FINAL JEE–MAIN EXAMINATION – JANUARY, 2024
(Held On Monday 29th January, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS
TEST PAPER WITH SOLUTION
SECTION-A
1. If in a G.P. of 64 terms, the sum of all the terms is
7 times the sum of the odd terms of the G.P, then
the common ratio of the G.P. is equal to
(1) 7 (2) 4
(3) 5 (4) 6
Ans. (4)
Sol.
2 3 63
a ar ar ar .... ar
=
2 4 62
7(a ar ar ..... ar )
64 64
2
a(1 r ) 7a(1 r )
1r 1r
r = 6
2. In an A.P., the sixth terms a6 = 2. If the
1 4 5
a a a
is
the greatest, then the common difference of the
A.P., is equal to
(1)
3
2
(2)
8
5
(3)
2
3
(4)
5
8
Ans. (2)
Sol.
6
a 2 a 5d 2
1 4 5
a a a a(a 3d)(a 4d)
=
(2 5d)(2 2d)(2 d)
2 2 3
f(d) 8 32d 34d 20d 30d 10d
f '(d) 2(5d 8)(3d 2)
+––
8/52/3
8
d5
3. If
2 2x , 1 x 0
f x ;
x
1 , 0 x 3
3
x , 3 x 0
gx x, 0 x 1
,
then range of (fog(x)) is
(1) (0, 1] (2) [0, 3)
(3) [0, 1] (4) [0, 1)
Ans. (3)
Sol.
2 2g(x) , 1 g(x) 0 .....(1)
f(g(x)) g(x)
1 , 0 g(x) 3 .....(2)
3
By (1)
x
And by (2)
x [ 3,0] and x [0,1]
(–3,3)
(1,1)
y=f(x)
–3 O 1
1
2/3 y=f(g(x))
Range of
f(g(x))
is [0, 1]
4. A fair die is thrown until 2 appears. Then the
probability, that 2 appears in even number of
throws , is
(1)
5
6
(2)
1
6
(3)
5
11
(4)
6
11
Ans. (3)
Sol. Required probability =
35
5 1 5 1 5 1 .....
6 6 6 6 6 6
=
5
16
25
6136
=
5
11
5. If
1
z 2i
2
, is such that
|z 1| z 1 i ,i 1
and
,R
, then
is equal to
(1) –4 (2) 3
(3) 2 (4) –1
Ans. (2)
Sol.
1
z 2i
2
z 1 z (1 i)
32i 2 i i
22
32i 2 i
22
9
2 and 4
24
3
6.
3
32
23
x
x2
11
lim cos dt
t
x2
is equal to
(1)
3
8
(2)
2
3
4
(3)
2
3
8
(4)
3
4
Ans. (3)
Sol. Using L’hopital rule
=
2
x2
0 cos x 3x
lim
2x 2
=
2
x2
sin x 3
2
lim 4
2x 2
=
2
3
8
7. In a
ABC,
suppose y = x is the equation of the
bisector of the angle B and the equation of the side
AC is 2x –y =2. If 2AB = BC and the point A and
B are respectively (4, 6) and
,
, then
2
is
equal to
(1) 42 (2) 39
(3) 48 (4) 45
Ans. (1)
Sol.
A(4,6)
B( , ) C(–2,–6)
A’(6,4)
y=x
(2,2)
D
AD : DC = 1 : 2
4 10
68
=
14
and = 14
8. Let
a, b
and
c
be three non-zero vectors such that
b
and
c
are non-collinear .if
a 5b
is collinear
with
c,b 6c
is collinear with
a
and
a b c 0
, then
is equal to
(1) 35 (2) 30
(3) – 30 (4)–25
Ans. (1)
Sol.
a 5b c
b 6c a
Eliminating
a
61
c 5b c b
1, 30
5
5, 30
9. Let
a
5, 4
, be the circumcenter of a triangle with
vertices
A a, 2 , B a, 6
and
a
C , –2
4
. Let
denote the circumradius,
denote the area and
denote the perimeter of the triangle. Then
is
(1) 60 (2) 53
(3) 62 (4) 30
Ans. (2)
Sol. A(a, –2), B(a, 6),
a
C , 2
4
,
a
O 5, 4
AO = BO
22
22
aa
(a 5) 2 (a 5) 6
44
a = 8
AB = 8, AC = 6, BC = 10
5, 24, 24
10. For
x,
22
, if
2
cosecx sinx
y x dx
cosecxsecx tanxsin x
and
x2
lim y x 0
then
y4
is equal to
(1)
11
tan 2
(2)
1
11
tan
22
(3)
1
11
tan
22
(4)
1
11
tan 2
2
Ans. (4)
Sol.
2
4
(1 sin x)cosx
y(x) dx
1 sin x
Put sinx = t
=
2
4
1tdt
t1
=
1
1
t
1t
tan C
22
x ,t 1
2
C = 0
1
11
y tan
42
2
11. If
,22
is the solution of
4cos 5sin 1
,
then the value of
tan
is
(1)
10 10
6
(2)
10 10
12
(3)
10 10
12
(4)
10 10
6
Ans. (3)
Sol.
4 5tan sec
Squaring :
2
24tan 40tan 15 0
10 10
tan 12
and
10 10
tan 12
is Rejected.
(3) is correct.
