FINAL JEE–MAIN EXAMINATION – JANUARY, 2024
(Held On Wednesday 31
st
January, 2024) TIME : 3 : 00 PM to 6 : 00 PM
MATHEMATICS
TEST PAPER WITH SOLUTION
SECTION-A
1. The number of ways in which 21 identical apples
can be distributed among three children such that
each child gets at least 2 apples, is
(1) 406
(2) 130
(3) 142
(4) 136
Ans. (4)
Sol. After giving 2 apples to each child 15 apples left
now 15 apples can be distributed in
15 3 1 17
22
CC
ways
17 16 136
2
2. Let A (a, b), B(3, 4) and (–6, –8) respectively
denote the centroid, circumcentre and orthocentre
of a triangle. Then, the distance of the point
P(2a + 3, 7b + 5) from the line 2x + 3y – 4 = 0
measured parallel to the line x – 2y – 1 = 0 is
(1)
15 5
7
(2)
17 5
6
(3)
17 5
7
(4)
5
17
Ans. (3)
Sol. A(a,b), B(3,4), C(-6, -8)
2 : 1
C
(-6, -8) AB
(3, 4)
(a, b)
a 0, b = 0 P 3,5
Distance from P measured along x – 2y – 1 = 0
x 3 rcos , y = 5+rsin
Where
1
tan 2
r 2cos 3sin 17
17 5 17 5
r77
3. Let z1 and z2 be two complex number such that z1
+ z2 = 5 and
33
12
z z 20 15i
. Then
44
12
zz
equals-
(1)
30 3
(2) 75
(3)
15 15
(4)
25 3
Ans. (2)
Sol.-
12
z z 5
33
12
z z 20 15i
3
33 1 2 1 2 1 2
12
z z z z 3z z z z
33 12
12
z z 125 3z .z 5
12
20 15i 125 15z z
12
3z z 25 4 3i
12
3z z 21 3i
12
z .z 7 i
2
12
z z 25
22
12
z z 25 2 7 i
11 2i
2
22
12
z z 121 4 44i
2
44
12
z z 2 7 i 117 44i
44
12
z z 117 44i 2 49 1 14i
44
12
z z 75
4. Let a variable line passing through the centre of the
circle x2 + y2 – 16x – 4y = 0, meet the positive
co-ordinate axes at the point A and B. Then the
minimum value of OA + OB, where O is the
origin, is equal to
(1) 12
(2) 18
(3) 20
(4) 24
Ans. (2)
Sol.-
y 2 m x 8
x-intercept
28
m
y-intercept
8m 2
2
OA OB 8 8m 2
m
2
2
f ' m 8 0
m
21
m4
1
m2
1
f 18
2
Minimum = 18
5. Let
f,g:(0, ) R
be two functions defined by
2
x2t
x
f(x) t t e dt
and
2
x1t
2
0
g(x) t e dt
.
Then the value of
ee
f log 9 g log 9
is
equal to
(1) 6
(2) 9
(3) 8
(4) 10
Ans. (3)
Sol.-
2
2
2
2
2 2 2
x2t
x
2x
x1t
2
0
x
x 2 x 2 x
f x t t e dt
f x 2. x x e ............ 1
g x t e dt
g x xe 2x 0
f x g x 2xe 2x e 2x e
Integrating both sides w.r.t.x
2
1
e
x
0
2
tt
0
0
log 9 1
f x g x 2xe dx
xt
e dt e
e
1
9 f(x) g(x) 1 9 8
9
6. Let
,,
be mirror image of the point (2, 3, 5)
in the line
x 1 y 2 z 3
2 3 4
.
Then
2 3 4
is equal to
(1) 32
(2) 33
(3) 31
(4) 34
Ans. (2)
Sol.
P(2,3,5)
R , ,
PR 2,3,4
PR. 2,3,4 0
2, 3, 5 . 2,3,4 0
2 3 4 4 9 20 33
7. Let P be a parabola with vertex (2, 3) and directrix
2x + y = 6. Let an ellipse
22
22
xy
E: 1,a b
ab
of
eccentricity
1
2
pass through the focus of the
parabola P. Then the square of the length of the
latus rectum of E, is
(1)
385
8
(2)
347
8
(3)
512
25
(4)
656
25
Ans. (4)
Sol.-
Focur axis
(2, 3)
(, )
(1.6, 2.8)
Slope of axis
1
2
1
y 3 x 2
2
2y 6 x 2
2y x 4 0
2x y 6 0
4x 2y 12 0
1.6 4 2.4
2.8 6 3.2
Ellipse passes through (2.4, 3.2)
22
22
24 32
10 10 1
ab
……….(1)
Also
22
22
b 1 b 1
1a 2 a 2
22
a 2b
Put in (1)
2328
b25
2
22
2
2
2b 4b 1 328 656
b4
a a 2 25 25
8. The temperature T(t) of a body at time t = 0 is 160o
F and it decreases continuously as per the
differential equation
dT K(T 80)
dt
, where K
is positive constant. If T(15) = 120oF, then T(45) is
equal to
(1) 85o F
(2) 95o F
(3) 90o F
(4) 80o F
Ans. (3)
