FINAL JEE–MAIN EXAMINATION – APRIL, 2024
(Held On Tuesday 09th April, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS
TEST PAPER WITH SOLUTION
SECTION-A
1. Let the line L intersect the lines
x – 2 = –y = z – 1, 2 (x + 1) = 2(y – 1) = z + 1
and be parallel to the line
x 2 y 1 z 2
3 1 2

.
Then which of the following points lies on L ?
(1)
1
– ,1,1
3



(2)
1
– ,1, 1
3



(3)
1
– , 1, 1
3




(4)
1
– , 1,1
3



Ans. (2)
Sol.
dr's of line
M(2+, –, 1+
N 1 ,1 , –1
22




L1
L2
1x 2 y z 1
L: 1 1 1

2x 1 y 1 z 1
L: 111
22
dr of line MN will be
3 , 1 ,2
22

& it will be
proportional to <3, 1, 2>
31
2
22
3 1 2

4 + = –6 4 + 3 = 0

42
&
33
Coordinate of M will be <
2 4 1
,,
3 3 3



and equation of required line will be.
2 4 1
x y z
3 3 3 k
3 1 2
So any point on this line will be
2 4 1
3k, k, 2k
3 3 3



21
3k
33

1
k3

Point lie on the line for
1
k3

is
1,1, 1
3




2. The parabola y2 = 4x divides the area of the circle
x2 + y2 = 5 in two parts. The area of the smaller
part is equal to :
(1)
1
22
5sin
35



(2)
1
12
5sin
35



(3)
1
12
5sin
35



(4)
1
22
5sin
35



Ans. (1)
Sol. y2 = 4x
x2 + y2 = 5
Area of shaded region as shown in the figure
will be
A(1,2)
B(1,–2)
C( 5, 0)
15
2
101
A 4x dx 5 x dx

15
321
2
1
0
4 x 5 x
x 5 x sin
3 2 2 5
 
 


1
1 5 5 1
sin
3 4 2 5



Required Area = 2 A1
1
2 5 1
5sin
32 5



1
21
5 sin
32 5



1
21
5cos
35

1
22
5sin
35

 

3. The solution curve, of the differential equation
dy dy
2y 3 5
dx dx

, passing through the point
(0, 1) is a conic, whose vertex lies on the line :
(1) 2x + 3y = 9 (2) 2x + 3y = –9
(3) 2x + 3y = –6 (4) 2x + 3y = 6
Ans. (1)
Sol.
dy
2y 5 3
dx
2y 5 dy 3dx
2
y
2 5y 3x
2
Curve passes through (0, 1)
 = – 4
Curve will be
2
53
y 3 x
24
Vertex of parabola will be
35
,
42



2x + 3y = 9
4. A ray of light coming from the point P (1, 2) gets
reflected from the point Q on the x-axis and then
passes through the point R (4, 3). If the point S (h,
k) is such that PQRS is a parallelogram, then hk2 is
equal to :
(1) 80 (2) 90
(3) 60 (4) 70
Ans. (4)
Sol.
Image of P wrt x-axis will be P'(1, –2) equation of
line joining P'R will be
5
y 3 x 4
3
Above line will meet x-axis at Q where
11
y 0 x 5
11
Q ,0
5



PQRS is parallelogram so their diagonals will
bisects each other

11 h
4 1 2 3 k 0
5&
2 2 2 2


14
h &k 5
5

22
14
hk 5 70
5
5. Let , R. If the system of equations
3x + 5y + z = 3
7x + 11y –9z = 2
97x + 155y – 189z =
has infinitely many solutions, then +2 is equal
to :
(1) 25 (2) 24
(3) 27 (4) 22
Ans. (1)
Sol. 3x + 5y + z = 3
7x + 11y –9z = 2
97x + 155y – 189z =
93x + 155y + 31z = 93
97x + 155y – 189z =
– – + –
–4x + (31 + 189)z = 93–
1085x + 1705y – 1395z = 310
1067x + 1705y – 2079z = 11
– – + –
18x + 684z = 310 – 11
–36x + 9(31 + 189)z = 9(93 – )
36x + 1368z = 2 (310 – 11 )
(279 + 3069)z = 1457 – 31 
for infinite solutions -
3069 341
279 31

1457
31

1457 682 775
2 25
31 31
6. The coefficient of x70 in x2(1 + x)98 + x3(1 + x)97 +
x4 (1 + x)96 + ........ +x54(1 + x)46 is 99Cp – 46Cq.
Then a possible value to p + q is :
(1) 55 (2) 61
(3) 68 (4) 83
Ans. (4)
Sol.
98 96
2 3 97 4
x 1 x x 1 x x 1 x .......
46
54
x 1 x
Coeff. of
70 98 97 96
68 67 66
x : C C C ..........
47 46
17 16
CC
46 47 98
30 30 30
C C ........... C
46 46 47 98 46
31 30 30 30 31
C C C ........... C – C
47 47 98 46
31 30 30 31
C C ........... C – C
......
99 46 99 46
31 31 p q
C C C C
Possible values of (p + q) are 62, 83, 99, 46
p + q = 83
7. Let
e
2 tan x 1
dx x log sin x cosx C
3 tan x 2
, where C is the constant of integration.
Then

is equal to :
(1) 3 (2) 1
(3) 4 (4) 7
Ans. (3)
Sol.
2 tan x dx
3 tan x
2cos x sin x dx
3cosx sin x
2 cosx – sinx = A(3cosx + sinx) + B(cosx – 3sinx)
3A + B = 2
A – 3B = –1
11
A ,B
22

2cos x sin x dx
3cosx sin x
x1
ln 3cosx sin x C
22
1x ln 3cosx sin x C
2
1x ln sin x cosx C
2
= 1, = 1, = 3
3
14
1
8. A variable line L passes through the point (3, 5)
and intersects the positive coordinate axes at the
points A and B. The minimum area of the triangle
OAB, where O is the origin, is :
(1) 30 (2) 25
(3) 40 (4) 35
Ans. (1)
Sol.
xy1
ab

351
ab


5a
b ,a 3
a3

(0, b)
O
A
(a, 0)
(3, 5)
B
1 1 5a
A ab a
2 2 a 3

5a
2 a 3

2
5 a 9 9
2 a 3




59
a3
2 a 3


59
a 3 6 30
2 a 3



9. Let
1
cos cos 60 cos 60 , 0,2
8
Then, the sum of all
0,2
, where cos 3
attains its maximum value, is :
(1) 9 (2) 18
(3) 6 (4) 15
Ans. (3)
Sol. We know that
(cos ) (cos (60° – ) (cos (60° + )
1cos3
4

