FINAL JEE–MAIN EXAMINATION – APRIL, 2024
(Held On Friday 05th April, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS
TEST PAPER WITH SOLUTION
SECTION-A
1. Let d be the distance of the point of intersection of
the lines
x 6 y z 1
3 2 1


and
x 7 y 9 z 4
4 3 2

from the point (7, 8, 9). Then
d2+ 6 is equal to :
(1) 72 (2) 69
(3) 75 (4) 78
Ans. (3)
Sol.
x 6 y z 1
3 2 1

…(1)
x = 3 – 6, y = 2, z = – 1
x 7 y 9 z 4 µ
4 3 2
…(2)
x = 4µ + 7, y = 3µ + 9, z = 2µ + 4
3 – 6 = 4µ + 7 3 – 4µ = 13 …(3) × 2
2 = 3µ + 9 2 – 3µ = 9 …(4) × 3
6 – 8µ = 26
6 – 9µ = 27
–+–.
µ = –1
3 – 4(–1) = 13
3 = 9
= 3
int. point (3, 6, 2) ; (7, 8, 9)
d2 = 16 + 4 + 49 = 69
Ans. d2 + 6 = 69 + 6 = 75
2. Let a rectangle ABCD of sides 2 and 4 be inscribed
in another rectangle PQRS such that the vertices of
the rectangle ABCD lie on the sides of the
rectangle PQRS. Let a and b be the sides of the
rectangle PQRS when its area is maximum. Then
(a + b)2 is equal to :
(1) 72 (2) 60
(3) 80 (4) 64
Ans. (1 )
Sol.
2cos(90–)
P
B
Q
R
S
A
4cos(90–)
2sin(90–)
D
90–
Area = (4cos + 2sin) (2cos + 4sin)
= 8cos2 + 16sincos + 4sincos + 8sin2
= 8 + 20 sincos
= 8 + 10 sin2
Max Area = 8 + 10 = 18 (sin2
(a + b)2 = (4cos + 2sin+ 2cos + 4sin)2
= (6cos + 6sin)2
= 36 (sin + cos)2

2
36( 2)

3. Let two straight lines drawn from the origin
O intersect the line 3x + 4y = 12 at the points P and
Q such that OPQ is an isosceles triangle and
POQ = 90°. If l = OP2 + PQ2 + QO2, then the
greatest integer less than or equal to l is :
(1) 44 (2) 48
(3) 46 (4) 42
Ans. (3)
Sol.
O
P(rcos, rsin)
Q(rcos(90+, rsin(90+) = (–rsinrcos)
3x + 4y = 12
3(rcos) + 4(rsin) = 12
r(3cos + 4sin) = 12 ...(1)
3(–rsin) + 4(rcos) = 12
r(–3sin + 4cos) = 12 ...(2)
22
22
12 12 (3cos 4sin ) ( 3sin 4cos )
rr
2
12
2 9 16
r




2
2 144 25
r
288 = 25r2

2
288 r
25
12
2r
5




= OP2 + PQ2 + QO2
= r2 + r2+ r2(cos + sin)2 + r2(sin + cos)2
= 2r2 + r2(1 + sin2 + 1 – 2sin2)
= 2r2 + 2r2
= 4r2
1152
288
4 46.08
25
25



[] = 46
4. If y = y(x) is the solution of the differential
equation
dy
dx
+ 2y = sin (2x), y(0) =
3
4
, then
y8



is equal to :
(1) e–/8 (2) e–/4
(3) e/4 (4) e/8
Ans. (2 )
Sol.
dy 2y sin 2x
dx 
,
3
y(0) 4
I.F =
2 dx
e
= e2x
2x 2x
y.e e sin 2x dx
2x
2x e (2sin 2x 2cos2x)
y.e C
44

x = 0, y =
3
4
3 1(0 2)
.1 C
48

31
C
44
1 = C
2x
2sin 2x 2cos2x
y 1.e
8

x8
,
28
1
y 2sin 2cos e
8 4 4







4
y 0 e

5. For the function
f(x) = sinx + 3x –
2
(x2 + x), where x
0, 2



,
consider the following two statements :
(I) f is increasing in
0, 2



.
(II) f is decreasing in
0, 2



.
Between the above two statements,
(1) only (I) is true.
(2) only (II) is true.
(3) neither (I) nor (II) is true.
(4) both (I) and (II) are true.
Ans. (4)
Sol. f(x) = sinx + 3x –
2
(x2 + x)
x 0, 2



f(x) = cosx + 3 –
2
(2x + 1) > 0 f(x)
f(x) = –sinx + 0 –
2
(2)
= –sinx –
4
< 0 f(x)
0 < x <
2
1
11
20 2x

3 3 3
2 2 2
(2x 1) ( 1)
( ve) ( ve)
2 2 2
3 3 (2x 1) 3 ( 1)

6. If the system of equations
11x + y + z = –5
2x + 3y + 5z = 3
8x – 19y – 39z = µ
has infinitely many solutions, then 4 – µ is equal
to :
(1) 49 (2) 45
(3) 47 (4) 51
Ans. (3)
Sol. 11x + y + z = –5
2x + 3y + 5z = 3
8x – 19y – 39z = µ
for infinite sol.
11 1
D 2 3 5 0
8 19 39


11(–117 + 95) – 1(–78 – 40) + (–38 – 24)
11(–22) + 118 – (62) = 0
62 = 118 – 242
124 2
62
1
5 1 2
D 3 3 5 0
µ 19 39



–5(–117 + 95) – 1(–117 – 5µ) – 2(–57 – 3µ) = 0
–5(–22) + 117 + 5µ + 114 + 6µ = 0
11µ = –110 – 231 = –341
µ = –31
4 – µ = (–2)4 – (–31) = 16 + 31 = 47
7. Let A = {1, 3, 7, 9, 11} and B = {2, 4, 5, 7, 8, 10, 12}.
Then the total number of one-one maps
f : A B, such that f (1) + f(3) = 14, is :
(1) 180 (2) 120
(3) 480 (4) 240
Ans. (4)
Sol.
1
3
7
9
11
(5)
2
(7)
12
4
5
7
8
10
A = {1, 3, 7, 9, 11}
B = {2, 4, 5, 7, 8, 10, 12}
f(1) + f(3) = 14
(i) 2 + 12
(ii) 4 + 10
2 × (2 × 5 × 4 × 3) = 240
8. If the function
3
sin3x sin x cos3x
f(x) x
,
x R, is continuous at x = 0, then f(0) is equal to :
(1) 2 (2) –2
(3) 4 (4) –4
Ans. (4)
Sol.
3
sin3x sin x cos3x
f(x) x
is continuous at x = 0
332
3
x0
(3x) x(3x)
3x ... x ... 1 ...
332
lim f(0)
x