Sol.-
Tt
160 0
T
160
kt
k.15
k15
k.45
3
k.15
dT k T 80
dt
dT Kdt
T 80
ln T 80 kt
ln T 80 ln80 kt
T 80
ln kt
80
T 80 80e
120 80 80e
40 1
e
80 2
T 45 80 80e
80 80 e
1
80 80 8
90
9. Let 2nd, 8th and 44th, terms of a non-constant A.P.
be respectively the 1st, 2nd and 3rd terms of G.P. If
the first term of A.P. is 1 then the sum of first
20 terms is equal to-
(1) 980 (2) 960
(3) 990 (4) 970
Ans. (4)
Sol.- 1 + d, 1 + 7d, 1 + 43d are in GP
(1 + 7d)2 = (1 + d) (1 + 43d)
1 + 49d2 + 14d = 1 + 44 d + 43d2
6d2 – 30d = 0
d = 5
20 20
S 2 1 20 1 5
2
=10 2 95
=970
10. Let
f : R (0, )
be strictly increasing
function such that
x
f(7x)
lim 1
f(x)
. Then, the value
of
x
f(5x)
lim 1
f(x)
is equal to
(1) 4
(2) 0
(3) 7/5
(4) 1
Ans. (2)
Sol.-
f :R (0, )
x
f 7x
lim 1
fx
f is increasing
f x f 5x f 7x
x
f x f 5x f 7x
f x f x f x
f 5x
1 lim 1
fx
f 5x 1
fx
1 1 0
11. The area of the region enclosed by the parabola
y = 4x – x2 and 3y = (x – 4)2 is equal to
(1)
32
9
(2) 4
(3) 6
(4)
14
3
Ans. (3)
Sol.-
Area
2
42
1
x4
4x x dx
3
Area
4
3
23
1
x4
4x x
2 3 9
64 64 4 1 27
2 3 2 3 9
27 21 6
12. Let the mean and the variance of 6 observation a,
b, 68, 44, 48, 60 be 55 and 194, respectively if
a > b, then a + 3b is
(1) 200
(2) 190
(3) 180
(4) 210
Ans. (3)
Sol.- a, b, 68, 44, 48, 60
Mean = 55 a > b
Variance = 194 a + 3b
a b 68 44 48 60 55
6
220 a b 330
a b 110......(1)
Also,
2
i
2 2 2 2
22
22
22
22
22
22
xx 194
n
a 55 b 55 68 55 44 55
48 55 60 55 194 6
a 55 b 55 169 121 49 25 1164
a 55 b 55 1164 364 800
a 3025 110a b 3025 110b 800
a b 800 6050 12100
a b 6850.......(2)
Solve (1) & (2);
a
=75,b=35
a 3b 75 3 35 75 105 180
13. If the function
f :( , 1] (a,b]
defined by
3
x 3x 1
f(x) e
is one-one and onto, then the
distance of the point P(2b + 4, a + 2) from the line
x + e–3y = 4 is :
(1)
6
2 1 e
(2)
6
4 1 e
(3)
6
3 1 e
(4)
6
1e
Ans. (1)
Sol.-
3
x 3x 1
f x e
3
x 3x 1 2
f ' x e . 3x 3
3
x 3x 1
e .3 x 1 x 1
For
f ' x 0
fx
is increasing function
a e 0 f
1 3 1 3
b e e f 1
P(2b + 4, a + 2)
3
P 2 e 4,2
x + e y = 4
-3
P
d
33 6
6
2e 4 2e 4
d 2 1 e
1e
14. Consider the function
f :(0, ) R
defined by
e
log x
f(x) e
. If m and n be respectively the
number of points at which f is not continuous and f
is not differentiable, then m + n is
(1) 0
(2) 3
(3) 1
(4) 2
Ans. (3)
Sol.-
e
log x
f : 0, R
f x e
lnx
lnx
lnx
1;0 x 1
1e
fx 1
e;x 1
e
1x;0 x 1
1
x
1,x 1
x
1
01
m = 0 (No point at which function is not continuous)
n = 1 (Not differentiable)
m + n = 1
15. The number of solutions, of the equation
sinx sinx
e 2e 2
is
(1) 2
(2) more than 2
(3) 1
(4) 0
Ans. (4)
Sol.- Take
sin x
e t t 0
2
t2
t
2
t2
2
t
2
t 2t 2 0
2
t 2t 1 3
2
t 1 3
t 1 3
t 1 1.73
t 2.73
or -0.73 (rejected as t > 0)
sin x
e 2.73
sinx
ee
log e log 2.73
e
sinx log 2.73 1
So no solution.
16. If
1
a sin sin 5
and
1
b cos cos 5
,
then
22
ab
is equal to
(1)
2
4 25
(2)
2
8 40 50
(3)
2
4 20 50
(4) 25
Ans. (2)
Sol.
1
a sin sin5 5 2
and
1
b cos cos5 2 5
22
22
a b 5 2 2 5
2
8 40 50
17. If for some m, n;
6 6 6 8
m m 1 m 2 3
C 2 C C C
and
n 1 n
34
P : P 1:8,
then
n n 1
m 1 m
PC
is equal
to
(1) 380
(2) 376
(3) 384
(4) 372
Ans. (4)
Sol.-
6 6 6 8
m m 1 m 2 3
C 2 C C C
7 7 8
m 1 m 2 3
C C C
88
m 2 3
CC
m2
And
n 1 n
34
P : P 1:8
n 1 n 2 n 3 1
n n 1 n 2 n 3 8
n8
n n 1 8 9
m 1 m 3 2
P C P C
98
8 7 6 2
= 372
18. A coin is based so that a head is twice as likely to
occur as a tail. If the coin is tossed 3 times, then
the probability of getting two tails and one head is-
(1)
2
9
(2)
1
9
(3)
2
27
(4)
1
27
Ans. (1)
Sol. Let probability of tail is
1
3
Probability of getting head =
2
3
Probability of getting 2 tails and 1 head
1 2 1 3
333
23
27
2
9
19. Let A be
a 3 3
real matrix such that
1 1 1 1 0 0
A 0 2 0 ,A 0 4 0 ,A 1 2 1 .
