FINAL JEE–MAIN EXAMINATION – JANUARY, 2024
(Held On Tuesday 30
th
January, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS
TEST PAPER WITH SOLUTION
SECTION-A
1. A line passing through the point A(9, 0) makes an angle
of 30º with the positive direction of x-axis. If this line is
rotated about A through an angle of 15º in the clockwise
direction, then its equation in the new position is
(1)
9
32
yx
(2)
9
32
xy
(3)
9
32
xy
(4)
9
32
yx
Ans. (1)
Sol.
Eqn : y – 0 = tan15° (x – 9) y = (2
3
(x 9)
2. Let Sa denote the sum of first n terms an arithmetic
progression. If S20 = 790 and S10 = 145, then S15 –
S5 is :
(1) 395
(2) 390
(3) 405
(4) 410
Ans. (1)
Sol.
20 20
S 2a 19d 790
2
2a + 19d = 79 …..(1)
10 10
S 2a 9d 145
2
2a + 9d = 29 …..(2)
From (1) and (2) a = -8, d = 5
15 5 15 5
S S 2a 14d 2a 4d
22
15 5
16 70 16 20
22
= 405-10
= 395
3. If z = x + iy, xy
0 , satisfies the equation
2
z i z 0
, then |z2| is equal to :
(1) 9
(2) 1
(3) 4
(4)
1
4
Ans. (2)
Sol.
2
2
z iz
z iz
|z2| = |z|
|z|2 – |z| = 0
|z|(|z| – 1) = 0
|z| = 0 (not acceptable)
|z| = 1
|z|2 = 1
4. Let
i 2 3
ˆ ˆ ˆ
a a i a j a k
and
1 2 3
ˆ ˆ ˆ
b b i b j b k
be
two vectors such that
a 1;
a.b 2
and
b 4.
If
c 2 a b 3b
, then the angle between
b
and
c
is equal to :
(1)
12
cos 3
(2)
11
cos 3
(3)
13
cos 2
(4)
12
cos 3
Ans. (3)
Sol. Given
a 1, b 4, a.b 2
c 2 a b 3b
Dot product with
a
on both sides
c.a 6
…..(1)
Dot product with
b
on both sides
b.c 48
…..(2)
22
c.c 4 a b 9 b
22
2 2 2
c 4 a b a.b 9 b
22
c 4 1 4 4 9 16
2
c 4 12 144
2
c 48 144
2
c 192
b.c
cos bc
48
cos 192.4
48
cos 8 3.4
3
cos 23
1
33
cos cos
22
5. The maximum area of a triangle whose one vertex
is at (0, 0) and the other two vertices lie on the
curve y = -2x2 + 54 at points (x, y) and (-x, y)
where y > 0 is :
(1) 88
(2) 122
(3) 92
(4) 108
Ans. (4)
Sol.
(x, y) (-x, y)
(0, 0)
Area of
0 0 1
1x y 1
2x y 1
1xy xy xy
2
2
Area xy x 2x 54
2
dd
6x 54 0
dx dx
at x = 3
Area = 3 (-2 × 9 + 54) = 108
6. The value of
3
n
2 2 2
nk1
n
lim n k n 3k
is :
(1)
2 3 3
24
(2)
13
8 4 3 3
(3)
13 2 3 3
8
(4)
8 2 3 3
Ans. (2)
Sol.
3
n
22
nk1 422
n
lim k 3k
n 1 1
nn
3
n
22
nk1 22
1n
lim nk 3k
11
nn
1
22
0
dx
1
3 1 x x
3
22
1
22
0
1
x 1 x
13 3dx
1
321 x x 3
1
22
2
0
1 1 1 dx
21x
1
x3
11
11
0
0
11
3tan 3x tan x
22
31
2 3 2 4 8
23
13
8. 4 3 3
7. Let g :
RR
be a non constant twice
differentiable such that
13
g' g'
22
. If a real
valued function f is defined as
1
f x g x g 2 x
2
, then
(1) f”(x) = 0 for atleast two x in (0, 2)
(2) f”(x) = 0 for exactly one x in (0, 1)
(3) f”(x) = 0 for no x in (0, 1)
(4)
31
f ' f ' 1
22
Ans. (1)
Sol.
31
g' g'
g' x g' 2 x 322
f ' x ,f ' 0
2 2 2
Also
13
g' g'
11
22
f ' 0, f ' 0
2 2 2
31
f ' f ' 0
22
13
rootsin ,1 and 1,
22
f " x
is zero at least twice in
13
,
22
8. The area (in square units) of the region bounded by
the parabola y2 = 4(x – 2) and the line y = 2x - 8
(1) 8
(2) 9
(3) 6
(4) 7
Ans. (2)
Sol. Let X = x – 2
y2 = 4x, y = 2 (x + 2) – 8
y2 = 4x, y = 2x – 4
42
2
y y 4
A42
-2
4
= 9
9. Let y = y (x) be the solution of the differential
equation sec x dy + {2(1 – x) tan x + x(2 – x)}
dx = 0 such that y(0) = 2.Then y(2) is equal to :
(1) 2
(2) 2{1 – sin (2)}
(3) 2{sin (2) + 1}
(4) 1
Ans. (1)
Sol.