12. A function y = f(x) satisfies
2
f x sin2x sinx 1 cos x f ' x 0
with condition
f(0) = 0 . Then
f2
is equal to
(1) 1 (2) 0 (3) –1 (4) 2
Ans. (1)
Sol.
2
dy sin 2x y sin x
dx 1 cos x
I.F. = 1 + cos2x
2
y 1 cos x sin x dx
= – cosx + C
x = 0, C = 1
y1
2
13. Let O be the origin and the position vector of A
and B be
ˆ ˆ ˆ
2i 2j k
and
ˆ ˆ ˆ
2i 4j 4k
respectively. If
the internal bisector of
AOB
meets the line AB
at C, then the length of OC is
(1)
231
3
(2)
234
3
(3)
334
4
(4)
331
2
Ans. (2)
Sol.
O
C
1 : 2
3
B
A(2, 4, 4)
(2, 2, 1)
6
length of
136 2 34
OC 33
14. Consider the function
1
f : ,1 R
2
defined by
3
f x 4 2x 3 2x 1
. Consider the statements
(I) The curve y = f(x) intersects the x-axis exactly
at one point
(II) The curve y = f(x) intersects the x-axis at
x cos12
Then
(1) Only (II) is correct
(2) Both (I) and (II) are incorrect
(3) Only (I) is correct
(4) Both (I) and (II) are correct
Ans. (4)
Sol.
2
f ' x 12 2x 3 2 0
for
1,1
2
1
f0
2
f(1) > 0 (A) is correct.
3
f x 2 4x 3x 1 0
Let cos = x,
cos 3 = cos
4
=
12
x cos12
(4) is correct.
15. Let
1 0 0
A0
0
and
3 21
|2A | 2
where
,Z
,
Then a value of
is
(1) 3 (2) 5
(3) 17 (4) 9
Ans. (2)
Sol.
22
A
321
2A 2
4
A2
22
16
16
4 or 5
16. Let PQR be a triangle with
R 1,4, 2
. Suppose
M(2, 1, 2) is the mid point of PQ. The distance of
the centroid of
PQR
from the point of
intersection of the line
x 2 y z 3 x 1 y 3 z 1
and
0 2 1 1 3 1
is
(1) 69 (2) 9
(3)
69
(4)
99
Ans. (3)
Sol. Centroid G divides MR in 1 : 2
G(1, 2, 2)
Point of intersection A of given lines is (2,–6, 0)
AG 69
17. Let R be a relation on Z × Z defined by
(a, b)R(c, d) if and only if ad – bc is divisible by 5.
Then R is
(1) Reflexive and symmetric but not transitive
(2) Reflexive but neither symmetric not transitive
(3) Reflexive, symmetric and transitive
(4) Reflexive and transitive but not symmetric
Ans. (1)
Sol. (a, b)R(a, b) as ab – ab = 0
Therefore reflexive
Let (a,b)R(c,d)ad – bc is divisible by 5
bc – ad is divisible by 5 (c,d)R(a,b)
Therefore symmetric
Relation not transitive as (3,1)R(10,5) and
(10,5)R(1,1) but (3,1) is not related to (1,1)
18. If the value of the integral
2023
22
2
xsinx
2
x cosx 1 sin x dx a 2
14
1e
,
then the value of a is
(1) 3 (2)
3
2
(3) 2 (4)
3
2
Ans. (1)
Sol.
2023
/2 22
xsin x
/2
x cos x 1 sin x
I dx
11e
2023
/2 22
xsin( x)
/2
x cos x 1 sin x
I dx
11e
On Adding, we get
/2
22
/2
2I x cos x 1 sin x dx
On solving
23
I2
44
a = 3
19. Suppose
x x 1 2
3
2
2 2 tanx tan x x 1
fx 7x 3x 1
,
Then the value of f '(0) is equal to
(1)
(2) 0
(3)
(4)
2
Ans. (3)
Sol.
h0
f(h) f(0)
f '(0) lim h
=
h h 1 2
23
h0
(2 2 ) tan h tan (h h 1) 0
lim
(7h 3h 1) h
=
20. Let A be a square matrix such that
T
AA I
. Then
22
TT
1A A A A A
2
is equal to
(1)
2
AI
(2)
3
AI
(3)
2T
AA
(4)
3T
AA
Ans. (4)
Sol.
TT
AA I A A
On solving given expression, we get
2 T 2 T 2 T 2 T
1A A (A ) 2AA A (A ) 2AA
2
=
2 T 2
A[A (A ) ]
=
3T
AA
SECTION-B
21. Equation of two diameters of a circle are
2x 3y 5
and
3x 4y 7
. The line joining the
points
22,4
7
and
1,3
7
intersects the circle
at only one point
P,
. Then
17
is equal to
Ans. (2)
Sol. Centre of circle is (1, 1)
C(1,–1)
A(–22/7,–4) B(–1/7,3)
P( , )
Equation of AB is 7x – 3y + 10 = 0 …(i)
Equation of CP is 3x + 7y + 4 = 0 …(ii)
Solving (i) and (ii)
41 1
,
29 29
17 2
22. All the letters of the word "GTWENTY" are
written in all possible ways with or without
meaning and these words are written as in a
dictionary. The serial number of the word
"GTWENTY" IS
Ans. (553)
Sol. Words starting with E = 360
Words starting with GE = 60
Words starting with GN = 60
Words starting with GTE = 24
Words starting with GTN = 24
Words starting with GTT = 24
GTWENTY = 1
Total = 553
23. Let
,
be the roots of the equation
2
x x 2 0
with
Im Im
. Then
6 4 4 2
5
is
equal to
Ans. (13)
Sol.
