2
2
2
2
2
a
2a
2  b  a 
  a  b2 
Ans (2)
Consider a Gaussian closed sphere of radius r. The net flux passing through if the electric field is E is
r

1
E  4r 2    (4x 2 )dx  Q 
0  a

r

1
E  4r 2   4A  xdx  Q 
a
r
a

 0
b
 4A 2 2
1
E  4r 2  
(r  a )  Q 
 2
 0
 A  a2 
Q 1
E   1  2  
2
 2  r  4r  0
Since the field has to be constant (i.e., independent of r) hence
Q
Aa 2
Q
 A

2
2
2a 2
2r
4r
51. A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is
90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean
time should be:
(1) 92  3s
(2) 92  2 s
(3) 92  5.0 s
(4) 92  1.8 s
Ans (2)
Given data
For 100 oscillations time period is record in four crials.
Time periods
T1 = 90 s
T2 = 91 s
T3 = 95 s
T4 = 92 s
Standard deviation ‘’ of measurements is given by
  (RMS value of Times periods) 2  Square of mean of Time periods 

RMS value of time period

T12  T22  T32  T42
4
 90 
2
2
2
  91   95    92 
4
= 92.019 s
Mean value of T’s = 92 s

 92.019 
2
2
  92   3.49  1.87  2
Reported mean is 92  2 s
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22
2
2
52. The temperature dependence of resistances of Cu and undoped Si in the temperature range 300–400 K, is
best described by:
(1) Linear decrease for Cu, linear decrease for Si.
(2) Linear increase for Cu, linear increase for Si
(3) Linear increase for Cu, exponential increase for Si
(4) Linear increase for Cu, exponential decreases for Si
Ans (4)
For a metallic conductor, R = R0 (1 +  (T – T0))
b
For an intrinsic semiconductor, R  ae T
53. Identify the semiconductor devices whose characteristics are given below, in the order (1), (2), (3), (4):
(1) Zener diode, Solar cell, Simple diode, Light dependent resistance
(2) Simple diode, Zener diode, Solar cell, Light dependent resistance
(3) Zener diode, Simple diode, Light dependent resistance, Solar cell
(4) Solar cell, Light dependent resistance, Zener diode, Simple diode
Ans (3)
Conceptual
54. A roller is made by joining together two cones at their vertices O. It is kept on two
rails AB and CD which are placed asymmetrically (see figure), with its axis
perpendicular to CD and its centre O at the centre of line joining AB and CD (see
figure). It is given a light push so that it starts rolling with its centre O moving
parallel to CD in the direction shown. As it moves, the roller will tend to:
(1) turn left and right alternately.
(2) turn left
(3) turn right
(4) go straight
Ans (4)
As such the net torque and force acting on the system is 0 hence no change in trajectory.
55. A pendulum clock loses 12 s a day if the temperature is 40C and gains 4s a day if the temperature is
20C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion
() of the metal of the pendulum shaft are respectively:
(1) 55C;  = 1.85 × 10–2/C
(2) 25C;  = 1.85 × 10–5/C
(3) 60C;  = 1.85 × 10–4/C
(4) 30C;  = 1.85 × 10–3/C
Ans (2)
1
t      86400 
2
1
12    40  t C  86400 
… (1)
2
1
4    20  t C  86400
… (2)
2
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23
40  t C
(1)
 3 
(2)
20  t C
60  3TC  40  t C
100  4TC  t C  25C
1
using (1), 12    40  25  86400 
2
24

15  86400
= 1.85 × 10–5/C
56. A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at
its lowest end. It starts moving up the string. The time taken to reach the support is: (take g = 10 ms–2)
(1) 2 s
(2) 2 2 s
(3) 2s
(4) 2 2 s
Ans (4)
Velocity of the pulse when it has travelled x from bottom
v  xg

dx
 xg
dt
So,
dx
 dt g
x
L

dx
0
x
t
  dt g
0
L
2  x   t g
0
2 Lt g
t2
L
g
20
10
2 2 s
2
 L  20m 
57. A point particle of mass m, moves along the uniformly rough track PQR as
shown in the figure. The coefficient of friction, between the particle and the
rough track equals . The particle is released, from rest, from the point P and
it comes to rest at a point r. the energies, lost by the ball, over the parts, PQ
and QR, of the track, are equal to each other, and no energy is lost when
particle changes direction from PQ to QR.
The values of the coefficient of friction  and the distance x (= QR), are, respectively close to:
(1) 0.29 and 6.5 m
(2) 0.2 and 6.5 m
(3) 0.2 and 3.5 m
(4) 0.29 and 3.5 m
Ans (4)
For the path PQ
 2 
Work done by friction WF = ( mg cos 30) 

 sin 30 
WF = 2 3 mg
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24
1
According to question WF  mgh
2
1
2 3 mg  mg  2 
2
1