1 1 1 1 0 0
Then, the system
x1
A 3I y 2
z3
has
(1) unique solution
(2) exactly two solutions
(3) no solution
(4) infinitely many solutions
Ans. (1)
Sol.- Let
1 1 1
2 2 2
3 3 3
x y z
A x y z
x y z
Given
12
A 0 0
12
…. (1)
11
22
33
xz 2
x z 0
2
xz
11
x z 2
…. (2)
22
x z 0
…. (3)
33
x z 0
…. (4)
Given
14
A 0 0
14
11
22
33
xz 4
x z 0
4
xz
11
x z 4
…. (5)
22
x x 0
…. (6)
33
x z 4
Given
00
A 1 2
00
1
2
3
y0
y2
0
y
1 2 3
y 0, y 2, y 0
from (2), (3), (4), (5), (6) and (7)
1 2 3
x 3x, x 0, x 1
1 2 3
y 0, y 2, y 0
1 2 3
z 1, z 0, z 3
3 0 1
A 0 2 0
1 0 3
x1
Now A 3l y 2
z3
x1
0 0 1
0 1 0 y 2
1 0 0 z3
z1
y2
x3
z 1 , y 2 , x 3
20. The shortest distance between lines L1 and L2,
where
1x 1 y 1 z 4
L: 2 3 2
and L2 is the line
passing through the points
A 4,4,3 .B 1,6,3
and perpendicular to the line
x 3 y z 1,
2 3 1
is
(1)
121
221
(2)
24
117
(3)
141
221
(4)
42
117
Ans. (3)
Sol.-
2
2 1 2 1 2 1
12
12
x 4 y 4 z 3
L3 2 0
x x y y z z
2 3 2
3 2 0
S.D nn
5 5 7
2 3 2
3 2 0
nn
141
ˆ ˆ ˆ
4i 6j 13k
141
16 36 169
141
221
SECTION-B
21.
2
3 4 4
0
120 x sin xcosx dx
sin x cos x
is equal to _________.
Ans. (15)
Sol.-
2
44
0
x sinx .cosx dx
sin x cos x
22
2
44
0
sin x . cosx x x dx
sin x cos x
22
44
0
.
sin x cosx 2 x
sin x cos x
22
2
4 4 4 4
00
xsin x cosx sinx cosx
2 dx dx
sin x cos x sin x cos x
22
2
4 4 4 4
00
sin x cosx sin x cosx
.
2 dx dx
4 sin x cos x sin x cos x
2
2
44
0
sin x cos x dx
2 sin x cos x
2
2
22
0
sin x cosxdx
2 1 2sin x cos x
2
2
2
0
sin2x dx
2 2 sin 2x
2
2
2
0
sin2x dx
2 1 cos 2x
Let
cos2x t
22. Let a, b, c be the length of three sides of a triangle
satisfying the condition (a2 + b2)x2 – 2b(a + c).
x + (b2 + c2) = 0. If the set of all possible values of
x is the interval
,
, then
22
12
is equal
to ______.
Ans. (36)
Sol.-
2 2 2 2 2
a b x 2b a c x b c 0
2 2 2 2 2 2
22
a x 2abx b b x 2bcx c 0
ax b bx c 0
ax b 0, bx c 0
a b c b + c > a c + a > b
1 5 1 5
x
22
1 5 1 5
x , or x
22
5 1 5 1
x
22
5 1 5 1
,
22
22
22 5 1 5 1
12 12 =36
4
23. Let A(–2, –1), B(1, 0),
C,
and
D,
be
the vertices of a parallelogram ABCD. If the point
C lies on 2x – y = 5 and the point D lies on
3x – 2y = 6, then the value of
is
equal to ________.
Ans. (32)
Sol.-
P
3x – 2y = 6
A(-2,-1) B(1,0) 2x – y = 5
C,
D,
2 1 1
P , ,
2 2 2 2
2 1 1
and
2 2 2 2
3.....(1), 1......(2)
Also,
,
lies on 3x – 2y = 6
3 2 6 ......(3)
and
,
lies on 2x – y = 5
2 5.........(4)
Solving (1), (2), (3), (4)
3, = 11, = 6, 12
32
24. Let the coefficient of xr in the expansion of
n 1 n 2
n 3 2 n 1
x 3 x 3 x 2
x 3 x 2 ....... x 2
be
r.
If
nnn
r
r0
, , N,
then the value
of
22
equals __________.
Ans. (25)
Sol.-
n 1 n 2 n 3
2 n 1
n 1 n 2 n 3 2 n 1
r
2 n 1
n1
n
n1
n n n n
22
x 3 x 3 x 2 x 3
x 2 ........ x 2
4 4 3 4 3 ....... 3
3 3 3
4 1 .....
4 4 4
3
14
43
14
43
4, 3
16 9 25
25. Let A be a
33
matrix and det (A) = 2. If
n det adj adj ...... adjA
2024-times
Then the remainder when n is divided by 9 is equal
to ___________.
Ans. (7)
Sol.-
A2
2024
n1
2024 times
adj adj adj..... a A
2024
2
A
2024
2
2
674 674
2024 2 2022
2 2 2 4 8 4 9 1
2024
2 4 mod 9
2024
2 9m 4, m even
3m
9m 4 3
.
2 16 2 16 mod9
7
26. Let
ˆ ˆ ˆ ˆ ˆ ˆ
a 3i 2j k,b 2i j 3k
and
c
be a
vector such that
ˆˆ
a b c 2 a b 24j 6k
and
ˆ
a b i .c 3.
Then
2
c
is equal to _____.
Ans. (38)
Sol.-
ˆˆ
a b c 2 a b 24j 6k
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
5i j 4k c 2 7i 7j 7k 24j 6k
ˆ ˆ ˆ
i j k
ˆ ˆ ˆ
5 1 4 14i 10j 20k
x y z
ˆ ˆ ˆ ˆ ˆ ˆ
i z 4y j 5z 4x k 5y x 14i 10j 20k
z 4y 14,4x 5z 10,5y x 20
2
a b i .c 3
ˆ ˆ ˆ
2i 3j 2k .c 3
2x 3y 2z 3
x 5, y 3,z 2
c 25 9 4 38
27. If
2 x x
e
2
x0
ax e blog 1 x cxe
lim 1,
x sin x
then 16(a2 + b2 + c2) is equal to ______.
Ans. (81)
Sol.-
2 3 2 3
2
23
x0 3
23
3
x
x x x x
ax 1 x ..... b x .......
2! 3! 2 3
xx
cx 1 x .......
x! 3!
lim sin x
x. x
b b c
c b x c a x a x ......
2 3 2
lim 1
x
2 2 2
2 2 2
b
c b 0, c a 0
2
b c 3 3
a 1 a= b c
3 2 4 2
9 9 9
a b c 16 4 4
16 a b c 81
28. A line passes through A(4, –6, –2) and B(16, –2,4).
The point P(a, b, c) where a, b, c are non-negative
integers, on the line AB lies at a distance of 21
units, from the point A. The distance between the
points P(a, b, c) and Q(4, –12, 3) is equal to ____.