6 4 4 2
5
=
4 4 2 2
( 2) 5 ( 2)
=
5 4 2 2
5 4 4
=
3 4 2
( 2) 5 2 4 4
=
32
2 5 3 2
=
2
2 ( 2) 5 3 2
=
2
7 4 3 2
=
7( 2) 4 3 2
=
3 3 16 3(1) 16 13
24. Let
x2
f x 2 x ,x R
. If m and n are
respectively the number of points at which the
curves y = f(x) and y = f '(x) intersects the x-axis,
then the value of m + n is
Ans. (5)
Sol.
2 4
y=x2
y=2x
m = 3
x
f '(x) 2 ln 2 2x 0
x
2 ln2 2x
1
y=2 ln2
x
y=2x
n = 2
m + n = 5
25. If the points of intersection of two distinct conics
22
x y 4b
and
22
2
xy1
16 b
lie on the curve
22
y 3x
, then
33
times the area of the rectangle
formed by the intersection points is __
Ans. (432)
Sol. Putting
22
y 3x
in both the conics
We get
2
xb
and
b3
1
16 b
b 4,12
(b = 4 is rejected because curves
coincide)
b = 12
Hence points of intersection are
12, 6
area of rectangle = 432
26. If the solution curve
y y x
of the differential
equation
2e
1 y 1 log x dx x
dy = 0, x > 0
passes through the point (1, 1) and
3
tan 2
y e ,
3
tan 2
then
2
is
Ans. (3)
Sol.
2
1 ln x dy
dx 0
xx 1y
21
(ln x)
ln x tan y C
2
Put x = y = 1
C4
21
(ln x)
ln x tan y
24
Put x = e
3
1 tan
32
y tan 3
42 1 tan 2
1, 1
+ 2 = 3
27. If the mean and variance of the data 65, 68, 58, 44,
48, 45, 60,
, ,60
where
are 56 and 66.2
respectively, then
22
is equal to
Ans. (6344)
Sol.
x 56
266.2
22
25678 (56) 66.2
10
22
6344
28. The area (in sq. units) of the part of circle
22
x y 169
which is below the line
5x y 13
is
1
65 12
sin
2 2 13
where
,
are coprime
numbers. Then
is equal to
Ans. (171)
Sol.
(5,12)
(0,–13)
Area =
12
2
13
1
169 y dy 25 5
2
=
1
169 65 169 12
sin
2 2 2 2 13
171
29. If
11 11 11
1 2 9
C C C n
.....
2 3 10 m
with gcd(n, m) = 1,
then n +m is equal to
Ans. (2041)
Sol.
11
9r
r1
C
r1
=
912 r1
r1
1C
12
=
12
1 2035
2 26
12 6
m + n = 2041
30. A line with direction ratios 2, 1, 2 meets the lines
x = y +2 = z and x + 2 = 2y = 2z respectively at
the point P and Q. if the length of the
perpendicular from the point (1, 2, 12) to the line
PQ is l, then l2 is
Ans. (65)
Sol. Let
P(t,t 2, t)
and
Q(2s 2,s,s)
D.R’s of PQ are 2, 1, 2
2s 2 t s t 2 s t
2 1 2
t 6 and s 2
P(6,4,6) and Q(2,2,2)
x 2 y 2 z 2
PQ : 2 1 2
Let
F(2 2, 2,2 2)
A(1,2,12)
AF·PQ 0
2
So F(6,4, 6) and
AF 65
A
PQ
F
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
31. In the given circuit, the breakdown voltage of the
Zener diode is 3.0 V. What is the value of Iz?
(1) 3.3 mA (2) 5.5 mA
(3) 10 mA (4) 7 mA
Ans. (2)
Sol.
Vz = 3V
Let potential at B = 0 V
Potential at E(VE) = 10 V
VC = VA = 3 V
Iz + I1 = I
10 – 3 7
IA
1000 1000
1
3
IA
2000
Therefore
z
7 –1.5
I 5.5mA
1000
32. The electric current through a wire varies with time
as I = I0 + t. where I0 = 20 A and = 3 A/s. The
amount of electric charge crossed through a section
of the wire in 20 s is :
(1) 80 C (2) 1000 C
(3) 800 C (4) 1600 C
Ans. (2)
Sol. Given that
Current I = I0 + t
0
I 20A
3A/s
I = 20 + 3t
dq 20 3t
dt
q20
00
dq 20 3t dt
20 20
00
q 20dt 3tdt
20
2
0
3t
q 20t 1000 C
2
33. Given below are two statements:
Statement I : If a capillary tube is immersed first
in cold water and then in hot water, the height of
capillary rise will be smaller in hot water.
Statement II : If a capillary tube is immersed first
in cold water and then in hot water, the height of
capillary rise will be smaller in cold water.
In the light of the above statements, choose the
most appropriate from the options given below
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Ans. (3)
Sol. Surface tension will be less as temperature
increases
2T cos
hgr
Height of capillary rise will be smaller in hot water
and larger in cold water.
