FINAL JEE–MAIN EXAMINATION – JANUARY, 2024
(Held On Tuesday 30th January, 2024) TIME : 3 : 00 PM to 6 : 00 PM
MATHEMATICS
TEST PAPER WITH SOLUTION
SECTION-A
1. Consider the system of linear equations
x + y + z = 5, x + 2y +2z = 9,
x + 3y +z = , where , R. Then, which of
the following statement is NOT correct?
(1) System has infinite number of solution if 1
and =13
(2) System is inconsistent if 1 and 13
(3) System is consistent if 1 and  13
(4) System has unique solution if 1 and 13
Ans. (4)
Sol.

2
1 1 1
1 2 0
13
22 – – 1 = 0
1
1, 2

2
1 1 5
2 9 0 13
3
Infinite solution = 1 & = 13
For unique soln  1
For no soln = 1 & 13
If 1 and  13
Considering the case when
1
2
and
this
will generate no solution case
2. For



, 0, 2
, let
3sin( ) 2sin( )
and a
real number k be such that
tan k tan
. Then the
value of k is equal to :
(1)
2
3
(2) –5
(3)
2
3
(4) 5
Ans. (2 )
Sol. 3sin cos + 3sin cos
= 2sin cos – 2sin cos
5sin cos = –sin cos
1
tan tan
5
tan = –5tan
3. Let A(, 0) and B(0, ) be the points on the line
5x + 7y = 50. Let the point P divide the line
segment AB internally in the ratio 7 : 3. Let 3x –
25 = 0 be a directrix of the ellipse

22
22
xy
E : 1
ab
and the corresponding focus be S. If from S, the
perpendicular on the x-axis passes through P, then
the length of the latus rectum of E is equal to
(1)
25
3
(2)
32
9
(3)
25
9
(4)
32
5
Ans. (4 )
Sol.
A (10, 0)
P (3, 5)
50
B 0, 7



x =
S
25
3
ae = 3
a 25
e3
a = 5
b = 4
2
2b 32
Length of LR a5

4. Let
ˆˆ ˆ
a i j k, , R
. Let a vector
b
be such
that the angle between
a
and
b
is
4
and
2
b6
,
If
a.b 3 2
, then the value of
2
22
ab
is
equal to
(1) 90 (2) 75
(3) 95 (4) 85
Ans. (1)
Sol.
2
| b | 6
;
 | a || b | cos 3 2
2 2 2
| a | | b | cos 18
 
2
| a | 6
Also 1 + 2 + 2 = 6
2 + 2 = 5
to find
(2 + 2)
2 2 2
| a | | b | sin

=



1
(5)(6)(6) 2
= 90
5. Let
23
f(x) (x 3) (x 2) , x [ 4, 4]
. If M and m are
the maximum and minimum values of f,
respectively in [–4, 4], then the value of M – m is :
(1) 600 (2) 392
(3) 608 (4) 108
Ans. (3)
Sol. f'(x) = (x + 3)2 . 3(x – 2)2 + (x –2)3 2(x + 3)
= 5(x + 3) (x – 2)2 (x + 1)
f'(x) = 0, x = –3, –1, 2
–3 –1 2
+–+ +
f(–4) = –216
f(–3) = 0, f(4) = 49 × 8 = 392
M = 392, m = –216
M – m = 392 + 216 = 608
Ans = '3'
6. Let a and b be be two distinct positive real
numbers. Let 11th term of a GP, whose first term is
a and third term is b, is equal to pth term of another
GP, whose first term is a and fifth term is b. Then p
is equal to
(1) 20 (2) 25
(3) 21 (4) 24
Ans. (3)
Sol. 1st GP t1 = a, t3 = b = ar2 r2 =
b
a
t11 = ar10 = a(r2)5 =



5
b
aa
2nd G.P. T1 = a, T5 = ar4 = b
1/ 4
4bb
rr
aa
Tp = arp –1
p1
4
b
aa



p1
54
11 p
bb
t T a a
aa
p1
5 p 21
4
7. If x2 – y2 + 2hxy + 2gx + 2fy + c = 0 is the locus of
a point, which moves such that it is always
equidistant from the lines x + 2y + 7 = 0 and 2x – y
+ 8 = 0, then the value of g + c + h – f equals
(1) 14 (2) 6
(3) 8 (4) 29
Ans. (1)
Sol. Cocus of point P(x, y) whose distance from
Gives
X + 2y + 7 = 0 & 2x – y + 8 = 0 are equal is

x 2y 7 2x y 8
55
(x + 2y + 7)2
– (2x – y + 8)2
= 0
Combined equation of lines
(x – 3y + 1) (3x + y + 15) = 0
3x2 – 3y2 – 8xy + 18x – 44y + 15 = 0
x2 – y2 –
8 44
xy 6x y 5 0
33
x2 – y2 + 2h xy + 2gx 2 + 2fy + c = 0
4 22
h , g 3, f , c 5
33
4 22
g c h f 3 5 8 6 14
33
8. Let
a and b
be two vectors such that
| b | 1 and | b a | 2
. Then
2
(b a) b
is equal
to
(1) 3
(2) 5
(3) 1
(4) 4
Ans. (2)
Sol.
| b | 1 & | b a | 2
b a b b b a 0

2 2 2
(b a ) b b a b
= 4 + 1 = 5
9. Let
y f(x)
be a thrice differentiable function in
(–5, 5). Let the tangents to the curve y=f(x) at
(1, f(1)) and (3, f(3)) make angles
6
and
4
,
respectively with positive x-axis. If
32
1
27 f (t) 1 f (t)dt 3
where , are
integers, then the value of + equals
(1) –14
(2) 26
(3) –16
(4) 36
Ans. (2)
Sol. y = f(x)
dy f '(x)
dx
(1,f (1))
dy 1 1
f '(1) tan f '(1)
dx 6 33
(3,f (3))
dy f '(3) tan 1 f '(3) 1
dx 4
32
1
27 f '(t) 1 f "(t)dt 3

