FINAL JEE–MAIN EXAMINATION – JANUARY, 2024
(Held On Monday 29th January, 2024) TIME : 3 : 00 PM to 6 : 00 PM
MATHEMATICS
TEST PAPER WITH SOLUTION
SECTION-A
1. Let A =
2 1 2
6 2 11
3 3 2
and P =
1 2 0
5 0 2
7 1 5
. The sum
of the prime factors of
1
P AP 2I
is equal to
(1) 26 (2) 27 (3) 66 (4) 23
Ans. (1)
Sol.
1 1 1
P AP 2I P AP 2P P
=
1
P (A 2I)P
=
1
P A 2I P
= |A-2I|
0 1 2
6 0 11
3 3 0
= 69
So, Prime factor of 69 is 3 & 23
So, sum = 26
2. Number of ways of arranging 8 identical books
into 4 identical shelves where any number of
shelves may remain empty is equal to
(1) 18 (2) 16 (3) 12 (4) 15
Ans. (4)
Sol. 3 Shelf empty : (8, 0, 0, 0) 1way
2 shelf empty :
(7,1,0,0)
(6,2,0,0) 4ways
(5,3,0,0)
(4,4,0,0)
1 shelf empty :
(6,1,1,0) (3,3,2,0)
(5,2,1,0) (4,2,2,0) 5ways
(4,3,1,0)
0 Shelf empty :
(1,2,3,2) (5,1,1,1)
(2,2,2,2) 5ways
(3,3,1,1)
(4,2,1,1)
Total = 15 ways
3. Let P(3, 2, 3), Q (4, 6, 2) and R (7, 3, 2) be the
vertices of PQR. Then, the angle QPR is
(1)
6
(2)
17
cos 18
(3)
11
cos 18
(4)
3
Ans. (4)
Sol.
P(3, 2, 3)
R(7, 3, 2)Q(4, 6, 2)
Direction ratio of PR = (4, 1, -1)
Direction ratio of PQ = (1, 4, -1)
Now,
4 4 1
cos .
18 18
3
4. If the mean and variance of five observations are
24
5
and
194
25
respectively and the mean of first
four observations is
7,
2
then the variance of the
first four observations in equal to
(1)
4
5
(2)
77
12
(3)
5
4
(4)
105
4
Ans. (3)
Sol.
2
24 194
X;
5 25
Let first four observation be x1, x2, x3, x4
Here,
12345
x x x x x 24......(1)
55
Also,
1 2 3 4
x x x x 7
42
1 2 3 4
x x x x 14
Now from eqn -1
x5 = 10
Now,
2194
25
22222
1 2 3 4 5
xxxxx576 194
5 25 25
2222
1 2 3 4
x x x x 54
Now, variance of first 4 observations
Var =
2
44
2
ii
i 1 i 1
xx
44
54 49
44
=
5
4
5. The function f(x) =
2
3
2x 3(x) ,x ,
has
(1) exactly one point of local minima and no
point of local maxima
(2) exactly one point of local maxima and no
point of local minima
(3) exactly one point of local maxima and
exactly one point of local minima
(4) exactly two points of local maxima and
exactly one point of local minima
Ans. (3)
Sol.
2
3
f(x) 2x 3(x)
1
3
f '(x) 2 2x
=
1
3
1
21
x
=
1
3
1
3
x1
2
x
+
–1
M
0
m
–+
So, maxima (M) at x = -1 & minima(m) at x = 0
6. Let r and
respectively be the modulus and
amplitude of the complex number
5
z 2 i 2 tan 8
, then
(r, )
is equal to
(1)
33
2sec ,
88
(2)
35
2sec ,
88
(3)
53
2sec ,
88
(4)
11 11
2sec ,
88
Ans. (1)
Sol. z =
5
2 i 2 tan 8
= x + iy (let)
r =
2 2 1 y
x y & tan x
r =
2
25
(2) 2 tan 8
=
53
2sec 2sec
88
= 2 sec
3
8
&
1
5
2 tan 8
tan 2
=
15
tan tan 8
=
3
8
7. The sum of the solutions
x
of the equation
3
66
3cos2x cos 2x
cos x sin x
= x3 – x2 + 6 is
(1) 0 (2) 1
(3) –1 (4) 3
Ans. (3)
Sol.
3
66
3cos2x cos 2x
cos x sin x
= x3 – x2 + 6
2
22
cos2x (3 cos 2x)
cos2x (1 sin xcos x)
= x3 – x2 + 6
2
2
4(3 cos 2x)
(4 sin 2x)
= x3 – x2 + 6
2
2
4(3 cos 2x)
(3 cos 2x)
= x3 – x2 + 6
x3 – x2 + 2 = 0 (x + 1)(x2 – 2x + 2) = 0
so, sum of real solutions = –1
8. Let
OA a,OB 12a 4b
and
OC b,
where O
is the origin. If S is the parallelogram with adjacent
sides OA and OC, then
area of the quadrilateral OABC
area of S
is equal to ___
(1) 6 (2) 10
(3) 7 (4) 8
Ans. (4)
Sol.
AB
CO
a
b
12a + 4b
Area of parallelogram,
S a b
Area of quadrilateral =Area(OAB)+Area (OBC)
=
1a (12a 4b) b (12a 4b)
2
=
8 (a b)
Ratio =
8 (a b)
(a b)
= 8
9. If loge a, loge b, loge c are in an A.P. and loge a –
loge2b, loge2b – loge3c, loge3c – loge a are also in
an A.P, then a : b : c is equal to
(1) 9 : 6 : 4 (2) 16 : 4 : 1
(3) 25 : 10 : 4 (4) 6 : 3 : 2
Ans. (1)
Sol. logea, logeb, logec are in A.P.
b2 = ac …..(i)
Also
eee
a 2b 3c
log ,log ,log
2b 3c a
are in A.P.
2
2b a 3c
3c 2b a
b3
c2
Putting in eq. (i) b2 = a ×
2b
3
a3
b2
a : b : c = 9 : 6 : 4
10. If
33
22
33
sin x cos x dx A cos tan x sin B cos sin cot x C,
sin x cos x sin(x )
where C is the integration constant, then AB is
equal to
(1)
4cosec(2 )
(2)
4sec
(3)
2sec
(4)
8cosec(2 )
Ans. (4)
Sol.