34. A convex mirror of radius of curvature 30 cm
forms an image that is half the size of the object.
The object distance is :
(1) –15 cm (2) 45 cm
(3) –45cm (4) 15 cm
Ans. (1)
Sol.
Given R = 30 cm
f = R/2 = +15 cm
Magnification (m) =
1
2
For convex mirror, virtual image is formed for real
object.
Therefore, m is +ve
1f
2 f u
u = – 15 cm
35. Two charges of 5Q and –2Q are situated at the
points (3a, 0) and (–5a, 0) respectively. The
electric flux through a sphere of radius '4a' having
center at origin is :
(1)
0
2Q
(2)
0
5Q
(3)
0
7Q
(4)
0
3Q
Ans. (2)
Sol.
5Q charge is inside the spherical region
flux through sphere =
0
5Q
36. A body starts moving from rest with constant
acceleration covers displacement S1 in first (p – 1)
seconds and S2 in first p seconds. The displacement
S1 + S2 will be made in time :
(1)
2p 1 s
(2)
2
2p 2p 1 s
(3)
2p 1 s
(4)
2
2p 2p 1 s
Ans. (2)
Sol. S1 in first (p – 1) sec
S2 in first p sec
2
1
1
S a p 1
2
2
2
1
S a p
2
2
12
1
S S at
2
(p – 1)2 + p2 = t2
2
t 2p 1 2p
37. The potential energy function (in J) of a particle in
a region of space is given as U = (2x2 + 3y3+ 2z).
Here x, y and z are in meter. The magnitude of
x - component of force (in N) acting on the particle
at point P (1, 2, 3) m is :
(1) 2 (2) 6
(3) 4 (4) 8
Ans. (3)
Sol. Given U = 2x2 + 3y3 + 2z
x
U
F 4x
x
At x = 1 magnitude of Fx is 4N
38. The resistance
V
RI
where
V 200 5 V
and
I 20 0.2 A
, the percentage error in the
measurement of R is :
(1) 3.5%
(2) 7%
(3) 3%
(4) 5.5%
Ans. (1)
Sol.
V
Rl
According to error analysis
dR dV dI
R V I
dR 5 0.2
R 200 20
dR 7
R 200
% error
dR 7
100 100 3.5%
R 200
39. A block of mass 100 kg slides over a distance of
10 m on a horizontal surface. If the co-efficient of
friction between the surfaces is 0.4, then the work
done against friction (in J) is :
(1) 4200
(2) 3900
(3) 4000
(4) 4500
Ans. (3)
Sol. Given m = 100 kg
s = 10 m
= 0.4
As f = mg = 0.4 × 100 × 10 = 400 N
Now W = f.s = 400 ×10 = 4000 J
40. Match List I with List II
List I
List II
A.
E
0 c 0 0
d
B.dl i dt
I.
Gauss’
law for
electricity
B.
B
d
E.dl dt
II.
Gauss'
law for
magnetism
C.
0
Q
E.dA
III.
Faraday
law
D.
B.dA 0
IV.
Ampere –
Maxwell
law
Chose the correct answer from the options given
below
(1) A-IV, B-I, C-III, D-II
(2) A-II, B-III, C-I, D-IV
(3) A-IV, B-III, C-I, D-II
(4) A-I, B-II, C-III, D-IV
Ans. (3)
Sol. Ampere – Maxwell law
E
0 c 0 0
d
B.dl i dt
Faraday law
B
d
E.dl dt
Gauss’ law for electricity
0
Q
E.dA
Gauss ‘ law for magnetism
B.dA 0
41. If the radius of curvature of the path of two
particles of same mass are in the ratio 3:4, then in
order to have constant centripetal force, their
velocities will be in the ratio of:
(1)
3 : 2
(2)
1: 3
(3)
3 :1
(4)
2 : 3
Ans. (1)
Sol. Given m1 = m2
and
1
2
r3
r4
As centripetal force
2
mv
Fr
In order to have constant (same in this question)
centripetal force
F1 = F2
22
1 1 2 2
12
m v m v
rr
11
22
vr3
v r 2
42. A galvanometer having coil resistance 10 shows
a full scale deflection for a current of 3mA. For it
to measure a current of 8A, the value of the shunt
should be:
(1) 3 × 10–3 (2) 4.85 × 10–3
(3) 3.75 × 10–3 (4) 2.75 × 10–3
Ans. (3)
Sol. Given G = 10
Ig = 3mA
I = 8A
In case of conversion of galvanometer into
ammeter.