Ans. (22)
Sol.-
x 4 x 6 z 2
12 4 6
x 4 y 6 z 2 21
6 2 3
7 7 7
6 2 3
21 4, 21 6, 21 2
7 7 7
22,0,7 a,b,c
324 144 16 22
29. Let y = y(x) be the solution of the differential
equation
2 2y 2
sec xdx e tan x tan x dy 0,
0 x ,y 0. If y ,
2 4 6
Then
8
e
is equal to ______.
Ans. (9)
Sol.-
2 2y 2
2
2y 2
2 2y
2y
2
2
dx
sec x e tan x tan x 0
dy
dx dt
Put tan x t sec x dy dy
dt e t t 0
dy
dt t t .e
dy
1 dt 1 e
t dy t
1 1 dt du
Put u
t t dy dy
2y
2y
dy y
y y 2y
yy
2
8
du ue
dy
du ue
dy
I.F. e e
ue e e dy
1e e c
tan x
x ,y 0,c 0
4
x , y =
6
3e e 0
e3
e9
30. Let A = {1, 2, 3, ………100}. Let R be a relation
on A defined by (x, y)
R if and only if 2x = 3y.
Let R1 be a symmetric relation on A such that
1
RR
and the number of elements in R1 is n.
Then, the minimum value of n is ___________.
Ans. (66)
Sol.-
R 3,2 , 6,4 , 9,6 , 12,8 ,......... 99,66
n(R) 33
66
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
31. A light string passing over a smooth light fixed
pulley connects two blocks of masses
1
m
and
2
m
.
If the acceleration of the system is g/8, then the
ratio of masses is
(1)
9
7
(2)
8
1
(3)
4
3
(4)
5
3
Ans. (1)
Sol.
12
12
m m g g
am m 8
1 2 1 2
8m 8m m m
12
7m 9m
1
2
m9
m7
32. A uniform magnetic field of
3
2 10 T
acts along
positive Y-direction. A rectangular loop of sides 20
cm and 10 cm with current of 5 A is Y-Z plane.
The current is in anticlockwise sense with
reference to negative X axis. Magnitude and
direction of the torque is :
(1)
4
2 10 N m
along positive Z –direction
(2)
4
2 10 N m
along negative Z-direction
(3)
4
2 10 N m
along positive X-direction
(4)
4
2 10 N m
along positive Y-direction
Ans. (2)
Sol.
M iA
=
ˆ
5 0.2 0.1 i
=
ˆ
0.1 i
3
ˆˆ
M B 0.1 i 2 10 j
=
4ˆ
2 10 k N m
33. The measured value of the length of a simple
pendulum is 20 cm with 2 mm accuracy. The time
for 50 oscillations was measured to be 40 seconds
with 1 second resolution. From these
measurements, the accuracy in the measurement of
acceleration due to gravity is N%. The value of N
is:
(1) 4 (2) 8
(3) 6 (4) 5
Ans. (3)
Sol.
T2 g
2
2
4
gT
g 2 T
gT
0.2 1
2
20 40
=
0.3
20
Percentage change =
0.3 100
20
= 6%
34. Force between two point charges
1
q
and
2
q
placed in vacuum at ‘r’ cm apart is F. Force
between them when placed in a medium having
dielectric K = 5 at ‘r/5’ cm apart will be:
(1) F/25 (2) 5F
(3) F/5 (4) 25F
Ans. (2)
Sol. In air
12
02
qq
1
F4r
In medium
1 2 1 2
22
00
q q q q
1 25
F' 5F
4 K 4 5
r' r
35. An AC voltage
V 20sin200 t
is applied to a
series LCR circuit which drives a current
I 10sin 200 t 3
. The average power
dissipated is:
(1) 21.6 W (2) 200 W
(3) 173.2 W (4) 50 W
Ans. (4)
Sol.
P IV cos
o
20 10 cos60
22
50 W
36. When unpolarized light is incident at an angle of
60° on a transparent medium from air. The
reflected ray is completely polarized. The angle of
refraction in the medium is
(1)
0
30
(2)
0
60
(3)
0
90
(4)
0
45
Ans. (1)
Sol. By Brewster’s law
At complete reflection refracted ray and reflected
ray are perpendicular.
37. The speed of sound in oxygen at S.T.P. will be
approximately:
(Given,
1
R 8.3 JK , 1.4)
(1)
310 m / s
(2) 333 m/s
(3) 341 m/s
(4) 325 m/s
Ans. (1)
Sol.
3
RT 1.4 8.3 273
vM 32 10
314.8541 315 m / s
38. A gas mixture consists of 8 moles of argon and 6
moles of oxygen at temperature T. Neglecting all
vibrational modes, the total internal energy of the
system is
(1) 29 RT
(2) 20 RT
(3) 27 RT
(4) 21 RT
Ans. (3)
Sol.
V
U nC T
12
1 V 2 V
U n C T n C T
3R 5R
8 T 6 T
22
= 27RT
39. The resistance per centimeter of a meter bridge
wire is r, with X
resistance in left gap. Balancing
length from left end is at 40 cm with 25
resistance in right gap. Now the wire is replaced by
another wire of 2r resistance per centimeter. The
new balancing length for same settings will be at
(1) 20 cm
(2) 10 cm
(3) 80 cm
(4) 40 cm
Ans. (4)
Sol.
G
X
25
1=40c
2=60cm
12
25 X
rr
.....(i)
12
25 X
2r ' 2r '
..... (ii)
From (i) and (ii)
22
' 40 cm
40. Given below are two statements:
Statement I: Electromagnetic waves carry energy
as they travel through space and this energy is
equally shared by the electric and magnetic fields.
Statement II: When electromagnetic waves strike
a surface, a pressure is exerted on the surface.
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. (2)
Sol.