33
22
33
sin x cos x dx
sin x cos x sin(x )
I =
33
22
33
sin x cos x dx
sin x cos x (sin x cos cosx sin )
=
33
22
33
22
22
sin x cos x
dx dx
sin x cos x tan xcos sin sin x cos x cos cot x sin
=
22
sec x cosec x
dx dx
tan xcos sin cos cot x sin
I = I1 + I2 …… {Let}
For I1, let tan x cos – sin = t2
22t dt
sec xdx cos
For I2 , let
2
cos cot xsin z
22z dz
cosec x dx sin
I = I1 + I2
=
2t dt 2z dz
cos t sin z
=
2t 2z
cos sin
=
2sec tan x cos sin 2cosec cos cot xsin
Comparing
AB =
8 cosec2
11. The distance of the point (2, 3) from the line 2x –
3y + 28 = 0, measured parallel to the line
3x y 1 0,
is equal to
(1)
42
(2)
63
(3)
3 4 2
(4)
4 6 3
Ans. (4)
Sol.
P
2x – 3y + 28 = 0
r
A
(2, 3)
Writing P in terms of parametric co-ordinates 2 + r
cos , 3 + r sin as tan =
3
r 3r
P(2 ,3 )
22
P must satisfy 2x – 3y + 28 = 0
So,
r 3r
2(2 ) 3(3 ) 28 0
22
We find r =
4 6 3
12. If sin
e
ylog | x |
x2
is the solution of the
differential equation x cos
y dy y
ycos x
x dx x
and y(1) =
,
3
then
2
is equal to
(1) 3 (2) 12
(3) 4 (4) 9
Ans. (1)
Sol. Differential equation :–
y dy y
xcos y cos x
x dx x
y dy
cos x y x
x dx
Divide both sides by x2
2
dy
xy
y1
dx
cos xx
x
Let
yt
x
dt 1
cos t dx x
1
cost dt dx
x
Integrating both sides
sin t = ln | x | + c
y
sin ln x c
x
Using y(1) =
,
3
we get c =
3
2
So,
2
33
13. If each term of a geometric progression a1, a2, a3,…
with
1
1
a8
and
21
a a ,
is the arithmetic mean of
the next two terms and Sn = a1 + a2 + …+an, then
S20 – S18 is equal to
(1) 215 (2) –218
(3) 218 (4) –215
Ans. (4)
Sol. Let r’th term of the GP be arn–1. Given,
2ar = ar+1 + ar+2
2arn–1 = arn + arn+1
21r
r
r2 + r – 2 = 0
Hence, we get, r = – 2 (as r 1)
So, S20 – S18 = (Sum upto 20 terms) – (Sum upto
18 terms) = T19 + T20
T19 + T20 = ar18 (1 + r)
Putting the values a =
1
8
and r = – 2;
we get T19 + T20 = –215
14. Let A be the point of intersection of the lines 3x +
2y = 14, 5x – y = 6 and B be the point of
intersection of the lines 4x + 3y = 8, 6x + y = 5.
The distance of the point P(5, –2) from the line
AB is
(1)
13
2
(2) 8 (3)
5
2
(4) 6
Ans. (4)
Sol. Solving lines L1 (3x + 2y = 14) and L2 (5x – y = 6)
to get A(2, 4) and solving lines L3 (4x + 3y = 8)
and L4 (6x + y = 5) to get B
1,2 .
2
Finding Eqn. of AB : 4x – 3y + 4 = 0
Calculate distance PM
4(5) 3( 2) 4 6
5
15. Let x =
m
n
(m, n are co-prime natural numbers) be
a solution of the equation
11
cos 2sin x 9
and let
, ( )
be the roots of the equation mx2 – nx –
m + n = 0. Then the point
( , )
lies on the line
(1) 3x + 2y = 2 (2) 5x – 8y = –9
(3) 3x – 2y = –2 (4) 5x + 8y = 9
Ans. (4)
Sol. Assume sin–1 x =
1
cos(2 ) 9
2
sin 3
as m and n are co-prime natural numbers,
2
x3
i.e. m = 2, n = 3
So, the quadratic equation becomes 2x2 – 3x + 1 =
0 whose roots are
1
1, 2
1
1, 2
lies on 5x + 8y = 9
16. The function f(x) =
2
x,x
x 6x 16
–{–2, 8}
(1) decreases in (–2, 8) and increases in
( , 2) (8, )
(2) decreases in
( , 2) ( 2,8) (8, )
(3) decreases in
( , 2)
and increases in
(8, )
(4) increases in
( , 2) ( 2,8) (8, )
Ans. (2)
Sol. f(x) =
2
x
x 6x 16
Now,
2
22
(x 16)
f '(x) (x 6x 16)
f '(x) 0
Thus f(x) is decreasing in
( ,–2) ( 2,8) (8, )
17. Let y =
2
e2
1x
log ,
1x
–1 < x < 1. Then at x =
1,
2
the value of
225(y' y")
is equal to
(1) 732 (2) 746
(3) 742 (4) 736
Ans. (4)
Sol. y =
2
e2
1x
log 1x
4
dy 4x
y'
dx 1x
Again,
24
2 4 2
d y 4(1 3x )
y"
dx (1 x )
Again
4
4 4 2
4x 4(1 3x )
y' y" 1 x (1 x )
at
1
x,
2
736
y' y" 225
Thus 225
(y' y")
= 225 ×
736
225
= 736
18. If R is the smallest equivalence relation on the set
{1, 2, 3, 4} such that {(1,2), (1,3)}
R, then the
number of elements in R is ______
(1) 10 (2) 12
(3) 8 (4) 15
Ans. (1)
Sol. Given set {1, 2, 3, 4}
Minimum order pairs are
(1, 1), (2, 2), (3, 3), (4, 4), (3, 1),(2, 1), (2, 3),(3,2),