We have IgG = (I – Ig)S
g
g
IG
SII
–3
3 10 10
S8 – 0.003
= 3.75 ×10–3
43. The de-Broglie wavelength of an electron is the
same as that of a photon. If velocity of electron is
25% of the velocity of light, then the ratio of K.E.
of electron and K.E. of photon will be:
(1)
1
1
(2)
1
8
(3)
8
1
(4)
1
4
Ans. (2)
Sol. For photon
Pp
pP
hc hc
EE
For electron
e
e
e e e
hv
h
m v 2K
Given ve = 0.25 c
e
ee
h 0.25c hc
2K 8K
Also
pe
pe
hc hc
E 8K
e
p
K1
E8
44. The deflection in moving coil galvanometer falls
from 25 divisions to 5 division when a shunt of
24 is applied. The resistance of galvanometer
coil will be :
(1) 12 (2) 96
(3) 48 (4) 100
Ans. (2)
Sol. Let x = current/division
After applying shunt
Now 5x × G = 20x × 24
G = 4 × 24
G = 96
45. A biconvex lens of refractive index 1.5 has a focal
length of 20 cm in air. Its focal length when
immersed in a liquid of refractive index 1.6 will
be:
(1) – 16 cm
(2) – 160 cm
(3) + 160 cm
(4) + 16 cm
Ans. (2)
Sol. l = 1.5
m = 1.6
fa = 20 cm
As
lm
m
a l m
1
f
f
m1.5 1 1.6
f
20 1.5 1.6
fm = –160 cm
46. A thermodynamic system is taken from an original
state A to an intermediate state B by a linear
process as shown in the figure. It's volume is then
reduced to the original value from B to C by an
isobaric process. The total work done by the gas
from A to B and B to C would be :
(1) 33800 J (2) 2200 J
(3) 600 J (4) 1200 J
Ans. (BONUS)
Sol.
Work done AB =
1
2
(8000 + 6000) Dyne/cm2 ×
4m3 = (6000Dyne/cm2) × 4m3
Work done BC = –(4000 Dyne/cm2) × 4m3
Total work done = 2000 Dyne/cm2 × 4m3
33
52
1N
2 10 4m
10 cm
23
–4 2
N
2 10 4m
10 m
= 2 × 102 × 4 Nm = 800 J
47. At what distance above and below the surface of
the earth a body will have same weight, (take
radius of earth as R.)
(1)
5R R
(2)
3R R
2
(3)
R
2
(4)
5R R
2
Ans. (4)
Sol.
2
p2
gR
g
Rh
q
h
g g 1 R
gp = gq
2
gh
g1 R
h
1R
2
2
hh
1 1 1
RR
Take
hx
R
So
x3 – x + x2 = 0
51
x2
R
h 5 1
2
48. A capacitor of capacitance 100 F is charged to a
potential of 12 V and connected to a 6.4 mH
inductor to produce oscillations. The maximum
current in the circuit would be :
(1) 3.2 A (2) 1.5 A
(3) 2.0 A (4) 1.2 A
Ans. (2)
Sol. By energy conservation
22
max
11
CV LI
22
max
C
I V
L
–6
–3
100 10 12
6.4 10
12 3 1.5 A
82
49. The explosive in a Hydrogen bomb is a mixture of
1H2, 1H3 and 3Li6 in some condensed form. The
chain reaction is given by
3Li6 + 0n12He4 + 1H3
1H2 + 1H32He4 + 0n1
During the explosion the energy released is
approximately
[Given : M(Li) = 6.01690 amu. M (1H2) = 2.01471
amu. M (2He4) = 4.00388amu, and 1 amu = 931.5
MeV]
(1) 28.12 MeV (2) 12.64 MeV
(3) 16.48 MeV (4) 22.22 MeV
Ans. (4)
Sol. 3Li6 + 0n12He4 + 1H3
1H2 + 1H32He4 + 0n1
6 2 4
3 1 2
Li H 2 He
Energy released in process
Q = mc2
Q = [M(Li)+ M (1H2) –2 × M(2He4)] × 931.5 MeV
Q = [6.01690+2.01471–2 × 4.00388] × 931.5 MeV
Q = 22.216 MeV
Q = 22.22 MeV
50. Two vessels A and B are of the same size and are
at same temperature. A contains 1g of hydrogen
and B contains 1g of oxygen. PA and PB are the
pressures of the gases in A and B respectively, then
A
B
P
P
is :
(1) 16 (2) 8 (3) 4 (4) 32
Ans. (1)
Sol.
A A A A
B B B B
P V n RT
P V n RT
Given VA =VB
And TA = TB
AA
BB
Pn
Pn
A
B
P1/ 2 16
P 1/ 32
SECTION-B
51. When a hydrogen atom going from n = 2 to n = 1
emits a photon, its recoil speed is
x
5
m/s. Where
x = ______ . (Use : mass of hydrogen atom
= 1.6 × 10–27 kg)
Ans. (17)
Sol.
–3.4 eV
–13.6 eV
n = 2
n = 1
E = 10.2 eV
Recoil speed(v) =
E
mc
=
–27 8
10.2eV
1.6 10 3 10
–19
–27 8
10.2 1.6 10
1.6 10 3 10
v = 3.4 m/s =
17
5
m/s
Therefore, x = 17
52. A ball rolls off the top of a stairway with
horizontal velocity u. The steps are 0.1 m high and
0.1 m wide. The minimum velocity u with which
that ball just hits the step 5 of the stairway will be
x
ms–l where x = ___________[use g = 10 m/s2].
Ans. (2)
Sol.
0
1
2
3
4
5
Just miss
(for minimum speed)
The ball needs to just cross 4 steps to just hit 5th
step
Therefore, horizontal range (R) = 0.4 m
R = u.t
Similarly, in vertical direction
2
1
h gt
2
2
1
0.4 gt
2
2
1 0.4
0.4 g
2u
u2 = 2
u =
2
m/s
Therefore, x = 2
53. A square loop of side 10 cm and resistance 0.7 is
placed vertically in east-west plane. A uniform
magnetic field of 0.20 T is set up across the plane
in north east direction. The magnetic field is
decreased to zero in 1 s at a steady rate. Then,
magnitude of induced emf is
x
× 10–3V. The
value of x is ________.