32
1
I f '(t) 1 f "(t)dt
f'(t) = z f"(t) dt = dz
z = f'(3) = 1
z = f'(1) =
1
3



1
13
2
1/ 3 1/ 3
z
I (z 1)dz z
3






1 1 1 1
1
33
3 3 3
4 10 4 10 3
3 3 27
93



4 10
3 27 3 36 10 3
3 27
= 36, = – 10
+ = 36 – 10 = 26
10. Let P be a point on the hyperbola

22
xy
H : 1
94
,
in the first quadrant such that the area of triangle
formed by P and the two foci of H is
2 13
. Then,
the square of the distance of P from the origin is
(1) 18
(2) 26
(3) 22
(4) 20
Ans. (3)
Sol.
yp
Os1
s2
x
(, )
22
xy
1
94

a2 = 9, b2 = 4
2
2 2 2 2 2
b
b a (e 1) e 1 a
24 13
e1
99
12
13 13
e s s 2ae 2 3 2 13
33
Area of
1 2 1 2
1
PS S s s 2 13
2
1(2 13) 2 13 2
2
2 2 2 2
1 1 1 18 3 2
9 4 9
Distance of P from origin =
22
=
18 4 22
11. Bag A contains 3 white, 7 red balls and bag B
contains 3 white, 2 red balls. One bag is selected at
random and a ball is drawn from it. The probability
of drawing the ball from the bag A, if the ball
drawn in white, is :
(1)
1
4
(2)
1
9
(3)
1
3
(4)
3
10
Ans. (3)
Sol. E1 : A is selected
A
3W
7R
B
3W
2R
E2 : B is selected
E : white ball is drawn
P (E1/E) =
1
1 1 2 2
1
P(E).P(E / E ) 2 10
1 3 1 3
P(E ). P(E / E ) P(E ). P(E / E )
2 10 2 5
=
31
3 6 3
12. Let f : R R be defined
2x x
f(x) ae be cx
. If
f(0) 1
,
e
f (log 2) 21
and
e
log 4
0
39
f(x) cx dx 2

, then the value of |a+b+c|
equals :
(1) 16 (2) 10
(3) 12 (4) 8
Ans. (4)
Sol. f(x) = ae2x + bex + cx f(0) = –1
a + b = –1
f(x) = 2ae2x + bex + c f (ln 2) = 21
8a + 2 b + c = 21

ln 4 2x x
0
39
(ae be )dx 2




ln 4
2x
x
0
ae 39
be
22
8a + 4b –

a 39
b
22
15a + 6b = 39
15 a – 6a – 6 = 39
9a = 45 a = 5
b = - 6
c = 21 – 40 + 12 = –7
a + b + c – 8
|a + b + c| = 8
13. Let

1ˆ ˆ ˆ ˆ
ˆˆ
L : r (i j 2k) (i j 2k), R
2ˆ ˆ ˆ
ˆˆ
L : r ( j k) (3i j pk), R
and

3ˆˆ
ˆ
L : r ( i mj nk) R
Be three lines such that L1 is perpendicular to L2
and L3 is perpendicular to both L1 and L2. Then the
point which lies on L3 is
(1) (–1, 7, 4) (2) (–1, –7, 4)
(3) (1, 7, –4) (4) (1, –7, 4)
Ans. (1)
Sol. L1 L2 L3 L1, L2
3 – 1 + 2 P = 0
P = – 1
ˆˆ ˆ
i j k
ˆˆˆ
1 1 2 i 7 j 4k
3 1 1
( , 7 , 4 ) 
will lie on L3
For = 1 the point will be (-1, 7, 4)
14. Let a and b be real constants such that the function
f defined by

2
x 3x a , x 1
f(x) bx 2 , x 1
be
differentiable on R. Then, the value of
2
2
f(x) dx
equals
(1)
15
6
(2)
19
6
(3) 21 (4) 17
Ans. (4)
Sol. f is continuous f(x) = 2x + 3 , k < 1
4 + a = b + 2 b , x > 1
a = b – 2 f is differentiable
b = 5
a = 3

12
2
21
(x 3x 3)dx (5x 2)dx
=
12
3 2 2
21
x 3x 5x
3x 2x
3 2 2
=
1 3 8 5
3 6 6 10 4 2
3 2 3 2
=
35
6 12 17
22
15. Let
f : {0}
be a function satisfying



x f(x)
fy f(y)
for all x, y, f(y) 0. If
f (1)
= 2024,
then
(1) xf (x) – 2024 f(x) = 0
(2) xf(x) + 2024f(x) = 0
(3) xf(x) +f(x) = 2024
(4) xf(x) –2023f(x) = 0
Ans. (1)
Sol.
x f(x)
fy f(y)



f (1) 2024
f(1) 1
Partially differentiating w. r. t. x
x 1 1
f . f (x)
y y f(y)




yx
1 f (x)
f (1). x f(x)
2024f(x) = xf(x)  xf(x) – 2024 f(x) = 0
16. If z is a complex number, then the number of
common roots of the equation
1985 100
z z 1 0
and
32
z 2z 2z 1 0
, is equal to :
(1) 1 (2) 2
(3) 0 (4) 3
Ans. (2)
Sol. z1985 + z100 + 1 = 0 & z3 + 2z2 + 2z + 1 = 0
(z + 1) (z2 – z + 1) + 2z(z + 1) = 0
(z + 1) (z2 + z + 1) = 0
z = – 1 , z = w, w2
Now putting z = –1 not satisfy
Now put z = w
w1985 + w100 + 1
w2 + w + 1 = 0
lso, z = w2
w3970 + w200 + 1
w + w2 + 1 = 0
Two common root
17. Suppose 2 – p, p, 2 – are the coefficient of
four consecutive terms in the expansion of (1+x)n.
Then the value of
22
p 6 2p
equals
(1) 4 (2) 10
(3) 8 (4) 6
Ans. (Bonus)
Sol. 2 – p, p, 2 – ,
Binomial coefficients are
nCr , nCr+1 , nCr+2 , nCr+3 respectively
nCr + nCr+1 = 2
n+1Cr+1 = 2 ……(1)
Also, nCr+2 + nCr+3 = 2
n+1Cr+3 = 2 …….(2)
From (1) and (2)
n+1Cr+1 = n+1Cr+3
2r + 4 = n + 1
n = 2r + 3
2r+4Cr+1 = 2
Data Inconsistent
18. If the domain of the function f(x) = loge