(1, 3), (1, 2)
Thus no. of elements = 10
19. An integer is chosen at random from the integers 1,
2, 3, …, 50. The probability that the chosen integer
is a multiple of atleast one of 4, 6 and 7 is
(1)
8
25
(2)
21
50
(3)
9
50
(4)
14
25
Ans. (2)
Sol. Given set = {1, 2, 3, …….. 50}
P(A) = Probability that number is multiple of 4
P(B) = Probability that number is multiple of 6
P(C) = Probability that number is multiple of 7
Now,
12 8 7
P(A) , P(B) , P(C)
50 50 50
again
4 1 1
P(A B) ,P(B C) ,P(A C)
50 50 50
P(A B C) 0
Thus
12 8 7 4 1 1
P(A B C) 0
50 50 50 50 50 50
=
21
50
20. Let a unit vector
ˆ ˆ ˆ
ˆ
u xi yj zk
make angles
,
23
and
2
3
with the vectors
1 1 1 1
ˆ ˆ ˆ ˆ
i k , j k
2 2 2 2
and
11
ˆˆ
ij
22
respectively. If
111
ˆ ˆ ˆ
v i j k ,
222
then
2
ˆ
uv
is equal to
(1)
11
2
(2)
5
2
(3) 9 (4) 7
Ans. (2)
Sol. Unit vector
ˆ ˆ ˆ
ˆ
u xi yj zk
12
1 1 1 1
ˆ ˆ ˆ ˆ
p i k,p j k
2 2 2 2
3
11
ˆˆ
p i j
22
Now angle between
1
ˆ
u and p 2
1
xz
ˆ
u p 0 0
22
x z 0
….(i)
Angle between
2
ˆ
u and p 3
22
ˆˆ
u p u p cos 3
y z 1 1
yz
2
2 2 2
….(ii)
Angle between
3
2
ˆ
uand p 3
33
2
ˆˆ
u p u p cos 3
x 4 1 1
xy
2
2 2 2
…. (iii)
from equation (i), (ii) and (iii) we get
1
x2
y = 0 z =
1
2
Thus
1 1 1 1 1
ˆ ˆ ˆ ˆ ˆ
ˆ
u v i k i j k
2 2 2 2 2
21
ˆˆ
ˆ
u v i j
22
2
24 1 5
ˆ
uv 2 2 2
SECTION-B
21. Let , be the roots of the equation
2
x 6x 3 0
such that Im () > Im (). Let a, b
be integers not divisible by 3 and n be a natural
number such that
99 98 n
3 (a ib),i 1
. Then
n + a + b is equal to _______.
Ans. 49
Sol.
2
x 6x 6 0
6 i 6 6
x (1 i)
22
i4
3(e )
,
i4
3(e )
99 98 98 1
=
98 ()
=
i99
49 4
3 e 2
= 349 (-1+i)
= 3n (a+ib)
n = 49, a = -1, b = 1
n + a + b = 49 – 1 + 1 = 49
22. Let for any three distinct consecutive terms a, b, c
of an A.P, the lines ax + by + c = 0 be concurrent
at the point P and Q (,) be a point such that the
system of equations
x + y + z = 6,
2x + 5y + z = and
x + 2y + 3z = 4, has infinitely many solutions.
Then (PQ)2 is equal to ________.
Ans. 113
Sol. a, b, c and in A.P
2b = a + c a – 2b + c = 0
ax + by + c passes through fixed point (1, -2)
P = (1, - 2)
For infinite solution,
D = D1 = D2 = D3 = 0
1 1 1
D: 2 5 0
1 2 3
= 8
1
6 1 1
D : 5 0
4 2 3
= 6
Q = (8,6)
PQ2 =113
23. Let P(,) be a point on the parabola y2 = 4x. If P
also lies on the chord of the parabola x2 = 8y
whose mid point is
5
1, 4
. Then (-28) (-8) is
equal to ___________.
Ans. 192
Sol. Parabola is x2 = 8y
Chord with mid point (x1,y1) is T = S1
xx1 – 4(y+y1) = x1
2-8y1
(x1, y1) =
5
1, 4
55
x 4 y 1 8 9
44
x-4y + 4 = 0 (i)
(,) lies on (i) & also on y2 = 4x
- 4 + 4 = 0 (ii)
& 2 = 4(iii)
Solving (ii) & (iii)
2 = 4(4-4) 2-16 + 16 = 0
= 8
43
and = 4 - 4 = 28
16 3
(, ) = (
28 16 3,8 4 3
) &
28 16 3,8 4 3
( 28)( 8) 16 3 4 3
= 192
24. If
3
6
1 sin2xdx 2 3
, where ,
and
are rational numbers, then 3+4-
is equal
to _______.
Ans. 6
Sol. =
3
6
1 sin2x dx
=
3
6
sinx cosx dx
=
3
4
64
cosx sinx dx sinx cosx dx
=
1 2 2 3
=
23
1, 2, 1
34
= 6
25. Let the area of the region {(x, y):
0 x 3,0 y
min{x2 + 2, 2x + 2}}be A. Then 12A is equal to
_____.
Ans. 164
Sol.
x
y(2, 6)
(3, 0)
y = x + 2
2
y = 2x + 2
(0, 2)
23
2
02
A (x 2)dx (2x 2)dx
A =
41
3
12A = 41 × 4 = 164
26. Let O be the origin, and M and N be the points on
the lines
x 5 y 4 z 5
4 1 3
and
x 8 y 2 z 11
12 5 9
respectively such that MN is
the shortest distance between the given lines. Then
.
OM ON
is equal to _____.
Ans. 9
Sol.
1x 5 y 4 z 5
L: 4 1 3
drs (4,1,3) = b1
M (4+5, + 4, 3 + 5)
2x 8 y 2 z 11
L:12 5 9
N(12 – 8, 5 -2, 9-11)
MN 4 12 13, 5 6,3 9 16
..(1)
Now
12
bb
ˆ ˆ ˆ
i j k
4 1 3
12 5 9
=
ˆˆ
6i 8k
…(2)
Equation (1) and (2)
4 12 13
6
=
56
0
=
3 9 16
8
I and II
– 5 + 6 = 0 ….(3)
I and III
– 3 + 4 = 0 ….(4)
Solve (3) and (4) we get
= –1, = 1
M (1, 3, 2)
N (4, 3, –2)
.