Ans. (2)
Sol.
2ˆ
A 0.1 j
0.2 0.2
ˆˆ
B i j
22
Magnitude of induced emf
3
B·A 0
e 2 10 V
t1
54. A cylinder is rolling down on an inclined plane of
inclination 60°. It's acceleration during rolling
down will be
x
3
m/s2, where x = _________.
(use g = 10 m/s2).
Ans. (10)
Sol.
For rolling motion,
cm
2
gsin
aI
1MR
gsin
a1
12
3
2 10 2
3
10
3
Therefore x = 10
55. The magnetic potential due to a magnetic dipole at
a point on its axis situated at a distance of 20 cm
from its center is 1.5 × 10–5Tm. The magnetic
moment of the dipole is__________Am2.
(Given :
0
4
= 10–7TmA–1)
Ans. (6)
Sol.
0
2
M
V4r
–5 –7
2
–2
M
1.5 10 10
20 10
–5 –4
–7
1.5 10 20 20 10
M10
M = 1.5 × 4 = 6
56. In a double slit experiment shown in figure, when
light of wavelength 400 nm is used, dark fringe is
observed at P. If D = 0.2 m. the minimum distance
between the slits S1 and S2 is ______ mm.
Ans. (0.20)
Sol. Path difference for minima at P
22
2 D d 2D 2
22
D d D 4
22
D d D
4
2
2 2 2 D
D d D 16 2
2
2D
d2 16
9 14
20.2 400 10 4 10
d24
2 10
d 400 10
5
d 20 10
d 0.20mm
57. A 16 wire is bend to form a square loop. A 9V
battery with internal resistance 1 is connected
across one of its sides. If a 4F capacitor is
connected across one of its diagonals, the energy
stored by the capacitor will be
x
2
J. where
x = _________.
Ans. (81)
Sol.
eq
V
IR
eq
V 9 9
I12 4
R4
112 4
19 4 9
I4 16 16
A B 1 99
V V I 8 8
16 2
V
1 81
U 4 J
24
81
UJ
2
x 81
58. When the displacement of a simple harmonic
oscillator is one third of its amplitude, the ratio of
total energy to the kinetic energy is
x
8
, where
x = _________.
Ans. (9)
Sol. Let total energy = E =
2
1KA
2
22
1 A KA E
UK
2 3 2 9 9
E 8E
KE E 99
Ratio
Total E
8E
KE 9
=
9
8
x = 9
59. An electron is moving under the influence of the
electric field of a uniformly charged infinite plane
sheet S having surface charge density +. The
electron at t = 0 is at a distance of 1 m from S and
has a speed of 1 m/s. The maximum value of if
the electron strikes S at t = l s is
0
2
mC
em
the value of is
Ans. (8)
Sol.
0
e
u 1 m / s; a 2m
t 1 s
S = –1 m
Using
2
1
S ut at
2
2
0
1e
1 1 1 1
2 2 m
0m
8e
8
60. In a test experiment on a model aeroplane in wind
tunnel, the flow speeds on the upper and lower
surfaces of the wings are 70 ms–1 and 65 ms–1
respectively. If the wing area is 2 m2 the lift of the
wing is _______ N.
(Given density of air = 1.2 kg m–3)
Ans. (810)
Sol.
22
12
1
F v v A
2
22
1
F 1.2 70 65 2
2
= 810 N
CHEMISTRY
TEST PAPER WITH SOLUTION
SECTION-A
61. Given below are two statements : one is labelled as
Assertion A and the other is labelled as Reason R:
Assertion A: The first ionisation enthalpy decreases
across a period.
Reason R: The increasing nuclear charge outweighs
the shielding across the period.
In the light of the above statements, choose the
most appropriate from the options given below:
(1) Both A and R are true and R is the correct
explanation of A
(2) A is true but R is false
(3) A is false but R is true
(4) Both A and R are true but R is NOT the
correct explanation of A
Ans. (3)
Sol. First ionisation energy increases along the period.
Along the period Z increases which outweighs the
shielding effect
62. Match List I with List II
LIST-I LIST-II
(Substances) (Element Present)
A.Ziegler catalyst I.Rhodium
B.Blood Pigment II. Cobalt
C.Wilkinson catalyst III.Iron
D.Vitamin B12 IV.Titanium
Choose the correct answer from the options given
below:
(1) A-II, B-IV, C-I, D-III
(2) A-II, B-III, C-IV, D-I
(3) A-III, B-II, C-IV, D-I
(4) A-IV, B-III, C-I, D-II
Ans. (4)
Sol. Ziegler catalyst Titanium
Blood pigment Iron
Wilkinson catalyst Rhodium
Vitamin B12Cobalt
63. In chromyl chloride test for confirmation of Cl–
ion, a yellow solution is obtained. Acidification of
the solution and addition of amyl alcohol and 10%
H2O2 turns organic layer blue indicating formation
of chromium pentoxide. The oxidation state of
chromium in that is
(1)+6 (2)+5
(3)+10 (4)+3
Ans. (1)
Sol.
