FINAL JEE–MAIN EXAMINATION – APRIL, 2024
(Held On Saturday 06th April, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS
TEST PAPER WITH SOLUTION
SECTION-A
1. If
31
x sin , x 0
f(x) = ,then
x
0 , x 0



(1)
f ''(0) 1
(2)
2
2 24 –
f '' 2




(3)
2
2 12 –
f '' 2




(4)
f ''(0) 0
Ans. (2)
Sol. f'(x) =
1
sin
1 1 1 x
f "(x) 6x sin 3cos cos
x x x x



2
2 12 24
f" 22



2. If A(3,1,–1),
5 7 1
B , ,
3 3 3



, C(2,2,1) and
10 2 –1
D , ,
3 3 3



are the vertices of a quadrilateral
ABCD, then its area is
(1)
42
3
(2)
52
3
(3)
22
(4)
22
3
Ans. (1)
Sol.
D
C
B
A
Area =
1BD AC
2
552
ˆˆ ˆ
BD i j k
3 3 3
ˆˆ ˆ
AC i j 2k
3.
/4 22
2
33
0
cos xsin x dx
cos x sin x
is equal to
(1) 1/12 (2) 1/9
(3) 1/6 (4) 1/3
Ans. (3)
Sol. Divide Nr & Dr by cosx
/4 22
2
3
0
tan x sec xdx dx
1 tan x
Let 1 + tan3x = t
tan2x sec2x dx =
dt
3
2
2
1
1 dt 1
36
t
4. The mean and standard deviation of 20 observations
are found to be 10 and 2, respectively. On
respectively, it was found that an observation by
mistake was taken 8 instead of 12. The correct
standard deviation is
(1)
3.86
(2) 1.8
(3)
3.96
(4) 1.94
Ans. (3)
Sol. Mean
(x) 10
i
x10
20

xi = 10×20 = 200
If 8 is replaced by 12, then xi = 200 – 8 + 12 = 204
Correct mean
i
x
(x) 20
204 10.2
20

Standard deviation = 2
Variance = (S.D.)2 = 22 = 4
2
2
ii
xx
4
20 20






22
i
x10 4
20

2
i
x104
20
xi
2 = 2080
Now, replaced '8' observations by '12'
Then,
2 2 2
i
x 2080 8 12 2160
Variance of removing observations
2
2
ii
xx
20 20






2
2160 10.2
20
108 – 104.04
3.96
Correct standard deviation
=
3.96
5. The function
2
2
x 2x – 15
f(x) x – 4x 9
, xR is
(1) both one-one and onto.
(2) onto but not one-one.
(3) neither one-one nor onto.
(4) one-one but not onto.
NTA Ans. (3)
Ans. Bonus
Sol.
2
(x 5)(x 3)
f(x) x 4x 9


Let g(x) = x2 – 4x + 9
D < 0
g(x) > 0 for x R
x
f( 5) 0
f(3) 0

So, f(x) is many-one.
again,
yx2 – 4xy + 9y = x2 + 2x – 15
x2 (y – 1) – 2x(2y + 1) + (9y + 15) = 0
for x R D 0
D = 4(2y + 1)2 – 4(y –1) (9y + 15) 0
5y2 + 2y + 16 0
(5y – 8) (y + 2)  0
–2
8/5
8
y –2, 5



range
Note : If function is defined from f : R R then
only correct answer is option (3)
Bonus
6. Let A = {n [100, 700] N : n is neither a
multiple of 3 nor a multiple of 4}. Then the
number of elements in A is
(1) 300 (2) 280
(3) 310 (4) 290
Ans. (1)
Sol. n(3) multiple of 3
102, 105, 108, ..... , 699
Tn= 699 = 102 + (n – 1)(3)
n = 200
n(3) = 200
n(4) multiple of 4
100, 104, 108, ...., 700
Tn = 700 = 100 + (n – 1) (4)
n = 151
n(4) = 151
n(34) multiple of 3 & 4 both
108, 120, 132, ....., 696
Tn = 696 = 108 + (n – 1)(12)
n = 50
n(3 4) = 50
n(3 4) = n(3) + n(4) – n(3 4)
= 200 + 151 – 50
= 301
n 3 4
= Total – n(3 4) = neither a multiple
of 3 nor a multiple of 4
= 601 – 301 = 300
7. Let C be the circle of minimum area touching the
parabola y = 6 – x2 and the lines
y = 3 x
. Then,
which one of the following points lies on the circle
C ?
(1) (2, 4) (2) (1, 2)
(3) (2, 2) (4) (1, 1)
Ans. (1)
Sol.
(0,6)
(0,6–r)
Equation of circle
x2 + (y – (6 – r))2 = r2
touches
3 x y 0
p = r
0 (6 r) r
2

|r – 6| = 2r
r = 2
Circle x2 + (y – 4)2 = 4
(2, 4) Satisfies this equation
8. For , R and a natural number n, let
2
2
r
n
r1
2
A 2r 2 n –
n(3n – 1)
3r – 2 3 2


. Then 2A10 – A8 is
(1) 4 + 2 (2) 2 + 4
(3) 2n (4) 0
Ans. (1)
Sol.
2
2
r
n
r1
2
A 2r 2 n –
n(3n – 1)
3r – 2 3 2


2
2
10 8
n
20 1 2
2A – A 40 2 n –
n(3n – 1)
56 3 2


2
2
n
81 2
– 16 2 n –
n 3n – 1
22 3 2

2
2
n
12 1 2
24 2 n –
n 3n – 1
34 3 2


2
2
n
01 2
0 2 n –
n(3n – 1)
–2 3 2


22
– 2 (n – ) – n 2
– 2(– – 2 )
42
9. The shortest distance between the lines
x – 3 y 15 z – 9
2 –7 5