OM ON
= 4 + 9 – 4 = 9
27. Let
22 f (r)
3r
22
rx
2r (f(r)) f(x)f(r)
f(x) lim r e
rx
be differentiable in
( ,0) (0, )
and f(1) = 1.
Then the value of ea, such that f(a) = 0, is equal to
_____.
Ans. 2
Sol. f(1)=1, f(a) = 0
f (r)
22
23
r
22
rx
2r (f (r) f(x)f(r))
f (x) Lim r e
rx
f (r)
23r
rx
2r f(r) (f(r) f(x))
Lim r e
r x r x
f (x)
2
23
x
2x f(x)
f (x) f '(x) x e
2x
y
23
x
dy
y xy x e
dx
y
2x
y dy x e
x dx y
Put
dy dv
y vx v x
dx dx
v
dv x
v v x e
dx v
vv
dv e e vdv dx
dx v
Integrating both side
ev (x + c) + 1 + v = 0
f(1) 1 x 1,y 1
2
c1
e
v2
e 1 x 1 v 0
e
y
x2y
e 1 x 1 0
ex
x = a, y = 0
2
ae
ae = 2
28. Remainder when
32
32
64
is divided by 9 is equal to
____.
Ans. 1
Sol. Let 3232 = t
32
32
64
= 64t = 82t = (9 – 1)2t
= 9k + 1
Hence remainder = 1
29. Let the set
2y
C x,y | x 2 2023,x,y
.
Then
(x,y) C
(x y)
is equal to _______.
Ans. 46
Sol. x2 – 2y = 2023
x 45,y 1
(x,y) C
(x y) 46.
30. Let the slope of the line 45x + 5y + 3 = 0 be
2
19r
27r 2
for some r1,
2
r R.
Then
x2
x3 23
2
321
8t
Lim dt
3r x r x r x 3x
2
is equal to ____.
Ans. 12
Sol. According to the question ,
2
19r
27r 9
2
x2
3
x3 23
221
8t dt
lim 3r x r x r x 3x
2
=
2
2
x3 2
221
8x
lim 3r 2r x 3r x 3
2
(using LH’ Rule)
=
221
72
3r 6r 27r 3
2
=
21
72
9r 27r 3
2
=
72 12
93
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
31. Two sources of light emit with a power of 200 W.
The ratio of number of photons of visible light
emitted by each source having wavelengths 300
nm and 500 nm respectively, will be :
(1) 1 : 5 (2) 1 : 3
(3) 5 : 3 (4) 3 : 5
Ans. (4)
Sol.
1
1
hc
n 200
2
2
hc
n 200
11
22
n300
n 500
1
2
n3
n5
32. The truth table for this given circuit is :
A
Y
B
(1)
A B Y
001
011
101
110
(2)
A B Y
000
011
100
111
(3)
A B Y
000
010
100
111
(4)
A B Y
001
010
101
110
Ans. (2)
Sol.
A
BA.B
Output = A.B A B+
A.B
A
B
A
A
B
Y = A.B +
A.B
=
A A .B
Y = 1.B
Y = B
33. A physical quantity Q is found to depend on
quantities a, b, c by the relation
43
2
ab
Qc
. The
percentage error in a, b and c are 3%, 4% and 5%
respectively. Then, the percentage error in Q is :
(1) 66% (2) 43%
(3) 34% (4) 14%
Ans. (3)
Sol.
43
2
ab
Qc
Q a b c
4 3 2
Q a b c
Q a b c
100 4 100 3 100 2 100
Q a b c
% error in Q = 4 × 3% + 3 × 4% + 2 × 5%
= 12% + 12% + 10%
= 34%
34. In an a.c. circuit, voltage and current are given by :
V = 100 sin (100 t) V and
I = 100 sin (100 t +
3
) mA respectively.
The average power dissipated in one cycle is :
(1) 5 W (2) 10 W
(3) 2.5 W (4) 25 W
Ans. (3)
Sol.
avg
P
rms
V
rms
I
cos()
3
100 100 10 cos 3
22
4
3
10 1 10
22
10 2.5W
4
35. The temperature of a gas having 2.0 × 1025
molecules per cubic meter at 1.38 atm (Given, k =
1.38 × 10–23 JK–1) is :
(1) 500 K (2) 200 K
(3) 100 K (4) 300 K
Ans. (1)
Sol. PV = nRT
A
N
PV RT
N
N = Total no. of molecules
N
P kT
V
1.38 × 1.01 × 105 = 2 × 1025 × 1.38 × 10–23 × T
1.01 × 105 = 2 × 102 × T
3
1.01 10
T 500 K
2
36. A stone of mass 900g is tied to a string and moved
in a vertical circle of radius 1m making 10 rpm.
The tension in the string, when the stone is at the
lowest point is (if 2 = 9.8 and g = 9.8 m/s2)
(1) 97 N (2) 9.8 N
(3) 8.82 N (4) 17.8 N
Ans. (2)
Sol. Given that
T
mg
v
m = 900 gm =
900 9
kg kg
1000 10
r = 1m
2 N 2 (10) rad / sec
60 60 3
T – mg = mr2
T = mg + mr2
=
2
99
9.8 1
10 10 3
=
2
9
8.82 10 9
= 8.82 + 0.98
= 9.80 N
37. The bob of a pendulum was released from a
horizontal position. The length of the pendulum is
10m. If it dissipates 10% of its initial energy
against air resistance, the speed with which the bob
arrives at the lowest point is : [Use, g : 10 ms–2]
(1)
65
ms–1 (2)
56
ms–1
(3)
55
ms–1 (4)
25
ms–1
Ans. (1)
Sol. = 10 m,
Initial energy = mg
So,
2
91
mg mv
10 2
2
91
10 10 v
10 2
v2 = 180
v 180 6 5 m / s
38. If the distance between object and its two times
magnified virtual image produced by a curved
mirror is 15 cm, the focal length of the mirror must
be :
(1) 15 cm (2) –12 cm
(3) –10 cm (4) 10/3 cm
Ans. (3)
Sol.