2
3
3
x0
27
9x
x(3 ) x ...
33
2
lim f(0)
x





for exist
= 0, 3+ = 0,
27 f(0)
33
= –3,
27 ( 3) f(0)
66
27 3
f(0) 4
6

9. The integral
4
0
136sin x dx
3sin x 5cos x
is equal to :
(1) 3 – 50 loge 2 + 20 loge 5
(2) 3 – 25 loge 2 + 10 loge 5
(3) 3 – 10 loge
22
+ 10 loge 5
(4) 3 – 30 loge 2 + 20 loge 5
Ans. (1)
Sol.
/4
0
136sin x
I dx
3sin x 5cos x
136sinx = A(3sinx + 5cosx) + B(3cosx – 5sinx)
136 = 3A – 5B …(1)
0 = 5A + 3B …(2)
3B = –5A
5
BA
3

5
136 3A 5 A
3



25
136 3A A
3

34A
136 3
136 3
A 12
34

5
B (12) –20
3

/4 /4
00
A(3sin x 5cosx) B(3cosx 5sin x)
I3sin x 5cosx 3sin x 5cosx





/4
/4
00
A x B n 3sin x 5cosx


35
12 20 n n 0 5
422




 
3 20 n4 2 20 n5
5
3 20 n2 20 n5
2
3 50 n2 20 n5
10. The coefficients a, b, c in the quadratic equation
ax2 + bx + c = 0 are chosen from the set
{1, 2, 3, 4, 5, 6, 7, 8}. The probability of this
equation having repeated roots is :
(1)
3
256
(2)
1
128
(3)
1
64
(4)
3
128
Ans. (3)
Sol. ax2 + bx + c = 0
a, b, c {1, 2, 3, 4, 5, 6,7, 8}
Repeated roots D = 0
b2 – 4ac = 0 b2 = 4ac
Prob =
81
8 8 8 64

(a, b, c)
(1, 2, 1) ; (2, 4, 2) ; (1, 4, 4) ; (4, 4, 1) ; (3, 6, 3) ;
(2, 8, 8) ; (8, 8, 2) ; (4, 8, 4)
8 case
11. Let A and B be two square matrices of order 3
such that |A| = 3 and |B| = 2.
Then |AT A(adj(2A))–1 (adj(4B))(adj(AB))–1AAT|
is equal to :
(1) 64 (2) 81
(3) 32 (4) 108
Ans. (1)
Sol. |A| = 3, |B| = 2
|ATA(adj(2A))–1 (adj(4B)) (adj(AB))–1AAT|
= 3×3×|(adj(2A)–1| × |adj(4B)| × |(adj(AB))–1|×3×3
6
62
1
adj(2A)
1
2adjA
1
2 ·3
212× 22
22
1
adj(AB)
1
adjB·adjA
1
2 ·3
4 12 2
6 2 2 2
11
3 · ·2 ·2 · 64
2 ·3 2 ·3

12. Let a circle C of radius 1 and closer to the origin be
such that the lines passing through the point (3, 2)
and parallel to the coordinate axes touch it. Then
the shortest distance of the circle C from the point
(5, 5) is :
(1)
22
(2) 5
(3)
42
(4) 4
Ans. (4)
Sol.
O
R
R
P(3, 2)
Q(5, 5)
C(2, 1)
Coordinates of the centre will be (2, 1)
Equation of circle will be
(x – 2)2 + (y – 1)2 = 1
QC =
22
(5 2) (5 1)
QC = 5
shortest distance
= RQ = CQ – CR
= 5 – 1
= 4
13. Let the line 2x + 3y – k = 0, k > 0, intersect the
x-axis and y-axis at the points A and B,
respectively. If the equation of the circle having the
line segment AB as a diameter is x2 + y2 – 3x – 2y = 0
and the length of the latus rectum of the ellipse
x2 + 9y2 = k2 is
m
n
, where m and n are coprime,
then 2m + n is equal to
(1) 10 (2) 11
(3) 13 (4) 12
Ans. (2)
Sol. Centre of the circle =
3,1
2



Equation of diameter = 2x + 3y – k = 0
3
2 3(1) – k 0
2




k = 6
Now, Equation of ellipse becomes
x2 + 9y2 = 36
22
22
xy
1
62

length of LR
22
2b 2.2 8 4 m
a 6 6 3 n
2m + n = 2(4) + 3 = 11
14. Consider the following two statements :
Statement I : For any two non-zero complex
numbers z1, z2
12
1 2 1 2
12
zz
z z 2 z z
zz
and
Statement II : If x, y, z are three distinct complex
numbers and a, b, c are three positive real numbers
such that
a b c
y z z x x y

, then
2 2 2
a b c 1
y z z x x y
.
Between the above two statements,
(1) both Statement I and Statement II are incorrect.
(2) Statement I is incorrect but Statement II is correct.
(3) Statement I is correct but Statement II is incorrect.
(4) both Statement I and Statement II are correct.
Ans. (3)
Sol. Statement I :
12
12
12
zz
zzzz

Since
1 2 1 2
1 2 1 2
z z z z
z z z z
12
12
1 2 1 2
zz
zz
z z z z
12
12
zz2
zz

12
1 2 1 2
12
zz
z z 2 z z
zz




statement I is correct
For Statement II :
a b c
y z z x x y

2 2 2
2 2 2
a b c
y z z x x y
a2 = (|y – z|2) = (y – z)
yz
b2 = (z – x)
zx
and c2 = (x – y)
xy
2 2 2
a b c y z z x x y 0
y z z x x y
Statement II is false
15. Suppose
0, 4



is a solution of 4 cos – 3 sin = 1.
Then cos is equal to :
(1)
4
3 6 2
(2)
66
3 6 2
(3)
66
3 6 2
(4)
4
3 6 2
Ans. (1)
Sol.
2
22
2 tan
1 tan / 2 2
4 3 1
1 tan / 2 1 tan 2











let
tan t
2
2
2
4 4t 6t 1
1t

4 – 4t2 – 6t = 1 + t2
5t2 + 6t – 3 = 0
6 36 4(5)( 3)
t2(5)
6 96
10

6 4 6
10

3 2 6
t5

2
2
22
2 6 3 24 9 12 6
11
5 25
1t
cos 1t 24 9 12 6
2 6 3 1
125
5










25 33 12 6 12 6 8 6 6 4 29 6 6
25 33 12 6 58 12 6 29 6 6 29 6 6
100 150 6 4 6 6 4 6 6
625 25 4 6 6
200 8 4
4 6 6 3 6 2
25 4 6 6


16. If
1 1 1
... m
1 2 2 3 99 100
and
1 1 1
... n
1 2 2 3 99 100
, then the point (m, n)
lies on the line
(1) 11(x – 1) – 100(y – 2) = 0
(2) 11(x – 2) – 100(y – 1) = 0
(3) 11(x – 1) – 100y = 0
(4) 11x – 100y = 0
Ans. (4)
Sol.
1 1 1
... m
1 2 2 3 99 100
1 2 2 3 99 100
... m
1 1 1