and
x 1 y – 1 z – 9
2 1 –3

is
(1)
63
(2)
43
(3)
53
(4)
83
Ans. (2)
Sol.
x – 3 y 15 z – 9
2 –7 5

&
x 1 y – 1 z – 9
2 1 –3

S.D =
2 1 1 2
12
a .a . b .b
bb
a1 = 3, –15, 9 b1 = 2, –7, 5
a2 = –1, 1, 9 b2 = 2, 1, –3
a2 – a1 = –4, 16, 0
12
ˆˆ ˆ
i j k
ˆˆ ˆ
b b 2 7 5 i(16) j( 16) k(16)
2 1 3
ˆˆˆ
16(i j k)
12
b b 16 3
2 1 1 2
a a . b b 16 4 16 (16)(12)
S.D. =
(16)(12) 43
16 3
10. A company has two plants A and B to manufacture
motorcycles. 60% motorcycles are manufactured
at plant A and the remaining are manufactured at
plant B. 80% of the motorcycles manufactured at
plant A are rated of the standard quality, while
90% of the motorcycles manufactured at plant B
are rated of the standard quality. A motorcycle
picked up randomly from the total production is
found to be of the standard quality. If p is the
probability that it was manufactured at plant B,
then 126p is
(1) 54 (2) 64
(3) 66 (4) 56
Ans. (1)
Sol.
A
B
Manufactured
60%
40%
Standard quality
80%
90%
P(Manufactured at B / found standard quality) = ?
A : Found S.Q
B : Manufacture B
C : Manufacture A
P(E1) =
40
100
P(E2) =
60
100
P(A/E1) =
90
100
P(A/E2) =
80
100
P(E1/A) =
11
1 1 2 2
P A / E P E
P A / E P E P A / E P E
3
7
126 P = 54
11. Let, , be the distinct roots of the equation
22
x – t – 5t 6 x 1 0,
t R and an = n
+ n
.
Then the minimum value of
2023 2025
2024
aa
a
is
(1) 1/4 (2) –1/2
(3) –1/4 (4) 1/2
Ans. (3)
Sol. by newton's theorem
an+2 – (t2 – 5t + 6)an+1 + an = 0
 a2025 + a2023 = (t2 – 5t + 6) a2024

2
2025 2023
2024
aa t 5t 6
a
2
21
5
t 5t 6 t4
2



 minimum value =
1
4
12. Let the relations R1 and R2 on the set
X = {1, 2, 3, ..., 20} be given by
R1 = {(x, y) : 2x – 3y = 2} and
R2 = {(x, y) : –5x + 4y = 0}. If M and N be the
minimum number of elements required to be added
in R1 and R2, respectively, in order to make the
relations symmetric, then M + N equals
(1) 8 (2) 16
(3) 12 (4) 10
Ans. (4)
Sol. x = {1, 2, 3, .......20}
R1 = {(x, y) : 2x – 3y = 2}
R2 = {(x, y) : –5x + 4y = 0}
R1 = {(4, 2), (7, 4), (10, 6), (13, 8), (16, 10), (19, 12)}
R2 = {(4, 5), (8, 10), (12, 15), (16, 20)}
in R1 6 element needed
in R2 4 element needed
So, total 6+4 = 10 element
13. Let a variable line of slope m > 0 passing through
the point (4, –9) intersect the coordinate axes at the
points A and B. the minimum value of the sum of
the distances of A and B from the origin is
(1) 25 (2) 30
(3) 15 (4) 10
Ans. (1)
Sol. equation of line is
y + 9 = m (x – 4)
9 4m
A,0
m



B = (0, – 9 – 4m)
OA + OB =
9 4m 9 4m
m

m0
9
13 4m
m
9
4m 9
m36 4m 12
2m
OA + OB 25
14. The interval in which the function f(x) = xx, x > 0,
is strictly increasing is
(1)
1
0, e


(2)
2
1,1
e


(3) (0, ) (4)
1,
e


Ans. (4)
Sol. f(x) = xx ; x > 0
ny = xnx
1 dy x nx
y dx x

x
dy x (1 nx)
dx 
for strictly increasing
dy 0
dx
xx (1 + nx) 0
nx –1
x e–1
x
1
e
1
x,
e



15. A circle in inscribed in an equilateral triangle of
side of length 12. If the area and perimeter of any
square inscribed in this circle are m and n,
respectively, then m + n2 is equal to
(1) 396 (2) 408
(3) 312 (4) 414
Ans. (2)
Sol. r =
s
=
2
3a a 12 23
3a 2 3 2 3
4. 2

A
A =
r2
=
26
Area = m = A2 = 24
Perimeter = n = 4A =
86
m + n2 = 24 + 384
= 408
16. The number of triangles whose vertices are at the
vertices of a regular octagon but none of whose
sides is a side of the octagon is
(1) 24 (2) 56
(3) 16 (4) 48
Ans. (3)
Sol. no. of triangles having no side common with a n
sided polygon =
n n 4
12
C . C
3
84
12
C . C 16
3

17. Let y = y(x) be the solution of the differential
equation
–1
2 tan x
dy
1 x y e ,
dx
y(1) = 0. Then
y(0) is
(1)
/2
1e – 1
4
(2)
/2
11 – e
2
(3)
/2
11 – e
4
(4)
/2
1e – 1
2
Ans. (2)
Sol.
1
tan x
22
dy y e
dx 1 x 1 x


I.F. =
1
2
1dx tan x
1x
ee
1
11
tan x
tan x tan x
2
e
y.e e .dx
1x





Let tan–1x = z
2
dx dz
1x
y.ez =
2z
2z e
e dz C
2

1
12 tan x
tan x e
y.e C
2

1
1
tan x
tan x
eC
y2e
y(1) = 0
/4
/4
eC
02e
/2
e
C2

1
1
tan x /2
tan x
ee
y22e

/2
1e
y0 2

18. Let y = y(x) be the solution of the differential
equation
ee
dy 3
2x log x 2y log x,
dx x

x > 0 and
y(e–1) = 0. Then, y(e) is equal to
(1)
3
–2e
(2)
2
–3e
(3)
3
–e
(4)
2
–e
Ans. (3)
Sol.
2
dy y 3
dx x n x 2x

I.F. =
1dx
x n x
e
= en(n(x)) = nx
ynx =
2
3 n x dx
2x
22
3 n x 3
x dx . x dx dx
2 2x




3 n x 1 3 1 dx
2 x 2x x
y.nx =
3 nx 3 C
2x 2x
y(e–1) = 0
0 (–1) =
3e 3e C
22

C = 0
y =
3 nx 3
2x 2x
3 3 3
y e –
2e 2e e

19. Let the area of the region enclosed by the curves
y = 3x, 2y = 27 – 3x and
y 3x – x x
be A. Then
10 A is equal to
(1) 184 (2) 154
(3) 172 (4) 162
Ans. (4)
Sol. y = 3x, 2y = 27 – 3x & y = 3x –
xx
y = 3x
(3,9)
x
(9,0)
(0,0)
y
2y = 27 – 3x
(4,4)
x=4
39
03
27 – 3x
A 3x – 3x – x x dx – 3x – x x dx
2





39
3/2 3/2
03
27 9x
A x dx – x dx
22

3 9 9
5/2 2 5/2
9
3
3
03
2x 27 9 x 2x
A x –
5 2 2 2 5

5/2 5/2 5/2
2 27 9 2
A 3 (6) – 72 9 – 3
5 2 4 5
5/2 5 5/2
2 2 2
A 3 81 – 162 3 – 3
5 5 5
486 81
A – 81
55

10A = 162
Ans. = 4
20. Let
f : (– , ) – {0} R
be a differentiable
function such that
2
a
1
f '(1) lim a f a