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
uv = 15 – u
f
15 cm
m = 2 =
v
u
(15 u)
2u
2u = 15 –u
3u = 15 u = 5 cm
v = 15 – u = 15 – 5 = 10 cm
1 1 1
f v u
1 1 1 2 1
10 ( 5) 10 10
f = –10 cm
39. Two particles X and Y having equal charges are
being accelerated through the same potential
difference. Thereafter they enter normally in a
region of uniform magnetic field and describes
circular paths of radii R1 and R2 respectively. The
mass ratio of X and Y is :
(1)
2
2
1
R
R
(2)
2
1
2
R
R
(3)
1
2
R
R
(4)
2
1
R
R
Ans. (2)
Sol.
2mqV
2m(KE)
mv p
RqB qB qB qB
Rm
2
mR
2
11
22
mR
mR
40. In Young’s double slit experiment, light from two
identical sources are superimposing on a screen.
The path difference between the two lights
reaching at a point on the screen is
7
4
. The ratio
of intensity of fringe at this point with respect to
the maximum intensity of the fringe is :
(1) 1/2 (2) 3/4 (3) 1/3 (4) 1/4
Ans. (1)
Sol.
7
x4
2 2 7 7
x42
2
max
I I cos 2
2 2 2
max
I 7 7
cos cos cos
I 2 2 2 4
2
cos 2 4
2
cos 4
1
2
41. A small liquid drop of radius R is divided into 27
identical liquid drops. If the surface tension is T,
then the work done in the process will be :
(1) 8R2T (2) 3R2T
(3)
2
1RT
8
(4) 4R2T
Ans. (1)
Sol. Volume constant
33
44
R 27 r
33
R3 = 27r3
R = 3r
R
r3
2
2R
r9
Work done = T.A
= 27 T(4r2) – T 4R2
=
2
2
R
27T4 4 R T
9
= 8R2T
42. A bob of mass ‘m’ is suspended by a light string of
length ‘L’. It is imparted a minimum horizontal
velocity at the lowest point A such that it just
completes half circle reaching the top most
position B. The ratio of kinetic energies
A
B
(K.E.)
(K.E.)
is :
O
mg
BVm
VH
VL
C
L
Amg
(1) 3 : 2 (2) 5 : 1
(3) 2 : 5 (4) 1 : 5
Ans. (2)
Sol. Apply energy conservation between A & B
22
LH
11
mV mV mg(2L)
22
L
V 5gL
So,
H
V gL
2
A
2
B
1m 5gL
(K.E) 5
2
1
(K.E) 1
m gL
2
43. A wire of length L and radius r is clamped at one
end. If its other end is pulled by a force F, its
length increases by l. If the radius of the wire and
the applied force both are reduced to half of their
original values keeping original length constant,
the increase in length will become.
(1) 3 times (2) 3/2 times
(3) 4 times (4) 2 times
Ans. (4)
Sol.
stress
Ystrain
2
F
r
Y
L
2
F Y r L
….(i)
2
F / 2
r / 4
Y
L
2
r
F Y 2
L4
From (i)
2
2r
Y r Y
L L 2
= 2
44. A planet takes 200 days to complete one revolution
around the Sun. If the distance of the planet from
Sun is reduced to one fourth of the original
distance, how many days will it take to complete
one revolution ?
(1) 25 (2) 50
(3) 100 (4) 20
Ans. (1)
Sol. T2 r3
22
12
33
12
TT
rr
2
2
2
3
3
T
(200)
rr
4
2
2
200 200 T
444
2
200
T42
T2 = 25 days
45. A plane electromagnetic wave of frequency
35 MHz travels in free space along the X-direction.
At a particular point (in space and time)
ˆ
E 9.6 jV / m
. The value of magnetic field at this
point is :
(1)
8ˆ
3.2 10 kT
(2)
8ˆ
3.2 10 iT
(3)
ˆ
9.6 jT
(4)
8ˆ
9.6 10 kT
Ans. (1)
Sol.
EC
B
8
E3 10
B
88
E 9.6
B3 10 3 10
B = 3.2 × 10–8 T
ˆˆ
ˆ
B v E
=
ˆˆ
ij
=
ˆ
k
So,
8ˆ
B 3.2 10 k T
46. In the given circuit, the current in resistance R3 is :
R2
R1
R4
R3
10 V
2
4
4 1
(1) 1 A (2) 1.5 A
(3) 2 A (4) 2.5 A
Ans. (1)
Sol.
R2
R1R4
R3
10 V
2
4
4 1
1A
1A
2A
2A
2A
Req = 2 + 2 + 1 = 5
eq
V 10
i 2A
R5
Current in resistance R3 =
4
244
4
28
= 1A
47. A particle is moving in a straight line. The
variation of position ‘x’ as a function of time ‘t’ is
given as x = (t3 – 6t2 + 20t + 15) m. The velocity of
the body when its acceleration becomes zero is :
(1) 4 m/s (2) 8 m/s
(3) 10 m/s (4) 6 m/s
Ans. (2)
Sol. x = t3 – 6t2 + 20t + 15
dx
dt
v = 3t2 – 12t + 20
dv
dt
a = 6t –12
When a = 0
6t – 12 = 0; t = 2 sec
At t = 2 sec
v = 3(2)2 – 12(2) + 20
v = 8 m/s
48. N moles of a polyatomic gas (f = 6) must be mixed
with two moles of a monoatomic gas so that the
mixture behaves as a diatomic gas. The value of N
is :
(1) 6 (2) 3 (3) 4 (4) 2
Ans. (3)
Sol.
1 1 2 2
eq
12
n f n f
fnn
For diatomic gas feq = 5
(N)(6) (2)(3)
5N2
5N + 10 = 6N + 6
N = 4
49. Given below are two statements :
Statement I : Most of the mass of the atom and all
its positive charge are concentrated in a tiny
nucleus and the electrons revolve around it, is
Rutherford’s model.
Statement II : An atom is a spherical cloud of
positive charges with electrons embedded in it, is a
special case of Rutherford’s model.
In the light of the above statements, choose the
most appropriate 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. According to Rutherford atomic model, most of
mass of atom and all its positive charge is
concentrated in tiny nucleus & electron revolve
around it.
According to Thomson atomic model, atom is
spherical cloud of positive charge with electron
embedded in it.
Hence,
Statement I is true but statement II false.
50. An electric field is given by
ˆ ˆ ˆ
(6i 5j 3k)
N/C.