100 1 m
m = 9
1 1 1
... n
1 2 2 3 99 100
1 1 1 1 1 1
... n
1 2 2 3 99 100
1
1n
100

99 n
100
(m, n) =
99
9, 100



11(9) – 100
99
100



= 99 – 99 = 0
Ans. option (4) 11x – 100y = 0
17. Let f(x)= x5 + 2x3 + 3x + 1, x R, and g(x) be a
function such that g(f(x))=x for all x R. Then
g(7)
g'(7)
is equal to :
(1) 7 (2) 42
(3) 1 (4) 14
Ans. (4)
Sol. f(x) = x5 + 2x3 + 3x + 1
f(x) = 5x4 + 6x2 + 3
f(1) = 5 + 6 + 3 = 14
g(f(x)) = x
g(f(x))f(x) = 1
for f(x) = 7
x5 + 2x3 + 3x + 1 = 7
x = 1
g(7) f(1) = 1
11
g'(7) f '(1) 14

x = 1, f(x) = 7 g(7) = 1
g(7) 1 14
g'(7) 1 / 14

18. If A(l, –1, 2), B(5, 7, –6), C(3, 4, –10) and
D(–l, –4, –2) are the vertices of a quadrilateral
ABCD, then its area is :
(1)
12 29
(2)
24 29
(3)
24 7
(4)
48 7
Ans. (1)
Sol. A(1, –1, 2)
B(5, 7, –6)
C(3, 4, –10)
D(–1, –4, –2)
Area =
11
ˆ ˆ ˆ ˆ
ˆˆ
AC BD 2i 5j 12k 6i 11j 4k
22
1ˆˆ
ˆ
112i 64 j 8k
2
ˆˆ
ˆ
4 14i 8j k
4 196 64 1
4 261
12 29
19. The value of
2
2y(1 sin y) dy
1 cos y

is :
(1) 2 (2)
2
2
(3)
2
(4) 22
Ans. (1)
Sol.
2
2y(1 sin y) dy
1 cos y

22
2y 2y sin y
dy dy
1 cos y 1 cos y




(Odd) (Even)
2
0
sin y
0 2.2 y dy
1 cos y

 

2
0
ysin y
I 4 dy
1 cos y
2
0
( y)sin y
I 4 dy
1 cos y

2
0
sin y
2I 4 dy
1 cos y
2
0
sin y
I 2 dy
1 cos y

1
0
2 tan cosy
244




2
2
24



20. If the line
2 x 3y 2 4z
3 4 1


makes a right
angle with the line
x 3 1 2y 5 z
3µ 6 7

, then
4 + 9µ is equal to :
(1) 13 (2) 4
(3) 5 (4) 6
Ans. (4)
Sol.
2 x 3y 2 4z
3 4 1


…(1)
2
y
x 2 z 4
3
41
( 3) ( 1)
3







x 3 1 2y 5 z
3µ 6 7

…(2)
1
y
x 3 z 5
2
3µ ( 3) ( 7)



Right angle
41
( 3)(3µ) ( 3) ( 1)( 7) 0
3




–9µ – 4 – 1 + 7 = 0
4 + 9µ = 6
SECTION-B
21. From a lot of 10 items, which include 3 defective
items, a sample of 5 items is drawn at random. Let
the random variable X denote the number of
defective items in the sample. If the variance of X
is 2, then 962 is equal to__________.
Ans. (56)
Sol. X = denotes number of defective
x
0
1
2
3
P(x)
7
15
5
12
5
12
1
12
x1
2
0
1
4
9
Pix1
2
0
5
12
20
12
9
12
pixi
0
5
12
10
12
3
12
ii
18
µ p x 12
2
i1
34
px 12

2 2 2
i1
p x (µ)
2
34 18 17 9
12 12 6 4



34 27 7
12 12
27
96 96 56
12
22. If the constant term in the expansion of
(l + 2x – 3x3)
9
2
31
x
2 3x



is p, then 108p is equal
to
Ans. (54)
Sol.
9
32
31
1 2x 3x x
2 3x



General term m
9
2
31
x
2 3x



9 2r
9 r 18 3r
r9r
3
C ( 1) x
2
Put r = 6 to get coeff. of
0 9 0 0
63
17
x C x x
18
6
Put r = 7 to get coeff. of
5
3 9 7 3
r2
3
x C ( 1) x
2

9 3 3
752
1
C x x
27
32


3 0 3
71
1 2x 3x x x
18 27



7 3 7 1 7 2 9 1
18 27 18 9 18 18 2
1
108 54
2

23. The area of the region enclosed by the parabolas
y = x2 – 5x and y = 7x – x2 is __________.
Ans. (72)
NTA Ans. (198)
Sol. y = x2 – 5x and y = 7x – x2
(6, 6)
7
5
O
9(x)
f(x)
6
0(g(x) f(x))dx
622
0(7x x ) (x 5x) dx
6
23
62
00
x 2x
12x 2x dx 12 23




23
2
6(6) (6)
3

= 216 – 144 = 72 unit2
24. The number of ways of getting a sum 16 on
throwing a dice four times is __________.
Ans. (125)
Sol. (x1 + x2 ....+ x6)4
4
6
41x
x·
1x



x4·(1–x6)4·(1–x)–4
x4[1–4x6 + 6x12....] [(1–x)–4]
(x4 – 4x10 + 6x16....) (1–x)–4
(x4 – 4x10 + 6x16) (1 + 15C12 x12 + 9C6x6 ....)
(15C12 – 49C6 + 6)x16
(15C3 – 49C6 + 6)
= 35 × 13 – 6 × 8 × 7 + 6
= 455 – 336 + 6
= 125
25. If S = {a R : |2a – 1| = 3[a]+2{a}}, where [t]
denotes the greatest integer less than or equal to t
and {t} represents the fractional part of t, then
aS
72 a
is equal to __________.
Ans. (18)
Sol. |2a – 1| = 3[a] + 2{a}
|2a – 1| = [a] + 2a
Case-1 :
1
a2
2a – 1 = [a] + 2a
[a] = –1 a [–1, 0) Reject
Case-2 :
1
a2
–2a + 1 = [a] + 2a
a = I + f
–2(I + f) + 1 = I + 2I + 2f
I = 0,
1
f4
1
a4
Hence
1
a4
aS
1
72 a 72 18
4
26. Let f be a differentiable function in the interval
(0, ) such that f(l) = 1 and
22
tx
t f(x) x f(t)
lim 1
tx
for each x > 0. Then 2 f(2) + 3 f(3) is equal to
__________.
Ans. (24)
Sol.
22
tx
t f(x) x f(t)
lim 1
tx
2
tx
2t.f(x) x f '(x)
lim 1
1
2x.f(x) – x2f(x) = 1
2
dy 2 1
y
dx x x
2dx
x
2
1
I.f. e x