2
dy 2 x 1 sinx x 2x cosx
dx
Now both side integrate
2
y x 2 x 1 sinxdx x 2x sinx 2x 2 sinx dx
2
y x x 2x sinx
y 0 0 2
2
y x x 2x sin x 2
y 2 2
be the foot of perpendicular from the
point (1, 2, 3) on the line
x 3 y 1 z 4
5 2 3
.
then
19
is equal to :
(1) 102
(2) 101
(3) 99
(4) 100
Ans. (2)
Sol.
(1, 2, 3)
P( )
Let foot P (5k – 3, 2k + 1, 3k – 4)
DR's AP:5k 4, 2k 1, 3k 7
DR's Line:5, 2, 3
Condition of perpendicular lines (25k-20) + (4k-2) + (9k – 21)=0
Then
43
k38
Then
19
=101
11. Two integers x and y are chosen with replacement
from the set {0, 1, 2, 3, ….., 10}. Then the
probability that
|x y| 5
is :
(1)
30
121
(2)
62
121
(3)
60
121
(4)
31
121
Ans. (1)
Sol. If x = 0, y = 6, 7, 8, 9, 10
If x = 1, y = 7, 8, 9, 10
If x = 2, y = 8, 9, 10
If x = 3, y = 9, 10
If x = 4, y = 10
If x = 5, y = no possible value
Total possible ways = (5 + 4 + 3 + 2 + 1) × 2
= 30
Required probability
30 30
11 11 121
12. If the domain of the function
1
1e
2x
f x cos log 3 x
4
is
[ , ) y
, then
is equal to :
(1) 12
(2) 9
(3) 11
(4) 8
Ans. (3)
Sol.
2x
11
4
2x 1
4
–4 2 – |x| 4
–6 – |x| 2
–2 |x| 6
|x| 6
x [–6, 6] …(1)
Now, 3 – x 1
And x 2 …(2)
and 3 – x > 0
x < 3 …(3)
From (1), (2) and (3)
x [–6, 3) – {2}
= 6
= 3
= 2
+ + = 11
13. Consider the system of linear equation x + y + z =
4, x + 2y +
2z
=
10
, x + 3y + 4
2z = 2 +15,
where
,
R
. Which one of the following
statements is NOT correct ?
(1) The system has unique solution if
1
2
and
1
, 15
(2) The system is inconsistent if
1
2
and
1
(3) The system has infinite number of solutions if
1
2
and
15
(4) The system is consistent if
1
2
Ans. (2)
Sol. x + y + z = 4, x + 2y +
2z
=
10
, x + 3y + 4
2z = 2 +15,
2
1 1 1
1 2 2
1 3 4
=
2
21
For unique solution
1
0, 2 1 0, 2
Let
1
0, 2
y x z 2
4 1 1
0, 10 2 1
15 3 1
15 1
For infinite solution
1, 1 or 15
2
14. If the circles
22
2
x 1 y 2 r
and
22
x y 4x 4y 4 0
intersect at exactly two
distinct points, then
(1) 5 < r < 9
(2) 0 < r < 7
(3) 3 < r < 7
(4)
1r7
2
Ans. (3)
Sol. If two circles intersect at two distinct points
1 2 1 2 1 2
r r C C r r
r 2 9 16 r 2
|r – 2 | < 5 and r + 2 > 5
-5 < r – 2 < 5 r > 3 ……….(2)
-3 < r < 7 ………(1)
From (1) and (2)
3 < r < 7
15. If the length of the minor axis of ellipse is equal to
half of the distance between the foci, then the
eccentricity of the ellipse is :
(1)
5
3
(2)
3
2
(3)
1
3
(4)
2
5
Ans. (4)
Sol. 2b = ae
be
a2
2
e
e1
4
2
e5
16. Let M denote the median of the following
frequency distribution.
Class
0-4
4-8
8-12
12-16
16-20
Frequency
3
9
10
8
6
Then 20 M is equal to :
(1) 416
(2) 104
(3) 52
(4) 208
Ans. (4)
Sol.
Class
Frequency
Cumulative
frequency
0-4
3
3
4-8
9
12
8-12
10
22
12-16
8
30
16-20
6
36
NC
2
M l h
f
18 12
M 8 4
10
M = 10.4
20M = 208
17. If
4 4 2
4 4 2
4 4 2
2cos x 2sin 3 sin 2
3 2cos 2sin sin 2
2cos 3 2sin sin 2
xx
f x x x x
x x x
then
1'0
5f
is equal to ______
(1) 0
(2) 1
(3) 2
(4) 6
Ans. (1)
Sol.
4 4 2
4 4 2
4 2 2
2cos x 2sin x 3 sin 2x
3 2cos x 2sin x sin 2x
2cos x 3 2sin 4x sin 2x
2 2 1 3 3 1
R R R , R R R
4 4 2
2cos x 2sin x 3 sin 2x
3 0 3
0 3 3
f(x) = 45
f ' x 0
18. Let A (2, 3, 5) and C(-3, 4, -2) be opposite vertices
of a parallelogram ABCD if the diagonal
ˆ
ˆˆ
23 BD i j k
then the area of the parallelogram
is equal to
(1)
1410
2
(2)
1474
2
(3)
1586
2
(4)
1306
2
Ans. (2)
Sol. Area =
AC BD
ˆ ˆ ˆ
i j k
5 1 7
1 2 3
11
ˆ ˆ ˆ
17i 8j 11k 474
22
19. If 2sin3x + sin 2x cos x + 4sinx – 4 = 0 has exactly
3 solutions in the interval
n
0, 2
,
nN
, then the
roots of the equation
2
x nx n 3 0
belong
to :
(1)
0,
(2)
,0
(3)
17 17
,
22
(4) Z
Ans. (2)
Sol.