The electric flux through a surface area
ˆ
30i
m2
lying in YZ-plane (in SI unit) is :
(1) 90 (2) 150
(3) 180 (4) 60
Ans. (3)
Sol.
ˆ ˆ ˆ
E 6i 5j 3k
ˆ
A 30i
E.A
ˆ ˆ ˆ ˆ
(6i 5j 3k).(30i)
6 30 180
SECTION-B
51. Two metallic wires P and Q have same volume and
are made up of same material. If their area of cross
sections are in the ratio 4 : 1 and force F1 is applied
to P, an extension of l is produced. The force
which is required to produce same extension in Q
is F2.
The value of
1
2
F
F
is_______ .
Ans. (16)
Sol.
Stress F / A F
YStrain / A
F
AY
V
VA A
2
FV
AY
Y & V is same for both the wires
2
F
A
2
1 1 2
2
22
1
FA
F
A
12
22
1 2 2 1
F A F A
2
2
11
2
22
FA 416
F1
A
52. A horizontal straight wire 5 m long extending from
east to west falling freely at right angle to
horizontal component of earth’s magnetic field
0.60 × 10–4 Wbm–2. The instantaneous value of emf
induced in the wire when its velocity is 10 ms–1 is
_________ × 10–3 V.
Ans. (3)
Sol. BH = 0.60 × 10–4 Wb/m2
Induced emf
H
e B v
= 0.60 × 10–4 × 10 × 5
= 3 × 10–3 V
53. Hydrogen atom is bombarded with electrons
accelerated through a potential different of V,
which causes excitation of hydrogen atoms. If the
experiment is being formed at T = 0 K. The
minimum potential difference needed to observe
any Balmer series lines in the emission spectra will
be
V
10
, where = ________.
Ans. (121)
Sol. For minimum potential difference electron has to
make transition from n = 3 to n = 2 state but first
electron has to reach to n = 3 state from ground
state. So, energy of bombarding electron should be
equal to energy difference of n = 3 and n = 1 state.
2
1
E 13.6 1 e eV
3
13.6 8 V
9
V 12.09 V 12.1 V
So,
121
54. A charge of 4.0 C is moving with a velocity of
4.0 × 106 ms–1 along the positive y-axis under a
magnetic field
B
of strength
ˆ
2k
T. The force
acting on the charge is
ˆ
x i N
. The value of x is __.
Ans. (32)
Sol. q = 4 C,
6ˆ
v 4 10 j
m/s
ˆ
B 2kT
F q v B
=
66
ˆˆ
4 10 4 10 j 2k
=
66
ˆ
4 10 8 10 i
ˆ
F 32i N
x = 32
55. A simple harmonic oscillator has an amplitude A
and time period 6 second. Assuming the
oscillation starts from its mean position, the time
required by it to travel from x = A to
3
xA
2
will be
x
s, where x = _______ :
Ans. (2)
Sol.
P
QA
A
From phasor diagram particle has to move from P
to Q in a circle of radius equal to amplitude of
SHM.
3A
3
2
cos A2
6
Now,
t
6
2t
6T
2t
66
t2
So, x = 2
56. In the given figure, the charge stored in 6F
capacitor, when points A and B are joined by a
connecting wire is _______C.
9V
66µF
AB
3µF 3
Ans. (36)
Sol. At steady state, capacitor behaves as an open
circuit and current flows in circuit as shown in the
diagram.
66
33
9V 9V
6 F
3F
A AB B
i
eq
R9
9V
i 1A
9
6
V 1 6
6 V
VA = 3 V
So, potential difference across 6F is 6 V.
Hence Q =
CV
= 6 × 6 × 10–6 C
36 C
57. In a single slit diffraction pattern, a light of
wavelength 6000 Å is used. The distance between
the first and third minima in the diffraction pattern
is found to be 3 mm when the screen in placed 50
cm away from slits. The width of the slit is ______
× 10–4 m.
Ans. (2)
Sol. For nth minima
bsin = n
( is small so sin is small, hence sin tan )
btan = n
y
bn
D
n
nD
yb
(Position of nth minima)
D
1
3
b
y1
y3
A
B
B 1st minima, A 3rd minima
31
3 D D
y , y
bb
31
2D
y y y b
10
–3 2 6000 10 0.5
3 10 b
10
3
2 6000 10 0.5
b3 10
b = 2 × 10–4 m
x = 2
58. In the given circuit, the current flowing through the
resistance 20 is 0.3 A, while the ammeter reads
0.9 A. The value of R1 is ______ .
A
20
15
R1
Ans. (30)
Sol.
BA A
20
15
R1
i3
i1
i2
Given, i1 = 0.3 A, i1 + i2 + i3 = 0.9 A
So, VAB = i1 × 20 = 20 × 0.3 V = 6 V
i2 =
6V 2 A
15 5
i1 + i2 + i3
9A
10
3
29
i
10 5 10
3
79
i
10 10
i3 = 0.2 A
So, i3 × R1 = 6 V
(0.2)R1 = 6
1
6
R 30
0.2
59. A particle is moving in a circle of radius 50 cm in
such a way that at any instant the normal and
tangential components of its acceleration are equal.
If its speed at t = 0 is 4 m/s, the time taken to
complete the first revolution will be
2
11e
s,
where = ______.
Ans. (8)
Sol.
Ct
aa
2
v dv
r dt
vt
2
40
dv dt
r
v
v
4
1t
vr
112t
v4
4 ds
v1 8t dt
ts
00
dt
4 ds
1 8t
(r = 0.5 m
s = 2r =
t
0
n 1 8t
48
n 1 8t 2
1 – 8t = e–2
21
t 1 e s
8
So, = 8
60. A body of mass 5 kg moving with a uniform speed
1
3 2 ms
in X – Y plane along the line y = x + 4.
The angular momentum of the particle about the
origin will be ______ kg m2s–1.