1
2
2x 3 2x 1
cos x2
4x x 3
is
( , ],
then the
value of 5 – 4 is equal to
(1) 10 (2) 12
(3) 11 (4) 9
Ans. (2)
Sol.

2
2x 3 0
4x x 3
and
2x – 1
11
x2


23 0
(4x 3)(x 1)



3x 1 x 3
0 & 0
x 2 x 2
–+–+
–3/2 –1 3/4



1
, 2 ,
3
…..(1)
2,3
…….(2)



1,3
3
……(3) (1)
(2)
(3)


3,3
4
33
4
5 4 15 3 12
19. Let f : R R be a function defined
4 1/ 4
x
f(x) (1 x )
and
g(x) f(f(f(f(x))))
then
25
2
0
18 x g(x)dx
(1) 33 (2) 36
(3) 42 (4) 39
Ans. (4)
Sol.
4 1/4
x
f(x) (1 x )
fof(x) =





4 1/ 4
4 1/ 4 1/4 4 1/ 4
4
4
x
f(x) x
(1 x )
(1 f(x) ) (1 2x )
x
11x
f(f(f(f(x)))) =
4 1/4
x
(1 4x )
18
25 3
4 1/ 4
0
x
(1 4x )
dx
Let 1 + 4x4 = t4
16x3 dx = 4t3 dt
33
1
18 t dt
4t
=



3
3
1
t
23
=
3[26] 39
2
20. Let R =





x00
0 y 0
0 0 z
be a non-zero 3 × 3 matrix,
where x sin = y sin

24
zsin
33
0, (0,2 )
. For a square matrix M, let trace
(M) denote the sum of all the diagonal entries of
M. Then, among the statements:
(I) Trace (R) = 0
(II) If trace (adj(adj(R)) = 0, then R has exactly one
non-zero entry.
(1) Both (I) and (II) are true
(2) Neither (I) nor (II) is true
(3) Only (II) is true
(4) Only (I) is true
Ans. (2)
Sol.
24
x sin y sin z sin 0
33

x, y, z 0
Also,
24
sin sin sin 0 R
33

1 1 1 0
x y z
xy + yz + zx = 0
(i) Trace (R) = x + y + z
If x + y + z = 0 and x y + yz + zx = 0
x = y = z = 0
Statement (i) is False
(ii) Adj(Adj(R)) = |R| R
Trace (Adj(Adj(R)))
= xyz (x + y + z) 0
Statement (ii) is also False
SECTION-B
21. Let Y = Y(X) be a curve lying in the first quadrant
such that the area enclosed by the line
Y – y = Y(x) (X – x) and the co-ordinate axes,
where (x, y) is any point on the curve, is always
2
y1
2Y x
, Y(x) 0. If Y(1) = 1, then 12Y(2)
equals ______ .
Ans. (20)
Sol.
1y
A x y xY / x
2 Y x




=
2
y1
2Y x
2
y xY x y xY x y 2Y x


2
22
y xyY x xyY x x Y x
=

2
y 2Y x
2xy – x2 Y(x) = 2
2
dy 2xy 2
dx x

2
dy 2 2
y
dx x x
I.F. =
2 ln x
2
1
ex
3
2
12
y x c
3
x
Put x = 1, y = 1

2
1c
3
1
c3
2
2 1 1
YX
3 X 3
5
12Y 2 12 20
3
22. Let a line passing through the point (–1, 2, 3)
intersect the lines

1
x 1 y 2 z 1
L: 3 2 2
at
M , ,
and


2
x 2 y 2 z 1
L: 3 2 4
at N(a, b,
c). Then the value of

2
2
a b c
equals ____.
Ans. (196)
Sol.
M 3 1,2 2, 2 1
32
N 3 2, 2 2,4 1
 a b c 1
(–1, 2, 3)
L1L2
N(a,b,c)M( )

3 2 2 2 4
3 1 2 4 2
  3 2 3
2
  22
 2
 2
0
14
a + b + c = –1

2
2196
a b c
23. Consider two circles C1 : x2 + y2 = 25 and C2 : (x –
)2 + y2 = 16, where  (5, 9). Let the angle
between the two radii (one to each circle) drawn
from one of the intersection points of C1 and C2 be



163
sin 8
. If the length of common chord of C1
and C2 is , then the value of ()2 equals _____ .
Ans. (1575)
Sol.