2
2
00
1B
E
22
00
1
E CBand C
41. In a photoelectric effect experiment a light of
frequency 1.5 times the threshold frequency is
made to fall on the surface of photosensitive
material. Now if the frequency is halved and
intensity is doubled, the number of photo electrons
emitted will be:
(1) Doubled (2) Quadrupled
(3) Zero (4) Halved
Ans. (3)
Sol. Since
0
ff
2
i.e. the incident frequency is less
than threshold frequency. Hence there will be no
emission of photoelectrons.
current 0
42. A block of mass 5 kg is placed on a rough inclined
surface as shown in the figure.
5kg
300
=0.1
If
1
F
is the force required to just move the block
up the inclined plane and
2
F
is the force required
to just prevent the block from sliding down, then
the value of
12
FF
is : [Use
2
g 10m / s
]
(1)
25 3 N
(2)
50 3 N
(3)
53
N
2
(4) 10 N
Ans. () BONUS
Sol.
K
f mg cos
50 3
0.1 2
2.5 3 N
1K
F mg sin f
25 2.5 3
2K
F mg sin f
25 2.5 3
12
F F 5 3 N
43. By what percentage will the illumination of the
lamp decrease if the current drops by 20%?
(1) 46% (2) 26%
(3) 36% (4) 56%
Ans. (3)
Sol.
2
P i R
2
int int
P I R
2
final int
P 0.8I R
% change in power =
final int
int
PP
100 (0.64 1) 100 36%
P
44. If two vectors
A
and
B
having equal magnitude
R are inclined at an angle
, then
(1)
A B 2 R sin 2
(2)
A B 2 R sin 2
(3)
A B 2 R cos 2
(4)
A B 2 R cos 2
Ans. (3)
Sol. The magnitude of resultant vector
22
R' a b 2ab cos
Here
a b R
Then
2 2 2
R' R R 2R cos
R 2 1 cos
2
2R 2cos 2
2Rcos 2
45. The mass number of nucleus having radius equal to
half of the radius of nucleus with mass number 192
is:
(1) 24 (2) 32
(3) 40 (4) 20
Ans. (1)
Sol.
2
1
R
R2
1/3 1/3
0
0 1 2
R
R A A
2
12
1
AA
8
1
192
A 24
8
46. The mass of the moon is 1/144 times the mass of a
planet and its diameter 1/16 times the diameter of a
planet. If the escape velocity on the planet is v, the
escape velocity on the moon will be:
(1)
v
3
(2)
v
4
(3)
v
12
(4)
v
6
Ans. (1)
Sol.
escape
2GM
VR
planet
2GM
VV
R
Moon
2GM 16 1 2GM
V144 R 3 R
Planet
Moon
V
V3
=
V
3
47. A small spherical ball of radius r, falling through a
viscous medium of negligible density has terminal
velocity 'v'. Another ball of the same mass but of
radius 2r, falling through the same viscous medium
will have terminal velocity:
(1)
v
2
(2)
v
4
(3) 4v (4) 2v
Ans. (1)
Sol. Since density is negligible hence Buoyancy force
will be negligible
At terminal velocity.
Mg 6 rv
1
Vr
(as mass is constant)
Now,
v r '
v ' r
r ' 2r
So,
v
v' 2
48. A body of mass 2 kg begins to move under the
action of a time dependent force given by
2
ˆˆ
F 6t i 6t j N
. The power developed by the
force at the time t is given by:
(1)
45
6t 9t W
(2)
35
3t 6t W
(3)
53
9t 6t W
(4)
35
9t 6t W
Ans. (4)
Sol.
2
ˆˆ
F 6t i 6t j N
2
ˆˆ
F ma 6ti 6t j
2
Fˆˆ
a 3ti 3t j
m
t2
3
0
3t ˆˆ
v adt i t j
2
35
P F.v 9t 6t W
49.
The output of the given circuit diagram is
(1)
(2)
(3) (4)
Ans. (3)
Sol.
If A = 0 ;
A1
A = 1 ;
A0
B = 0 ;
B1
B = 1 ;
B0
Y =
A B A B
=
11
= 0
50. Consider two physical quantities A and B related
to each other as
2
Bx
EAt
where E, x and t have
dimensions of energy, length and time
respectively. The dimension of AB is
(1)
2 1 0
L M T
(2)
2 1 1
L M T
(3)
2 1 1
L M T
(4)
0 1 1
L M T
Ans. (2)
Sol.
2
BL
22
2 2 1
x L 1
AtE TML T MT
1
A M T
2 1 1
AB L M T
SECTION-B
51. In the following circuit, the battery has an emf of 2
V and an internal resistance of
2
3
. The power
consumption in the entire circuit is ______ W.
Ans. (3)
Sol.
eq
4
R3
2
eq
V4
P 3 W
R 4 / 3
52. Light from a point source in air falls on a convex
curved surface of radius 20 cm and refractive index
1.5. If the source is located at 100 cm from the
convex surface, the image will be formed at____
cm from the object.
Ans. (200)
Sol.
2 1 2 1
v u R
1.5 1 1.5 1
v 100 20
v 100 cm
Distance from object
= 100+100
= 200 cm
53. The magnetic flux
(in weber) linked with a
closed circuit of resistance 8 Ω varies with time (in
seconds) as
2
5t 36t 1
. The induced current
in the circuit at t = 2s is _______ A.
Ans. (2)
Sol.
d10t 36
dt
at t 2, 16 V
16
i 2 A
R8
54. Two blocks of mass 2 kg and 4 kg are connected
by a metal wire going over a smooth pulley as
shown in figure. The radius of wire is
5
4.0 10
m and Young's modulus of the metal is
11 2
2.0 10 N / m
. The longitudinal strain
developed in the wire is
1
. The value of
is____. [Use
2
g 10 m / s
)
Ans. (12)
Sol.
12
12
2m m
Tg
mm
=
80 N
3
2 10 2
A r 16 10 m
Strain =
FT
AY AY
=
10 11
80 / 3 1
12
16 10 2 10
12
55. A body of mass 'm' is projected with a speed ‘u’
making an angle of 45° with the ground. The
angular momentum of the body about the point of
projection, at the highest point is expressed as
3
2 mu
Xg
. The value of 'X' is_______.
Ans. (8)
Sol.