So equation reduces to
11
cos3
48


1
cos3 2


11
cos3
22
maximum value of cos3 =
1
2
, here
3 2n 3
2n
39

As [0, 2] possible values are
5 7 11 13 17
, , , , ,
9 9 9 9 9 9



Whose sum is
5 7 11 13 17
9 9 9 9 9 9
54 6
9

10. Let
OA 2a,OB 6a 5b
and
OC 3b
,
where O is the origin. If the area of the
parallelogram with adjacent sides
OA
and
OC
is
15 sq. units, then the area (in sq. units) of the
quadrilateral OABC is equal to :
(1) 38 (2) 40
(3) 32 (4) 35
Ans. (4)
Sol.
C
(3b)
O
A
(2a)
B
(6aT5b)
Area of parallelogram having sides
OA&OC
OA OC 2a 3b 15
6 a b 15
5
ab 2

........(1)
Area of quadrilateral
12
1
OABC d d
2

1AC OB
2

13b 2a 6a 5b
2
118b a –10a b 14 a b
2
5
14 35
2
11. If the domain of the function
1x1
f x sin 2x 3



is R – (, )
then 12 is equal to :
(1) 36 (2) 24
(3) 40 (4) 32
Ans. (4)
Sol. Domain of
1x1
f(x) sin 2x 3



is
x1
3
2x 3 0 &x and 1
2 2x 3
x 1 2x 3
y = |2x + 3|
y = |x – 1|
x = 1
x = – 4
For
2x 3 x 1
2
x – , 4 ,
3



2
4& :12 32
3

12. If the sum of series
1 1 1
.......
1 1 d 1 d 1 2d 1 9d 1 10d
is equal to 5, then 50d is equal to :
(1) 20 (2) 5
(3) 15 (4) 10
Ans. (2)
Sol.
11
.............
1 1 d 1 d 1 2d

15
1 9d 1 10d

1 d 1 1 2d 1 d
1..............
d 1 1 d 1 d 1 2d




1 10d 1 9d 5
1 9d 1 10d

1 1 1 1
1 ..............
d 1 d 1 d 1 2d
115
1 9d 1 10d





11
15
d 1 10d




10d 5d
1 10d
50d 5
13. Let
32
f x ax bx ex 41
be such that
f(1) = 40, f'(1) = 2 and f''(1) = 4.
Then a2 + b2 + c2 is equal to :
(1) 62 (2) 73
(3) 54 (4) 51
Ans. (4)
Sol.
32
f(x) ax bx cx 41
2
f '(x) 3ax 2bx cx

f '(1) 3a 2b c 2
......... (1)
f ''(n) 6ax 2b

f ''(1) 6a 2b 4
3a + b = 2 ......... (2)
(1) – (2)
b + c = 0 ......... (3)
f(1) = 40
a + b + c + 41 = 40
use (3)
a + 41 = 40
by (2)
–3 + b = 2 b = 5 & c = –5
a2 + b2 +c2 = 1 + 25 + 25 = 51
14. Let a circle passing through (2, 0) have its centre at
the point (h, k). Let (xc, yc) be the point of
intersection of the lines 3x + 5y = 1 and (2 + c) x +
5c2y = 1. If
c
c1
h limx
and
c
c1
k lim y
, then the
equation of the circle is :
(1) 25x2 + 25y2 – 20x + 2y – 60 = 0
(2) 5x2 + 5y2 – 4x – 2y – 12 = 0
(3) 25x2 + 25y2 – 2x + 2y – 60 = 0
(4) 5x2 + 5y2 – 4x + 2y – 12 = 0
Ans. (1)
Sol.
21 3x
2 c x 5c 1
5



2
22
1 c 1 3x c 1
x ,y
2 c 3c 5 5 2 c 3c
 
c1
1 c 1 c 2
h lim 1 c 2 3c 5



c1
c 1 1
K lim 5 c 1 3c 2 25
21
Centre , ,
25 25



22
2 1 64 1
r 2 0
5 25 25 625
161
r25
22
2 1 161
xy
5 25 125

22
25x 25y 20x 2y 60 0
15. The shortest distance between the line
x 3 y 7 z 1
4 11 5


and
x 5 y 9 z 2
3 6 1

is :
(1)
187
563
(2)
178
563
(3)
185
563
(4)
179
563
Ans. (1)
Sol.
N
M
A(3, – 7, 1)
B
(5, 9, – 2)
n p q
ˆ ˆ ˆ
i j k
ˆ ˆ ˆ
n 4 11 5 19i 11j 9k
3 6 1
S.d. = projection of
ABonn
ˆ ˆ ˆ ˆ ˆ ˆ
2i 16j 3k . 19i 11j 9k
AB n
n361 121 81

38 176 27
563

187
S.d. 563
16. The frequency distribution of the age of students in
a class of 40 students is given below.
Age
15
16
17
18
19
20
No. of
Students
5
8
5
12
x
y
If the mean deviation about the median is 1.25,
then 4x + 5y is equal to :
(1) 43 (2) 44
(3) 47 (4) 46
Ans. (2)
Sol. x + y = 10 .........(1)
Median = 18 = M
ii
i
f x M
M.D. f

36 x 2y
1.25 40

x + 2y = 14 .........(1)
by (1) & (2)
x = 6, y = 4
4x + 5y = 24 + 20 = 44
Age(xi)
f
|xi – M|
fi|xi – M|
15
5
3
15
16
8
2
16
17
5
1
5
18
12
0
0
19
x
1
x
20
y
2
2y
17. The solution of the differential equation
(x2 + y2)dx – 5xy dy = 0, y(1) = 0, is :
(1)
5
2 2 2
x 4y x
(2)
6
22
x 2y x
(3)
6
22
x 4y x
(4)
5
2 2 2
x 2y x
Ans. (1)
Sol. (x2 + y2) dx = 5xydy
22
dy x y
dx 5xy

Put y = Vx
2
dv 1 V
Vx
dx 5V
2
xdv 1 4V
dx 5V

2
V dx
dV
1 4V 5x


Let 1 – 4 V2 = t
– 8V dV = dt
dt dx
8 t 5x


11
ln t ln x lnC
85
5ln t 8ln x ln K
85
ln x ln t lnK 0
85
x t C
5
82
x 1– 4V C
5
22
8
2
x 4y
xC
x