.
Then
–1 2
e
a
a(a 1) 1
lim tan a – 2 log a
2a




is equal
to
(1)
3
24
(2)
3
84
(3)
5
28
(4)
3
48
Ans. (3)
Sol. f : (–,) – {0} R
f (1) =
2
a
1
lim a f a




12
a
a a 1 1
lim tan a 2 n a
2a





21
2
a
1
112
a
lim a tan 1 n a
2a
a










f(x) =
1
2
(1 + x) tan–1(x) + 1 – 2x2 n(x)
f (x) =
1
2
1
2
1x tan x 4x n x 2x
1x

1
f 1 1 2
24



5
f1 28

Ans. (3)
SECTION-B
21. Let  = 45 ; , R. If x(, 1, 2) + y(1, , 2)
+ z(2, 3, ) = (0, 0, 0) for some x, y, z R, xyz
0, then 6 + 4 + is equal to_______
Ans. (55)
Sol.  = 45,  R
x(,1,2) + y(1,,2) + z(2,3,) = (0,0,0)
x, y, z R, xyz 0
x + y + 2z = 0
x + y + 3z = 0
2x + 2y + z = 0
xyz 0 non-trivial
12
1 3 0
22

( – 6) – 1( – 6) + 2(2 – 2) = 0
 – 6 – + 6 + 4 – 4 = 0
6 + 4 + = 55
22. Let a conic C pass through the point (4, –2) and
P(x, y), x 3, be any point on C. Let the slope of
the line touching the conic C only at a single point
P be half the slope of the line joining the points
P and (3, –5). If the focal distance of the point
(7, 1) on C is d, then 12d equals_______.
Ans. (75)
Sol. P(x, y) & x 3
Slope of line at P(x, y) will be
dy 1 y 5
dx 2 x 3



dy 1
2 dx
(y 5) (x 3)

2n(y + 5) = n(x – 3) + C
Passes through (4, –2)
2n(3) = n(1) + C
C = 2n(3)
2n(y + 5) = n(x – 3) + 2n(3)
y5
2 n n(x 3)
3






2
y5 (x 3)
3




(y + 5)2 = 9(x – 3)
Parabola
4a = 9
9
a4


(3,–5)
(7,11)
d
2
2
7
d6
4




625
d4
25
d4
12d = 75
23. Let
k
17
0
kk1
17
0
1 – x dx
r , k N
1 – x dx

. Then the value of
10
k1 k
1
7 r – 1
is equal to _______.
Ans. (65)
Sol.
7K
K
I 1.(1 x ) dx
1
1
7 K 7 K 1 6
K00
I (1 x ) x 7K (1 x ) x .xdx
1
7 K 1 7
K
0
I 7K (1 x ) ((1 x ) 1)dx
IK = –7K IK + 7K IK–1
K
K1
7K 8
I 7K 7
K
7K 8
r7K 7
K
1
r1
7(K 1)

K
1
7(r 1) K1

10
K1
(K 1) 11(6) 1 65
24. Let x1, x2, x3, x4 be the solution of the equation
4x4 + 8x3 – 17x2 – 12x + 9 = 0 and
2 2 2 2
1 2 3 4
125
4 x 4 x 4 x 4 x m
16
.
Then the value of m is ________.
Ans. (221)
Sol. 4x4 + 8x3 – 17x2 – 12x + 9
= 4(x – x1) (x – x2) (x – x3) (x – x4)
Put x = 2i & –2i
64 – 64i + 68 – 24i + 9 = (2i – x1) (2i – x2) (2i – x3)
(2i – x4)
= 141 – 88i ......(1)
64 + 64i + 68 + 24i + 9 = 4(–2i – x1) (–2i – x2) (–2i
– x3) (–2i – x4)
= 141 + 88i .......(2)
22
125 141 88
m
16 16
m = 221
25. Let L1, L2 be the lines passing through the point
P(0, 1) and touching the parabola
9x2 + 12x + 18y – 14 = 0. Let Q and R be the
points on the lines L1 and L2 such that the PQR
is an isosceles triangle with base QR. If the slopes
of the lines QR are m1 and m2. then
22
12
16 m m
is equal to _______.
Ans. (68)
Sol. 9x2 + 12x + 4 = – 18(y – 1)
(3x + 2)2 = –18(y – 1)
2
22(y 1)
x3



R
Q
P(0,1)
(0, 1)
y = mx + 1
2
22(y 1)
x3



(3x + 2)2 = –18mx
9x2 + (12 + 18m)x + 4 = 0
4(6 + 9m)2 = 4(36)
6 + 9m = 6, – 6
m = 0,
4
3
R
Q
P
4
tan 3
2
2 tan 4
2
3
1 tan 2
tan – 2 2 tan 1 0
22


–1
tan 2,
22
QR
m tan 90 2




– cot 2
1
–1
m2
2
–1
m2
–1 / 2

22
12
1
16 m m 16 4
4



= 4 + 64 = 68
26. If the second, third and fourth terms in the
expansion of (x + y)n are 135, 30 and
10
3
,
respectively, then
32
6 n x y
is equal to
_____.
Ans. (806)
Sol.
n n–1
1
C x y 135
....(i)
n n–2 2
2
C x y 30
....(ii)
n n –3 3
3
10
C x y 3
....(iii)
By
i
ii
n
1
n
2
Cx9
y2
C
.....(iv)
By
ii
iii
n
2
n
3
Cx9
y
C
.....(v)
By
iv
v
nn
13
nn
22
CC 1
2
CC
2
2n n – 1 n – 2 n n – 1 n(n – 1)
6 2 2
4n – 8 = 3n – 3
n5
put in (v)
x9
y
x = 9y
put in (i)
54
1
x
C x 135
9



x5 = 27 × 9
x = 3,
1
y3
32
6 n x y
1
6 125 9 3



= 806
27. Let the first term of a series be T1 = 6 and its rth
term Tr = 3 Tr–1 + 6r, r = 2, 3, ....., n. If the sum of
the first n terms of this series is
2
1n – 12n 39
5
nn
4.6 – 5.3 1
. Then n is equal to ______.
Ans. (6)
Sol. Tr = 3Tr–1 + 6r , r = 2, 3, 4, … n
T2 = 3.T1 + 62
T2 = 3.6 + 62 …(1)
T3 = 3T2 + 63
T3 = 3T2 + 63
T3 = 3(3.6 + 62) + 63
T3 = 32.6 + 3.62 + 63 …(2)
Tr = 3r–1.6 + 3r–2.62 + … + 6r
2 r 1
r1
r
6 6 6
T 3 6 1 ...
3 3 3