32
2sin x 2sinx.cos x 4sinx 4 0
32
2sin x 2sin x. 1 sin x 4sin x 4 0
6sinx 4 0
2
sinx 3
n = 5 (in the given interval)
2
x 5x 2 0
5 17
x2
Required interval
,0
20. Let
f : , R
22
be a differentiable function
such that
1
f 0 ,
2
If the
2
x
0x
x0
x f t dt
lim e1
,
then
2
8
is equal to :
(1) 16
(2) 2
(3) 1
(4) 4
Ans. (2)
Sol.
2
x
0
x
x0 2
2
x f t dt
lim e1
x
x
2
xx
02
x 0 x 0
f t dt e1
lim lim 1
xx
=
x0
fx
lim 1
(using L Hospital)
1
f0 2
2
1
2
82
SECTION-B
21. A group of 40 students appeared in an examination
of 3 subjects – Mathematics, Physics & Chemistry.
It was found that all students passed in at least one
of the subjects, 20 students passed in Mathematics,
25 students passed in Physics, 16 students passed
in Chemistry, at most 11 students passed in both
Mathematics and Physics, at most 15 students
passed in both Physics and Chemistry, at most 15
students passed in both Mathematics and
Chemistry. The maximum number of students
passed in all the three subjects is ________.
Ans. (10)
Sol.
MP
C
x
11 – x 0 (Maths and Physics)
x 11
x =11 does not satisfy the data.
For x = 10
Hence maximum number of students passed in all
the three subjects is 10.
22. If d1 is the shortest distance between the lines
x + 1 = 2y = -12z, x = y + 2 = 6z – 6 and d2 is the
shortest distance between the lines
x 1 y 8 z 4 x 1 y 2 z 6
,
2 7 5 2 1 3
, then the
value of
1
2
32 3d
d
is :
Ans. (16)
Sol.
21 x 1 y z x y 2 z – 1
, L : 1
1 1/ 2 1/12
:11 6
L
d1 = shortest distance between L1 & L2
=
2 1 1 2
12
a a . b b
bb
1
d2
34
x 1 y 8 z 4 x 1 y 2 z 6
L : , L :
2 7 5 2 1 3
d2 = shortest distance between L3 & L4
212
d Hence
3
1
2
32 3d 32 3 2 16
12
d3
23. Let the latus rectum of the hyperbola
22
2
xy
1
9b
subtend an angle of
3
at the centre of the
hyperbola. If b2 is equal to
1n
m
l
, where l
and m are co-prime numbers, then l2 + m2 + n2 is
equal to __________
Ans. (182)
Sol. LR subtends 60º at centre
30º
22
2
b / a b 1
tan30 ae ae 3
2
3b
e9
Also,
2 2 4
2b b 3b
e 1 1
9 9 81
b4 = 3b2 + 27
b4 - 3b2 – 27 = 0
23
b (1 13)
2
3,m 2,n 13
2 + m2 + n2 = 182
24. Let A = {1, 2, 3,….7} and let P(1) denote the
power set of A. If the number of functions
f :A P A
such that
a f a , a A
is mn, m
and
nN
and m is least, then m + n is equal to
_______.
Ans. (44)
Sol. f : A P(A)
a f(a)
That means ‘a’ will connect with subset which
contain element ‘a’.
Total options for 1 will be 26. (Because 26 subsets
contains 1)
Similarly, for every other element
Hence, total is 26 × 26 × 26× 26× 26× 26× 26 = 242
Ans. 2+42 = 44
25. The value
9
0
10x
9 dx
x1
, where [t] denotes the
greatest integer less than or equal to t, is _____.
Ans. (155)
Sol.
10x 1
x1
1
x9
10x 4
x1
2
x3
10x 9
x1
x9
1/9 2/3 9
0 1/9 2/3
I 9 0dx 1.dx 2dx
= 155
26. Number of integral terms in the expansion of
824
11
26
7 11
is equal to ________.
Ans. (138)
Sol. General term in expansion of
824
1/2 1/6
(7) (11)
is
tr + 1 = 824Cr
824 r r/6
2
(7) (11)
For integral term, r must be multiple of 6.
Hence r = 0, 6, 12, ……….822
27. Let y = y(x) be the solution of the differential
equation (1 – x2) dy =
32
xy x 2 3 1 x dx,
1 x 1,y 0
=0. If
1m
y2n
, m and n are co-
prime numbers, then m + n is equal to ________.,
Ans. (97)
Sol.
32
22
(x 2) 3(1 x )
dy xy
dx 1 x 1 x
IF =
2
2
x1
dx ln(1 x )
1 x 2
ee
2
1x
23
y 1 x 3 (x 2)dx
4
2x
y 1 x 3 2x c
4
y(0) 0
c0
1 65
y2 32
m
n
m n 97
28. Let
,N
be roots of equation x2 – 70x + = 0,
where
,N
23
. If
assumes the minimum
possible value, then
1 1 35
is
equal to :
Ans. (60)
Sol.
2
x 70x 0
= 70
=
(70 – ) =
Since, 2 and 3 does not divide
= 5, = 65, = 325
By putting value of we get the required
value 60.
29. If the function
2
1, x 2
x
fx
ax 2b, x 2
is
differentiable on R, then 48 (a + b) is equal to
_____.
Ans. (15)
Sol.