5
2 2 2
x 4y Cx
given y(1) = 0
5
1C
C1
5
2 2 2
x 4y x
18. Let three vectors
ˆ ˆ ˆ
a i 4j 2k
,
ˆ ˆ ˆ
b 5i 3j 4k
,
ˆ ˆ ˆ
c xi yj zk
from a triangle
such that
c a – b
and the area of the triangle is
56
. if is a positive real number, then
2
c
is :
(1) 16 (2) 14
(3) 12 (4) 10
Ans. (2)
Sol.
c a b
(x, y, z) = (–5, 1, –2)
x = –5, y = 1, z = – 2 ........(1)
Area of = 5
6
(given)
156
ac
2
ˆ ˆ ˆ
i j k
10 6
42
x 1 2
ˆ ˆ ˆ 10 6
10i j k
2 2x 4x
(2 + 2 – 10)2 + ( – 4 + 20)2 = 500
(4 – 10)2 + (20 – 3)2 = 500
252 – 80 – 120 = 0
(25 – 200) = 0
= 8 (given is +ve number)
x = – 5 = 3
22 2 2
x y z
c
= 9 + 1 + 4
= 14
19. Let , be the roots of the equation
2
x 2 2 x –1 0
. The quadratic equation,
whose roots are 4 + 4 and
6
1
10
, is :
(1) x2 – 190x + 9466 = 0
(2) x2 – 195x + 9466 = 0
(3) x2 – 195x + 9506 = 0
(4) x2 – 180x + 9506 = 0
Ans. (3)
Sol. x2 + 2
2
x – 1 = 0
+ = –2
2
 = – 1
4 + 4 = (2 + 2)2 – 222
= (( + )2 – 2)2 – 2()2
= (8 + 2)2 – 2(–1)2
= 100 – 2 = 98
6 + 6 = (3 + 3)2 –233
= (( + ) (( + )2 –3)2 – 2()3
= (–2
2
(8 + 3))2 + 2
= (8) (121) + 2 = 970
1
10
(6 + 6) = 97
x2 – (98 + 97)x + (98) (97) = 0
x2 – 195x + 9506 = 0
20. Let
2
f x x 9
,
x
gx x9
and
a = fog(10), b = gof(3). If e and l denote the
eccentricity and the length of the latus rectum of
the ellipse
22
xy1
ab

, then 8e2 + l2 is equal to.
(1) 16 (2) 8
(3) 6 (4) 12
Ans. (2)
Sol. f(x) = x2 + 9 g(x) =
x
x9
a = f(g(10)) = f
10
10 9



= f(10) = 109
b = g (f(3)) = g (9 + 9)
= g(18) =
18
9
= 2
E :
22
xy
1
109 2

e2 = 1 –
2 107
109 109
24
2
109 109
22
8 16
107
8e 109 109
= 8
SECTION-B
21. Let a, b and c denote the outcome of three
independent rolls of a fair tetrahedral die, whose
four faces are marked 1, 2, 3, 4. If the probability
that ax2 + bx + c = 0 has all real roots is
m
n
,
gcd(m, n) = 1, then m + n is equal to ________.
Ans. (19)
Sol. a, b, c {1, 2, 3, 4}
1
2
3
4
Tetrahedral dice
ax2 + bx + c = 0
has all real roots

D0
2
b 4ac 0
Let b = 1 
1 4ac 0
(Not feasible)
b = 2
4 4ac 0
1 ac
a = 1, c = 1,
b = 3
9 4ac 0
9ac
4
a = 1, c = 1
a = 1, c = 2
a = 2, c = 1
b = 4
16 4ac 0
4 ac
a = 1, c = 1
a = 1, c = 2 a = 2, c = 1
a = 1, c = 3 a = 3, c = 1
a = 1, c = 4 a = 4, c = 1
a = 2, c = 2
Probability
12 3 m
4 4 4 16 m
m + n = 19
22. The sum of the square of the modulus of the
elements in the set
z a ib :a,b Z,z C, z 1 1, z 5 z 5i
is ________.
Ans. (9)
Sol.
z 1 1
x 1 iy 1


22
x 1 y 1

22
x 1 y 1
............ (1)
Also
z 5 z 5i
22
22
x 5 y x y 5
10x 10y

xy
............ (2)
Solving (1) and (2)

22
x 1 x 1

2
2x 2x 0

x x 1 0

x 0orx 1
y 0or y 1
(2, 0)
(1, 0)
1
y = x
Given x, y I
Points (0, 0), (1, 0), (2, 0), (1, 1), (1, –1)
to find
2 2 2 2 2
1 2 3 4 5
z z z z z
= 0 + 1 + 4 + 1 + 1 + 1 + 1 = 9
23. Let the set of all positive values of , for which the
point of local minimum of the function
(1 + x (2–x2)) satisfies
2
2
x x 2 0
x 5x 6


, be (, ).
Then 2 + 2 is equal to ________.
Ans. (39)
Sol.
2
2
x x 2 0
x 5x 6



10
x 2 x 3

+
–2
–
–3
+
x (–3, –2) ..........(1)
22
f(x) 1 x x
Finding local minima
22
f '(x) x 2x .x
Put f'(x) = 0
= 3x2

x3

+
–
–
3

3
Local min Local max
We want local min

x3


from (1)
x 3, 2
3 –2
3

3 3 2 3
2 3, 3 3
22
12 27 39
24. Let
4 2 4 4 2 4
n
n 2n n 8n
lim n 1 n 1 n 1 n 16 n 4 n 16

2
4 4 2 2 4 4
n 2n n
....... n n n n n n
be
k
,
using only the principal values of the inverse
trigonometric functions. Then k2 is equal to _____.
Ans. (32)
Sol.
2
4 4 2 2 4 4
r1
n 2nr
n r n r n r
2
4 2 4
r1
1r
12nn
n
r r r
1 1 1
n n n





12
4 2 4
0
dx 2x dx
1 x 1 x 1 x
12
24
0
1x dx
1 x 1 x

12
2
02
11
xdx
11
xx
xx





12
2
0
1
1xdx
11
x x 2
xx
1
xt
x


2
1
1 dx dt
x

2
2
dt
t t 2

2
2
tdt
t t 2
take t2 – 2 = 2
t dt = d
2
2
d
2


2
2
d
2


2
1
1tan
22

11
11
tan 1 tan
22


1
24
2




K
42

So
K 4 2
K2 = 32
25. The remainder when 4282024 is divided by 21 is
________.
Ans. (1)
Sol. (428)2024 = (420 + 8)2024
= (21 × 20 + 8)2024
= 21m + 82024
Now 82024 = (82)1012
= (64)1012
= (63 + 1)1012
= (21 × 3 + 1)1012
= 21n + 1
Remainder is 1.
26. Lef f:(0, )R be a function given by
tan8x
tan7x
btanx
a
8, 0 x
72
f x a 8, x 2
(1 cot x ) , x
2




Where a, b Z. If f is continuous at
x2
, then
a2 + b2 is equal to ________.
Ans. (81)
Sol. LHL at x =
2
tan8x
tan7x
x2
8
lim 7



=
0
81
7



RHL at x =
2
btanx
a
x2
lim 1cot x
=
x2
b
lim cotx tanx b
aa
ee
1 = a – 8 =
b
a
e
a = 9, b = 0
a2 + b2 = 81
27. Let A be a non-singular matrix of order 3. If
det(3adj(2adj((detA)A))) = 3–13
2–10 and det
(3adj(2A)) = 2m
3n, then
3m 2n
is equal to
________.
Ans. (14)
Sol. |3 adj(2adj (|A|A))| = |3adj (2|A|2 adj(A)|
= |3.22|A|4 adj(adj (A)| = 2633 |A|12 |A|4
= 26 33 |A|16 = 2–10 3–13
|A|16 = 2–16 3–16 |A| = 2–1 3–1
Now |3adj (2A)| = |3.22 adj(A)|
= 26 33 |A|2 = 2–m 3–n
26 33 2–2 3–2 = 2–m 3–n
2–m 3–n = 24 31
m = – 4, n = – 1
|3m + 2n| = |–12 – 2| = 14
28. Let the centre of a circle, passing through the point
(0, 0), (1, 0) and touching the circle x2 + y2 = 9, be
(h, k). Then for all possible values of the
coordinates of the centre (h, k), 4(h2 + k2) is equal
to ________.
Ans. (9)
Sol.
y
0, (0, 0)
(1, 0)
(3, 0)
x
x2 + y2 = 9
(x – h)2 + (y – k)2 = h2 + k2
x2 + y2 – 2hx – 2ky = 0
passes through (1, 0)
1 + 0 – 2h = 0
h = 1/2
OC =
OP
2
223
1k2
2