Tr = 3r–1.6(1 + 2 + 22 + … + 2r–1)
r
r1
r
(1 2 )
T 6 3 1. ( 1)

Tr = 6.3r–1.(2r – 1)
r
r
r
63
T .(2 1)
3

Tr = 2.(6r – 3r)
rr
n
S 2 6 3
nn
n
6.(6 1) 3.(3 1)
S 2. 52





nn
n
12(6 1) 15(3 1)
S2 10



4n
n
3
S 4.6 5.3 1
5

n2 – 12n + 39 = 3
n2 – 12n + 36 = 0
n = 6
28. For n N, if cot–13 + cot–14 + cot–15 +
1
cot n 4
,
then n is equal to ________.
Ans. (47)
Sol. cot–13 + cot–14 + cot–15 +
1
cot n 4
1 1 1 1
1 1 1 1
tan tan tan tan
3 4 5 n 4
11
46 1
tan tan
48 n 4





11
23 1
tan tan
24 n 4





1 1 1
1 23
tan tan 1 tan
n 24

11
23
1
124
tan tan 23
n124







11
1
124
tan tan 47
n
24







11
11
tan tan
n 47

n = 47
29. Let P be the point (10, –2, –1) and Q be the foot of
the perpendicular drawn from the point R(1, 7, 6)
on the line passing through the points (2, –5, 11)
and (–6, 7, –5). Then the length of the line segment
PQ is equal to ________.
Ans. (13)
Sol.
P(10,–2,–1)
Q
R(1,7,6)
(2,–5,11)
(–6, 7,–5)
Line :
x 6 y 7 z 5
8 12 16


x 6 y 7 z 5
2 3 4
Q(2 – 6, 7 – 3, 4 – 5)
QR 2 7, 3 ,4 11
QR
· dr’s of line = 0
4 – 14 + 9 + 16 – 44 = 0
29 = 58 = 2
Q(–2, 1, 3)
PQ 144 9 16 169 13
30. Let
ˆˆ ˆ
a 2i – 3j 4k,
ˆˆˆ
b 3i 4j – 5k,
and a vector
c
be such that
ˆˆ ˆ
a b c b c i 8j 13k
.
If
a c 13
, then
(24 – b c)
is equal to ______.
Ans. (46)
Sol.
a b a c b c (1,8,13)
a a b a a c a b c
ˆˆ ˆ
a i 8j 13k
22 ˆˆ ˆ
a b a a b a c a a c a c b a b c a i 8j 13k

ˆˆ ˆ
26a 29b 13a 29c 13b 26c a i 8j 13k
ˆˆ ˆ
13a 16b 3c a i 8j 13k
2ˆˆ ˆ
13a b 16b 3b c a i 8j 13k b
2 3 4
13 26 16(50) 3b c 1 8 13
3 4 5

462 3b c 396

b c 22
Hence
24 – b c 46
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
31. To find the spring constant (k) of a spring
experimentally, a student commits 2% positive
error in the measurement of time and 1% negative
error in measurement of mass. The percentage
error in determining value of k is :
(1) 3% (2) 1%
(3) 4% (4) 5%
Ans. (4)
Sol.
m
T2 k

2m
Tk
2 T m k
% % %
T m k

k m 2 T
% % %
k m T

k% ( 1)% 2(2)% | 5% | 5%
k
32. A bullet of mass 50 g is fired with a speed 100 m/s
on a plywood and emerges with 40 m/s. The
percentage loss of kinetic energy is :
(1) 32% (2) 44%
(3) 16% (4) 84%
Ans. (4)
Sol.
2
i
1
K m(100)
2
2
f
1
K m(40)
2
%loss
fi
i
| K K |
K
× 100
=
22
2
11
m(40) m(100)
22 100
1m(100)
2
=
|1600 100 100 |
100

= 84%
33. The ratio of the shortest wavelength of Balmer
series to the shortest wavelength of Lyman series
for hydrogen atom is :
(1) 4 : 1 (2) 1 : 2
(3) 1 : 4 (4) 2 : 1
Ans. (1)
Sol
n = 1
Balmer
n = 2
Lyman
n =
2
22
12
1 1 1
Rz nn





2
2
L
2
2
B
11
Rz 1
11
Rz 2






B
L
4 :1
34. To project a body of mass m from earth's surface
to infinity, the required kinetic energy is (assume,
the radius of earth is RE, g = acceleration due to
gravity on the surface of earth) :
(1) 2mgRE (2) mgRE
(3)
1
2
mgRE (4) 4mgRE
Ans. (2)
Sol.
2
e
E
1 GMm
mv
2R
2
E
GM
gR
E
K mgR
35. Electromagnetic waves travel in a medium with
speed of 1.5 × 108 ms–1. The relative permeability
of the medium is 2.0. The relative permittivity will
be :
(1) 5 (2) 1
(3) 4 (4) 2
Ans. (4)
Sol.
2
mm
00 2
1
v
1
c
2
rr 2
c
v
82
r82
(3 10 )
2(1.5 10 )
r24
r2
36. Which of the following phenomena does not
explain by wave nature of light.
(A) reflection (B) diffraction
(C) photoelectric effect (D) interference
(E) polarization
Choose the most appropriate answer from the
options given below :
(1) E only (2) C only
(3) B, D only (4) A, C only
Ans. (2)
Sol. (Theory)
Photoelectric effect prove particle nature of light.
37. While measuring diameter of wire using screw
gauge the following readings were noted. Main
scale reading is 1 mm and circular scale reading
is equal to 42 divisions. Pitch of screw gauge is
1 mm and it has 100 divisions on circular scale. The
diameter of the wire is
xmm
50
. The value of x is :
(1) 142 (2) 71
(3) 42 (4) 21
Ans. (2)
Sol. MSR = 1mm, CSR = 42, pitch = 1 mm
pitch 1
LC 0.01mm
No. of CSD 100



Diameter = MSR + LC × CSD
Diameter = 1 + (0.01) × 42 mm
Diameter = 1.42 mm =
x
50
x = 71
38. is the uniform surface charge density of a thin
spherical shell of radius R. The electric field at any
point on the surface of the spherical shell is :
(1) /0R (2) /20
(3) /0 (4) /40
Ans. (3)
Sol.
Gaussin Surface
dA
E
E=0
By Gauss law
in
0
q
E ·dA
0
dA
EdA 
0
E
39. The value of unknown resistance (x) for which the
potential difference between B and D will be zero
in the arrangement shown, is :
24
12
12
1
1
B
C
A
0.5
12
12
x
14.5V
D
(1) 3 (2) 9
(3) 6 (4) 42
Ans. (3)
Sol.
24
12
12
1
1
B
C
A
0.5
12
12
x
14.5V
D
0.5
B
C
A
14.5V
D
0.5
In case of balanced Wheatstone Bridge
BC
AB
AD CD
V
V
VV
12 0.5
6 x 0.5
x = 6
40. The specific heat at constant pressure of a real gas
obeying PV2 = RT equation is :
(1)
V
CR
(2)
V
RC
3
(3) R (4)
V
R
C2V
Ans. (4)
Sol. dQ = du + dW
CdT = CVdT + PdV …..(1)
 PV2 = RT
P = constant
P(2VdV) = RdT
PdV =
RdT
2V
Put in equation (1)
V
R
CC 2V