2
1; x 2
x
f(x) ax 2b; 2 x 2
1; x 2
x
Continuous at x = 2
1a2b
24
Continuous at x = -2
1a2b
24
Since, it is differentiable at x = 2
2
12ax
x
Differentiable at x = 2
14a
4
1
a16
, b
=
3
8
30. Let
2 2 2 2 2 2
1 4 8 13 19 26 ......
upto
10 terms and
10 4
n1
n
. If
4 55k 40
,
then k is equal to __________.
Ans. (353)
Sol. = 12 + 42 + 82 ….
tn = an2 + bn + c
1 = a + b + c
4 = 4a + 2b + c
8 = 9a + 3b + c
On solving we get, a =
1
2
, b =
3
2
, c = - 1
2
2
10
n1
n 3n 1
22
10 2
2
n1
4 n 3n 2
,
10 4
n1
n
10 32
n1
4 (6n 5n 12n 4)
= 55(353) + 40
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
31. Match List-I with List-II.
List-I
List-II
A.
Coefficient of viscosity
I.
[M L2T–2]
B.
Surface Tension
II.
[M L2T–1]
C.
Angular momentum
III.
[M L–1T–1]
D.
Rotational kinetic energy
IV.
[M L0T–2]
(1) A–II, B–I, C–IV, D–III
(2) A–I, B–II, C–III, D–IV
(3) A–III, B–IV, C–II, D–I
(4) A–IV, B–III, C–II, D–I
Ans. (3)
Sol.
dv
FA
dy
2 2 1
MLT L T
11
ML T
2
02
.MLT
F
S T ML T
L
21
L mvr ML T
2 2 2
1
.2
K E I ML T
32. All surfaces shown in figure are assumed to be
frictionless and the pulleys and the string are light.
The acceleration of the block of mass 2 kg is :
(1) g (2)
3
g
(3)
2
g
(4)
4
g
Ans. (2)
Sol.
T
Fixed 4 kg
2 kg
30º
2a
a
2T
40 – 2T = 4a
T – 10 = 4a
20 = 12 a
a = 5/3
2a =
3
g
33. A potential divider circuit is shown in figure. The
output voltage V0 is
V0
3.3k100100100100100100100
4V
(1) 4V (2) 2 mV
(3) 0.5 V (4) 12 mV
Ans. (3)
Sol.
4000
eq
R
41
4000 1000
iA
01
. 500 0.5
1000
V i R V
34. Young’s modules of material of a wire of length
‘L’ and cross-sectional area A is Y. If the length of
the wire is doubled and cross-sectional area is
halved then Young’s modules will be :
(1)
4
Y
(2)
4Y
(3) Y (4) 2Y
Ans. (3)
Sol. Young’s modulus depends on the material not
length and cross sectional area. So young’s
modulus remains same.
35. The work function of a substance is 3.0 eV. The
longest wavelength of light that can cause the
emission of photoelectrons from this substance is
approximately:
(1) 215 nm (2) 414 nm
(3) 400 nm (4) 200 nm
Ans. (2)
Sol. For P.E.E. :
e
hc
W
1240
3
nm eV
eV
413.33nm
max 414nm
for P.E.E.
36. The ratio of the magnitude of the kinetic energy to
the potential energy of an electron in the 5th excited
state of a hydrogen atom is :
(1) 4 (2)
1
4
(3)
1
2
(4) 1
Ans. (3)
Sol.
1
2PE KE
for each value of n (orbit)
1
2
KE
PE
37. A particle is placed at the point A of a frictionless
track ABC as shown in figure. It is gently pushed
toward right. The speed of the particle when it
reaches the point B is : (Take g = 10 m/s2).
1m
A
B
0.5m
C
(1) 20 m/s (2)
10 /ms
(3)
2 10 /ms
(4) 10 m/s
Ans. (2)
Sol. By COME
KEA + UA = KEB + UB
0 + mg(1) =
2
1mv mg 0.5
2
10 /v g m s
38. The electric field of an electromagnetic wave in
free space is represented as
0ˆ
cosE E t kz i
.
The corresponding magnetic induction vector will
be :
(1)
0ˆ
cosB E C t kz j
(2)
0ˆ
cos
E
B t kz j
C
(3)
0ˆ
CcosB E t kz j
(4)
0ˆ
cos
E
B t kz j
C
Ans. (2)
Sol. Given
0ˆ
cosE E t kz i
0ˆ
cos
E
B t kz j
C
ˆˆˆ
C E B
39. Two insulated circular loop A and B radius ‘a’
carrying a current of ‘I’ in the anti clockwise
direction as shown in figure. The magnitude of the
magnetic induction at the centre will be :
O
A
B
(1)
0
2I
a
(2)
0
2
I
a
(3)
0
2
I
a
(4)
0
2I
a
Ans. (3)
Sol.
Bnet
0
2
2
net I
Ba
40. The diffraction pattern of a light of wavelength 400
nm diffracting from a slit of width 0.2 mm is
focused on the focal plane of a convex lens of focal
length 100 cm. The width of the 1st secondary
maxima will be :
(1) 2 mm (2) 2 cm
(3) 0.02 mm (4) 0.2 mm
Ans. (1)
Sol. Width of 1st secondary maxima =
.D
a
Here
3
9
2
0.2 10
400 10
100 10
am
m
D
Width of 1st secondary maxima
92
3
400 10 100 10
0.2 10
2mm
41. Primary coil of a transformer is connected to
220 V ac. Primary and secondary turns of the
transforms are 100 and 10 respectively. Secondary
coil of transformer is connected to two series
resistance shown in shown in figure. The output
voltage (V0) is :
~
220V
7k
15k
V0
(1) 7 V (2) 15 V
(3) 44 V (4) 22 V
Ans. (1)
Sol.