24
y1
dx C
xx
23
y1
C
x 3x

Put f(1) = 1
2
C3
2
1 2x
y3x 3

3
2x 1
y3x
17
f(2) 6
55
f(3) 9
2f(2) + 3f(3) =
17 55 72 24
3 3 3
27. Let a1, a2, a3, ... be in an arithmetic progression of
positive terms.
Let Ak = a1
2 – a2
2 + a3
2 – a4
2 + ... + a2k–1
2 – a2k
2.
If A3 = –153, A5 = –435 and a1
2 + a2
2 + a3
2 = 66,
then a17 – A7 is equal to __________.
Ans. (910)
Sol. d common diff.
Ak = –kd[2a + (2k – 1)d]
A3 = –153
153 = 13d[2a + 5d]
51 = d[2a + 5d] …(1)
A5 = –435
435 = 5d[2a + 9d]
87 = d[2a + 9d]
(2) – (1)
36 = 4d2
d = 3, a = 1
a17 – A7 = 49 – [–7.3[2 + 39]] = 910
28. Let
ˆˆ ˆ
a i 3j 7k
,
ˆˆˆ
b 2i j k
and
c
be a
vector such that
a 2b c 3 c a
. If
a c 130
, then
bc
is equal to __________.
Ans. (30)
Sol.
a 2b c 3 c a
2b 4a c 0
ˆˆ ˆ
c 4a 2b 8i 14 j 30k
a c 130
8 + 42 + 210 = 130
1
2

ˆˆ ˆ
c 4i 7j 15k
bc
= 8 + 7 + 15 = 30
29. The number of distinct real roots of the equation
|x| |x + 2| – 5|x + l| – 1 = 0 is __________.
Ans. (3)
Sol.
–2
–1
0
Case-1
x 0
x2 + 2x – 5x – 5 – 1 = 0
x2 – 3x – 6 = 0
3 9 24 3 33
x22

One positive root
Case-2
–1 x < 0
–x2 – 2x – 5x – 5 – 1 = 0
x2 + 7x + 6 = 0
(x + 6) (x + 1) = 0
x = –1
one root in range
Case-3
–2 x < –1
x2 – 2x + 5x + 5 – 1 = 0
x2 – 3x – 4 = 0
(x – 4) (x + 1) = 0
No root in range
Case-4
x < –2
x2 + 7x + 4 = 0
7 49 16 7 33
x22

one root in range
Total number of distinct roots are 3
30. Suppose AB is a focal chord of the parabola
y2 = 12x of length l and slope
m3
. If the
distance of the chord AB from the origin is d, then
ld2 is equal to __________.
Ans. (108)
Sol.
S
3
d
= 4a cosec2
2
9
12 d

d2 = 108
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
31. Light emerges out of a convex lens when a source
of light kept at its focus. The shape of wavefront of
the light is:
(1) Both spherical and cylindrical
(2) Cylindrical
(3) Spherical
(4) Plane
Ans. (4)
Sol. Light emerges parallel
planor wavefront
F
32. Following gates section is connected in a complete
suitable circuit.
A
R
B
C
D
For which of the following combination, bulb will
glow (ON):
(1) A = 0, B = 1, C = 1, D = 1
(2) A = 1, B = 0, C = 0, D = 0
(3) A = 0, B = 0, C = 0, D = 1
(4) A = 1, B = 1, C = 1, D = 0
Ans. (2)
Sol. Bulb will glow if bulb have potential drop on it.
One end of bulb must be at high (1) and other must
be at low (0).
Option (2) satisfy this condition
A
R
B
C
D
0
1
1
0
1
0
0
0
33. If G be the gravitational constant and u be the
energy density then which of the following
quantity have the dimension as that the
uG
:
(1) Pressure gradient per unit mass
(2) Force per unit mass
(3) Gravitational potential
(4) Energy per unit mass
Ans. (2)
Sol. [uG] = [(M1L–1T–2) (M–1L3T–2)]
[uG] = [M0L2T–4]
[ uG]
= [L1T–2]
Option (2) is correct
34. Given below are two statements :
Statement-I: When a capillary tube is dipped into
a liquid, the liquid neither rises nor falls in the
capillary. The contact angle may be 0°.
Statement-II: The contact angle between a solid
and a liquid is a property of the material of the
solid and liquid as well :
In the light of above statement, 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 true.
(3) Both Statement-I and Statement-II are false.
(4) Statement-I is true and Statement-II is false.
Ans. (1)
Sol. Capillary rise
2T cos
hgr
;
If = 0° then rise is non-zero
Statement-1 is incorrect.
Option(1) is correct
35. Given below are two statements:
V0
v
M1
M2
Statement-I: Figure shows the variation of
stopping potential with frequency () for the two
photosensitive materials M1 and M2. The slope
gives value of
h
e
, where h is Planck's constant, e is
the charge of electron.
Statement-II: M2 will emit photoelectrons of
greater kinetic energy for the incident radiation
having same frequency.
In the light of the above statements, choose the
most appropriate answer from the options given
below.
(1) Statement-I is correct and Statement-II is
incorrect.
(2) Statement-I is incorrect but Statement-II is
correct.
(3) Both Statement-I and Statement-II are
incorrect.
(4) Both Statement-I and Statement-II are
correct.
Ans. (1)
Sol.
0
eV hv
0
h
Vv
ee

M2 material has higher work function, so
statement-(II) is incorrect.
Option (1) is correct.
36. The angle between vector
Q
and the resultant of
(2Q 2P)
and
(2Q 2P)
is:
(1) 0°
(2)
1(2Q 2P)
tan 2Q 2P
(3)
1P
tan Q



(4)
12Q
tan P



Ans. (1)
Sol.
R (2Q 2P) (2Q 2P)
R 4Q
Angle between
Q
and
R
is zero
Option (1) is correct
37. In hydrogen like system the ratio of coulombian
force and gravitational force between an electron
and a proton is in the order of:
(1) 1039 (2) 1019
(3) 1029 (4) 1036
Ans. (1)
Sol.
9 19 19
12
e22
kQ Q 9 10 1.6 10 1.6 10
Frr