2
19
k
44

k2 = 2
k = ±
2
Possible coordinate of
c (h, k)
1,2
2



1,2
2



4(h2 + k2) = 4
12
4



= 4
9
4



= 9
29. If a function f satisfies f(m + n) = f(m) + f(n) for
all m, n N and f(1) = 1, then the largest natural
number such that
2022 2
k1
f k 2022
is
equal to ________.
Ans. (1010)
Sol. f (m + n) = f(m) + f(n)
f(x) = kx
f (1) = 1
k = 1
f (x) = x
Now
2022 2
k1
f (2022)
k


2022 2
k1
(2022)
k

2022 +
2022 2023
2
(2022)2
 2022 –
2023
2
 1010.5
largest natural no. is 1010.
30. Let A = {2, 3, 6, 7} and B = {4, 5, 6, 8}. Let R be a
relation defined on A × B by (a1, b1) R (a2, b2) is
and only if a1 + a2 = b1 + b2. Then the number of
elements in R is ________.
Ans. (25)
Sol. A = {2, 3, 6, 7}
B = { 2, 5, 6, 8}
(a1, b1) R (a2, b2)
a1 + a2 = b1 + b2
1. (2, 4) R (6, 4)
2. (2, 4) R (7, 5)
3. (2,5) R (7, 4)
4. (3, 4) R (6, 5)
5. (3, 5) R (6, 4)
6. (3, 5) R (7, 5)
7. (3,6) R (7, 4)
8. (3, 4) R (7, 6)
9. (6, 5) R (7, 8)
10. (6, 8) R (7, 5)
11. (7,8) R (7, 6)
12. (6, 8) R (6, 4)
13. (6,6) R (6, 6)
× 2
Total 24 + 1 = 25
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
31. A proton, an electron and an alpha particle have
the same energies. Their de-Broglie wavelengths
will be compared as:
(1) e > > p (2) < p < e
(3) p < e < (4) p > e >
Ans. (2)
Sol. DB =
hh
p2mk
DB
1
m
a < p < e
32. A particle moving in a straight line covers half the
distance with speed 6 m/s. The other half is
covered in two equal time intervals with speeds 9
m/s and 15 m/s respectively. The average speed of
the particle during the motion is :
(1) 8.8 m/s (2) 10 m/s
(3) 9.2 m/s (4) 8 m/s
Ans. (4)
Sol.
A
t1
S
B
t
S
C
D
t
BD S = 9t + 15t = 24t
AB S = 6t1 = 24t t1 = 4t
< speed > =
1
dist. 48t
time 2t t
=
48t 48t
2t 4t 6t
8 m/s
33. A plane EM wave is propagating along x direction.
It has a wavelength of 4 mm. If electric field is in y
direction with the maximum magnitude of
60 Vm–1, the equation for magnetic field is:
(1)
8
zˆ
B 60sin x 3 10 t kT
2



(2)
–7 3 8
zˆ
B 2 10 sin 10 x 3 10 t kT
2



(3)
8
xˆ
B 60sin x 3 10 t iT
2



(4)
–7 8
zˆ
B 2 10 sin x 3 10 t kT
2



Ans. (2)
Sol. E = BC 60 = B × 3 × 108
B = 2 × 10–7
Also C = f
3 × 108 = f × 4 × 10–3
f =
11
310
4
= 2f =
11
32 10
4
=
3
10 C
2
Electric field y direction
Propagation x direction
Magnetic field z-direction
34. Given below are two statements:
Statement (I): When an object is placed at the
centre of curvature of a concave lens, image is
formed at the centre of curvature of the lens on the
other side.
Statement (II): Concave lens always forms a
virtual and erect image.
In the light of the above statements, choose the
correct answer from the options given below:
(1) Statement I is false but Statement II is true.
(2) Both Statement I and Statement II are false.
(3) Statement I is true but Statement II is false.
(4) Both Statement I and Statement II are true.
NTA Ans. (1)
Allen Ans. (2)
Sol.
1 1 1
v u f

1 1 1
v 2f f


11
v 2f
v = –2f
1 1 1
v u f

Virtual image of Real object.
In statement II, it is not mentioned that object is
real or virtual hence Statement II is false.
35. A light emitting diode (LED) is fabricated using
GaAs semiconducting material whose band gap is
1.42 eV. The wavelength of light emitted from the
LED is:
(1) 650 nm (2) 1243 nm
(3) 875 nm (4) 1400 nm
Ans. (3)
Sol.
1240
1.42

= 875 nm (Approx)
36. A sphere of relative density and diameter D has
concentric cavity of diameter d. The ratio of
D
d
, if
it just floats on water in a tank is:
(1)
1
3
1




(2)
1
3
1
1





(3)
1
3
1



(4)
1
3
2
2





Ans. (1)
Sol. weight (w) =
33
4 D d g
38





Buoyant force (Fb) =
3
4D
1g
38




For Just Float w = Fb
(D3 – d3) = D3
3
3
d1
1
D

3
1d
1D




1
3D
1d

37. A capacitor is made of a flat plate of area A and a
second plate having a stair-like structure as shown
in figure. If the area of each stair is
A
3
and the
height is d, the capacitance of the arrangement is:
d
d
d
A/3
A/3
A/3
A
(1)
0
11 A
18d
(2)
0
13 A
17d
(3)
0
11 A
20d
(4)
0
18 A
11d
Ans. (1)
Sol. All capacitor are in parallel combination.
Also effective area is common area only
Ceq = C1 + C2 + C3
Ceq =
0 0 0
A A A
3d 3(2d) 3(3d)

q0
e
A11
3 6d
C


q0
e
11A
18d
C

38. A light unstretchable string passing over a smooth
light pulley connects two blocks of masses m1 and
m2. If the acceleration of the system is
g
8
, then the
ratio of the masses
2
1
m
m
is:
(1) 9 : 7 (2) 4 : 3
(3) 5 : 3 (4) 8 : 1
Ans. (1)
Sol.
21
sys
12
mm g
ag
m m 8