41. Match List I with List II
LIST I
LIST II
A.
Torque
I.
1 1 2 2
[M L T A ]

B.
Magnetic field
II.
21
[L A ]
C.
Magnetic moment
III.
1 2 1
[M T A ]

D.
Permeability of
free space
IV.
1 2 2
[M L T ]
Choose the correct answer from the options given
below :
(1) A-I, B-III, C-II, D-IV
(2) A-IV, B-III, C-II, D-I
(3) A-III, B-I, C-II, D-IV
(4) A-IV, B-II, C-III, D-I
Ans. (2)
Sol.
22
[ ] [r F] [ML T ]
[F] [qVB]

212
1
F MLT
B [MA T ]
qV ATLT







[M] = [I × A] = [AL2]
0
2
Idlsin
B4r

2
Br
[] Idl




22
[MLT A ]

42. Given below are two statements :
Statement I : In an LCR series circuit, current is
maximum at resonance.
Statement II : Current in a purely resistive circuit
can never be less than that in a series LCR circuit
when connected to same voltage source.
In the light of the above statements, choose the
correct from the options given below :
(1) Statement I is true but Statement II is false
(2) Statement I is false but Statement II is true
(3) Both Statement I and Statement II are true
(4) Both Statement I and Statement II are false
Ans. (3)
Sol. Statement-I
m
m22
LC
V
I
R (X X )

at resonance XL = XC
Thus,
m
m
V
IR
Impendence is minimum therefore I is
maximum at resonance.
Statement-II
V
IR



in purely resistive circuit.
2 1 2
MT A L
AL




43. The correct truth table for the following logic
circuit is :
A
B
Y
Options :
(1)
A B Y
0 0 0
0 1 1
1 0 0
1 1 1
(2)
A B Y
0 0 1
0 1 1
1 0 0
1 1 1
(3)
A B Y
0 0 1
0 1 1
1 0 0
1 1 0
(4)
A B Y
0 0 0
0 1 0
1 0 0
1 1 1
Ans. (2)
Sol.
A
B
Y
NOT
AND
OR
44. A sample contains mixture of helium and oxygen
gas. The ratio of root mean square speed of helium
and oxygen in the sample, is :
(1)
1
32
(2)
22
1
(3)
1
4
(4)
1
22
Ans. (2)
Sol.
rms
w
3RT
VM

2
2
Ow,He
He w,O
VM
VM
=
41
32 22
2
He
O
V22
V1
45. A light string passing over a smooth light
pulley connects two blocks of masses m1 and m2
(where m2 > m1). If the acceleration of the system
is
g
2
, then the ratio of the masses
1
2
m
m
is :
(1)
21
21
(2)
15
51
(3)
15
21
(4)
31
21
Ans. (1)
Sol.
21
12
MM
ag
MM



21
12
MM
gg
MM
2



1 2 2 1
M M 2M 2M
1
2
M21
M21




46. Four particles A, B, C, D of mass
m
2
, m, 2m, 4m,
have same momentum, respectively. The particle
with maximum kinetic energy is :
(1) D (2) C
(3) A (4) B
Ans. (3)
Sol.
2
p
KE 2m
Same momentum, so less mass means more KE.
So
m
2
will have max. KE.
47. A train starting from rest first accelerates
uniformly up to a speed of 80 km/h for time t, then
it moves with a constant speed for time 3t. The
average speed of the train for this duration of
journey will be (in km/h) :
(1) 80 (2) 70
(3) 30 (4) 40
Ans. (2)
M1
a
M2
a
Sol. Average speed =
total distance
time taken
=
80 t 80 3t
2
4t

= 70 km/hr.
48. An element
ˆ
xi l
is placed at the origin and
carries a large current I = 10A. The magnetic field
on the y-axis at a distance of 0.5 m from the
elements x of 1 cm length is :
y
P
0.5 m
x
x
(1)
8
4 10 T
(2)
8
8 10 T
(3)
8
12 10 T
(4)
8
10 10 T
Ans. (1)
Sol.
y
P
x
0
3
I(d r)
dB 4r
l
(Tesla)
7
3
11 ˆ
10 10 ( k)
2 100
1
2






=
8ˆ
4 10 T ( k)

49. A small ball of mass m and density is dropped in
a viscous liquid of density
0
. After sometime, the
ball falls with constant velocity. The viscous force
on the ball is :
(1)
0
mg 1



(2)
0
mg 1



(3)
0
mg 1
(4)
0
mg 1



Ans. (4)
Sol. mg – FB – Fv = ma
a = 0 for constant velocity
mg – FB = Fv
Fv = mg – v 0 g = mg –
0
0
mg mg 1




50. In photoelectric experiment energy of 2.48 eV
irradiates a photo sensitive material. The stopping
potential was measured to be 0.5 V. Work function
of the photo sensitive material is :
(1) 0.5 eV (2) 1.68 eV
(3) 2.48 eV (4) 1.98 eV
Ans. (4)
Sol. eVs = h–
0.5 V = 2.48 –
work function () = 2.48 V – 0.5 V = 1.98 V
SECTION-B
51. If the radius of earth is reduced to three-fourth
of its present value without change in its mass
then value of duration of the day of earth will be
______ hours 30 minutes.
Ans. (13)
Sol. By conservation of angular momentum
1 1 2 2
II
2
2
12
2 2 2 3 2
MR M R
5 T 5 4 T

12
19
T 16T
1
2
19
T
T 16

924hr
16

27 hr
2
= 13 hr 30 mins.
52. Three infinitely long charged thin sheets are placed
as shown in figure. The magnitude of electric field
at the point P is
0
x
. The value of x is _________
(all quantities are measured in SI units).
–
Y
–2
X
–X
Z
X = –a
X = a
X = 3a
P
O
Ans. (2)
Sol.
–
Y
–2
X
–X
Z
X = –a
X = a
X = 3a
P
O
p
000
2ˆ
E ( i)
222