11 31 27
12
g22
Gm m 6.67 10 9.1 10 1.6 10
Frr

40 39
e
g
F0.23 10 2.3 10
F
Option (1)
38. In a co-axial straight cable, the central conductor
and the outer conductor carry equal currents in
opposite directions. The magnetic field is zero.
(1) inside the outer conductor
(2) in between the two conductors
(3) outside the cable
(4) inside the inner conductor
Ans. (3)
Sol.
I
r
I
0 enc
B.d i 0
B = 0 outside the cable
39. An electron rotates in a circle around a nucleus
having positive charge Ze. Correct relation
between total energy (E) of electron to its potential
energy (U) is:
(1) E = 2U (2) 2E = 3U
(3) E = U (4) 2E = U
Ans. (4)
Sol.
2
2
k(Ze)(e) mv
Frr

2
1 1 K(Ze)(e)
KE mv
2 2 r

K(Ze)(e)
PE r

K(Ze)(e) K(Ze)(e) K(Ze)(e)
TE 2r r 2r

PE
TE 2
2TE = PE
Option (4)
40. If the collision frequency of hydrogen molecules in
a closed chamber at 27°C is Z, then the collision
frequency of the same system at 127° C is :
(1)
3Z
2
(2)
4Z
3
(3)
2Z
3
(4)
3Z
4
Ans. (3)
Sol. Assuming mean free path constant.
f v T
11
22
fT300
f T 400

21
42
f f Z
33
41. Ratio of radius of gyration of a hollow sphere to
that of a solid cylinder of equal mass, for moment
of Inertia about their diameter axis AB as shown in
figure is
8
x
. The value of x is:
A
B
R
M
2R
4R
B
A
Diameter
(1) 34 (2) 17
(3) 67 (4) 51
Ans. (3)
Sol.
22
sphere 1
2
I MR Mk
3

2 2 2
cylinder
11
I M(4R ) MR M(2R)
12 4
22
2
67 MR Mk
12

1
2
k2 12 8
.
k 3 67 67

42. Two conducting circular loops A and B are placed
in the same plane with their centres coinciding as
shown in figure. The mutual inductance between
them is:
O
A
a
B
b
b >> a
(1)
2
0a
2b

(2)
2
0b
.
2a
(3)
2
0b
2a

(4)
2
0a
.
2b
Ans. (1)
Sol. = Mi = BA
2
0i
Mi a
2b
2
0a
M2b


43. Match list-I with list-II:
List-I
List-II
(A)
Kinetic energy of planet
(I)
GMm
a
(B)
Gravitation Potential
energy of Sun-planet
system.
(II)
GMm
2a
(C)
Total mechanical energy
of planet
(III)
Gm
r
(D)
Escape energy at the
surface of planet for unit
mass object
(IV)
GMm
2a
(Where a = radius of planet orbit, r = radius of
planet, M = mass of Sun, m = mass of planet)
Choose the correct answer from the options given
below:
(1) (A) – II, (B) – I, (C) – IV, (D) – III
(2) (A) – III, (B) – IV, (C) – I, (D) – II
(3) (A) – I, (B) – IV, (C) – II, (D) – III
(4) (A) – I, (B) – II, (C) – III, (D) – IV
Ans. (1)
Sol.
2
1 GMm
KE mv
2 2a

PE = –2KE
TE = –KE
44. A wooden block of mass 5kg rests on soft
horizontal floor. When an iron cylinder of mass 25
kg is placed on the top of the block, the floor yields
and the block and the cylinder together go down
with an acceleration of 0.1 ms–2. The action force
of the system on the floor is equal to:
(1) 297 N (2) 294 N
(3) 291 N (4) 196 N
Ans. (3)
Sol. Taking g = 9.8 m/s2
0.1 m/s2
25 kg
5 kg
30 × 9.8= 294
N
294 – N = 30 × 0.1
N = 291
45. A simple pendulum doing small oscillations at a
place R height above earth surface has time period
of T1 = 4 s. T2 would be it's time period if it is
brought to a point which is at a height 2R from
earth surface. Choose the correct relation [R =
radius of Earth]:
(1) T1 = T2 (2) 2T1 = 3T2
(3) 3T1 = 2T2 (4) 2T1 = T2
Ans. (3)
Sol.
2
1
T 2 (2R)
GM

2
2
T 2 (3R)
GM

1
2
T2
T3

46. A body of mass 50 kg is lifted to a height of 20 m
from the ground in the two different ways as
shown in the figures. The ratio of work done
against the gravity in both the respective cases,
will be:
Case-1: Pulled straight up
M=50kg
h =20m
30°
Case-2: Along the ramp
h =20m
30°
M=50k
(1) 1 : 1 (2) 2 : 1
(3)
3 : 2
(4) 1 : 2
Ans. (1)
Sol. Work done by gravity is independent of path. It
depends only on vertical displacement so work
done in both cases will be same.
Option (1) is correct
47. Time periods of oscillation of the same simple
pendulum measured using four different measuring
clocks were recorded as 4.62 s, 4.632 s, 4.6 s and
4.64 s. The arithmetic mean of these reading in
correct significant figure is.
(1) 4.623 s (2) 4.62 s
(3) 4.6 s (4) 5 s
Ans. (3)
Sol. Sum of number by considering significant digit
sum = 4.6 + 4.6 + 4.6 + 4.6 = 18.4
Arithmetic Mean =
sum 18.4
44
= 4.6
48. The heat absorbed by a system in going through
the given cyclic process is :
P(in cc)
V
(in kPa)
340
60
0
60
340
(1) 61.6 J (2) 431.2 J
(3) 616 J (4) 19.6 J
Ans. (1)
Sol. U = 0 (Cyclic process)
Q = W = area of P-V curve.
= × (140 ×103 Pa) × (140 ×10–6 m3)
Q = 61.6 J
49. In the given figure R1 = 10, R2 = 8, R3 = 4
and R4 = 8. Battery is ideal with emf 12V.
Equivalent resistant of the circuit and current
supplied by battery are respectively.
12 V
R1
R2
R4
R3
+
–
(1) 12 and 11.4 A (2) 10.5 and 1.14 A
(3) 10.5 and 1 A (4) 12 and 1 A
Ans. (4)
Sol. Here R2, R3, R4 are in parallel
234 2 3 4
1 1 1 1
R R R R
R234 = 2
R234 is in series with R1 so
Req = R234 + R1 = 2 + 10 = 12
12
i12
= 1Amp
50. An alternating voltage of amplitude 40 V and
frequency 4 kHz is applied directly across the
capacitor of 12 µF. The maximum displacement
current between the plates of the capacitor is
nearly:
(1) 13 A (2) 8 A
(3) 10 A (4) 12 A
Ans. (4)
Sol. Displacement current is same as conduction
current in capacitor.
C
11
XC 2 fC