450H
uucos450
O
22
u sin
L mu cos 2g
31
mu x 8
4 2 g
56. Two circular coils P and Q of 100 turns each have
same radius of
cm. The currents in P and R are
1 A and 2 A respectively. P and Q are placed with
their planes mutually perpendicular with their
centers coincide. The resultant magnetic field
induction at the center of the coils is
x mT
,
where x =_________.
[Use
71
04 10 TmA
]
Ans. (20)
Sol.
3
0 1 0
P
Ni 1 100
B 2 10 T
2r 2
3
0 2 0
Q
Ni 2 100
B 4 10 T
2r 2
22
net P Q
B B B
=
20 mT
x = 20
57. The distance between charges +q and –q is
2l
and
between +2 q and -2 q is
4l
. The electrostatic
potential at point P at a distance r from centre O is
9
210
ql V
r
, where the value of
is
______. (Use
9 2 2
0
19 10
4
Nm C
)
Ans. (27)
Sol.
1
P 2q
2
P 8q
P
r
600
1200
net
P 6q
9
0
32
9 10 6q
K p.r
V cos 120
rr
=
9 2 2
2
q
27 10 Nm c
r
27
58. Two identical spheres each of mass 2 kg and radius
50 cm are fixed at the ends of a light rod so that the
separation between the centers is 150 cm. Then,
moment of inertia of the system about an axis
perpendicular to the rod and passing through its
middle point is
2
20
xkg m
, where the value of x is
____.
Ans. (53)
Sol.
I
50 cm
50 cm
2 kg 2 kg
75 cm
150 cm
22
2
I mR md 2
5
22
2
2 1 3 53
I 2 2 2 kg m
5 2 4 20
X = 53
59. The time period of simple harmonic motion of
mass M in the given figure is
5
M
K
, where the
value of
is _______.
Ans. (12)
Sol.
eq 2k . k 5k
kk
3k 3
Angular frequency of oscillation
eq
k
m
5k
3m
Period of oscillation
2 3m
25k
12m
5k
60. A nucleus has mass number
1
A
and volume
1
V
.
Another nucleus has mass number
2
A
and volume
2
V
. If relation between mass number is
21
4AA
,
then
2
1
V
V
________.
Ans. (4)
Sol. For a nucleus
Volume:
3
4
VR
3
1/3
0
R R A
3
0
4
V R A
3
22
11
VA4
VA
CHEMISTRY
TEST PAPER WITH SOLUTION
SECTION-A
61. Match List I with List II
LIST – I
(Complex ion)
LIST – II
(Electronic
Configuration
A.
3
26
Cr H O
I.
0
2
2g g
te
B.
3
26
Fe H O
II.
30
2g g
te
C.
2
26
Ni H O
III.
32
2g g
te
D.
3
26
V H O
IV.
62
2g g
te
Choose the correct answer from the options given
below :
(1) A-III, B-II, C-IV, D-I
(2) A-IV, B-I, C-II, D-III
(3) A-IV, B-III, C-I, D-II
(4) A-II, B-III, C-IV, D-I
Ans. (4)
Sol:-
3
26
Cr H O
Contains
3
3 3 o
g
2g
Cr : Ar 3d : t e
3
26
Fe H O
Contains
3
3 5 2
g
2g
Fe : Ar 3d :t e
2
26
Ni H O
Contains
6
2 8 2
g
2g
Ni : Ar 3d :t e
3
26
V H O
Contains
2
3 2 o
g
2g
V : Ar 3d : t e
62. A sample of
3
CaCO
and
3
MgCO
weighed 2.21 g
is ignited to constant weight of 1.152 g. The
composition of mixture is :
(Given molar mass in g
1
mol
33
CaCO :100, MgCO :84
)
(1)
33
1.187 g CaCO 1.023g MgCO
(2)
33
1.023g CaCO 1.023g MgCO
(3)
33
1.187 g CaCO 1.187 g MgCO
(4)
33
1.023g CaCO 1.187 g MgCO
Ans. (1)
Sol:-
32
CaCO s CaO s CO g
32
MgCO s MgO s CO g
Let the weight of
3
CaCO
be x gm
weight of
3
MgCO 2.21 x gm
Moles of
3
CaCO
decomposed = moles of CaO
formed
x
100
moles of CaO formed
weight of CaO formed
x56
100
Moles of
3
MgCO
decomposed = moles of MgO
formed
2.21 x
84
moles of MgO formed
weight of MgO formed
2.21 x 40
84
2.21 x x
40 56 1.152
84 100
3
x 1.1886g weight of CaCO
& weight of
3
MgCO 1.0214g
63. Identify A and B in the following reaction sequence.
Conc. HNO3ANaOH B
Br
HCl
(i)
(ii)
A=
Br
NO2
NO2
,B=
(1) NO2
OH
NO2
NO2
NO2
(2) A=
Br
NO2
,B=
NO2
OH
Br
(3) A=
Br
NO2
,B=
OH
Br
NO2
OH
(4) A= ,B=
OH
NO2NO2
Ans. (1)
NO2
(i)NaOH NO2
NO2
OH
NO2
NaOH (Acid base
reaction)
NO2
NO2
O
NO2
Na+
NO2(ii) HCl
NO2
NO2
Br
Br
Con. HNO3
NO2
OH
NO2
Sol:-
64. Given below are two statements :
Statement I:
8
S
solid undergoes
disproportionation reaction under alkaline
conditions to form
2
223
S and S O
Statement II:
4
ClO
can undergo
disproportionation reaction under acidic condition.
In the light of the above statements, choose the
most appropriate answer from the options given
below :
(1) Statement I is correct but 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:-
1 8
S : S 12 OH2
22 2
3
4S 2S O 6H O
S : ClO
2 4
cannot undergo disproportionation
reaction as chlorine is present in it’s highest
oxidation state.
65. Identify major product ‘P’ formed in the following
reaction.
C
O
Cl
+
Anhydrous
AlCl3‘P’
(Major Product)
OC
O
Cl
(1)
COCH3
(2)
C
O
H
(3)
C
O
(4)
Ans. (4)
C
O
Cl +AlCl3
C
O
+C
O
Sol:-
..