11 2
22
100 22
10
NV
N
0
3
22 1 , 7
22 10
I mA V V
42. The gravitational potential at a point above the
surface of earth is
7
5.12 10 /J kg
and the
acceleration due to gravity at that point is 6.4 m/s2.
Assume that the mean radius of earth to be
6400 km. The height of this point above the earth’s
surface is :
(1) 1600 km
(2) 540 km
(3) 1200 km
(4) 1000 km
Ans. (1)
Sol.
7
5.12 10
E
E
GM
Rh
…. (i)
26.4
E
E
GM
Rh
…. (ii)
By (i) and (ii)
5
16 10 1600h m km
43. An electric toaster has resistance of 60
at room
temperature (27ºC). The toaster is connected to a
220 V supply. If the current flowing through it
reaches 2.75 A, the temperature attained by toaster
is around : (if
4
2 10 /ºC
)
(1) 694ºC
(2) 1235ºC
(3) 1694ºC
(4) 1667ºC
Ans. (3)
Sol. RT=27 = 60,
220 80
2.75
T
R
R = R0 (1+ T)
80 = 60 [1+ 2 × 10–4(T–27)]
T
1694ºC
44. A Zener diode of breakdown voltage 10V is used
as a voltage regulator as shown in the figure. The
current through the Zener diode is
20V
200500
(1) 50 mA (2) 0
(3) 30 mA (4) 20 mA
Ans. (3)
Sol.
20V
200500
I1I3
I2
10V
Zener is in breakdown region.
3
1
10 1
500 50
10 1
200 20
I
I
2 1 3
I I I
21 1 3 30
20 50 100
I mA
45. Two thermodynamical process are shown in the
figure. The molar heat capacity for process A and
B are CA and CB. The molar heat capacity at
constant pressure and constant volume are
represented by CP and CV, respectively. Choose the
correct statement.
45º
tan–1
A
B
log P
log V
O
(1)
,0
BA
CC
(2)
0
AB
C and C
(3)
P V A B
C C C C
(4)
A P V
CCC
Ans. (Bonus)
Sol. For process A
log logV P V , 1
Constant
P
PV
1
AVR
CC
…. (i)
Likewise for process B
1tanPV Cons t
11
Bv
R
CC
2
Bv
R
CC
… (ii)
Pv
C C R
…. (iii)
By (i), (ii) & (iii)
P B A v
CCCC
[No answer matching]
46. The electrostatic potential due to an electric dipole
at a distance ‘r’ varies as :
(1) r (2)
2
1
r
(3)
3
1
r
(4)
1
r
Ans. (2)
Sol.
2
coskP
Vr
& can also checked dimensionally
47. A spherical body of mass 100 g is dropped from a
height of 10 m from the ground. After hitting the
ground, the body rebounds to a height of 5m. The
impulse of force imparted by the ground to the
body is given by : (given g = 9.8 m/s2)
(1) 4.32 kg ms–1 (2) 43.2 kg ms–1
(3) 23.9 kg ms–1 (4) 2.39 kg ms–1
Ans. (4)
Sol.
fi
I P P P
M = 0.1 kg
0.1 2 9.8 5 2 9.8 10 IP
0.1 14 7 2 2.39
kg ms–1
48. A particle of mass m projected with a velocity ‘u’
making an angle of 30º with the horizontal. The
magnitude of angular momentum of the projectile
about the point of projection when the particle is at
its maximum height h is :
(1)
3
3
16
mu
g
(2)
2
3
2
mu
g
(3)
3
2
mu
g
(4) zero
Ans. (1)
Sol. L =
cosmu H
22
2
33
sin
cos 2
3 1 3
2 2 2 16
u
mu g
mu mu
gg
49. At which temperature the r.m.s. velocity of a
hydrogen molecule equal to that of an oxygen
molecule at 47ºC?
(1) 80 K (2) –73 K
(3) 4 K (4) 20 K
Ans. (4)
Sol.
3 320
3
2 32
R
RT
320 20
16
TK
50. A series L,R circuit connected with an ac source
E = (25 sin 1000 t) V has a power factor of
1
2
. If
the source of emf is changed to E = (20 sin 2000
t)V, the new power factor of the circuit will be :
(1)
1
2
(2)
1
3
(3)
1
5
(4)
1
7
Ans. (3)
Sol. E = 25 sin (1000 t)
1
cos 2
LR circuit
XL
R
Initially
1
11
1
tan tan45º
R
L
1
L
XL
21
2
, given
21
2
tan ' LL
RR
'2tan
1
cos ' 5
SECTION-B
51. The horizontal component of earth’s magnetic field
at a place is
5
3.5 10 T
. A very long straight
conductor carrying current of
2A
in the
direction from South east to North West is placed.
The force per unit length experienced by the
conductor is ……… × 10–6 N/m.
Ans. (35)
Sol.