 0.29
2 3
For the path QR
Work done by friction Wf =  mg x
1
According to problem WF  mgh
2
1
mgx  mg  2 
2
1
x   2 3 = 3.5 m

58. A pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water so
that half of it is in water. The fundamental frequency of the air column is now:
f
3f
(1) f
(2)
(3)
(4) 2f
2
4
Ans (1)
Let length of pipe = L
v
Fundamental frequency when the pipe is open, f 
2L
L
v
v
v
When dipped in water, new length L 
f closed 


f
2
4L
 L  2L
4 
2
59. A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is
2A
at a distance
from equilibrium position. The new amplitude of the motion is:
3
7A
A
(1)
(2)
41
(3) 3A
(4) A 3
3
3
Ans (1)
Let initial amplitude be A and new amplitude be A and angular frequency be . Hence maximum initial
speed = A
2
Now from phasor cos  
3
A
Hence speed at the given instant

5
2A
v = A sin  
A
3
3
Tripled speed v  5 A
2
Hence applying energy conservation
1  2A  1
1
2
K
  m v 2  K A
2  3  2
2

1
4 1
m2 A 2    m2  5A 2   m2 A2
2
2
9 2
7
4

A 2    5  A 2 A  A
3
9

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60. A arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms),
50 Hz AC supply, the series inductor needed for it to work is close to:
(1) 0.065 H
(2) 80 H
(3) 0.08 H
(4) 0.044 H
Ans (1)
V
R  8 
L
8
I
Z  64  2 L2
220
64  2 L2 
10
2 2
64   L  484
L = 0.065 H
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bulb
~
220 V, 50 Hz
26
Chemistry
61. Which one of the following statements about water is FALSE?
(1) Ice formed by heavy water sinks in normal water
(2) Water is oxidized to oxygen during photosynthesis
(3) Water can act both as an acid and as a base
(4) There is extensive intramolecular hydrogen bonding in the condensed phase.
Ans (4)
In the condensed phase, there is extensive intermolecular hydrogen bonding and not intramolecular
hydrogen bonding.
62. The concentration of fluoride, lead, nitrate and iron in a water sample from an underground lake was
found to be 1000 ppb, 40ppb, 100 ppm and 0.2 ppm, respectively. This water is unsuitable for drinking
due to high concentration of
(1) Iron
(2) Fluoride
(3) lead
(4) Nitrate
Ans (4)
Nitrate  100 ppm  Higher concentration.
63. Galvanization is applying a coating of
(1) Zn
(2) Pb
(3) Cr
(4) Cu
Ans (1)
Zinc is used for galvanization
64. Which one of the following complexes shows optical isomerism?
(1) [Co(NH3)4Cl2]Cl
(2) [Co(NH3)3Cl3]
(3) cis[Co(en)2Cl2]Cl (4) trans[Co(en)2Cl2]Cl
(en = ethylenediamine)
Ans (3)
Cl
en
Cl
Cl
Cl
Co
en
Cl
Cl
Co
en
en
Contains didentate symmetrical ligand and has no element of symmetry.
65. Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure pi and temperature T1
are connected through a narrow tube of negligible volume as shown in the figure below. The temperature
of one of the bulbs is then raised to T2. The final pressure pf is
 TT 
(1) 2pi  1 2 
 T1  T2 
 TT 
(2) pi  1 2 
 T1  T2 
 T1 
(3) 2pi 

 T1  T2 
Ans (4)
At constant volume
2pi
At I set
T1
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27
 T2 
(4) 2pi 