Ans. (60)
Sol. y – x – 4 = 0
dl is perpendicular distance of given line from
origin.
l22
4
d 2 2 m
11
So,
l
L mvd 5 3 2 2 2
kg m2/s
= 60 kg m2/s
CHEMISTRY
TEST PAPER WITH SOLUTION
SECTION-A
61. The ascending acidity order of the following H
atoms is
HC C
A
HC
2CH
H
B
CH
HC
3
HC
3
HC
3
C
HC
3CH2H
D
H
(1) C < D < B < A
(2) A < B < C < D
(3) A < B < D < C
(4) D < C < B < A
Ans. (1)
Sol.
CH C
>
2
CH CH
>
32
H C – CH
>
C
CH3
CH3
CH3
Stability of conjugate base acidic strength
C < D < B < A
62. Match List I with List II
List I (Bio Polymer)
List II (Monomer)
A.
Starch
I.
nucleotide
B.
Cellulose
II.
-glucose
C.
Nucleic acid
III.
-glucose
D.
Protein
IV.
-amino acid
Choose the correct answer from the options given
below :-
(1) A-II, B-I, C-III, D-IV
(2) A-IV, B-II, C-I, D-III
(3) A-I, B-III, C-IV, D-II
(4) A-II, B-III, C-I, D-IV
Ans. (4)
Sol. A-II, B-III, C-I, D-IV
Fact based.
63. Match List I with List II
List I
(Compound)
List II
(pKa value)
A.
Ethanol
I.
10.0
B.
Phenol
II.
15.9
C.
m-Nitrophenol
III.
7.1
D.
p-Nitrophenol
IV.
8.3
Choose the correct answer from the options given
below :-
(1) A-I, B-II, C-III, D-IV
(2) A-IV, B-I, C-II, D-III
(3) A-III, B-IV, C-I, D-II
(4) A-II, B-I, C-IV, D-III
Ans. (4)
Sol. Ethanol 15.9
Phenol 10
M-Nitrophenol 8.3
P-Nitrophenol 7.1
64. Which of the following reaction is correct ?
(1)
2
2
HNO ,0ºC
3 2 2 2 3 2 2
HO
CH CH CH NH CH CH OH N HCl
(2)
CH3
+ HI
CH3
I
(3)
+ Br2
UV light Br
Br
(4)
2 5 2 2
2 5 2 2 2 3 2
C H CONH Br NaOH
C H CH NH Na CO NaBr H O
Ans. (2)
Sol.
CH3
HI
Markovnikov addition
CH3
I
65. According to IUPAC system, the compound
OH
is named as
(1) Cyclohex-1-en-2-ol (2) 1-Hydroxyhex-2-ene
(3) Cyclohex-1-en-3-ol (4) Cyclohex-2-en-1-ol
Ans. (4)
Cyclohex-2-en-1-ol
66. The correct IUPAC name of K2MnO4 is
(1) Potassium tetraoxopermanganate (VI)
(2) Potassium tetraoxidomanganate (VI)
(3) Dipotassium tetraoxidomanganate (VII)
(4) Potassium tetraoxidomanganese (VI)
Ans. (2)
Sol. K2MnO4
2 + x – 8 = 0
x = +6
O.S. of Mn = +6
IUPAC Name =
Potassium tetraoxidomanganate(VI)
67. A reagent which gives brilliant red precipitate with
Nickel ions in basic medium is
(1) sodium nitroprusside
(2) neutral FeCl3
(3) meta-dinitrobenzene
(4) dimethyl glyoxime
Ans. (4)
Sol. Ni2+ + 2dmg– [Ni(dmg)2]
Rosy red/Bright Red precipitate
68. Phenol treated with chloroform in presence of
sodium hydroxide, which further hydrolysed in
presence of an acid results
(1) Salicyclic acid
(2) Benzene-1,2-diol
(3) Benzene-1, 3-diol
(4) 2-Hydroxybenzaldehyde
Ans. (4)
Sol.
OH OH
+ CHCl + NaOH
3 CHO
-hydroxy-benzaldehyde
It is Reimer Tiemann Reaction
69. Match List I with List II
List I
(Spectral Series
for Hydrogen)
List II
(Spectral Region/Higher
Energy State)
A.
Lyman
I.
Infrared region
B.
Balmer
II.
UV region
C.
Paschen
III.
Infrared region
D.
Pfund
IV.
Visible region
Choose the correct answer from the options given
below :-
(1) A-II, B-III, C-I, D-IV
(2) A-I, B-III, C-II, D-IV
(3) A-II, B-IV, C-III, D-I
(4) A-I, B-II, C-III, D-IV
Ans. (3)
Sol. A – II, B – IV, C – III, D – I
Fact based.
70. On passing a gas, ‘X’, through Nessler’s reagent, a
brown precipitate is obtained. The gas ‘X’ is
(1) H2S (2) CO2
(3) NH3 (4) Cl2
Ans. (3)
Sol. Nessler’s Reagent Reaction :
2 4 3 2 2
Nessler's Reagent Iodine of Millon's base
Brown precipitate
2K HgI NH 3KOH HgO. Hg NH I 7KI 2H O
71. The product A formed in the following reaction is:
NH2NaNO , HCl, 0ºC
2
then Cu Cl
22 A
(1)
NH2
Cl
(2)
NH Cl
3
+–
(3)
Cl
(4)
Cl
Cl
Ans. (3)
Sol.
NH2NaNO HCl
2
0° C Cu Cl
22
N2
+Cl–Cl
72. Identify the reagents used for the following conversion
O CH3
CHO
A
OH OH
CHO
CHO B
OH
CHO
C
O
(1) A = LiAlH4, B = NaOH(aq), C = NH2–NH2/KOH,
ethylene glycol
(2) A = LiAlH4, B = NaOH(alc), C =Zn/HCl
(3) A = DIBAL-H, B= NaOH(aq),
C = NH2–NH2/KOH, ethylene glycol
(4) A = DIBAL-H, B = NaOH(alc), C = Zn/HCl
Ans. (4)
Sol.
OCH3
O
CHO
HO
CHO
CHO
HO
(A) DIBAL–H
{Selective
reduction
of ester} (B) NaOH
(alc.)