=
0
2ˆ
i
53. A big drop is formed by coalescing 1000 small
droplets of water. The ratio of surface energy of
1000 droplets to that of energy of big drop is
10
x
.
The value of x is _________.
Ans. (1)
Sol.
1000 drops
Big drop
33
44
1000 r R
33
10r = R
R = 10r
2
2
S.E. of 1000 drops 1000(4 r )T
S.E. of Big drop 4 R T
2
2
1000 r 10
10 x
(10r)
x = 1
54. When a dc voltage of 100V is applied to an
inductor, a dc current of 5A flows through it.
When an ac voltage of 200V peak value is
connected to inductor, its inductive reactance is
found to be
20 3
. The power dissipated in the
circuit is _________W.
Ans. (250)
Sol. For DC voltage
V 100
R 20
I5
for AC voltage
L
X 20 3
R = 20
22
L
Z X R
3 400 400
= 40
Power =
2
rms
iR
=
2
2
rms
200
V2
R 20 250W
Z 40


 
 
 

55. The refractive index of prism is
3
and the
ratio of the angle of minimum deviation to the
angle of prism is one. The value of angle of prism
is _________°.
Ans. (60)
Sol. For min
i = e
12
A
rr 2

min 1
A
2i A 1
A
2i = 2A
i = A
A
i
r1
r2
e
Snell's law
1 × sin i = sin r
A
sin i sin 2

 

A
sin A sin 2

 

A A A
2sin cos 3sin
2 2 2



A3
cos 22




A30
2
A = 60°
56. A wire of resistance R and radius r is stretched till
its radius became r/2. If new resistance of the
stretched wire is x R, then value of x is _________.
Ans. (16)
Sol. We know
RA
l
,
2
Rr
l
As we starch the wire, its length will increase but
its radius will decrease keeping the volume
constant
if
VV
2
2
f
r
r4
ll
f4ll
2
new
2
old
R4r 16
Rr
4







l
l
Rnew = 16R
 x = 16
57. Radius of a certain orbit of hydrogen atom is
8.48 Å. If energy of electron in this orbit is E/x,
then x = _________.
(Given a0 = 0.529Å, E = energy of electron in
ground state)
Ans. (16)
Sol. We know
2
n
r 0.529 Z

2
n
8.48 0.529 1
n2 = 16 n = 4
We know
2
1
En
th
n
E
E16
x = 16
58. A circular coil having 200 turns, 2.5 × 10– 4 m2 area
and carrying 100 A current is placed in a uniform
magnetic field of 1 T. Initially the magnetic dipole
moment
(M)
was directed along
B
. Amount of
work, required to rotate the coil through 90° from
its initial orientation such that
M
becomes
perpendicular to
B
, is _________ J.
Ans. (5)
Sol.
initial
final
B
M
B
M
We know
Wext
= U + KE
P.E. M ·B
fi
M ·B M ·B 0
= –MB cos 90 + MB cos 0
= MB
= NIAB
64
5
200 100 10 10 1
2

= 5J
59. A particle is doing simple harmonic motion of
amplitude 0.06 m and time period 3.14 s. The
maximum velocity of the particle is _______ cm/s.
Ans. (12)
Sol. We know
vmax = A at mean position
=
2A
T
=
20.06
= 0.12 m/sec
vmax = 12 cm/sec
60. For three vectors
ˆˆˆ
A ( xi 6 j 2k)
,
ˆˆˆ
B ( i 4j 3k)
and
ˆˆ ˆ
C ( 8i j 3k)
, if
A·(B C) 0
, them value of x is _________.
Ans. (4)
Sol.
ˆˆ
ˆ
i j k
B C 1 4 3
8 1 3