Basic medium 2
2 2 7 2 4 2 2 4
yellow solution
Cl K Cr O H SO CrO Cl CrO Cl
22
1.Acidification
2. Amyl alcohol
2
45
3.10% H O
yellow solution blue compound
CrO CrO
Cr
O
O
O–2
O–1
O–1
–1
–1
+6
64. The difference in energy between the actual
structure and the lowest energy resonance structure
for the given compound is
(1) electromeric energy
(2) resonance energy
(3) ionization energy
(4) hyperconjugation energy
Ans. (2)
Sol. The difference in energy between the actual structure
and the lowest energy resonance structure for the
given compound is known as resonance energy.
65. Given below are two statements :
Statement I : The electronegativity of group 14
elements from Si to Pb gradually decreases.
Statement II : Group 14 contains non-metallic,
metallic, as well as metalloid elements.
In the light of the above statements, choose the
most appropriate from the options given below :
(1) Statement I is false but Statement II is true
(2) Statement I is true but Statement II is false
(3) Both Statement I and Statement II are true
(4) Both Statement I and Statement II are false
Ans. (1)
Sol. Gr-14 EN
C 2.5
Si 1.8
Ge 1.8
Sn 1.8
Pb 1.9
The electronegativity values for elements from Si
to Pb are almost same. So Statement I is false.
66. The correct set of four quantum numbers for the
valence electron of rubidium atom (Z = 37) is:
(1)
1
5,0,0, 2
(2)
1
5,0,1, 2
(3)
1
5,1,0, 2
(4)
1
5,1,1, 2
Ans. (1)
Sol. Rb = [Kr]5s1
n = 5
l = 0
m = 0
s = +½ or –½
67. The major product(P) in the following reaction is
(1) (2)
(3) (4)
Ans. (4)
Sol.
C H = C H 2
O C H C H
23
C o n c H B r (e x c e s s) (P )
68. The arenium ion which is not involved in the
bromination of Aniline is .
(1) (2)
(3) (4)
Ans. (3)
Sol. Since
2
H
N
group is o/p directing hence arenium
ion will not be formed by attack at meta position
i.e.
NH2
H
Br
Hence Answer is (3)
69. Appearance of blood red colour, on treatment of
the sodium fusion extract of an organic compound
with FeSO4 in presence of concentrated H2SO4
indicates the presence of element/s
(1) Br (2) N
(3) N and S (4) S
Ans. (3)
Sol.
24
H
23
Conc.H SO
Fe Fe
SCN
3
3
Fe Fe SCN blood red colour
Appearance of blood red colour indicates presence
of both nitrogen and sulphur.
70. Given below are two statements : one is labelled as
Assertion A and the other is labelled as Reason R :
Assertion A : Aryl halides cannot be prepared by
replacement of hydroxyl group of phenol by
halogen atom.
Reason R : Phenols react with halogen acids violently.
In the light of the above statements, choose the
most appropriate from the options given below:
(1) Both A and R are true but R is NOT the
correct explanation of A
(2) A is false but R is true
(3) A is true but R is false
(4) Both A and R are true and R is the correct
explanation of A
Ans. (3)
Sol. Assertion (A): Given statement is correct because
in phenol hydroxyl group cannot be replaced by
halogen atom.
Reason (R) :
OH
HX N o R eac tio n
Given reason is false
Hence Assertion (A) is correct but Reason (R) is false
71. Identify product A and product B :
(1)
(2)
(3)
(4)
Ans. (4)
Sol.
+Cl2
Cl
(product A )
(Formed by free radical mechanism)
Cl
CCl4Cl
Cl (Formed by electrophilic
addition reaction on alkene)
(Product B)
hv
Hence correct Ans. (4)
72. Identify the incorrect pair from the following :
(1) Fluorspar- BF3
(2) Cryolite-Na3AlF6
(3) Fluoroapatite-3Ca3(PO4)2.CaF2
(4) Carnallite-KCl.MgCl2.6H2O
Ans. (1)
Sol. (1) Fluorspar is CaF2
73. The interaction between bond and lone pair of
electrons present on an adjacent atom is
responsible for
(1) Hyperconjugation
(2) Inductive effect
(3) Electromeric effect
(4) Resonance effect
Ans. (4)
Sol. It is a type of conjugation responsible for resonance.
74. KMnO4 decomposes on heating at 513K to form
O2 along with
(1) MnO2 & K2O2
(2) K2MnO4 & Mn
(3) Mn & KO2
(4) K2MnO4 & MnO2
Ans. (4)
Sol.
4 2 4 2 2
KMnO K MnO MnO O
75. In which one of the following metal carbonyls, CO
forms a bridge between metal atoms?
(1) [Co2(CO)8] (2) [Mn2(CO)10]
(3) [Os3(CO)12] (4) [Ru3(CO)12]
Ans. (1)
Sol. (1)
Co CoOC
OC
OC
CO
CO
CO
CO
CO
(2)
Mn Mn
CO
CO
CO
CO
CO
OC
OC
OC
OC
OC
(3)
Os CO
CO
CO
CO
Os
OC
OC
OC
OC Os
OC CO
COCO
(4)
Ru CO
CO
CO
CO
Ru
OC
OC
OC
OC Ru
OC CO
COCO
76. Type of amino acids obtained by hydrolysis of
proteins is :
(1)
(2)
(3)
(4)
Ans. (2)
Sol. Proteins are natural polymers composed of -amino
acids which are connected by peptide linkages.