..
..
C
O
Electrophile
C
O
H
AlCl4
C
O
AlCl4
+
66. Major product of the following reaction is –
CH3
DCl
_?
CH3
H
Cl D
(1)
CH3
H
Cl
D
(2)
CH3
HCl
D
(3)
CH3
H
Cl
D
(4)
Ans. (3 or 4)
Cl
D
HCH3
D
H
H C
3
Cl
D
HH C
3
Cl
+
CH3
DSol:-
(±) (±)
67. Identify structure of 2,3-dibromo-1-phenylpentane.
Br Br
1.
Br
Br
2.
Br
Br
3.
Br
Br
4.
Ans. (3)
Br
Br
1234 5
2, 3-dibromo -1-phenylpentane
Sol:-
68. Select the option with correct property -
(1)
2
4
4
Ni CO and NiCl
both diamagnetic
(2)
2
4
4
Ni CO and NiCl
both paramagnetic
(3)
2
4
NiCl
diamagnetic,
4
Ni CO
paramagnetic
(4)
4
Ni CO
diamagnetic,
2
4
NiCl
paramagnetic
Ans. (4)
Sol:-
4
Ni CO
diamagnetic,
3
sp
hybridisation,
number of unpaired electrons = 0
2
4
NiCl ,
paramagnetic,
3
sp
hybridisation,
number of unpaired electrons = 2
69. The azo-dye (Y) formed in the following reactions
is Sulphanilic
23
acid NaNO CH COOH X
X + Y
NH2
HSO3N = N
N = N SO H
3
1.
HO S
3N = N
HO S
3N = N
NH2
2.
HSO3N = N
NH2
3.
N = N NH2
4. HSO3
Ans. (4)
2
NH
3
SO H 3
SO H
2 3
NaNO CH COOH
3
N N O C CH
||
O
X
3
SO H
3
N N O C CH
||
O
+
2
NH
3
HO S N N2
NH
Y
Red azo-dye
Sol:-
This is known as Griess-Ilosvay test.
70. Given below are two statements :
Statement I: Aniline reacts with con.
24
H SO
followed by heating at 453-473 K gives p-
aminobenzene sulphonic acid, which gives blood
red colour in the 'Lassaigne's test'.
Statement II: In Friedel - Craft's alkylation and
acylation reactions, aniline forms salt with the
3
AlCl
catalyst. Due to this, nitrogen of aniline
aquires a positive charge and acts as deactivating
group.
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
Ans. (4)
Conc. H SO
2 4 453-473K
NH2NH2
43
NH HSO
Sol:-
SO H
3
Lassaigne’s test 2
Fe SCN
Blood red colour
71.
g
gg
C
AB
2
The correct relationship
between
P
K,
and equilibrium pressure P is
(1)
11
22
P12
P
K2
(2)
31
22
P12
P
K21
(3)
3
122
P32
P
K2
(4)
11
22
P32
P
K2
Ans. (2)
gg g C
A B 2
1 2
t = teq
Sol:-
B A C
12
P .P, P . P , P .P
1 1 1
2 2 2
1
2
BC
PA
31
22
1
2
P .P
KP
P
12
72. Choose the correct statements from the following
A. All group 16 elements form oxides of general
formula
23
EO and EO
where E = S, Se, Te and
Po. Both the types of oxides are acidic in nature.
B.
2
TeO
is an oxidising agent while
2
SO
is
reducing in nature.
C. The reducing property decreases from
2
HS
to
2
H Te
down the group.
D. The ozone molecule contains five lone pairs of
electrons.
Choose the correct answer from the options given
below:
1. A and D only 2. B and C only
3. C and D only 4. A and B only
Ans. (4)
Sol:- (A) All group 16 elements form oxides of the
23
EO and EO
type where
E S,Se,Teor Po
.
(B)
2
SO
is reducing while
2
TeO
is an oxidising
agent.
(C) The reducing property increases from
2
HS
to
2
H Te
down the group.
(D)
O
.. O
O
. .
..
..
..
..
have six lone pairs
73. Identify the name reaction.
CO, HCl CHO
Anhyd. AlCl /CuCl
3
(1) Stephen reaction
(2) Etard reaction
(3) Gatterman-koch reaction
(4) Rosenmund reduction
Ans. (3)
Sol:-
CO, HCl CHO
Anhyd. AlCl /CuCl
3
Gatterman-Koch reaction
74. Which of the following is least ionic ?
(1)
2
BaCl
(2) AgCl
(3) KCl (4)
2
CoCl
Ans. (2)
Sol:-
22
AgCl CoCl BaCl KCl
(ionic character)
Reason :
Ag
has pseudo inert gas configuration.
75. The fragrance of flowers is due to the presence of
some steam volatile organic compounds called
essential oils. These are generally insoluble in
water at room temperature but are miscible with
water vapour in vapour phase. A suitable method
for the extraction of these oils from the flowers is -
1. crystallisation
2. distillation under reduced pressure
3. distillation
4. steam distillation
Ans. (4)
Sol:- Steam distillation technique is applied to separate
substances which are steam volatile and are
immiscible with water.
76. Given below are two statements :
Statement I: Group 13 trivalent halides get easily
hydrolyzed by water due to their covalent nature.
Statement II:
3
AlCl
upon hydrolysis in acidified
aqueous solution forms octahedral
3
26
Al H O
ion.
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. Statement I is false but statement II is true
3. Both statement I and statement II are false
4. Both statement I and statement II are true
Ans. (4)
Sol:- In trivalent state most of the compounds being
covalent are hydrolysed in water. Trichlorides on
hydrolysis in water form tetrahedral
4
M OH
species, the hybridisation state of element M is
3
sp .
In case of aluminium, acidified aqueous solution
forms octahedral
3
26
Al H O
ion.