5
3.5 10
H
BT
5
6
sin , i 2
1
sin 2 3.5 10 2
35 10 /
F i B A
FiB
Nm
52. Two cells are connected in opposition as shown.
Cell E1 is of 8 V emf and 2
internal resistance;
the cell E2 is of 2 V emf and 4
internal
resistance. The terminal potential difference of cell
E2 is:
B
A C
E1E2
Ans. (6)
Sol.
4
C
B
A
2
8V
2V
I
8 2 6 1
2 4 6
IA
Applying Kirchhoff from C to B
2 4 1
6
CB
CB
VV
V V V
= 6V
53. A electron of hydrogen atom on an excited state is
having energy En = – 0.85 eV. The maximum
number of allowed transitions to lower energy
level is ….. . .
Ans. (6)
Sol.
2
13.6 0.85
4
n
En
n
No of transition
1 4 4 1 6
22
nn
54. Each of three blocks P, Q and R shown in figure
has a mass of 3 kg. Each of the wire A and B has
cross-sectional area 0.005 cm2 and Young’s
modulus 2 × 1011 N m–2. Neglecting friction, the
longitudinal strain on wire B is _____ × 10–4.
(Take g = 10 m/s2)
R
QP A B
Ans. (2)
Sol.
R
QP T2T1
2
10 /
3
a m s
1
1
30 3
20
Ta
TN
stress
strain Y
4
2 10
55. The distance between object and its two times
magnified real image as produced by a convex lens
is 45 cm. The focal length of the lens used is
______ cm.
Ans. (10)
Sol.
2
v
u
2vu
…(i)
45 ...( )
15
v u ii
u cm
30
1 1 1
v cm
f v u
10f cm
56. The displacement and the increase in the
velocity of a moving particle in the time interval
of t to (t + 1) s are 125 m and 50 m/s,
respectively. The distance travelled by the
particle in (t + 2)th s is ____ m.
Ans. (175)
Sol. Considering acceleration is constant
2
2
50 50 /
1
125 2
125 2
100 /
v u at
u u a a m s
ut at
a
u
u m s
21
2
th
n
a
S u n
= 175 m
57. A capacitor of capacitance C and potential V has
energy E. It is connected to another capacitor of
capacitance 2 C and potential 2V. Then the loss of
energy is
,
3
xE
where x is _____ .
Ans. (2)
Sol. Energy loss =
2
12 12
12
1
2
CC VV
CC
2.
32
E
x
58. Consider a Disc of mass 5 kg, radius 2m, rotating
with angular velocity of 10 rad/s about an axis
perpendicular to the plane of rotation. An identical
disc is kept gently over the rotating disc along the
same axis. The energy dissipated so that both the
discs continue to rotate together without slipping is
_____ J.
2m
Mass = 5kg
= 10 rad/sec
Ans. (250)
Sol.
22
. 100 /
2
ii
MR
L I kgm s
22
1. . 500
22
iMR
E
J
if
LL
100 2f
I
f = 5 rad/sec
22
52
1
2 . . 5 250
22
f
EJ
250EJ
59. In a closed organ pipe, the frequency of
fundamental note is 30 Hz. A certain amount of
water is now poured in the organ pipe so that the
fundamental frequency is increased to 110 Hz. If
the organ pipe has a cross-sectional area of 2 cm2,
the amount of water poured in the organ tube is
_______ g. (Take speed of sound in air is 330 m/s)
Ans. (400)
Sol.
1
1
11
30
44
Vm
2
2
3
110
44
Vm
2,m
Change in volume =
3
400A cm
M = 400 g ;
3
1/g cm
60. A ceiling fan having 3 blades of length 80 cm each
is rotating with an angular velocity of 1200 rpm.
The magnetic field of earth in that region is 0.5 G
and angle of dip is 30º. The emf induced across the
blades is
5
10NV
. The value of N is _____ .
Ans. (32)
Sol.
4
1
sin30 10
4
v
BB
2
2 1200 / s
60
f rad
2
1
2V
B
5
32 10
V
CHEMISTRY
TEST PAPER WITH SOLUTION
SECTION-A
61. Given below are two statements:
Statement-I: The gas liberated on warming a salt
with dil H2SO4, turns a piece of paper dipped in
lead acetate into black, it is a confirmatory test for
sulphide ion.
Statement-II: In statement-I the colour of paper
turns black because of formation of lead sulphite.
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 false
(2) Statement-I is false but Statement-II is true
(3) Statement-I is true but Statement-II is false
(4) Both Statement-I and Statement-II are true.
Ans. (3)
Sol. Na2S + H2SO4 Na2SO4 + H2S
(CH3COO)2Pb + H2S PbS + 2CH3COOH
Black lead
sulphide
62.
C
Cl
O
H2
Pd-BaSO4
CHO
This reduction reaction is known as:
(1) Rosenmund reduction
(2) Wolff-Kishner reduction
(3) Stephen reduction
(4) Etard reduction
Ans. (1)
Sol.
Cl
O
H2
Pd-BaSO4
CHO
O
It is known as rosenmund reduction that is the
partial reduction of acid chloride to aldehyde
63. Sugar which does not give reddish brown precipitate
with Fehling‟s reagent is:
(1) Sucrose (2) Lactose
(3) Glucose (4) Maltose
Ans. (1)
Sol. Sucrose do not contain hemiacetal group.
Hence it does not give test with Fehling solution.
While all other give positive test with Fehling
solution
64. Given below are the two statements: one is labeled as
Assertion (A) and the other is labeled as Reason (R).
Assertion (A): There is a considerable increase in
covalent radius from N to P. However from As to
Bi only a small increase in covalent radius is
observed.