2
1
m9
m7
39. The dimensional formula of latent heat is:
(1) [M0LT–2] (2) [MLT–2]
(3) [M0L2T–2] (4) [ML2T–2]
Ans. (3)
Sol. Latent heat is specific heat
22
ML T
M
= M0L2T–2
40. The volume of an ideal gas ( = 1.5) is changed
adiabatically from 5 litres to 4 litres. The ratio of
initial pressure to final pressure is:
(1)
4
5
(2)
16
25
(3)
8
55
(4)
2
5
Ans. (3)
Sol. For Adiabatic process
i i f f
P V P V
Pi (5)1.5 = Pf (4)1.5
31
22
i
f
P4 4 4
P 5 5 5
8
55
41. The energy equivalent of 1g of substance is:
(1) 11.2 × 1024 MeV (2) 5.6 × 1012 MeV
(3) 5.6 eV (4) 5.6 × 1026 MeV
Ans. (4)
Sol. E = mC2
E = (1 × 10–3) × (3 × 108)2 J
E = (10–3) (9 × 1016) (6.241 × 1018) eV
E = 56.169 × 1031 eV
E 5.6 × 1026 MeV
42. An astronaut takes a ball of mass m from earth to
space. He throws the ball into a circular orbit about
earth at an altitude of 318.5 km. From earth’s
surface to the orbit, the change in total mechanical
energy of the ball is
e
e
GM m
x21R
. The value of x is
(take Re = 6370 km):
(1) 11 (2) 9
(3) 12 (4) 10
Ans. (1)
Sol. h = 318.5 
e
R
20



TEi =
e
e
GM m
R
TEf =
ee
e
ee
GM m GM m
R
2(R h) 2R 20




TEf =
e
e
10GM m
21R
Change in total mechanical energy
= TEf – TEi
=
ee
GM m 11GM m
10
1
Re 21 21Re




43. Given below are two statements:
Statement (I) : When currents vary with time,
Newton’s third law is valid only if momentum
carried by the electromagnetic field is taken into
account.
Statement (II) : Ampere’s circuital law does not
depend on Biot-Savart’s law.
In the light of the above statements, choose the
correct answer from the options given below:
(1) Both Statement I and Statement II are false.
(2) Statement I is true but Statement II is false.
(3) Statement I is false but Statement II is true.
(4) Both Statement I and Statement II are true.
Ans. (2)
Sol. Conceptual.
44. A particle of mass m moves on a straight line with
its velocity increasing with distance according to
the equation
vx
, where is a constant. The
total work done by all the forces applied on the
particle during its displacement from x = 0 to
x = d, will be:
(1)
2
m
2d
(2)
2
md
2
(3)
2
md
2
(4) 2m2d
Ans. (3)
Sol.
vx
at x = 0 : v = 0
& at x = d ;
vd
W.D = Kf – Ki
W.D =
22
11
m d m(0)
22

W.D =
2
md
2
45. A galvanmeter has a coil of resistance 200 with
a full scale deflection at 20 A. The value of
resistance to be added to use it as an ammeter of
range (0–20) mA is:
(1) 0.40 (2) 0.20
(3) 0.50 (4) 0.10
Ans. (2)
Sol. G = 200
ig = 20 A
i =
g
G
i1
S



20 × 10–3 = 20 × 10–6
200 1
S



200 999
S
S  0.2 
46. A heavy iron bar, of weight W is having its one end
on the ground and the other on the shoulder of a
person. The bar makes an angle with the horizontal.
The weight experienced by the person is:
(1)
W
2
(2) W
(3) W cos (4) W sin
Ans. (1)
Sol.
W
Ng
R
R = net reaction force by shoulder
Balancing torque about pt of contact on ground:
L
W cos R Lcos
2



W
R2
47. One main scale division of a vernier caliper is
equal to m units. If nth division of main scale
coincides with (n + 1)th division of vernier scale,
the least count of the vernier caliper is:
(1)
n
(n 1)
(2)
m
(n 1)
(3)
1
(n 1)
(4)
m
n(n 1)
Ans. (2)
Sol. n MSD = (n + 1) VSD
1 VSD =
n
n1
MSD
LC = 1 MSD – 1 VSD
LC =
n
mm
n1



LC =
n 1 n
mn1




LC =
m
n1



48. A bulb and a capacitor are connected in series
across an ac supply. A dielectric is then placed
between the plates of the capacitor. The glow of
the bulb:
(1) increases (2) remains same
(3) becomes zero (4) decreases
Ans. (1)
Sol.
R
~
C
22
C
Z R X
&
C
1
XWC
due to dielectric
C XC Z
So, current increases & thus bulb will glow more
brighter.
49. The equivalent resistance between A and B is:
10
6
8
4
11
8
5
7
A
B
(1) 18 (2) 25
(3) 27 (4) 19
Ans. (4)
Sol.
10
6
8
4
11
8
5
7
A
B
D
D
C
C
A
C
15
15
15
D
8
B
6
A
C
5
D
B
6
8
Req = 6 + 5 + 8 = 19
50. A sample of 1 mole gas at temperature T is
adiabatically expanded to double its volume. If
adiabatic constant for the gas is
3
2

, then the
work done by the gas in the process is:
(1)
RT 2 2


(2)
R22
T


(3)
RT 2 2


(4)
T22
R


Ans. (1)
Sol. TV– 1
= constant
33
11
22
f
T(V) T 2V

1 1 1
2 2 2
f
TV T (2) (V)

f
T
T2



Now, W.D. =
T
1 R T
nR T 2
3
112





W.D. =
1
2RT 1 2



W.D. =
RT 2 2


SECTION-B
51. If
a
and
b
makes an angle
15
cos 9



with each
other, then
| a b | 2 | a b |
for
| a | n | b |
The integer value of n is _______.
Ans. (3)
Sol.
5
cos 9

a b 5
ab 9
.......(1)
| a b | 2 | a b |
2 2 2 2
a b 2a b 2a 2b 4a b
22
6a b a b
22
5
6 ab a b
9
22
10 ab a b
3
& a = nb
2 2 2 2
10 nb n b b
3
3n2 – 10n + 3 = 0
1
n3
and n = 3
integer value n = 3
52. At the centre of a half ring of radius R = 10 cm and
linear charge density 4n C m–1, the potential is
x V. The value of x is ________.
Ans. (36)
Sol. Potential at centre of half ring
V =
KQ
R
KR
VR

V = K V = 9 × 109 × 4 × 10–9
V = 36
53. A star has 100% helium composition. It starts to
convert three 4He into one 12C via triple alpha
process as 4He + 4He + 4He 12C + Q. The mass of
the star is 2.0 × 1032 kg and it generates energy at
the rate of 5.808 × 1030 W. The rate of converting
these 4He to 12C is n × 1042 s–1, where n is _______.
[Take, mass of 4He = 4.0026 u, mass of 12
4He + 4He + 4He 12C + Q
power generated =
NQ
t
where, N No. of reaction/sec.
Q =
2
He C
3m m C
Q = (3 × 4.0026 – 12) (3 × 108)2
Q = 7.266 MeV
30
6 19
N power 5.808 10
tQ
7.266 10 1.6 10