 T1  T2 
In second set
pf pf

T2 T2
When we equalise
2pi pf pf
 
T1 T1 T2
2pi T2 pf  T1pf

T1
T1T2
p (T  T2 )T1
2pi  f 1
T1T2
2pi [T2 ]
pf 
[T1  T2 ]
66. The heats of combustion of carbon and carbon monoxide are 393.5 and 283.5 kJ mol1, respectively.
The heat of formation (in kJ) of carbon monoxide per mole is:
(1) 110.5
(2) 110.5
(3) 676.5
(4) 676.5
Ans (1)
C + O2  CO2
H = 393.5 kJ mol1
… (1)
1
CO  O2  CO2 H = 283.5 kJ mol1
… (2)
2
(1)  (2)  Hf (CO) = 393.5
+283.5
110.0 kJ mol1
67. At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon required 375 mL air containing 20 % O2 by
volume for complete combustion. After combustion the gases occupy 330 mL. assuming that the water
formed is in liquid form and the volumes were measured at the same temperature and pressure, the
formula of the hydrocarbon is
(1) C4H10
(2) C3H6
(3) C3H8
(4) C4H8
Ans (3)
Volume of air
= 375 mL
Volume of O2
= 75 mL
Volume of N2 remaining = 300 mL
Volume of O2 consumed = 75 mL
C3H8(g) + 5O2(g)  3CO2 (g) + 4H2O(l)
1 ml of C3H8 consumed 5 mL of O2
15 ml of C3H8 consumed 75 mL of O2
68. Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2
decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches
0.05 M, the rate of formation of O2 will be
(1) 1.34  102 mol min1
(2) 6.93  102 mol min1
(3) 6.93  102 mol min1
(4) 2.66 L min1 at STP
Ans (3)
25min
25min
0.5M 
 0.25M 
 0.125M
50 min
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28
0.693
 0.27 min 1
25
Rate = k [H2O2]
= 0.27  0.05
= 1.38  103
rate of formation of O2 = 1.38  103/2
= 0.69  103 = 6.9  104
69. The pair having the same magnetic moment is
[At. No. Cr = 24, Mn = 25, Fe = 26, Co = 27]
(1) [CoCl4]2 and [Fe(H2O)6]2+
(2) [Cr(H2O)6]2+ and [CoCl4]2
(3) [Cr(H2O)6]2+ and [Fe(H2O)6]2+
(4) [Mn(H2O)6]2+ and [Cr(H2O)6]2+
Ans (3)
1. Co2+ : [Ar]3d7
No of unpaired es = 3
Fe2+ : [Ar]3d6
No. of unpaired es = 4
2. Cr2+ : [Ar]3d4
No. of unpaired es = 4
Co2+ : [Ar]3d7
No. of unpaired es = 3
3. Cr2+ : [Ar]3d4
No of unpaired es = 4
Fe2+ : [Ar]3d6
No of unpaired es = 4
4. Mn2+ : [Ar]3d5
No of unpaired es = 5
Cr2+ : [Ar]3d4
No of unpaired es = 4
If number of unpaired electrons are same then magnetic moment is also same
70. The species in which the N atom is in a state of sp hybridization is
(1) NO2
(2) NO2
(3) NO2
(4) NO3
k
Ans (2)

O  N  O sp hybridised
71. Thiol group is present in
(1) Methionine
Ans (4)
H2N
CH
COOH
CH2
SH
(2) Cytosine
(3) Cystine
COOH
or
HS
NH2
Thiol Group
72. The pair in which phosphorous atoms have a formal oxidation state of +3 is
(1) Pyrophosphorous and pyrophosphoric acids
(2) Orthophosphorous and pyrophosphorous acids
(3) Pyrophosphrous and hypophosporic acids
(4) Orthophosphorous and hypophosphoric acids
Ans (2)
3
3
H 3 P O3
H 4 P2 O5
O
O
H
P
OH
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OH
H
P
OH
O
O
P
H
OH
29
(4) Cysteine
73. The distillation technique most suited for separating glycerol from spent-lye in the soap industry is
(1) Distillation under reduced pressure
(2) Simple distillation
(3) Fractional distillation
(4) Steam distillation
Ans (1)
Glycerol has a high boiling point which decomposes at its boiling point. Hence, it is separated by
distillation under reduced pressure.
74. Which one of the following ores is best concentrated by froth floatation method?
(1) Malachite
(2) Magnetite
(3) Siderite
(4) Galena
Ans (4)
Galena, sulphide ore of Pb is concentrated by froth floatation process.
75. Which of the following atoms has the highest first ionization energy?
(1) Sc
(2) Rb
(3) Na
(4) K
Ans (1)
Rb, Na, K belong to the 1st group and have a low iH.
1 2
21Sc: [Ar]3d 4s has a high iH since electron has to be removed from completely filled s-orbital.
76. In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole
of amine produced are
(1) Four moles of NaOH and one mole of Br2
(2) One mole of NaOH and one mole of Br2
(3) Four moles of NaOH and two moles of Br2
(4) Two moles of NaOH and two moles of Br2
Ans (1)
For one mole of amine formed, four moles of NaOH and one mole of Br2 are used
O
R
C
NH 2 + 4NaOH + Br2
RNH2 + 2NaBr + Na2CO3 + 2H2O
77. Which of the following compounds is metallic and ferromagnetic?
(1) MnO2
(2) TiO2
(3) CrO2
(4) VO2
Ans (3)
CrO2 is ferromagnetic and metallic.
78. Which of the following statements about low density polythene is FALSE?
(1) It is used in the manufacture of buckets, dust-bins etc
(2) Its synthesis requires high pressure
(3) It is a poor conductor of electricity
(4) Its synthesis requires dioxygen or a peroxide initiator as a catalyst
Ans (1)
LDPE (Low density polythene) is generally used in the manufacture of polythene bags
79. 2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields.
CH3
(a) C2H5CH2C
CH3
(1) (a) and (b)
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OCH3
(b) C2H5CH2C
CH2
(c) C2H5CH
CH3
(2) all of these
C
CH3
CH3
(3) (a) and (c)
30
(4) (c) only
Ans (2)
Cl
O
SN
+ NaOCH3
+ NaCl
NaOCH3
NaCl
+
major
80. A stream of electrons from a heated filament was passed between two charged plates kept at a potential
h
difference V esu. If e and m are charge and mass of an electron, respectively, then the value of