Hence proteins upon acidic hydrolysis produce
-amino acids.
77. The final product A formed in the following
multistep reaction sequence is
(1)
(2)
(3)
(4)
Ans. (1)
Sol.
78. Which of the following is not correct?
(1)
G
is negative for a spontaneous reaction
(2)
G
is positive for a spontaneous reaction
(3)
G
is zero for a reversible reaction
(4)
G
is positive for a non-spontaneous reaction
Ans. (2)
Sol.
P,T
( G) ve
for non-spontaneous process
79. Chlorine undergoes disproportionation in alkaline
medium as shown below :
a Cl2(g) + b OH–(aq)
c ClO–(aq) + d Cl–(aq)
+ e H2O(l)
The values of a, b, c and d in a balanced redox
reaction are respectively :
(1) 1, 2, 1 and 1 (2) 2, 2, 1 and 3
(3) 3, 4, 4 and 2 (4) 2, 4, 1 and 3
Ans. (1)
Sol.
22
Cl 2OH Cl ClO H O
80. In alkaline medium.
4
MnO
oxidises I– to
(1)
4
IO
(2)IO–
(3) I2 (4)
3
IO
Ans. (4)
Sol.
alkaline medium
4 2 2 3
2MnO H O I 2MnO 2OH IO
SECTION-B
81. Number of compounds with one lone pair of
electrons on central atom amongst following is _
O3, H2O, SF4, ClF3, NH3, BrF5, XeF4
Ans. (4)
Sol.
N
H H
H,
Br
F
F
F
F
F
82. The mass of zinc produced by the electrolysis of
zinc sulphate solution with a steady current of
0.015 A for 15 minutes is ____ × 10–4 g.
(Atomic mass of zinc = 65.4 amu)
Ans. (45.75) or (46)
Sol.
2
Zn 2e Zn
W = Z × i × t
65.4 0.015 15 60
2 96500
4
45 75 10 gm
83. For a reaction taking place in three steps at same
temperature, overall rate constant
12
3
KK
KK
. If
Ea1, Ea2 and Ea3 are 40, 50 and 60 kJ/mol
respectively, the overall Ea is____kJ/mol.
Ans. (30)
Sol.
a a a
1 2 3
E E E
1 2 1 2 RT
33
K K A A
Ke
KA
a a a
1 2 3
a
E E E
E /RT 12 RT
3
AA
A e e
A
1 2 3
a a a a
E E E E 40 50 60 30 kJ / mole.
84. For the reaction
2 4 2
N O (g) 2NO (g)
,
Kp= 0.492 atm at 300K. Kc for the reaction at same
temperature is___ × 10–2 .
(Given : R = 0.082 L atm mol–1 K–1)
Ans. (2)
Sol.
g
n
PC
K K RT
ng = 1
2
P
c
K 0.492
K 2 10
RT 0.082 300
85. A solution of H2SO4 is 31.4% H2SO4 by mass and
has a density of 1.25g/mL.The molarity of the
H2SO4 solution is____M (nearest integer)
[Given molar mass of H2SO4 = 98g mol–1]
Ans. (4)
Sol.
solute
n
M 1000
V
31.4
98 1000
100
1.25
4.005 4
86. The osmotic pressure of a dilute solution is
7 × 105 Pa at 273K. Osmotic pressure of the same
solution at 283K is____× 104 Nm–2.
Ans. (72.56) or (73)
Sol. = CRT
11
22
T
T
5
12
2
1
T 7 10 283
T 273
42
72.56 10 Nm
Cl2Cl+ ClO
0+1
1
+e
e
87. Number of compounds among the following which
contain sulphur as heteroatom is____.
Furan,Thiophene, Pyridine, Pyrrole, Cysteine,
Tyrosine
Ans. (2)
Sol.
S
Thiophene
,
SH
OH
O
HN
2
Cysteine
88. The number of species from the following which
are paramagnetic and with bond order equal to one
is____.
22
2 2 2 2 2 2 2 2
H ,He ,O ,N ,O ,F ,Ne ,B
Ans. (1)
Sol. Magnetic behaviour Bond order
H2 Diamagnetic 1
2
He
Paramagnetic 0.5
2
O
Paramagnetic 2.5
2
2
N
Paramagnetic 2
2
2
O
Diamagnetic 1
F2 Diamagnetic 1
2
Ne
Paramagnetic 0.5
B2 Paramagnetic 1
89. From the compounds given below, number of
compounds which give positive Fehling's test is____.
Benzaldehyde, Acetaldehyde, Acetone,
Acetophenone,Methanal, 4-nitrobenzaldehyde,
cyclohexane carbaldehyde.
Ans. (3)
Sol. Acetaldehyde (CH3CHO),Methanal(HCHO), and
cyclohexane carbaldehyde
CHO
.
90.
C=C
CH3
HCH3
H(i) O3
(ii) Zn/H O
2
(P)
Consider the given reaction. The total number of
oxygen atoms present per molecule of the product
(P) is____.
Ans. (1)
Sol.
C=C
CH3
HCH3
H(i) O3
(ii) Zn/H O
2
2C=O
CH3
H
Hence total number of oxygen atom present per
molecule
C=O
CH3
H
is 1.