22
1
C : x y 25
,
22
2
C : x y 16
5 < < 9

 

163
sin 8
 63
sin 8
Area of OAP =



11
5 4sin
2 2 2
 63
40 8
5 63

225 63
= 1575
24. Let


 

2
n
nk
k0
C
k1
and

 

nn
n1
k k 1
k0
CC
k2
.
If 5 = 6, then n equals ___________ .
Ans. (10)
Sol.

nn
n
kk
k0
CC
n1
k 1 n 1
=

n
n 1 n
k 1 n k
k0
1CC
n1
2n 1
n1
1C
n1

n
n1
nk1
k
k0
Cn1
Ck 2 n 1

n1
n n 1
n k k 2
k0
1CC
n1
=
2n 1
n2
1C
n1


2n 1
n2
2n 1
n1
2n 1 n 2 1
C
n2
C


n5
n 2 6
n = 10
25. Let Sn be the sum to n-terms of an arithmetic
progression 3, 7, 11, ...... .
If




n
k
k1
6
40 S 42
n n 1
, then n equals ___ .
Ans. (9)
Sol. Sn = 3 + 7 + 11 + ...... n terms
=
2
n6 n 1 4 3n 2n 2n
2
= 2n2 + n

n n n
2
k
k 1 k 1 k 1
S 2 K K
=

n n 1 2n 1 n n 1
262
=




2n 1 1
n n 1 32
=
n n 1 4n 5
6

n
k
k1
6
40 S 42
n n 1
40 < 4n + 5 < 42
35 < 4n < 37
n = 9
26. In an examination of Mathematics paper, there are
20 questions of equal marks and the question paper
is divided into three sections : A, B and C . A
student is required to attempt total 15 questions
taking at least 4 questions from each section. If
section A has 8 questions, section B has 6
questions and section C has 6 questions, then the
total number of ways a student can select 15
questions is _______ .
Ans. (11376)
Sol. If 4 questions from each section are selected
Remaining 3 questions can be selected either in (1,
1, 1) or (3, 0, 0) or (2, 1, 0)
Total ways =
8 6 6 8 6 6
5 5 5 6 5 4
c c c c c c 2
+
8 6 6 8 6 6
5 6 4 4 6 5
c c c 2 c c c 2
+

8 6 6
7 4 4
c c c
= 56 . 6 . 6 + 28 . 6 . 15 . 2 + 56 . 15 . 2 + 70 . 6 . 2
+ 8 . 15 . 15
= 2016 + 5040 + 1680 + 840 + 1800 = 11376
27. The number of symmetric relations defined on the
set {1, 2, 3, 4} which are not reflexive is _____ .
Ans. (960)
Sol. Total number of relation both symmetric and
reflexive =
2
nn
2
2
Total number of symmetric relation =



2
nn
2
2
Then number of symmetric relation which are
not reflexive

n n 1 n n 1
22
22
210 – 26
1024 – 64
= 960
28. The number of real solutions of the equation
2
x x 3 x 5 x 1 6 x 2 0
is ______ .
Ans. (1)
Sol. x = 0 and
2
x 3 x 5 x 1 6 x 2 0
Here all terms are +ve except at x = 0
So there is no value of x
Satisfies this equation
Only solution x = 0
No of solution 1.
29. The area of the region enclosed by the parabola
(y – 2)2 = x – 1, the line x – 2y + 4 = 0 and the
positive coordinate axes is ______.
Ans. (5)
Sol. Solving the equations
(y – 2)2 = x – 1 and x – 2y + 4 = 0
X = 2(y – 2)

2
xx1
4
x2 – 4x + 4 = 0
(x – 2)2 = 0
x = 2
Exclose area (w.r.t. y-axis) =
3
0
x dy
– Area of .
=
32
0
1
y 2 1 dy 1 2
2
=
3
2
0
y 4y 5 dy 1
=



3
3
2
0
y2y 5y 1
3
= 9 – 18 + 15 – 1 = 5
30. The variance 2 of the data
xi
0
1
5
6
10
12
17
fi
3
2
3
2
6
3
3
Is _______ .
Ans. (29)
Sol.
xi
fi
fixi
fixi
2
0
3
0
0
1
2
2
2
5
3
15
75
6
2
12
72
10
6
60
600
12
3
36
432
17
3
51
867
fi = 22
fixi
2 = 2048
fixi = 176
So
ii
i
fx 176
x8
22
f
for
2
22
ii
1f x x
N
=
2
12048 8
22 
= 93.090964
= 29.0909
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
31. If 50 Vernier divisions are equal to 49 main scale
divisions of a travelling microscope and one
smallest reading of main scale is 0.5 mm, the
Vernier constant of travelling microscope is:
(1) 0.1 mm
(2) 0.1 cm
(3) 0.01 cm
(4) 0.01 mm
Ans. (4)
Sol. 50 V+S = 49S + S
S = 50 (S – V)
.5 = 50 (S – V)
0.5 1
S V 0.01 mm
50 100
32. A block of mass 1 kg is pushed up a surface
inclined to horizontal at an angle of 60° by a force
of 10 N parallel to the inclined surface as shown in
figure. When the block is pushed up by 10 m along
inclined surface, the work done against frictional
force is : [g = 10 m/s2]
10 N
M
s = 0.1
60°
(1)
5 3 J
(2) 5 J
(3) 5 × 103 J (4) 10 J
Ans. (2)
Sol. Work done again frictional force
N 10
0.1 5 10 5J
33. For the photoelectric effect, the maximum kinetic
energy (Ek) of the photoelectrons is plotted against
the frequency (v) of the incident photons as shown
in figure. The slope of the graph gives
Ek
v
(1) Ratio of Planck’s constant to electric charge
(2) Work function of the metal
(3) Charge of electron
(4) Planck’s constant
Ans. (4)
Sol. K.E. = hf –
tan = h
34. A block of ice at –10°C is slowly heated and
converted to steam at 100°C. Which of the
following curves represent the phenomenon
qualitatively:
(1)
Temperature
Heat supplied
(2)
Temperature
Heat supplied
(3)
Temperature
Heat supplied
(4)
Temperature
Heat supplied
Ans. (4)
35. In a nuclear fission reaction of an isotope of mass
M, three similar daughter nuclei of same mass are
formed. The speed of a daughter nuclei in terms of
mass defect M will be :
(1)
2c M
M
(2)
2
Mc
3
(3)
2M
cM
(4)
3M
cM
Ans. (3)
Sol. (X) (Y) + (Z) + (P)
M M/3 M/3 M/3
2 2 2 2
1 M 1 M 1 M
Mc V V V
2 3 2 3 2 3
2M
Vc M
36. Choose the correct statement for processes A & B
shown in figure.
A
B
P
V
(1) PV = k for process B and PV = k for process A.
(2) PV = k for process B and A.
(3)