ˆˆ ˆ
15i 21j 33k
ˆ ˆ ˆ ˆ
ˆˆ
A ·(B C) ( xi 6j 2k) ·(15i 21j 33k)
0 = –15x + 126 – 66
15x = 60
x = 4
CHEMISTRY
TEST PAPER WITH SOLUTION
SECTION-A
61. Functional group present in sulphonic acid is :
(1) SO4H (2) SO3H
(3)
– S – OH
O
(4) – SO2
Ans. (2)
Sol.
S
OH
O
O
Group present in sulphonic acids
62. Match List I with List II :
List I
(Molecule / Species)
List II
(Property / Shape)
A.
SO2Cl2
I.
Paramagnetic
B.
NO
II.
Diamagnetic
C.
–
2
NO
III.
Tetrahedral
D.
–
3
I
IV.
Linear
Choose the correct answer from the options given
below :
(1) A-IV, B-I, C-III, D-II
(2) A-III, B-I, C-II, D-IV
(3) A-II, B-III, C-I, D-IV
(4) A-III, B-IV, C-II, D-I
Ans. (2)
Sol.
(A)
SO2Cl2
sp3
(B)
NO
Paramagnetic
(C)
2
NO
Diamagnetic
(D)
3
I
sp3d
63. Given below are two statements :
Statement I : Picric acid is 2, 4, 6-trinitrotoluene.
Statement II : Phenol-2, 4-disulphuric acid is
treated with conc. HNO3 to get picric acid.
In the light of the above statement, 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
incorrect.
(3) Statement I is correct but Statement II is
incorrect.
(4) Both Statement I and Statement II are correct.
Ans. (1)
Sol.
NO2
OH
O2N
NO2
picric acid
(2, 4, 6 – trinitrophenol)
OH
Conc. H2SO4
OH
SO3H
SO3H
Conc. HNO3
NO2
OH
O2N
NO2
Picric acid
S
O
O
Cl
Cl
Tetrahedral
I
I
I
–
Linear
64. Which of the following is metamer of the given
compound (X) ?
NH – C
O
(X)
(1)
NH – C
O
(2)
NH
OHC
(3)
NH – C
O
(4)
NH – C
O
Ans. (4)
Sol. Metamer Isomer having same molecular
formula, same functional group but different
alkyl/aryl groups on either side of functional
group.
65. DNA molecule contains 4 bases whoes structure
are shown below. One of the structure is not
correct, identify the incorrect base structure.
(1)
N
HC
N
H
C
C
C
N
N
CH
NH2
(2)
HC
C
N
H
NH
C
H3C – C
O
O
(3)
N
H3C – C
N
H
C
C
C
N
NH
C
NH2
O
(4)
HC
C
N
H
N
C
HC
O
NH2
Ans. (3)
Sol.
N
HC
N
H
C
C
C
N
N
CH
NH2
Adenine
HC
C
N
H
NH
C
H3C – C
O
O
Thymine
HC
C
N
H
N
C
HC
O
NH2
Cytosine
Are bases of DNA molecule. As DNA contain four
bases, which are adenine, guanine, cytosine and
thymine.
66. Match List I with List II :
LIST I
(Hybridization)
LIST II
(Orientation in
Space)
A.
sp3
I.
Trigonal
bipyramidal
B.
dsp2
II.
Octahedral
C.
sp3d
III.
Tetrahedral
D.
sp3d2
IV.
Square planar
Choose the correct answer from the options given
below :
(1) A-III, B-I, C-IV, D-II
(2) A-II, B-I, C-IV, D-III
(3) A-IV, B-III, C-I, D-II
(4) A-III, B-IV, C-I, D-II
Ans. (4)
Sol. sp3 Tetrahedral
dsp2 Square planar
sp3d Trigonal Bipyramidal
sp3d2 Octahedral
67. Given below are two statements :
Statement I : Gallium is used in the
manufacturing of thermometers.
Statement II : A thermometer containing gallium
is useful for measuring the freezing point (256 K)
of brine solution.
In the light of the above statement, choose the
correct answer from the options given below :
(1) Both Statement I and Statement II are false.
(2) Statement I is false but Statement II is true.
(3) Both Statement I and Statement II are true.
(4) Statement I is true but Statement II is false.
Ans. (4)
Sol. Statement - I Correct
Statement - II False
Ga is used to measure high temperature
68. Which of the following statements are correct ?
A. Glycerol is purified by vacuum distillation
because it decomposes at its normal boiling
point.
B. Aniline can be purified by steam distillation as
aniline is miscible in water.
C. Ethanol can be separated from ethanol water
mixture by azeotropic distillation because it
forms azeotrope.
D. An organic compound is pure, if mixed M.P. is
remained same.
Choose the most appropriate answer from the
options given below :
(1) A, B, C only
(2) A, C, D only
(3) B, C, D only
(4) A, B, D only
Ans. (2)
Sol. Option (B) is incorrect because aniline is
immisible in water.
69. Match List I with List II :
LIST I
(Compound /
Species)
LIST II
(Shape / Geometry)
A.
SF4
I.
Tetrahedral
B.
BrF3
II.
Pyramidal
C.
–
3
BrO
III.
See saw
D.
4
NH
IV.
Bent T-shape
Choose the correct answer from the options given
below :
(1) A-II, B-III, C-I, D-IV
(2) A-III, B-IV, C-II, D-I
(3) A-II, B-IV, C-III, D-I
(4) A-III, B-II, C-IV, D-I
Ans. (2)
Sol.
(A)
SF4
sp3d
hybridisation
(B)
BrF3
sp3d
hybridisation
(C)
3
BrO
sp3
hybridisation
(D)
4
NH
sp3
hybridisation
S
F
F
F
F
Br
F
F
F
Bent T-Shape
Br
O
O–
Pyramidal
N
H
Tetrahedral
H
+
70. In Reimer - Tiemann reaction, phenol is converted
into salicylaldehyde through an intermediate. The
structure of intermediate is ______.
(1)
ONa+
CH3
–
(2)
CHCl2
OH
(3)
ONa+
CHO
–
(4)
ONa+
CHCl2
–
Ans. (4)
Sol.
OH
CHCl3+ aq NaOH
O
CHCl2
NaOH
Na
+
Intermediate
O
C–H
Na
+
O
H
+
OH
C–H
O
71. Which of the following material is not a
semiconductor.
(1) Germanium
(2) Graphite
(3) Silicon
(4) Copper oxide
Ans. (2)
Sol. Graphite is conductor
72. Consider the following complexes.
[CoCl(NH3)5]2+, [Co(CN)6]3–,
(A) (B)
[Co(NH3)5(H2O)]3+, [Cu(H2O)4]2+
(C) (D)
The correct order of A, B, C and D in terms of
wavenumber of light absorbed is :
(1) C < D < A < B
(2) D < A < C < B
(3) A < C < B < D
(4) B < C < A < D
Ans. (2)
Sol. As ligand field increases, light of more energy is
absorbed
Energy wave number
73. Match List I with List II :
LIST I
(Precipitating reagent and
conditions)
LIST II
(Cation)
A.
NH4Cl + NH4OH
I.
Mn2+
B.
NH4OH + Na2CO3
II.
Pb2+
C.
NH4OH + NH4Cl + H2S gas
III.
Al3+
D.
dilute HCl
IV.
Sr2+
Choose the correct answer from the options given
below :
(1) A-IV, B-III, C-II, D-I
(2) A-IV, B-III, C-I, D-II
(3) A-III, B-IV, C-I, D-II
(4) A-III, B-IV, C-II, D-I
Ans. (3)
Sol. Theory based question
74. The electron affinity value are negative for :
A. Be Be–
B. N N–
C. O O2–
D. Na Na–
E. Al Al–
Choose the most appropriate answer from the
options given below :
(1) D and E only (2) A, B, D and E only
(3) A and D only (4) A, B and C only
Allen Ans. (4)
NTA Ans. (1)
Sol. (A) Be + e– Be– , E.A = –ive
(B) N + e– N– E.A = –ive
(C) O + e– O–
O– + e– O–2 E.A = –ive
(D) Na + e– Na– E.A = +ive
(E) A + e– A– E.A = +ive
Ans. A,B and C only
75. The number of element from the following that
do not belong to lanthanoids is :
Eu, Cm, Er, Tb, Yb and Lu
(1) 3 (2) 4
(3) 1 (4) 5
Ans. (3)
Sol. Cm is Actinide
76. The density of 'x' M solution ('x' molar) of NaOH
is 1.12 g mL–1. while in molality, the concentration
of the solution is 3 m (3 molal). Then x is
(Given : Molar mass of NaOH is 40 g/mol)
(1) 3.5 (2) 3.0
(3) 3.8 (4) 2.8
Ans. (2)
Sol. Molality =
solute
1000 M
1000 d M (Mw)
3 =
1000 x
1000 1.12 (x 40)
x = 3
77. Which among the following aldehydes is most
reactive towards nucleophilic addition reactions?
(1)
H – C – H
O
(2)
C2H5 – C – H
O
(3)
CH3 – C – H
O
(4)
C3H7 – C – H
O
Ans. (1)
Sol.
H – C – H
O
has low steric hindrance at carbonyl
carbon and high partial positive charge at carbonyl
carbon.
78. At –20 °C and 1 atm pressure, a cylinder is filled
with equal number of H2. I2 and HI molecules for
the reaction
H2(g) + I2(g) 2HI(g), the KP for the process is
x × 10–1. x = _________.
[Given : R = 0.082 L atm K–1 mol–1]
(1) 2 (2) 1
(3) 10 (4) 0.01
Ans. (3)
Sol. ng = 0 Kp =
g
22
n
2
HI T
H I T
(n ) P
n n n