Intramolecular
Aldol
CHO
OH
Zn/HCl(C)
(Clemmensen
reduction)
73. Which of the following acts as a strong reducing
agent? (Atomic number : Ce = 58, Eu = 63,
Gd = 64, Lu = 71)
(1) Lu3+ (2) Gd3+
(3) Eu2+ (4) Ce4+
Ans. (3)
Sol. Eu+2 Eu+3 + 1e–
[Xe]4f76s0 [Xe] 4f66s0
74. Chromatographic technique/s based on the
principle of differential adsorption is/are
A. Column chromatography
B. Thin layer chromatography
C. Paper chromatography
Choose the most appropriate answer from the
options given below:
(1) B only (2) A only
(3) A & B only (4) C only
Ans. (3)
Sol. Memory Based
75. Which of the following statements are correct
about Zn, Cd and Hg?
A. They exhibit high enthalpy of atomization as
the d-subshell is full.
B. Zn and Cd do not show variable oxidation
state while Hg shows +I and +II.
C. Compounds of Zn, Cd and Hg are
paramagnetic in nature.
D. Zn, Cd and Hg are called soft metals.
Choose the most appropriate from the options
given below:
(1) B, D only (2) B, C only
(3) A, D only (4) C, D only
Sol. Ans. (1)
(A) Zn, Cd, Hg exhibit lowest enthalpy of
atomization in respective transition series.
(C) Compounds of Zn, Cd and Hg are diamagnetic
in nature.
76. The element having the highest first ionization
enthalpy is
(1) Si (2) Al
(3) N (4) C
Ans. (3)
Sol. Al < Si < C < N ; IE1 order.
77. Alkyl halide is converted into alkyl isocyanide by
reaction with
(1) NaCN (2) NH4CN
(3) KCN (4) AgCN
Ans. (4)
Sol. Covalent character of AgCN.
78. Which one of the following will show geometrical
isomerism?
(1)
CHBr
Br
(2)
CH2
Br
(3)
CHBr
Br
(4)
CH2
Br
Ans. (3)
Sol. Due to unsymmetrical.
79. Given below are two statements:
Statement I: Fluorine has most negative electron
gain enthalpy in its group.
Statement II: Oxygen has least negative electron
gain enthalpy in its group.
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) Statement I is true but Statement II is false
(3) Both Statement I and Statement II are false
(4) Statement I is false but Statement II is true
Ans. (4)
Sol. Statement-1 is false because chlorine has most
negative electron gain enthalpy in its group.
80. Anomalous behaviour of oxygen is due to its
(1) Large size and high electronegativity
(2) Small size and low electronegativity
(3) Small size and high electronegativity
(4) Large size and low electronegativity
Ans. (3)
Sol. Fact Based.
SECTION-B
81. The total number of anti bonding molecular
orbitals, formed from 2s and 2p atomic orbitals in a
diatomic molecule is _____________.
Ans. (4)
Sol. Antibonding molecular orbital from 2s = 1
Antibonding molecular orbital from 2p = 3
Total = 4
82. The oxidation number of iron in the compound
formed during brown ring test for
3
NO
ion is _____.
Ans. (1)
Sol. [Fe(H2O)5(NO)]2+,
Oxidation no. of Fe = +1
83. The following concentrations were observed at
500 K for the formation of NH3 from N2 and H2. At
equilibrium :[N2] = 2 × 10–2 M, [H2] = 3 ×10–2 M and
[NH3] = 1.5 ×10–2M. Equilibrium constant for the
reaction is _______.
Ans. (417)
Sol.
2
3
C3
22
NH
KNH
2
2
C3
22
1.5 10
K2 10 3 10
KC = 417
84. Molality of 0.8 M H2SO4 solution (density 1.06 g cm–3)
is _____×10–3 m.
Ans. (815)
Sol.
sol solute
M 1000
md 1000 M Molar mass
815 × 10–3 m
85. If 50 mL of 0.5 M oxalic acid is required to
neutralise 25 mL of NaOH solution, the amount of
NaOH in 50 mL of given NaOH solution is
_____g.
Ans. (4)
Sol. Equivalent of Oxalic acid = Equivalents of NaOH
50 × 0.5 × 2 = 25 × M × 1
MNaOH = 2M
WNaOH in 50ml = 2 × 50 × 40 × 10–3 g = 4g
86. The total number of ‘Sigma’ and Pi bonds in 2-
formylhex-4-enoic acid is ____.
Ans. (22)
Sol.
C C C C C C
C
O
H
OH H
O HH
H
HHHH
22 bonds
87. The half-life of radioisotopic bromine - 82 is 36
hours. The fraction which remains after one day is
___________ ×10–2.
(Given antilog 0.2006 = 1.587)
Ans. (63)
Sol. Half life of bromine – 82 = 36 hours
1/2 0.693
tK
0.693
K36
= 0.01925 hr–1
1st order rxn kinetic equation
2.303 a
t log
K a x
a t K
log a x 2.303
(t = 1day = 24hr)
1
a 24hr 0.01925hr
log a x 2.303
a
log 0.2006
ax
aantilog 0.2006
ax
a1.587
ax
If a = 1
11.587
1x
1 – x = 0.6301 = Fraction remain
after one day
88. Standard enthalpy of vapourisation for CCl4 is
30.5 kJ mol–1. Heat required for vapourisation of
284g of CCl4 at constant temperature is ____kJ.
(Given molar mass in g mol–1 ; C = 12, Cl = 35.5)
Ans. (56)
Sol.
0
vap 4
H CCl 30.5 kJ / mol
Mass of CCl4 = 284 gm
Molar mass of CCl4 = 154 g/mol
Moles of CCl4 =
284
154
= 1.844 mol
Hvapº for 1 mole = 30.5 kJ/mol
Hvapº for 1.844 mol = 30.5 × 1.844
= 56.242 kJ
89. A constant current was passed through a solution
of
4
AuCl
ion between gold electrodes. After a
period of 10.0 minutes, the increase in mass of
cathode was 1.314 g. The total charge passed
through the solution is _________ × 10-2 F.
(Given atomic mass of Au = 197)
Ans. (2)
Sol.
W charge
E 1F
1.314 Q
197 1F
3
Q = 2 × 10–2 F
90. The total number of molecules with zero dipole
moment among CH4, BF3, H2O, HF, NH3, CO2 and
SO2 is _________.
Ans. (3)
Sol. Molecules with zero dipole moment = CO2, CH4,
BF3