1
Pk
T
for process B and T = k for process A.
(4)

1
Tk
P
for process A and PV = k for process B.
Ans. (1 & 3)
Sol. Steeper curve (B) is adiabatic
Adiabatic PVv = const.
Or



T
P const.
P
v
1
Tconst.
P
v
v
Curve (A) is isothermal
T = const.
PV = const.
37. An electron revolving in nth Bohr orbit has
magnetic moment
n
. If
x
nn
, the value of x is:
(1) 2 (2) 1
(3) 3 (4) 0
Ans. (2)
Sol. Magnetic moment = ir2
evr
2



2
1n
n
n
x = 1
38. An alternating voltage V(t) = 220 sin 100 t volt is
applied to a purely resistive load of 50 . The time
taken for the current to rise from half of the peak
value to the peak value is:
(1) 5 ms
(2) 3.3 ms
(3) 7.2 ms
(4) 2.2 ms
Ans. (2)
Sol. Rising half to peak
t = T/6
21
t 3.33ms
6 3 300 300
39. A block of mass m is placed on a surface having
vertical cross section given by y = x2/4. If
coefficient of friction is 0.5, the maximum height
above the ground at which block can be placed
without slipping is:
(1) 1/4 m (2) 1/2 m
(3) 1/6 m (4) 1/3 m
Ans. (1)
Sol.
dy x
tan
dx 2 2
x = 1 , y = 1/4
40. If the total energy transferred to a surface in time t
is 6.48 × 105 J, then the magnitude of the total
momentum delivered to this surface for complete
absorption will be :
(1) 2.46 × 10–3 kg m/s
(2) 2.16 × 10–3 kg m/s
(3) 1.58 × 10–3 kg m/s
(4) 4.32 × 10–3 kg m/s
Ans. (2)
Sol.
53
8
E 6.48 10
p 2.16 10
C3 10
41. A beam of unpolarised light of intensity I0 is
passed through a polaroid A and then through
another polaroid B which is oriented so that its
principal plane makes an angle of 45° relative to
that of A. The intensity of emergent light is :
(1) I0/4 (2) I0
(3) I0/2 (4) I0/8
Ans. (1)
Sol. Intensity of emergent light
=

2
00
II
cos 45
24
42. Escape velocity of a body from earth is 11.2 km/s.
If the radius of a planet be one-third the radius of
earth and mass be one-sixth that of earth, the
escape velocity from the plate is:
(1) 11.2 km/s (2) 8.4 km/s
(3) 4.2 km/s (4) 7.9 km/s
Ans. (4)
Sol.

EE
PP
RM
R , M
36
e
e
e
2GM
VR
…(i)
P
P
P
2GM
VR
…(ii)
e
p
V2
V
e
P
V11.2
V 7.9 km/sec
22
43. A particle of charge ‘–q’ and mass ‘m’ moves in a
circle of radius ‘r’ around an infinitely long line
charge of linear density ‘+’. Then time period
will be given as:
(Consider k as Coulomb’s constant)
(1)
2
23
4m
Tr
2k q
(2)

m
T 2 r 2k q
(3)

1m
T2 r 2k q
(4)
1 2k q
T2m
Ans. (2)
Sol.

2
2k q mr
r

2
2
2k q
mr




2
2
2 2k q
Tmr

m
T 2 r 2k q
44. If mass is written as m = k cP G–1/2 h1/2 then the
value of P will be : (Constants have their usual
meaning with k a dimensionless constant)
(1) 1/2
(2) 1/3
(3) 2
(4) –1/3
Ans. (1)
Sol. m = k cP G–1/2 h1/2
1 0 0 –1 P –1 3 –2 –1/2 2 –1 1/2
M L T [LT ] [M L T ] [ML T ]
By comparing P = 1/2
45. In the given circuit, the voltage across load
resistance (RL) is:
D1
(Ge) D2
(Si) RL2.5 k
1.5 k
15V
(1) 8.75 V
(2) 9.00 V
(3) 8.50 V
(4) 14.00 V
Ans. (1)
Sol.
Ge
(0.3) Si
(0.7) RL2.5 k
1.5 k
15V
D1D2
i

14
i 3.5mA
4
LL
V iR 3.5 2.5 volt
8.75volt
46. If three moles of monoatomic gas




5
3
is
mixed with two moles of a diatomic gas
7
5




,
the value of adiabatic exponent for the mixture is:
(1) 1.75 (2) 1.40
(3) 1.52 (3) 1.35
Ans. (3)
Sol. f1 = 3, f2 = 5
n1 = 3, n2 = 2
1 1 2 2
mixture
12
n f n f 9 10 19
fn n f 5
mixture
2 5 29
1 1.52
19 19
47. Three blocks A, B and C are pulled on a horizontal
smooth surface by a force of 80 N as shown in
figure
5 kg
AF=80N
3 kg 2 kg
B C
T1T2
The tensions T1 and T2 in the string are
respectively:
(1) 40N, 64N
(2) 60N, 80N
(3) 88N, 96N
(4) 80N, 100N
Ans. (1)
Sol. aA = aB = aC

2
F 80 8m / s
5 3 2 10
5 kg
AT1
T1 = 5 × 8 = 40
3 kg
BT2
T1
T2 – T1 = 3 × 8 T2 = 64
48. When a potential difference V is applied across a
wire of resistance R, it dissipates energy at a rate
W. If the wire is cut into two halves and these
halves are connected mutually parallel across the
same supply, the same supply, the energy
dissipation rate will become:
(1) 1/4W (2) 1/2W
(3) 2W (4) 4W
Ans. (4)
Sol.
2
vW
R
….(i)



2
vW'
1R
22
….(ii)
From (i) & (ii), we get
W ' 4W
49. Match List I with List II
List-I
List-II
A.
Gauss’s law of
magnetostatics
I.