22
HI H I
n n n
so KP = 1
1 = x × 10–1 x = 10
79. Match List I with List II :
LIST I
(Compound)
LIST II
(Uses)
A.
Iodoform
I.
Fire extinguisher
B.
Carbon
tetrachloride
II.
Insecticide
C.
CFC
III.
Antiseptic
D.
DDT
IV.
Refrigerants
Choose the correct answer from the options given
below :
(1) A-I, B-II, C-III, D-IV
(2) A-III, B-II, C-IV, D-I
(3) A-III, B-I, C-IV, D-II
(4) A-II, B-IV, C-I, D-III
Ans. (3)
Sol. Iodoform – Antiseptic
CCl4 – Fire extinguisher
CFC – Refrigerants
DDT – Insecticide
80. A conductivity cell with two electrodes (dark side)
are half filled with infinitely dilute aqueous solution
of a weak electrolyte. If volume is doubled by
adding more water at constant temperature, the
molar conductivity of the cell will -
1cm
1cm
1cm
(+)
(–)
(1) increase sharply
(2) remain same or can not be measured accurately
(3) decrease sharply
(4) depend upon type of electrolyte
Ans. (2)
Sol. Solution is already infinitely dilute, hence no
change in molar conductivity upon addition of
water
SECTION-B
81. Consider the dissociation of the weak acid HX as
given below
HX(aq) H+(aq) + X– (aq), Ka = 1.2 × 10–5
[Ka : dissociation constant]
The osmotic pressure of 0.03 M aqueous solution
of HX at 300 K is _____ × 10–2 bar (nearest
integer).
[Given : R = 0.083 L bar Mol–1 K–1]
Ans. (76)
Sol. HX H+ + X– Ka = 1.2 × 10–5
0.03M
0.03 – x x x
Ka = 1.2 × 10–5 =
2
x
0.03 x
0.03 – x 0.03 (Ka is very small)
2
x
0.03
= 1.2 × 10–5
x = 6 × 10–4
Final solution : 0.03 – x + x + x
= 0.03 + x = 0.03 + 6 × 10–4
= (0.03 + (6 × 10–4)) × 0.083 × 300
= 76.19 × 10–2 76 × 10–2
82. The difference in the 'spin-only' magnetic moment
values of KMnO4 and the manganese product
formed during titration of KMnO4 against oxalic
acid in acidic medium is _____ BM. (nearest
integer)
Ans. (6)
Sol. Spin only magnetic moment of Mn in KMnO4 = 0
Spin only value of manganese product fromed
during titration of KMnO4 aganist oxalic acid in
acidic medium is = 6
Ans. 6
83. Time required for 99.9% completion of a first
order reaction is _____ time the time required for
completion of 90% reaction.(nearest integer).
Ans. (3)
Sol. K =
99.9% 90%
1 100 1 100
nn
t 0.1 t 10
t99.9% = t90%
3
n(10 )
n10
t99.9% = t90% × 3
84. Number of molecules from the following which
can exhibit hydrogen bonding is ______. (nearest
integer)
CH3OH, H2O, C2H6, C6H6,
NO2
OH
HF, NH3
Ans. (5)
Sol. CH3OH, H2O,
NO2
OH
HF, NH3
Can show H–bonding.
85. 9.3 g of pure aniline upon diazotisation followed
by coupling with phenol gives an orange dye. The
mass of orange dye produced (assume 100%
yield/ conversion) is ________g. (nearest integer)
Ans. (20)
Sol.
NH2
NaNO2 + HCl
N2
Cl
+
T < 5°C
OH
N = N
HO
Orange dye
Reaction suggests that 1 mole of aniline give 1
mole of orange dye.
so (mol)aniline = (mole)orange dye
11
9.3g mass of orange dye
93g mol 199g mol

mass of orange dye = 19.9 g 20 g
86. The major product of the following reaction is P.
CH3C C – CH3
3
4
(i) Na/liq.NH
(ii)dil.KMnO
273K

'P'
Number of oxygen atoms present in product 'P' is
_______ (nearest integer).
Ans. (2)
Sol. CH3 – C
C–CH3
3
Na/liq.NH

H
C=C
CH3
CH3
H
dil. KMnO4
C
C
CH3
CH3
OH
OH
H
H
(Product P)
87. Frequency of the de-Broglie wave of election in
Bohr's first orbit of hydrogen atom is ____ × 1013 Hz
(nearest integer).
[Given : RH (Rydberg constant) = 2.18 × 10–18 J.
h (Plank's constant) = 6.6 × 10–34 J.s.]
Allen Ans. (661)
NTA Ans. (658)
Sol. =
h
mv
=
2
hv
mv
2
mv v
h
(frequency)
Given
1
2
mv2 = 2.18 × 10–18 J
h = 6.6 × 10–34
=
18
34
4.36 10
6.6 10
= 660.60 × 1013 Hz
661 × 1013 Hz
88. The major products from the following reaction
sequence are product A and product B.
B
(i) Br2
(ii) alc. KOH (3 eq.)
(i) Br2
(ii) O– Na+ (1.0 eq.)
A
The total sum of electrons in product A and
product B are ______ (nearest integer)
Ans. (8)
Sol.
Br2
Br
Br
HCC–CH2–O
Na
+
Br
O–CH2–CCH
(A)
Br
Br
alc KOH
(3 eq)
(B)
89. Among CrO, Cr2O3 and CrO3, the sum of spin-only
magnetic moment values of basic and amphoteric
oxides is ______ 10–2 BM (nearest integer).
(Given atomic number of Cr is 24)
Ans. (877)
Sol. CrO Basic oxide
Cr2O3 Amphoteric oxide
In CrO, Cr exist as Cr+2 and have only = 4.90
In Cr2O3, Cr exist as Cr+3 and have only = 3.87
Sum of spin only magnetic moment
= 4.90 + 3.87 = 8.77
only = 877 × 10–2
Ans. 877
90. An ideal gas,
V
5
CR
2
, is expanded adiabatically
against a constant pressure of 1 atm untill it
doubles in volume. If the initial temperature and
pressure is 298 K and 5 atm, respectively then the
final temperature is ______ K (nearest integer).
[
V
C
is the molar heat capacity at constant volume]
Ans. (274)
Sol. U = q + w (q = 0)
nCVT = –Pext (V2 – V1)
V2 = 2V1
21
21
nRT 2nRT
PP
P1 = 5, T1 = 298
P2 =
2
5T
2 298
5
n2
R(T2 – T1) = – 1
21
11
nRT nRT
PP



Put T1 = 298
and P2 =
2
5T
2 298
Solve and we get T2 = 274.16 K
T2 274 K