36
13.317
2 4 10 12 10
C
V 40
I 12A
X 3.317
SECTION-B
51. In Young's double slit experiment, carried out with
light of wavelength 5000Å, the distance between
the slits is 0.3 mm and the screen is at 200 cm from
the slits. The central maximum is at x = 0 cm. The
value of x for third maxima is ............. mm.
Ans. (10)
Sol.
73
4
D 5 10 2 10 10
d 3 10 3

m
For 3rd maxima y3 = 310 ×10–3 m = 10 mm
52. A 2A current carrying straight metal wire of
resistance 1 , resistivity 2 × 10–6 m, area of
cross-section 10 mm2 and mass 500 g is suspended
horizontally in mid air by applying a uniform
magnetic field
B
. The magnitude of B is .............
× 10–1 T (given, g = 10 m/s2)
Ans. (5)
Sol.
6
5
2 10
R1
A 10
5
mg = Bi
mg 5
B 0.5
i 2 5
= 5 ×10–1 Tesla
53. The electric field between the two parallel plates of
a capacitor of 1.5 µF capacitance drops to one third
of its initial value in 6.6 µs when the plates are
connected by a thin wire. The resistance of this
wire is .............. . (Given, log 3 = 1.1)
Ans. (4)
Sol.
00
EV
EV
33
t
0
0
VVe
3
t n 3
6.6 × 10–6 = R (1.5 × 10–6)(1.1)
6
R4
1.5
54. Three blocks M1, M2, M3 having masses 4 kg, 6 kg
and 10 kg respectively are hanging from a smooth
pully using rope 1, 2 and 3 as shown in figure. The
tension in the rope 1, T1 when they are moving
upward with acceleration of 2ms–2 is ............... N
(if g = 10 m/s2)
1
M1
M2
M3
2
3
2m/s2
T1
T1
4g
6g
10g
T2
T2
T3
T3
6kg
10kg
Ans. (240)
Sol. FBD of M1 :
2 m/s2
20
T1
200
T1 – 200 = (4 + 6 + 10) × 2
T1 = 240
55. The density and breaking stress of a wire are 6 ×
104 kg /m3 and 1.2 × 108 N/m2 respectively. The
wire is suspended from a rigid support on a planet
where acceleration due to gravity is
rd
1
3
of the
value on the surface of earth. The maximum length
of the wire with breaking is ............ m (take, g =
10 m/s2)
Ans. (600)
Sol.
T
mg
T = mg
T mg
AA
( A )g
A
8
4
1.2 10 3 600
g 6 10 10
56. A body moves on a frictionless plane starting from
rest. If Sn is distance moved between t = n – 1 and t
= n and Sn – 1 is distance moved between t = n –2
and t = n – 1, then the ratio
n1
n
S
S
is
2
1x



for n
= 10. The value of x is ............
Ans. (19)
Sol.
n
1 19a
S a(2n 1)
22
n1
1 17a
S a(2n 3)
22
n1
n
S17 2
1 x 19
S 19 x
57. If three helium nuclei combine to form a carbon
nucleus then the energy released in this reaction is
.......... × 10–2 MeV. (Given 1 u = 931 MeV/c2,
atomic mass of helium = 4.002603 u)
Ans. (727)
Sol. Reaction :
4 12
26
3 He C rays
Mass defect = m = (3mHe – mC)
= (3 × 4.002603 – 12) = 0.007809 u
Energy released
= 931 m MeV
= 7.27 MeV = 727 × 10–2 MeV
58. An ac source is connected in given series LCR
circuit. The rms potential difference across the
capacitor of 20 µF is .............. V.
R= 300
L = 1H
~
V = sin 100t volt
C = 20 µF
Ans. (50)
Sol. XL = L = 100 × 1 = 100
C6
11
X 500
C 100 20 10
22
LC
Z (X X ) R
22
(100 500) 300
Z = 500
rms
rms
V50
i 0.1A
Z 500
rms voltage across capacitor
Vrms = XC irms
=500 × 0.1 = 50V
59. In the experiment to determine the galvanometer
resistance by half-deflection method, the plot of
1
vs the resistance (R) of the resistance box is shown
in the figure. The figure of merit of the
galvanometer is .............. ×10–1 A/division. [The
source has emf 2V]
R()
0
2
4
6
1/2
2
1
Ans. (5)
Sol. i = K
2K
GR

1 (G R)K K KG
R
2 2 2



Slope =
1
K1 K 0.5 5 10 A
24
60. Three capacitors of capacitances 25 µF, 30 µF and
45 µF are connected in parallel to a supply of 100
V. Energy stored in the above combination is E.
When these capacitors are connected in series to
the same supply, the stored energy is
9E
x
. The
value of x is .................
Ans. (86)
Sol. In parallel combination : Potential difference is
same across all
Energy =
2
1 2 3
1(C C C )V
2
26
1(25 30 45) (100) 10
2
= 0.5 = E
In series combination: Charge is same on all.
equ 1 2 3
1 1 1 1 1 1 1
C C C C 25 30 45
equ
1 (18 15 10) 43
C 450 450