(where  is wavelength associated with electron wave) is given by
(1) 2 meV
(2) meV
(3) 2meV
(4) meV
Ans (1)

h
h
 2 MeV
2meV 
81. 18 g glucose (C6H12O6) is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous
solution is
(1) 759.0
(2) 7.6
(3) 76.0
(4) 752.4
Ans (4)
According to Roault’s law
po  p
n2

o
p
n1  n 2
760  p
18 / 180
0.1
0.1



178.2
18
760
9.9  0.1 10

18
180
76
760  p =
 7.6
10
p = 760  7.6 = 752.4 Torr
or
o
p  p A
10
 760 
 752.4
10.1
82. The product of the reaction given below is
1. NBS/h
X
2. H2O/K2CO3
(1)
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O
OH
CO2H
(2)
(3)
31
(4)
Ans (3)
NBS/h
Br
K2CO3/H2O
HO
83. The hottest region of Bunsen flame shown in the figure below is
(1) region 4
(2) region 1
(3) region 2
(4) region 3
Ans (3)
The hottest region of Bunsen flame is at the tip of the inner cone.
84. The reaction of zinc with dilute and concentrated nitric acid, respectively, produces
(1) NO2 and N2O
(2) N2O and NO2
(3) NO2 and NO
(4) NO and N2O
Ans (2)
In dilute nitric acid
2HNO3  H 2O  N 2O  4[O]
Zn  2HNO 3  [O]  Zn(NO 3 ) 2  H 2O}  4
4Zn  10 HNO3  4Zn(NO 3 ) 2  5H 2O  N 2O
In concentrated nitric acid
2HNO3  H 2O  2NO 2  [O]
Zn  2HNO 3  [O]  Zn(NO3 ) 2  H 2O
Zn  4HNO 3  Zn(NO3 ) 2  2H 2O  2NO 2
85. Which of the following is an anionic detergent?
(1) Glyceryl oleate
(2) Sodium stearate
(3) Sodium lauryl sulphate
(4) Cetyltrimethyl ammonium bromide
Ans (3)
Sodium lauryl sulphate is anionic detergent
Glyceryl oleate is fat
Sodium stearate is a anionic soap
Cetyltrimethyl ammonium bromide is cationic detergents which is powerful germicide.
86. The reaction of propene with HOCl (Cl2 + H2O) proceeds through the intermediate
(1) CH 3  CHCl  CH 2
(2) CH3CH+ CH2OH
(3) CH3CH+CH2Cl
(4) CH 3  CH(OH)  CH 2
Ans (3)
The addition of HOCl to propene follows electrophilic addition path in which carbocation is formed.

..
HO
..

HO + Cl
Cl
+
+
+ Cl+
Cl
(more stable)
Cl
+
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87. For a linear plot of log (x/m) versus log p in a Freundlich adsorption isotherm, which of the following
statements is correct? (k and n are constants)
(1) log (1/n) appears as the intercept.
(2) Both k and 1/n appear in the slope term
(3) 1/n appears as the intercept
(4) only 1/n appears as the slope
Ans (4)
x
 kp1/n
1
Slope =
m
n
x
1
log  log k  log p
m
n
log p 
88. The main oxides formed on combustion of Li, Na and K in excess of air respectively are
(1) Li2O, Na2O2 and KO2
(2) Li2O, Na2O and KO2
(3) LiO2, Na2O2 and K2O
(4) Li2O2, Na2O2 and KO2
Ans (1)
Lithium forms monoxide
Sodium forms peroxide
Potassium forms superoxide
 C  D is 100. If the initial concentration of
89. The equilibrium constant at 297 K for a reaction A  B 
all the four species were 1 M each, then equilibrium concentration of D(in mol L1) will be
(1) 1.182
(2) 0.182
(3) 0.818
(4) 1.818
Ans (4)
A
B
C
D
Initial conc. of moles
1
1
1
1
Reacted
x
x


Remaining at equilibrium
(1  x) (1  x) (1 + x) (1 + x)
[C][D] (1  x) 2

 100
[A][B] (1  x) 2
1 x
Taking root :
 10
1 x
1 + x = 10  10 x  11x = 9  x = 0.81
Concentrated of (D) = 1 + x = 1 + 0.818 = 1.818
K eq 
90. The absolute configuration of
H
CO2H
OH
Cl
H
is
CH3
(1) (2R 3R)
(2) (2R 3S)
(3) (2S 3R)
Ans (3)
The absolute configuration of given isomer is 2S 3R.
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(4) (2S 3S)