Reason (R): covalent and ionic radii in a particular
oxidation state increases down the group.
In the light of the above statement, choose the
most appropriate answer from the options given
below:
(1) (A) is false but (R) is true
(2) Both (A) and (R) are true but (R) is not the
correct explanation of (A)
(3) (A) is true but (R) is false
(4) Both (A) and (R) are true and (R) is the correct
explanation of (A)
Ans. (2)
Sol. According to NCERT,
Statement-I : Factual data,
Statement-II is true.
But correct explanation is presence of completely
filled d and f-orbitals of heavier members
65. Which of the following molecule/species is most
stable?
(1)
(2)
(3)
(4)
Ans. (1)
Sol.
it is aromatic species
66. Diamagnetic Lanthanoid ions are:
(1) Nd3+ and Eu3+ (2) La3+ and Ce4+
(3) Nd3+ and Ce4+ (4) Lu3+ and Eu3+
Ans. (2)
Sol. Ce : [Xe] 4f15d16s2 ; Ce4+ diamagnetic
La : [Xe] 4f05d16s2 ; La3+ diamagnetic
67. Aluminium chloride in acidified aqueous solution
forms an ion having geometry
(1) Octahedral
(2) Square Planar
(3) Tetrahedral
(4) Trigonal bipyramidal
Ans. (1)
Sol. AlCl3 in acidified aqueous solution forms
octahedral geometry [Al(H2O)6]3+
68. Given below are two statements:
Statement-I: The orbitals having same energy are
called as degenerate orbitals.
Statement-II: In hydrogen atom, 3p and 3d
orbitals are not degenerate orbitals.
In the light of the above statements, choose the
most appropriate answer from the options given
(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. (1)
Sol. For single electron species the energy depends upon
principal quantum number „n‟ only. So, statement II
is false.
Statement I is correct definition of degenerate orbitals.
69. Example of vinylic halide is
(1)
X
(2)
CH X
2
(3)
X
(4)
X
Ans. (1)
Sol. Vinyl carbon is sp2 hybridized aliphatic carbon
x
is vinyl halide
While
x
is aryl halide
and
CH -X
2
X
are allyl halide
70. Structure of 4-Methylpent-2-enal is
(1)
3
22
CH O
| ||
H C C C CH C H
| |
H H
(2)
32
3
O
||
CH CH C CH C H
|
CH
(3)
32
3
O
||
CH CH CH C C H
|
CH
(4)
3
3
O
||
CH CH CH CH C H
|
CH
Ans. (4)
Sol.
CH - CH - CH = CH - C - H
3
54
3 2 1
O
CH34-methylpent-2-enal
M
71. Match List-I with List-II
List-I List-II
Molecule Shape
(A) BrF5 (I) T-shape
(B) H2O (II) See saw
(C) ClF3 (III) Bent
(D) SF4 (IV) Square pyramidal
(1) (A)-I, (B)-II, (C)-IV, (D)-III
(2) (A) –II, (B)-I, (C)-III, (D)-IV
(3) (A)-III, (B)-IV, (C)-I, (D)-II
(4) (A)-IV, (B)-III, (C)-I, (D)-II
Ans. (4)
Sol. BrF5
Br
F
F
FF
F
Square pyramidal
H2O
O
H H
Bent
ClF3
F
F
F
.
.
.
.
Cl
T-shape
SF4,
F
F
FS
F
See-saw
72. The final product A, formed in the following
multistep reaction sequence is:
Br
A
(i) Mg, ether
then CO, H
2
+
(ii) NH ,
3
(iii) Br, NaOH
2
(1)
NH2
O
(2)
NH2
(3)
Br
O
(4)
OH
O
Ans. (2)
Sol.
Br Mg, ether MgBr O = C = O C–OMgBr
O
H+
COOH
NH3
Br / NaOH
2
C–NH2
O
Hoffmann bromamide
reaction
NH2
73. In the given reactions identify the reagent A and
reagent B
CH3“A” + (CH CO) O
3 2
273-283K
“B” + CS2
[Intermediate]
[Intermediate]
CHO
H O
3
+
H O
3+
(1) A-CrO3 B-CrO3
(2) A-CrO3 B-CrO2Cl2
(3) A-CrO2Cl2 B-CrO2Cl2
(4) A-CrO2Cl2 B-CrO3
Ans. (2)
Sol.
CH3CH(OCOCH )
3 2
CHO
CH[OCrCl (OH)]
2 2
CrO /(CH CO) O
3 3 2
CrO Cl
2 2
CS2
H O
3
+
H O
3
+
Etard reaction
74. Given below are two statement one is labeled as
Assertion (A) and the other is labeled as Reason (R).
Assertion (A): CH2 = CH – CH2 – Cl is an
example of allyl halide
Reason (R): Allyl halides are the compounds in
which the halogen atom is attached to sp2
hybridised carbon atom.
In the light of the two above statements, choose the
most appropriate answer from the options given
below:
(1) (A) is true but (R) is false
(2) Both (A) and (R) are true but (R) is not the
correct explanation of (A)
(3) (A) is false but (R) is true
(4) Both (A) and (R) are true and (R) is the correct
explanation of (A)
Ans. (1)
Sol.
22
CH CH CH Cl
It is allyl carbon and sp3 hybridized
75. What happens to freezing point of benzene when
small quantity of napthalene is added to benzene?