0
1
o E da dV
B.
Faraday’s law of
electro magnetic
induction
II.
o B da 0
C.
Ampere’s law
III.

d
o E dl B da
dt
D.
Gauss’s law of
electrostatics
IV.

0
o B dl I
Choose the correct answer from the options given
below:
(1) A-I, B-III, C-IV, D-II
(2) A-III, B-IV, C-I, D-II
(3) A-IV, B-II, C-III, D-I
(4) A-II, B-III, C-IV, D-I
Ans. (4)
Sol. Maxwell’s equation
50. Projectiles A and B are thrown at angles of 45° and
60° with vertical respectively from top of a 400 m high
tower. If their ranges and times of flight are same, the
ratio of their speeds of projection vA : vB is :
(1)
1: 3
(2)
2 :1
(3) 1 : 2 (4)
1: 2
Ans. (Bonus)
Sol.
45° 60°
uA
uB
For uA & uB time of flight and range can not be
same. So above options are incorrect.
SECTION-B
51. A power transmission line feeds input power at 2.3
kV to a step down transformer with its primary
winding having 3000 turns. The output power is
delivered at 230 V by the transformer. The current in
the primary of the transformer is 5A and its
efficiency is 90%. The winding of transformer is
made of copper. The output current of transformer
is_____A.
Ans. (45)
Sol. Pi = 2300 ×5 watt
P0 = 2300 × 5 × 0.9 = 230 × I2
I2 = 45A
52. A big drop is formed by coalescing 1000 small
identical drops of water. If E1 be the total surface
energy of 1000 small drops of water and E2 be the
surface energy of single big drop of water, the E1 :
E2 is x : 1 where x = ________.
Ans. (10)
Sol.
4 3 4 3
33
r 1000 R
R = 10r
E1 = 1000 × 4r2 × S
E2 = 4 (10r)2 S

1
2
E10 , x 10
E1
53. Two discs of moment of inertia I1
= 4 kg m2 and
I2 = 2 kg m2 about their central axes & normal to
their planes, rotating with angular speeds 10 rad/s
& 4 rad/s respectively are brought into contact face
to face with their axe of rotation coincident. The
loss in kinetic energy of the system in the process
is_____J.
Ans. (24)
Sol. I11 + I22 = (I1 + I2)0 (C.O.A.M.)
gives 0 = 8 rad/s
22
1 1 1 2 2
11
E I I 216J
22
2
2 1 2 0
1
E I I 192J
2
E = 24J
54. In an experiment to measure the focal length (f) of
a convex lens, the magnitude of object distance (x)
and the image distance (y) are measured with
reference to the focal point of the lens. The y-x
plot is shown in figure.
The focal length of the lens is_____cm.
403020
10
y
(cm)
A
B
C
D E
x (cm)
10
20
30
40
Ans. (20)
Sol.
1 1 1
f 20 (f 20) f

21
f 20cm
f 20 f

Or x1x2 = f2 gives f = 20 cm
55. A vector has magnitude same as that of
ˆˆ
A 3j 4j
and is parallel to
ˆˆ
B 4i 3j
. The x
and y components of this vector in first quadrant
are x and 3 respectively where x = _______.
Ans. (4)
Sol.
ˆˆ
5(4i 3j) ˆˆ
ˆ
N |A|B 4i 3j
5
x = 4
56. The current of 5A flows in a square loop of sides 1
m is placed in air. The magnetic field at the centre
of the loop is
7
X 2 10 T
. The value of X
is_____.
Ans. (40)
Sol.
0i11
B4 122
42



=
7
4 10 5 2 2
=
7
40 2 10 T
57. Two identical charged spheres are suspended by
string of equal lengths. The string make an angle of
37° with each other. When suspended in a liquid of
density 0.7 g/cm3, the angle remains same. If
density of material of the sphere is 1.4 g/cm3, the
dielectric constant of the liquid is _________