equ
450
C43

Energy =
2 2 2
1 2 3
Q Q Q
2C 2C 2C

2
1 2 3
Q 1 1 1
2 C C C



22
equ equ
equ
(V C ) V C
1
2 C 2

2
6
(100) 450 10
2 43

4.5 9 9
E 0.5
86 x x
x = 86
CHEMISTRY
TEST PAPER WITH SOLUTION
SECTION-A
61. The incorrect postulates of the Dalton's atomic
theory are :
(A) Atoms of different elements differ in mass.
(B) Matter consists of divisible atoms.
(C) Compounds are formed when atoms of
different element combine in a fixed ratio.
(D) All the atoms of given element have different
properties including mass.
(E) Chemical reactions involve reorganisation of
atoms.
Choose the correct answer from the options given
below :
(1) (B), (D), (E) only
(2) (A), (B), (D) only
(3) (C), (D), (E) only
(4) (B), (D) only
Ans. (4)
Sol. B, D
62. The following reaction occurs in the Blast furnance
where iron ore is reduced to iron metal
2 (g)
3s
Fe O 3CO
2(g)
l
Fe 3CO
Using the Le-chatelier's principle, predict which
one of the following will not disturb the
equilibrium.
(1) Addition of Fe2O3
(2) Addition of CO2
(3) Removal of CO
(4) Removal of CO2
Ans. (1)
Sol. When solid added no effect on equilibrium.
63. Identify compound (Z) in the following reaction
sequence.
Cl
+ NaOH
623 K
300 atm
X
HCl
Conc. HNO3
Y
Z
(1)
OH
NO2
(2)
OH
NO2
NO2
(3)
OH
NO2
O2N
NO2
(4)
OH
NO2
Ans. (3)
Sol.
Cl
+ NaOH
ONa
(X)
OH
(Y)
OH
NO2
NO2
NO2
(Z)
64. Given below are two statements : One is labelled
as Assertion (A) and the other is labelled as
Reason (R)
Assertion (A): Enthalpy of neutralisation of strong
monobasic acid with strong monoacidic base is
always –57 kJ mol–1
Reason (R): Enthalpy of neutralisation is the
amount of heat liberated when one mole of H+ ions
furnished by acid combine with one mole of –OH
ions furnished by base to form one mole of water.
In the light of the above statements, choose the
correct answer from the options given below.
(1) (A) is true but (R) is false
(2) Both (A) and (R) are true and (R) is the
correct explanation of (A)
(3) (A) is false but (R) is true
(4) Both (A) and (R) are true but (R) is not the
correct explanation of (A)
Ans. (2)
Sol. Enthalpy of neutralization of SA & SB is always
–57 kJ / mol because strong monoacid gives one
mole of H+ and strong mono base gives one mole
of OH– which form one mole of water.
65. The statement(s) that are correct about the species
O2– , F –, Na+ and Mg2+.
(A) All are isoelectronic
(B) All have the same nuclear charge
(C) O2– has the largest ionic radii
(D) Mg2+ has the smallest ionic radii
Choose the most appropriate answer from the
options given below :
(1) (B), (C) and (D) only
(2) (A), (B), (C) and (D)
(3) (C) and (D) only
(4) (A), (C) and (D) only
Ans. (4)
Sol. O–2 F– Na+ Mg+2
(No. of e–) 10 10 10 10
(Ionic radius) O–2 > F– > Na+ > Mg+2
Zeff O–2 < F– < Na+ < Mg+2
66. For the compounds:
(A) H3C–CH2–O–CH2–CH2–CH3
(B) H3C–CH2–CH2–CH2–CH3
(C)
CH3–CH2–C–CH2–CH3
O
(D)
H
3
C–CH– CH
2
– CH
2
–CH3
OH
The increasing order of boiling point is :
Choose the correct answer from the options given
below :
(1) (A) < (B) < (C) < (D)
(2) (B) < (A) < (C) < (D)
(3) (D) < (C) < (A) < (B)
(4) (B) < (A) < (D) < (C)
Ans. (2)
Sol. Compounds having same number of carbon atoms
follow the boiling point order as:
(Boiling point)Hydrogen bonding >(Boiling point)high polarity >
(Boiling point)low polarity > (Boiling point)non polar
67. Given below are two statements :
Statement I: In group 13, the stability of +1
oxidation state increases down the group.
Statement II: The atomic size of gallium is greater
than that of aluminium.
In the light of the above statements, choose the
most appropriate answer from the options given
below:
(1) Statement I is incorrect but Statement II is
correct
(2) Both Statement I and Statement II are correct
(3) Both Statement I and Statement II are
incorrect
(4) Statement I is correct but Statement II is
incorrect
Ans. (4)
Sol. Statement I : Number of d & f electrons, increases
down the group and due to poor shielding of d & f
e–, stability of lower oxidation states increases
down the group
Statement II : The atomic size of aluminium is
greater than that of gallium.
68. Number of and bonds present in ethylene
molecule is respectively :
(1) 3 and 1 (2) 5 and 2
(3) 4 and 1 (4) 5 and 1
Ans. (4)
Sol. ethylene is
C = C
H
H
H
H
, it has 5 bonds and
1 bond.
69. Identify 'A' in the following reaction :
O
(i) N2H4
'A'
C
CH3
CH3
(ii) ethylene glycol / KOH
(1)
OH
CH3
CH3
(2)
CH3
CH3
(3)
C=N–NH2
CH3
H5C2
(4)
C=N–NH2
CH3
CH3
Ans. (2)
Sol.
O
CH3
CH3
24
(i) N H
(ii)ethylene glycol/KOH

CH3
CH3
– Wolf kishner reduction.
70. The reaction at cathode in the cells commonly used
in clocks involves.
(1) reduction of Mn from +4 to +3
(2) oxidation of Mn from +3 to +4
(3) reduction of Mn from + 7 to +2
(4) oxidation of Mn from + 2 to +7
Ans. (1)
Sol. In the cathode reaction manganese (Mn) is reduced
from the +4 oxidation state to the +3 state.
71. Which one of the following complexes will exhibit
the least paramagnetic behaviour ?
[Atomic number, Cr = 24, Mn = 25, Fe = 26, Co = 27]
(1) [Co(H2O)6]2+ (2) [Fe(H2O)6]2+
(3) [Mn(H2O)6]2+ (4) [Cr(H2O)6]2+
Ans. (1)
Sol.
Number of
unpaired e–
n(n 2) B.M.
[Co(H2O)6]2+
3
3.87
[Fe(H2O)6]2+
4
4.89
[Mn(H2O)6]2+
5
5.92
[Cr(H2O)6]2+
4
4.89
Least paramagnetic behaviour = [Co(H2O)6]2+
72. Given below are two statements : one is labelled
as Assertion (A) and the other is labelled as
Reason (R).
Assertion (A): Cis form of alkene is found to be
more polar than the trans form
Reason (R): Dipole moment of trans isomer of
2-butene is zero.
In the light of the above statements, choose the
correct answer from the options given below :
(1) Both (A) and (R) are true but (R) is NOT the
correct explanation of (A)
(2) (A) is true but (R) is false
(3) Both (A) and (R) are true and (R) is the
correct explanation of (A)
(4) (A) is false but (R) is true
Ans. (3)
Sol. Dipole moment is a vector quantity and for
compound net dipole moment is the vector sum of
all dipoles hence dipole moment of cis form is
greater than trans form.
:
C = C
CH3
H
CH3
H
Cis
( > 0)
>
C = C
H
CH3
CH3
H
trans
( = 0)
73. Given below are two statements :
Statement I: Nitration of benzene involves the
following step –
H – O – NO2
H
•
•
H2O + NO2
Statement II: Use of Lewis base promotes the
electrophilic substitution of benzene.
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. (2)
Sol. In nitration of benzene concentrated H2SO4 and
HNO3 is used as reagent which generates
electrophile
2
NO
in following steps:
H2SO4 + HNO3
H
Lewis acids can promote the formation of
electrophiles not Lewis base
74. The correct order of ligands arranged in increasing
field strength.
(1) Cl– < –OH < Br– < CN–
(2) F– < Br– < I– < NH3
(3) Br– < F– < H2O < NH3
(4) H2O < –OH < CN– < NH3
Ans. (3)
Sol. Experimental order Br– < F– < H2O < NH3
75. Which of the following gives a positive test with
ninhydrin ?
(1) Cellulose (2) Starch
(3) Polyvinyl chloride (4) Egg albumin
Ans. (4)
Sol. Ninhydrin test is a test of amino acids. Egg
albumin contains protein which is a natural
polymer of amino acids which will show positive
ninhydrin test
76. The metal that shows highest and maximum
number of oxidation state is:
(1) Fe (2) Mn
(3) Ti (4) Co
Ans. (2)
Sol. Mn shows highest oxidation state (Mn+7) in 3d
series metals.
77. Ail organic compound has 42.1% carbon, 6.4%
hydrogen and remainder is oxygen. If its molecular
weight is 342, then its molecular formula is :
(1) C11H18O12 (2) C12H20O12
(3) C14H20O10 (4)
C12H22O11
Ans. (4)
Sol. only C12H22O11 has 42.1% carbon, 6.4% hydrogen
& 51.5 percent oxygen.
78. Given below are two statement :
Statement I : Bromination of phenol in solvent
with low polarity such as CHCl3 or CS2 requires
Lewis acid catalyst.
Statement II : The lewis acid catalyst polarises the
bromine to generate Br+.
In the light of the above statements, choose the
correct answer from the options given below :
(1) Statement I is true but Statement II is false.
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false.
(4) Statement I is false but Statement II is true.
Ans. (4)
Sol. Phenol is a highly activated compound which can
undergo bromination directly with Bromine
without any lewis acid.
79. Molar ionic conductivities of divalent cation and
anion are 57 S cm2 mol–1 and 73 S cm2 mol–1
respectively. The molar conductivity of solution of
an electrolyte with the above cation and anion will
be :
(1) 65 S cm2 mol–1 (2) 130 S cm2 mol–1
(3) 187 S cm2 mol–1 (4) 260 S cm2 mol–1
Ans. (2)
Sol.
2 2 –1
C57Scm mol