(1) Increases
(2) Remains unchanged
(3) First decreases and then increases
(4) Decreases
Ans. (4)
Sol. On addition of naphthalene to benzene there is
depression in freezing point of benzene.
76. Match List-I with List-II
List-I List-II
Species Electronic distribution
(A) Cr+2 (I) 3d8
(B) Mn+ (II) 3d34s1
(C) Ni+2 (III) 3d4
(D) V+ (IV) 3d54s1
Choose the correct answer from the options given
below:
(1) (A)-I, (B)-II, (C)-III, (D)-IV
(2) (A)-III, (B) – IV, (C) – I, (D)-II
(3) (A)-IV, (B)-III, (C)-I, (D)-II
(4) (A)-II, (B)-I, (C)-IV, (D)-III
Ans. (2)
Sol. 24Cr [Ar] 3d54s1; Cr2+ [Ar] 3d4
25Mn [Ar] 3d54s2; Mn+ [Ar] 3d54s1
28Ni [Ar] 3d84s2; Ni2+ [Ar] 3d8
23V [Ar] 3d34s2; V+ [Ar] 3d34s1
77. Compound A formed in the following reaction reacts
with B gives the product C. Find out A and B.
B
3 3 2 2
3
CH C CH Na A CH C C CH CH NaBr
(C) |
CH
(1) A=CH3 –C
CNa
, B = CH3 – CH2 – CH2 – Br
(2) A=CH3–CH= CH2, B = CH3 – CH2 – CH2 – Br
(3) A = CH3 – CH2 – CH3, B = CH3 – C CH
(4) A = CH3 – C
CNa
, B = CH3 – CH2 – CH3
Ans. (1)
Sol.
Na
3
CH C CH 3
CH C C Na
3 2 2
CH CH CH Br
NaBr + 3 2 2 3
CH C C CH CH CH
78. Following is a confirmatory test for aromatic
primary amines. Identify reagent (A) and (B)
NH2
ANCl
2
B
NaOH Scarlet red dye
(1) A = HNO3/H2SO4;
(2) A= NaNO2 + HCl, 0 – 5°C;
(3) A=NaNO2 + HCl, 0 – 5°C;
(4) A = NaNO2 + HCl, 0 – 5°C;
OH
B =
Ans. (4)
OH
B =
NH2
B =
OH
B =
83.
30
20
10
0
V
(dm )
3
P(kPa)
10 20 30
A
B
C
An ideal gas undergoes a cyclic transformation
starting from the point A and coming back to the
same point by tracing the path A B C A
as shown in the diagram. The total work done in
the process is _____ J.
Ans. (200)
Sol. Work done is given by area enclosed in the P vs V
cyclic graph or V vs P cyclic graph.
Sign of work is positive for clockwise cyclic
process for V vs P graph.
W =
1
2
× (30 – 10) × (30 - 10) = 200 kPa – dm3
= 200 × 1000 Pa – L = 2 L-bar = 200 J
84. if IUPAC name of an element is “Unununnium”
then the element belongs to nth group of periodic
table. The value of n is______
Ans. (11)
Sol. 111 belongs to 11th group
85. The total number of molecular orbitals formed from
2s and 2p atomic orbitals of a diatomic molecule
Ans. (08)
Sol. Two molecular orbitals 2s and 2s.
Six molecular orbitals 2pz and 2pz.
2px , 2py and 2px , *2py
86. On a thin layer chromatographic plate, an organic
compound moved by 3.5 cm, while the solvent
moved by 5 cm. The retardation factor of the
organic compound is _____ × 10-1
Ans. (07)
Sol. Retardation factor =
Distance travelled by
sample/organiccompound
Distance travelled by solvent
1
3.5 7 10
5
87. The compound formed by the reaction of ethanal
with semicarbazide contains _____number of
nitrogen atoms.
Ans. (03)
Sol.
CH -C = O + HN – NH – C – NH
3 2 2
HSemicarbazide
O
CH – CH = N – NH – C – NH
3 2
O
88. 0.05 cm thick coating of silver is deposited on a
plate of 0.05 m2 area. The number of silver atoms
deposited on plate are _____ × 1023. (At mass Ag
= 108, d = 7.9 g cm-3)
Ans. (11)
Sol. Volume of silver coating = 0.05 × 0.05 × 10000
= 25 cm3
Mass of silver deposited = 25 × 7.9 g
Moles of silver atoms =
25 7.9
108
Number of silver atoms =
23
25 7.9 6.023 10
108
= 11.01 × 1023
Ans. 11
89.
4 2 2 2
2MnO bI cH O xI yMnO zOH
If the above equation is balanced with integer
coefficients, the value of z is _______
Ans. (08)
Sol. Reduction Half Oxidation Half
42
2MnO 2MnO
2
2I I 2e
4 2 2
2MnO 4H O 6e 2MnO 8OH
2
6I 3I 6e
Adding oxidation half and reduction half, net
reaction is
4 2 2 2
2MnO 6I 4H O 3I 2MnO 8OH
z = 8
Ans 8
90. The mass of sodium acetate (CH3COONa) required
to prepare 250 mL of 0.35 M aqueous solution is
______ g. (Molar mass of CH3COONa is 82.02 g mol-1)
Ans. (7)
Sol. Moles = Molarity × Volume in litres
= 0.35 × 0.25
Mass = moles × molar mass
= 0.35 × 0.25 × 82.02 = 7.18 g
Ans. 7