77. The four quantum numbers for the electron in the
outer most orbital of potassium (atomic no. 19) are
(1)
n 4,
l
= 2,
m 1,
1
2
s
(2)
n 4,
l
= 0,
m0
,
1
2
s
(3)
n 3,
l
= 0,
m1
,
1
2
s
(4)
n2
,
l
= 0,
m0
,
1
2
s
Ans. (2)
Sol:-
19 K
2 2 6 2 6 1
1s , 2s , 2p , 3s , 3p , 4s
.
Outermost orbital of potassium is 4s orbital
l1
n 4, l 0, m 0, s 2
.
78. Choose the correct statements from the following
A.
27
Mn O
is an oil at room temperature
B.
24
VO
reacts with acid to give
2
2
VO
C. CrO is a basic oxide
D.
25
VO
does not react with acid
Choose the correct answer from the options given
below :
1. A, B and D only
2. A and C only
3. A, B and C only
4. B and C only
Ans. (2)
Sol:- (A)
27
Mn O
is green oil at room temperature.
(B)
24
VO
dissolve in acids to give VO2+ salts.
(C) CrO is basic oxide
(D)
25
VO
is amphoteric it reacts with acid as well
as base.
79. The correct order of reactivity in electrophilic
substitution reaction of the following compounds
is :
CH3Cl NO2
A B C D
1.
B C A D
2.
D C B A
3.
A B C D
4.
B A C D
Ans. (4)
Sol:-
3
CH
shows
Mand I
.
Cl
shows
Mand I
but inductive effect
dominates.
2
NO
shows
Mand I
.
Electrophilic substitution
1
M and I
Mand I
Hence, order is
B A C D
.
80. Consider the following elements.
Group Period
A'B'
C'D'
Which of the following is/are true about
A', B', C' and D'
?
A. Order of atomic radii:
B' A' D' C'
B. Order of metallic character :
B' A' D' C'
C. Size of the element :
D' C' B' A'
D. Order of ionic radii :
B' A' D' C'
Choose the correct answer from the options given
below :
1. A only 2. A, B and D only
3. A and B only 4. B, C and D only
Ans. (2)
Sol:- In general along the period from left to right, size
decreases and metallic character decrease.
In general down the group, size increases and
metallic character increases.
B' A' size C' A' size
D' C' size D' B' size
B' A' metallic character
D' C' metallic character
''
B A size
''
D C size
C statement is incorrect.
SECTION-B
81. A diatomic molecule has a dipole moment of
1.2 D. If the bond distance is
1Å
, then fractional
charge on each atom is ______
1
10 esu
.
(Given
18
1D 10
esu cm)
Ans. (0)
Sol:-
1.2D q d
10
1.2 10 esu Å q 1Å
10
q 1.2 10 esu
82.
r k A
for a reaction, 50% of A is decomposed
in 120 minutes. The time taken for 90%
decomposition of A is ______ minutes.
Ans. (399)
Sol:-
r k A
So, order of reaction = 1
1/2
t 120min
For 90% completion of reaction
2.303 a
k log
t a x
1/2
0.693 2.303 100
log
t t 10
t 399min.
83. A compound (x) with molar mass
1
108g mol
undergoes acetylation to give product with molar
mass
1
192g mol
. The number of amino groups in
the compound (x) is ______.
Ans. (2)
2 3 3
R NH CH C Cl R NH C CH
OO
Sol:-
Gain in molecular weight after acylation with one
2
NH
group is 42.
Total increase in molecular weight = 84
Number of amino group in
84
x42
2
84. Number of isomeric products formed by mono-
chlorination of 2-methylbutane in presence of
sunlight is _______.
Ans. (6)
Cl /h
2
Cl +Cl
Cl Cl
+
*
*
Sol:-
Number of isomeric products = 6
85. Number of moles of
H
ions required by 1 mole
of
4
MnO
to oxidise oxalate ion to
2
CO
is ____.
Ans. (8)
Sol:-
22
4 2 2 2
4
2MnO 5C O 16H 2Mn 10CO 8H O
Number of moles of
H
ions required by 1
mole of
4
MnO
to oxidise oxalate ion to
2
CO
is 8
86. In the reaction of potassium dichromate, potassium
chloride and sulfuric acid (conc.), the oxidation
state of the chromium in the product is
(+)_______.
Ans. (6)
Sol:-
2 2 7 2 4
K Cr O s 4KCl s 6H SO conc.
2 2 4
2CrO Cl g 6KHSO
+ 3H2O
This reaction is called chromyl chloride test.
Here oxidation state of Cr is +6.
87. The molarity of 1L orthophosphoric acid
34
H PO
having 70% purity by weight (specific gravity
3
1.54 g cm
) is ______M.
(Molar mass of
1
34
H PO 98g mol
)
Ans. (11)
Sol:- Specific gravity (density) = 1.54 g/cc.
Volume
1L 1000ml
Mass of solution
1.54 1000
1540g
% purity of
24
H SO
is 70%
So weight of
34
H PO 0.7 1540 1078g
Mole of
34
1078
H PO 98
= 11
Molarity
11 11
1L
88. The values of conductivity of some materials at
298.15 K in
13
Sm are 2.1 10
,
16
1.0 10 ,1.2 10,
2
3.91,1.5 10 ,
73
1 10 ,1.0 10
. The number of conductors
among the materials is ______.
Ans. (4)
Sol:-
3
3
16
2
7
2.1 10
1.2 10 conductors at 298.15K
3.91
1 10
1 10 Insulator at 298.15 K
1.5 10 Semiconductor at 298.15 K
1 10
-1
Conductivity S m
Therefore number of conductors is 4.
89. From the vitamins A,
1 6 12
B , B , B ,
C, D, E and K,
the number vitamins that can be stored in our body
is ______.
Ans. (5)
Sol:- Vitamins A, D, E, K and
12
B
are stored in liver
and adipose tissue.
90. If 5 moles of an ideal gas expands from 10 L to a
volume of 100 L at 300 K under isothermal and
reversible condition then work, w, is
x
J. The
value of
x
is ______.
(Given
11
R 8.314J K mol
)
Ans. (28721)
Sol:- It is isothermal reversible expansion, so work done
negative
2
1
V
W 2.303 nRT log V
100
2.303 5 8.314 300log 10
28720.713
J
28721J