2 2 –1
A73Scm mol

22
Solution C A

= 57 + 73 = 130
80. The number of neutrons present in the more abundant
isotope of boron is 'x'. Amorphous boron upon
heating with air forms a product, in which the
oxidation state of boron is 'y'. The value of x + y is …
(1) 4 (2) 6
(3) 3 (4) 9
Ans. (4)
Sol. More abundant isotope = B11
[Number of neutrons = 6]
x = 6
B + O2 B2O3
Oxidation state of B in B2O3 = +3
So, y = 3
Hence x + y = 9
SECTION-B
81. The value of Rydberg constant (RH) is 2.18×10–18 J.
The velocity of electron having mass 9.1×10–31 kg
in Bohr's first orbit of hydrogen atom
= ……… ×105 ms–1 (nearest integer)
Ans. (22)
Sol. V = 2.18 × 106 ×
Z
n
= 21.8 × 105 ×
1
1
22 × 105 (nearest)
82.
A
B
In a borax bead test under hot condition, a metal
salt (one from the given) is heated at point B of the
flame, resulted in green colour salt bead. The
spin-only magnetic moment value of the salt is
…………. BM (Nearest integer)
[Given atomic number of Cu = 29, Ni = 28,
Mn = 25, Fe = 26]
Ans. (6)
Sol. Fe+3 will give green coloured bead when heated at
point B.
Number of unpaired e– in Fe+3 = 5
= 5.92
Nearest integer = 6
83. The heat of combustion of solid benzoic acid at
constant volume is –321.30 kJ at 27°C. The heat of
combustion at constant pressure is (–321.30 – xR)
kJ, the value of x is …………….
Ans. (150)
Sol. C6H5COOH(S) +
15
2
O2(g) 7CO2(g) + 3H2O()
H = U + ngRT
1R
321.30 300
2 100
= (–321.30 – 150R) kJ
84. Consider the given chemical reaction sequence :
OH
Conc. H2SO4
Product A
Conc. HNO3
Product B
Total sum of oxygen atoms in Product A and
Product B are ………
Ans. (14)
Sol. Picric acid is prepared by treating phenol first with
concentrated sulphuric acid which converts it to
phenol-2,4-disulphonic acid and then with
concentrated nitric acid to get 2, 4, 6 trinitrophenol.
85. The spin only magnetic moment value of the ion
among Ti2+, V2+, Co3+ and Cr2+, that acts as
strong oxidising agent in aqueous solution is
……….. BM (Near integer).
(Given atomic numbers : Ti : 22, V : 23, Cr : 24,
Co : 27)
Ans. (5)
Sol. Strong oxidising agent = Co+3
No. of unpaired e– in Co+3[3d6] = 4
Hence
n(n 2) 24 BM
Nearest integer = 5
86. During Kinetic study of reaction 2A + B C + D,
the following results were obtained :
A[M]
B[M]
initial rate of
formation of D
I
0.1
0.1
6.0 × 10–3
II
0.3
0.2
7.2 × 10–2
III
0.3
0.4
2.88 × 10–1
IV
0.4
0.1
2.40 × 10–2
Based on above data, overall order of the reaction
is ……….
Ans. (3)
Sol. r = K[A]x[B]y
(I) 6 × 10–3 = K[0.1]x[0.1]y
(IV) 2.4 × 10–2 = K[0.4]x[0.1]y
(IV)/(I)
4 = (4)x
x = 1
r = K[A]x[B]y
(III) 2.88 × 10–1 = K[0.3]x[0.4]y
(II) 7.2 × 10–2 = K[0.3]x[0.2]y
(III)/(II)
4 = 2y
y = 2
Overall order = x + y = 1 + 2 = 3
87. An artificial cell is made by encapsulating 0.2 M
glucose solution within a semipermeable
membrane. The osmotic pressure developed when
the artificial cell is placed within a 0.05 M solution
of NaCl at 300 K is _______ × 10–1 bar.
(Nearest Integer)
[Given : R = 0.083 L bar mol–1 K–1]
Assume complete dissociation of NaCl
Ans. (25)
Sol.
0.05 M
NaCl Sol.
0.2 M
Glucose
NaCl  Na+ + Cl–
0.05M 0.05M 0.05M
Total C1 = 0.05 + 0.05 = 0.1 M (NaCl)
C2 = 0.2 M (glucose)
= (C2 – C1) RT
= (0.2 – 0.1) × 0.083 × 300
= 2.49 bar
= 24.9 × 10–1 bar
88. The number of halobenzenes from the following
that can be prepared by Sandmeyer's reaction
is .......
F
I
Cl
II
Br
III
I
IV
At
V
Ans. (2)
Sol. In Sandmayer reaction only bromobenzene &
chlorobenzene are prepared
89. In the lewis dot structure for NO2
–, total number of
valence electrons around nitrogen is …….
Ans. (8)
Sol.
N
O
O–
Number of valence e– around N-atom = 8
90. 9.3 g of pure aniline is treated with bromine water
at room temperature to give a white precipitate of
the product 'P'. The mass of product 'P' obtained is
26.4 g. The percentage yield is ……… %.
Ans. (80)
Sol.
NH2
NH2
Br
Br
Br
(white ppt)
93 g of aniline produces 330 g of 2, 4, 6-
tribromoaniline. Hence 9.3 g of aniline should
produce 33g of 2, 4, 6-tribromoaniline. Hence
percentage yield
26.4 100 80%
33