3
tan37 4
.
Ans. (2)
Sol.
Tcos = mg
Tsin = Fe
tan =
e
F
mg
e
B
F
tan Vg

….(i)
e
BL
F
tan k
( )Vg

….(ii)
From Eq. (i) & (ii)
B B L
Vg ( )kVg
1.4 = 0.7 k
k = 2
58. A simple pendulum is placed at a place where its
distance from the earth’s surface is equal to the
radius of the earth. If the length of the string is 4m,
then the time period of small oscillations will be
_______s. [take g = 2 ms–2]
Ans. (8)
Sol. Acceleration due to gravity g’ =
g
4
4
T2 g
44
T2 g
4
T 2 8s
59. A point source is emitting sound waves of intensity
16 × 10–8 Wm–2 at the origin. The difference in
intensity (magnitude only) at two points located at
a distances of 2m and 4m from the origin
respectively will be _______ × 10–8 Wm–2.
Ans. (Bonus)
Sol. Question is wrong as data is incomplete.
60. Two resistance of 100 and 200 are connected
in series with a battery of 4 V and negligible
internal resistance. A voltmeter is used to measure
voltage across 100 resistance, which gives
reading as 1 V. The resistance of voltmeter must
be______.
Ans. (200)
Sol.
100200
4V
RV
v
v
R 100 200
R 100 3
3Rv = 2Rv + 200
Rv = 200
CHEMISTRY
TEST PAPER WITH SOLUTION
SECTION-A
61. Which among the following purification methods
is based on the principle of “Solubility” in two
different solvents?
(1) Column Chromatography
(2) Sublimation
(3) Distillation
(4) Differential Extraction
Ans. (4)
Sol. Different Extraction
Different layers are formed which can be separated
in funnel. (Theory based).
62. Salicylaldehyde is synthesized from phenol, when
reacted with
(1)
O
HCl
, NaOH
(2) CO2, NaOH
(3) CCl4, NaOH
(4) HCCl3, NaOH
Ans. (4)
Sol.
CHCl /NaOH
3
OH OH
CHO
63. Given below are two statements:
Statement – I: High concentration of strong
nucleophilic reagent with secondary alkyl halides
which do not have bulky substituents will follow
SN2 mechanism.
Statement – II: A secondary alkyl halide when
treated with a large excess of ethanol follows SN1
mechanism.
In the the light of the above statements, choose the
most appropriate from the questions given below:
(1) Statement I is true but Statement II is false.
(3) Statement I is false but Statement II is true.
(3) Both statement I and Statement II are false.
(4) Both statement I and Statement II are true.
Ans. (4)
Sol. Statement – I: Rate of SN2 [R–X][Nu–]
SN2 reaction is favoured by high concentration of
nucleophile (Nu–) & less crowding in the substrate
molecule.
Statement – II: Solvolysis follows SN1 path.
Both are correct Statements.
64. m–chlorobenzaldehyde on treatment with 50%
KOH solution yields
(1)
Cl
CH C
OH O
Cl
(2)
Cl
COO
+
Cl
CH2OH
(3)
OH
C H
O
OH
C H
O
+
(4)
OH
COO
+
OH
CH2OH
Ans. (2)
Sol. Meta–chlorobenzaldehyde will undergo
Cannizzaro reaction with 50% KOH to give m–
chlorobenzoate ion and m–chlorobenzyl alcohol.
OH
COO
+
OH
CH2OH
50% KOH
CHO
Cl
2
65. Given below are two statements: One is labelled as
Assertion A and the other is labelled as Reason R.
Assertion A : H2Te is more acidic than H2S.
Reason R: Bond dissociation enthalpy of H2Te is
lower than H2S.
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) Both A and R are true and R is the correct
explanation of A.
(3) A is false but R is true.
(4) A is true but R is false.
Ans. (2)
Sol. Due to lower Bond dissociation enthalpy of H2Te it
ionizes to give H+ more easily than H2S.
66. Product A and B formed in the following set of
reactions are:
B
B2H6
H2O2,NaOH(aq.)
CH3
A
H+/H2O
(1)
CH2OH
A =
CH2OH
B =
OH
(2)
CH3
OH
A =
CH3
B =
OH
(3)
CH2OH
A =
CH3
B =
OH
(4)
CH3
A =
CH3
B =
OH
OH
Ans. (2)
Sol.
OH
CH3B2H6
H2O2,NaOH
CH3
H+/H2O
CH3
OH
(A)
(B)
67. IUPAC name of following compound is
CH3—CH—CH2—CN
NH2
(1) 2–Aminopentanenitrile
(2) 2–Aminobutanenitrile
(3) 3–Aminobutanenitrile
(4) 3–Aminopropanenitrile
Ans. (3)
Sol.
CH3—CH—CH2—CN
NH2
4 3 21
3–Aminobutanenitrile
68. The products A and B formed in the following
reaction scheme are respectively
A
(i) conc.HNO3/conc.H2SO4
323–333 K
(ii) Sn/HCl
(i) NaNO2, HCl, 273–278 K
(ii) Phenol
B
(1)
Cl NO2
NO2
HO
,
(2)
NH2
,
NH2
OH
(3)
NH2
,N NHO
(4)
NH2
,
OH
N N
Ans. (3)
Sol.
N NHO
conc.HNO3
conc.H2SO4
NO2
Sn/HCl
NH2
NaNO2+HCl
0–5ºC
NNCl–
+
OH
69. The molecule/ion with square pyramidal shape is:
(1) [Ni(CN)4]2– (2) PCl5
(3) BrF5 (4) PF5
Ans. (3)
Sol. BrF5
Br
F
FF
F
F
Square Pyramidal.
70. The orange colour of K2Cr2O7 and purple colour of
KMnO4 is due to
(1) Charge transfer transition in both.
(2) dd transition in KMnO4 and charge transfer
transitions in K2Cr2O7.
(3) dd transition in K2Cr2O7 and charge transfer
transitions in KMnO4.
(4) dd transition in both.
Ans. (1)
Sol.
6
2 2 7
7
4
K Cr O Cr No d d transition Charge transfer
KMnO Mn No d d transition
71. Alkaline oxidative fusion of MnO2 gives “A”
which on electrolytic oxidation in alkaline solution
produces B. A and B respectively are:
(1) Mn2O7 and MnO4
–
(2) MnO4
2– and MnO4
–
(3) Mn2O3 and MnO4
2–
(4) MnO4
2– and Mn2O7
Ans. (2)
Sol. Alkaline oxidative fusion of MnO2:
–2
2 2 4 2
2MnO 4OH O 2MnO 2H O
Electrolytic oxidation of MnO4
2– in alkaline medium.
2
44
MnO MnO e