42
N5 10
t
rate of conversion of 4He into 12C = 15 × 1042
Hence, n = 15
54. In a Young’s double slit experiment, the intensity
at a point is
th
1
4



of the maximum intensity, the
minimum distance of the point from the central
maximum is ________ m.
(Given : = 600 nm, d = 1.0 mm, D = 1.0 m)
Ans. (200)
Sol.
2
0
I I cos 2




2
0
Icos
42




2
3

2 yd 2
D3




9
3
D 600 10 1
y3d 3 10

= 2 × 10–4 m
55. A string is wrapped around the rim of a wheel of
moment of inertia 0.40 kgm2 and radius 10 cm. The
wheel is free to rotate about its axis. Initially the
wheel is at rest. The string is now pulled by a force
of 40 N. The angular velocity of the wheel after 10
s is x rad/s, where x is _______.
Ans. (100)
Sol. = FR = I 40 × 0.1 = 0.4
= 10 rad/s2
Wf = 10 × 10 = 100 rad/s
56. A square loop of edge length 2 m carrying current
of 2 A is placed with its edges parallel to the x-y
axis. A magnetic field is passing through the x-y
plane and expressed as
0ˆ
B B (1 4x)k
, where
B0 = 5 T. The net magnetic force experienced by
the loop is _______ N.
Ans. (160)
C = 12 u]
NTA Ans. (5)
Sol.
Sol.
y
x
F1
F2
i = 2A
x = 0
x = 2
= 2m
B(x = 0) = B0, B(x = 2) = 9B0
Also, F = iB
F1 = iB0 & F2 = 9iB0
F = F2 – F1 = 8iB0 = 8 × 2 × 2 × 5
F = 160 N
57. Two persons pull a wire towards themselves. Each
person exerts a force of 200 N on the wire.
Young’s modulus of the material of wire is
1 × 1011 N m–2. Original length of the wire is 2 m
and the area of cross section is 2 cm2. The wire will
extend in length by ______ m.
Ans. (20)
Sol.
200 N
200 N
FF
Y=
A AY
=
–4 11
200 2
2 10 10

= 2 × 10–5 = 20m
58. When a coil is connected across a 20 V dc supply,
it draws a current of 5 A. When it is connected
across 20 V, 50 Hz ac supply, it draws a current of
4 A. The self inductance of the coil is ______ mH.
(Take = 3)
Ans. (10)
Sol. Case-I:
L,R
5 A
20 V
i =
20
R
R = 4
Case-II:
L,R
5 A
20 V, 50 Hz
~
i =
20
Z
4 =
22
L
22
L
20 R X 5
RX
R2 +
2
L
X
= 25 XL = 3
L =
3 1 1000 mH
2 f 2 50 100


L = 10 mH
59. The position, velocity and acceleration of a particle
executing simple harmonic motion are found to
have magnitudes of 4 m, 2 ms–1 and 16 ms–2 at a
certain instant. The amplitude of the motion is
x
m where x is _______.
Ans. (17)
Sol. x = 4 m, V = 2 m/s, a = 16 m/s2
|a| = 2 x
16 = 2(4)
= 2 rad/s
v =
22
A – x
A =
22
2
vx
A =
416
4
A =
17 m
60. The current flowing through the 1 resistor is
n
10
A. The value of n is ________.
B
E
4
2
10V
I1
1
C
2
5V
4
D
A
Ans. (25)
Sol.
5V
4
2
10V
1
yV
2
5V
4
0V
xV
I1
(x–10)V
y – 5 y – 0 y – x 10 0
2 2 1
y – 5 + y + 2y – 2x + 20 = 0
4y – 2x + 15 = 0 ....(i)
x – 5 x – 0 x –10 – y 0
4 4 1
x – 5 + x + 4x – 40 – 4y = 0
6x – 4y – 45 0 ..(i)
–2x 4y 15 0 ..(ii)
4x – 30 0
x =
15
2
& 4y – 15 + 15 = 0
y = 0
i =
y – x 10
1
i =
0 – 7.5 10
1
i = 2.5A =
nA
10
n = 25
CHEMISTRY
TEST PAPER WITH SOLUTION
SECTION-A
61. The molar conductivity for electrolytes A and B
are plotted against C1/2 as shown below.
Electrolytes A and B respectively are :
C1/2 (mol L–1)1/2
400
200
0
A
B
0.2
0.4
m(Scm2 mol–1)
0
A B
(1) Weak electrolyte weak electrolyte
(2) Strong electrolyte strong electrolyte
(3) Weak electrolyte strong electrolyte
(4) Strong electrolyte weak electrolyte
Ans. (3)
Sol. A Weak electrolyte
B Strong electrolyte
62. Methods used for purification of organic
compounds are based on :
(1) neither on nature of compound nor on the
impurity present.
(2) nature of compound only.
(3) nature of compound and presence of impurity.
(4) presence of impurity only.
Ans. (3)
Sol. Organic compounds are purified based on their
nature and impruity present in it.
63. In the following sequence of reaction, the major
products B and C respectively are :
Na/Et2O
(i) Mg/Et2O
(ii) D2O
A
B
CoF2
C
Br
Cl
(1)
D
D
F
F
and
(2)
D
D
F
F
and
(3)
D
D
F
F
and
(4)
F
F
and
Ans. (1)
Sol.
F
F
Cl
Cl
(C)
Swart
Reaction
Na/Et2O
Cl
Br
Wurtz
Reaction
(A)
D
D
(B)
(i) Mg/Et2O
(ii) D2O
CoF2
64. Correct order of basic strength of Pyrrole
N
H
,
Pyridine
N
and Piperidine
N
H
is:
(1) Piperidine > Pyridine > Pyrrole
(2) Pyrrole > Pyridine > Piperidine
(3) Pyridine > Piperidine > Pyrrole
(4) Pyrrole > Piperidine >Pyridine
Ans. (1)
Sol. Order of basic strength is
N(sp3, localized lone pair) > N(sp2, localized lone
pair) > N(sp2, delocalized lone pair, aromatic)
Piperidine > Pyridine > Pyrrole
65. In which one of the following pairs the central
atoms exhibit sp2 hybridization ?
(1) BF3 and
2
NO
(2)
2
NH
and H2O
(3) H2O and NO2
(4)
2
NH
and BF3
Ans. (1)
Sol. BF3 sp2
NO2
– sp2
H2O sp3
NO2 sp2
NH2
– sp3
66. The F– ions make the enamel on teeth much harder
by converting hydroxyapatite (the enamel on the
surface of teeth) into much harder fluoroapatite
having the formula.
(1) [3(Ca3(PO4)2).CaF2]
(2) [3(Ca2(PO4)2).Ca(OH)2]
(3) [3(Ca3(PO4)3).CaF2]
(4) [3(Ca3(PO4)2).Ca(OH)2]
Ans. (1)
Sol. Fluoroapatite [3Ca3(PO4)2.CaF2]
67. Relative stability of the contributing structures is :
CH2=CH–C–H
O
(I)
CH2–CH=C–H
O
(II)
–
+
CH2–CH=C–H
O
(III)
+
–
(1) (I) > (III) > (II)
(2) (I) > (II) > (III)
(3) (II) > (I) > (III)
(4) (III) > (II) > (I)
Ans. (2)
Sol. (1) Neutral structures are more stable than
charged ones. Therefore I is more stable than
II and III.
(2) +ve charge on less electronegative atom is
more stable i.e., Cis more stable than O
Order is I > II > III
68. Given below are two statements :
Statement (I) : The oxidation state of an element
in a particular compound is the charge acquired by
its atom on the basis of electron gain enthalpy
consideration from other atoms in the molecule.
Statement (II) : p-p bond formation is more
prevalent in second period elements over other
periods.
In the light of the above statements, choose the
most appropriate answer from the options given
below :
(1) Both Statement I and Statement II are
incorrect
(2) Statement I is correct but Statement II is
incorrect
(3) Both Statement I and Statement II are
correct
(4) Statement I is incorrect but Statement II is
correct
Ans. (4)
Sol. Oxidation state of an element in a particular
compound is defined by the charge acquired by its
atom on the basis of electronegativity
consideration from other atoms in molecule.
69. Given below are two statements : one is labelled
as Assertion (A) and the other is labelled as
Reason (R) :
Assertion (A) : SN2 reaction of C6H5CH2Br occurs
more readily than the SN2 reaction of CH3CH2Br.
Reason (R) : The partially bonded unhybridized
p-orbital that develops in the trigonal bipyramidal
transition state is stabilized by conjugation with the
phenyl ring.
In the light of the above statements, choose the
most appropriate answer from the options given
below :
(1) (A) is not correct but (R) is correct
(2) Both (A) and (R) are correct but (R) is not the
correct explanation of (A)
(3) Both (A) and (R) are correct and (R) is the
correct explanation of (A)
(4) (A) is correct but (R) is not correct
Ans. (3)
Sol. The benzyl group acts in much the same way using
the -system of the benzene ring for conjugation
with the p-orbital in the transition state.
Br
RO
benzyl bromide
OR
70. For the given compounds, the correct order of
increasing pKa value :
(A)
OH
(B)
OH
O2N
(C)
OCH3
HO
(D)
NO2
OH
(E)
OCH3
HO
(1) (E) < (D) < (C) < (B) < (A)
(2) (D) < (E) < (C) < (B) < (A)
(3) (E) < (D) < (B) < (A) < (C)
(4) (B) < (D) < (A) < (C) < (E)
Ans. BONUS
NTA Ans. (4)
Sol. Acidic strength order :-
B > D > C > A > E
Correct pKa Order :
B < D < C < A < E
All options are incorrect.
71. Given below are two statements : one is labelled as
Assertion (A) : and the other is labelled as Reason (R).
Assertion (A) : Both rhombic and monoclinic
sulphur exist as S8 while oxygen exists as O2.
Reason (R) : Oxygen forms p-p multiple bonds
with itself and other elements having small size
and high electronegativity like C, N, which is not
possible for sulphur.
In the light of the above statements, choose the
most appropriate answer from the options given
below :
(1) Both (A) and (R) are correct and (R) is the
correct explanation of (A).
(2) Both (A) and (R) are correct but (R) is not the
correct explanation of (A).
(3) (A) is correct but (R) is not correct.
(4) (A) is not correct but (R) is correct.
Ans. (3)
Sol. Oxygen can form 2p-2p multiple bond with
itself due to its small size while sulphur cannot
form multiple bond with itself as 3p-3p bond
will be unstable due to large size of sulphur, but
sulphur can form multiple bond with small size
atom like C and N.
eg. S=C=S
S=C=N–
S C N
72. Given below are two statements : one is labelled as
Assertion (A) and the other is labelled as Reason (R).
Assertion (A): The total number of geometrical
isomers shown by [Co(en)2Cl2]+ complex ion is three
Reason (R): [Co(en)2Cl2]+ complex ion has an
octahedral geometry.
In the light of the above statements, choose the most
appropriate answer from the options given below :
(1) Both (A) and (R) are correct and (R) is the
correct explanation of (A).
(2) (A) is correct but (R) is not correct.
(3) (A) is not correct but (R) is correct.
(4) Both (A) and (R) are correct but (R) is not the
correct explanation of (A).
Ans. (3)
Sol. [Co(en)2Cl2]+ has octahedral geometry with two
geometrical isomers.
N
Co
C
N
C
N
N
+
Cl
Co
N
Cl
N
N
N
+
cis
trans
73. The electronic configuration of Cu(II) is 3d9
whereas that of Cu(I) is 3d10. Which of the
following is correct ?
(1) Cu(II) is less stable
(2) Stability of Cu(I) and Cu(II) depends on nature
of copper salts
(3) Cu(II) is more stable
(4) Cu(I) and Cu(II) are equally stable
Ans. (3)
Sol. Cu(II) is more stable than Cu(I) because hydration
energy of Cu+2 ion compensate IE2 of Cu.
74.
AlCl3
O
O
O
+
A
Zn-Hg
HCl
B
Conc.H2SO4
C
What is the structure of C ?
(1)
O
(2)
O
C
C
OH
O
(3)
CH2
C
OH
O
(4)
O
Ans. (1)
Sol
+
O
O
O
AlCl3
O
HO
Zn-Hg
HCl
(A)
O
(C)
Conc.H2SO4
O
HO
(B)
O
75. Compare the energies of following sets of quantum
numbers for multielectron system.
(A) n = 4, 1 = 1 (B) n = 4, l = 2
(C) n = 3, l = 1 (D) n = 3, l = 2
(E) n = 4, 1 = 0
Choose the correct answer from the options given
below :
(1) (B) > (A) > (C) > (E) > (D)
(2) (E) > (C) < (D) < (A) < (B)
(3) (E) > (C) > (A) > (D) > (B)
(4) (C) < (E) < (D) < (A) < (B)
Ans. (4)
Sol. Energy level can be determined by comparing (n + )
values
(A) n = 4, = 1 (n + ) = 5
(B) n = 4, = 2 (n + ) = 6
(C) n = 3, = 1 (n + ) = 4
(D) n = 3, = 2 (n + ) = 5
(E) n = 4, = 0 (n + ) = 4
For same value of (n + ), orbital having higher
value of n, will have more energy.
(B) > (A) > (D) > (E) > (C)
76. Identify major product "X" formed in the
following reaction :
CO, HCl
X
(Major product)
Anhydrous
AlCl3/CuCl
(1)
O
C
Cl
(2)
O
C
(3)
CHO
(4)
CH2Cl
Ans. (3)
Sol. This is Gattermann-Koch reaction
+ CO + HCl
AlCl3
CHO
CuCl
77. Identify the product A and product B in the
following set of reactions.
H2O, H+
H2O, H2O2, OH
(BH3)2
Major
product A
Major
product B
CH3–CH=CH2
(1) A-CH3CH2CH2–OH, B-CH3CH2CH2–OH
(2) A-CH3CH2CH2–OH, B-
CH3CH–CH3
OH
(3) A-
CH3–CH–CH3
OH
, B-CH3CH2CH2–OH
(4) A-CH3CH2CH3, B-CH3CH2CH3
Ans. (3)
Sol. (1) Hydration Reaction :
3 2 3 3
(More stable)
CH CH CH H CH CH CH
CH3–CH–CH3 + H2O
+
(CH3–CH–CH3) + H+
OH
(A)
(2) Hydroboration Oxidation Reaction :
3CH3–CH=CH2 + B2H6
THF
2(CH3CH2CH2)3B
OH
3 2 2 3 2 2
(CH CH CH ) B 3H O

3 2 2 3 3
(B)
3CH CH CH OH H BO
78. On reaction of Lead Sulphide with dilute nitric
acid which of the following is not formed ?
(1) Lead nitrate (2) Sulphur
(3) Nitric oxide (4) Nitrous oxide
Ans. (4)
Sol. PbS + HNO3 Pb(NO3)2 + NO + S + H2O
Nitrous oxide (N2O) is not formed during the
reaction.
79. Identify the incorrect statements regarding
primary standard of titrimetric analysis
(A) It should be purely available in dry form.
(B) It should not undergo chemical change in air.
(C) It should be hygroscopic and should react
with another chemical instantaneously and
stoichiometrically.
(D) It should be readily soluble in water.
(E) KMnO4 & NaOH can be used as primary
standard.
Choose the correct answer from the options given
below :
(1) (C) and (D) only (2) (B) and (E) only
(3) (A) and (B) only (4) (C) and (E) only
Ans. (4)
Sol. KMnO4 & NaOH Secondary standard.
Primary standard should not be Hygroscopic.
80. 0.05M CuSO4 when treated with 0.01M K2Cr2O7
gives green colour solution of Cu2Cr2O7. The
[SPM : Semi Permeable Membrane]
2 2 7 4
K Cr O CuSO
Side X SPM SideY
Due to osmosis :
(1) Green colour formation observed on side Y.
(2) Green colour formation observed on side X.
(3) Molarity of K2Cr2O7 solution is lowered.
(4) Molarity of CuSO4 solution is lowered.
Ans. (4)
Sol. Only solvent Molecules are allowed to pass
through the SPM.
K2Cr2O
CuSo4
V

V

SECTION-B
81. The heat of solution of anhydrous CuSO4 and
CuSO45H2O are –70 kJ mol–1 and +12 kJ mol–1
respectively.
The heat of hydration of CuSO4 to CuSO45H2O is
–x kJ. The value of x is_____.
Ans. (82)
Sol.
x
4 2 4 2
12
4 2 2 4
70
4 2 4
(1) CuSO (s) 5H O CuSO .5H O
(2) CuSO .5H O H O CuSO (aq)
CuSO H O CuSO (aq)



from (1) & (2)
–70 = x + 12
x = –82
82. Given below are two statements :
Statement I : The rate law for the reaction
A + B C is rate (r) = k[A]2[B]. When the
concentration of both A and B is doubled, the
reaction rate is increased “x” times.
Statement II :
Time
Contration of R
o
[Ro]
–K = Slope
The figure is showing “the variation in
concentration against time plot” for a “y” order
reaction.
The value of x + y is ______________.
Ans. (8)
Sol. r = K[A]2|B|
if conc. are doubled
r' = K[2A]2[2B]1
r' = 8r x = 8
C
t
–K
Zero order, y = 0
x + y = 8
83. How many compounds among the following
compounds show inductive, mesomeric as well as
hyperconjugation effects?
OCH3
O
,
,
NO2
,
CH3
CH3
,
COCH3
CH3
,
Cl
NO2
,
,
Ans. (4)
Sol.
O
,
NO2
,
,
COCH3
CH3
84. The standard reduction potentials at 298 K for the
following half cells are given below :
Cr2O7
2– + 14H+ + 6e– 2Cr3+ + 7H2O, E° = 1.33V
Fe3+ (aq) + 3e– Fe E° = –0.04V
Ni2+ (aq) + 2e– Ni E° = –0.25V
Ag+ (aq) + e– Ag E° = 0.80V
Au3+ (aq) + 3e– Au E° = 1.40V
Consider the given electrochemical reactions,
The number of metal(s) which will be oxidized be
Cr2O7
2–, in aqueous solution is __________.
Ans. (3)
Sol. Fe, Ni, Ag will be oxidized due to lower S.R.P.
85. When equal volume of 1M HCl and 1M H2SO4 are
separately neutralised by excess volume of 1M
NaOH solution. X and y kJ of heat is liberated
respectively. The value of y/x is _________.
of anhyd. CuSO4 in 500 mL solution at 32 °C is
2 × 10–1 M. Its molality will be _______ × 10–3 m.
n n n
sol sol sol
M v d
500 1.25 625g
Mass of solute (x) = 0.2 × 0.5 × 159.5
= 15.95
nsolute = 0.1,
Mass of solvent = Mass of solution – Mass of solute
= 625 – 15.95
= 609.05
0.1
m609.05
1000
3
m 0.164 164 10
NO2
(nearest integer).
[Given density of the solution = 1.25 g/mL.]
NTA Ans. (81) BONUS
Sol.
Ans. (2)
Sol. H+ + OH– H2O x
2H+ + 2OH– 2H2O 2x = y
y/x = 2
86. Molarity (M) of an aqueous solution containing x g
87. The total number of species from the following in
which one unpaired electron is present, is _______.
2
2 2 2 2 2 2 2
N ,O ,C ,O ,O , H ,CN ,He
Ans. (4)
Sol. One unpaired e– is present in :
2 2 2 2
C ; O ; H ; He
88. Number of ambidentate ligands among the
following is ________.
22
2 2 4 3 4 2
NO ,SCN , C O , NH ,CN ,SO ,H O.
Ans. (3)
Sol. Ligands which have two different donor sites but at
a time connects with only one donor site to central
metal are ambidentate ligands.
Ambidentate ligands are NO2
– ; SCN– ; CN–
89. Total number of essential amino acid among the
given list of amino acids is _______.
Arginine, Phenylalanine, Aspartic acid, Cysteine,
Histidine, Valine, Proline
Ans. (4)
Sol. Essential Amino acids are :-
Arginine, Phenylalanine, Histidine, Valine
90. Number of colourless lanthanoid ions among the
following is _________.
Eu3+, Lu3+, Nd3+, La3+, Sm3+
Ans. (2)
Sol. La+3 – [Xe]4f0
Nd+3 – [Xe]4f3
Sm+3 – [Xe]4f5
Eu+3 – [Xe]4f6
Lu+3 – [Xe]4f14
La+3 and Lu+3 do not show any colour because no
unpaired electron is present.