72. If a substance ‘A’ dissolves in solution of a mixture
of ‘B’ and ‘C’ with their respective number of
moles as nA, nB and nC, mole fraction of C in the
solution is:
(1)
C
A B C
n
n n n
(2)
C
A B C
n
n n n
(3)
C
A B C
n
n – n – n
(4)
B
AB
n
nn
Ans. (2)
Sol. Mole fraction of C =
C
A B C
n
n n n
73. Given below are two statements:
Statement – I: Along the period, the chemical
reactivity of the element gradually increases from
group 1 to group 18.
Statement – II: The nature of oxides formed by
group 1 element is basic while that of group 17
elements is acidic.
In the the light above statements, choose the most
appropriate from the questions given below:
(1) Both statement I and Statement II are true.
(2) Statement I is true but Statement II is False.
(3) Statement I is false but Statement II is true.
(4) Both Statement I and Statement II is false.
Ans. (3)
Sol. Chemical reactivity of elements decreases along
the period therefore statement – I is false.
Group – 1 elements from basic nature oxides
while group – 17 elements form acidic oxides
therefore statement – II is true.
74. The coordination geometry around the manganese
in decacarbonyldimanganese(0)
(1) Octahedral (2) Trigonal bipyramidal
(3) Square pyramidal (4) Square planar
Ans. (1)
Sol. Mn2(CO)10
CO Mn Mn
CO
CO
CO
CO
CO
CO CO
CO CO
Octahedral around Mn
75. Given below are two statements:
Statement-I: Since fluorine is more electronegative
than nitrogen, the net dipole moment of NF3 is greater
than NH3.
Statement-II: In NH3, the orbital dipole due to
lone pair and the dipole moment of NH bonds are
in opposite direction, but in NF3 the orbital dipole
due to lone pair and dipole moments of N-F bonds
are in same direction.
In the light of the above statements. Choose the
most appropriate from the options given below.
(1) Statement I is true but Statement II is false.
(2) Both Statement I and Statement II are false.
(3) Both statement I and Statement II is are true.
(4) Statement I is false but Statement II is are true.
Ans. (2)
Sol.
FFF
N
HHH
N
76. The correct stability order of carbocations is
(1)
3 3 3 2 3 2 3
(CH ) C CH C H (CH ) C H C H
(2)
3 3 2 3 2 3 3
C H (CH ) C H CH C H (CH ) C
(3)
3 3 3 2 3 2 3
(CH ) C (CH ) C H CH C H C H
(4)
3 3 2 3 3
3
C H CH C H CH C H (CH )C
|
CH
Ans. (3)
Sol. More no. of hyperconjugable Hydrogens, more
stable is the carbocations.
77. The solution from the following with highest
depression in freezing point/lowest freezing point is
(1) 180 g of acetic acid dissolved in water
(2) 180 g of acetic acid dissolved in benzene
(3) 180 g of benzoic acid dissolved in benzene
(4) 180 g of glucose dissolved in water
Ans. (1)
Sol.
f
T
is maximum when i × m is maximum.
1)

1
180
m3
60
, i = 1 +
Hence
ff
T (1 ) k 3 1.86 5.58 C ( 1)
2)
2 f f
180 3
m 3, i 0.5, T k ' 7.68 C
60 2
3)
3 f f
180 1.48
m 1.48, i 0.5, T k ' 3.8 C
122 2
4)
4 f f
180
m 1, i 1, T 1 k ' 1.86 C
180
As per NCERT,
 1
f2
k ' (H O) 1.86 k kg mol
 1
f
k ' (Benzene) 5.12 k kg mol
78. A and B formed in the following reactions are:
2 2 2
2 2 2
CrO Cl 4NaOH A 2NaCl 2H O
A 2HCl 2H O B 3H O
(1) A = Na2CrO4, B = CrO5
(2) A = Na2Cr2O4, B = CrO4
(3) A = Na2Cr2O7, B = CrO3
(4) A = Na2Cr2O7, B = CrO5
Ans. (1)
Sol.
2 2 2 4 2
(A)
CrO Cl 4NaOH Na CrO 2NaCl 2H O
2 4 2 2 5 2
Missing from
(B) balanced eqaution
Na CrO 2H O 2HCl CrO 2NaCl 3H O
79. Choose the correct statements about the hydrides
of group 15 elements.
A. The stability of the hydrides decreases in the
order NH3 > PH3 > AsH3 > SbH3 > BiH3
B. The reducing ability of the hydrides increases
in the order NH3 < PH3 < AsH3 < SbH3 < BiH3
C. Among the hydrides, NH3 is strong reducing
agent while BiH3 is mild reducing agent.
D. The basicity of the hydrides increases in the
order NH3 < PH3 < AsH3 < SbH3 < BiH3
Choose the most appropriate from the option given
below:
(1) B and C only (2) C and D only
(3) A and B only (4) A and D only
Ans. (3)
Sol. On moving down the group, bond strength of M–H
bond decreases, which reduces the thermal stability
but increases reducing nature of hydrides, hence A
and B are correct statements.
80. Reduction potential of ions are given below:
4 4 4
ClO IO BrO
E 1.19V E 1.65V E 1.74V
The correct order of their oxidising power is:
(1)

4 4 4
ClO IO BrO
(2)

4 4 4
BrO IO ClO
(3)

4 4 4
BrO ClO IO
(4)

4 4 4
IO BrO ClO
Ans. (2)
Sol. Higher the value of ve SRP (Std. reduction
potential) more is tendency to undergo reduction,
so better is oxidising power of reactant.
Hence, ox. Power:-

4 4 4
BrO IO ClO
SECTION-B
81. Number of complexes which show optical
isomerism among the following is ________.
3
22
–
cis [Cr(ox) Cl ] ,
3
3
[Co(en) ] ,
2
22
cis [Pt(en) Cl ] ,
22
cis [Co(en) Cl ] ,
2
22
trans [Pt(en) Cl ] ,
3
22
–
trans [Cr(ox) Cl ]
Ans. (4)
Sol.
3
22
–
cis [Cr(ox) Cl ]
can show optical isomerism
(no POS & COS)
3
3
[Co(en) ]
can show (no POS & COS)
2
22
cis [Pt(en) Cl ]
can show (no POS & COS)
22
cis [Co(en) Cl ]
can show (no POS & COS)
2
22
trans [Pt(en) Cl ]
can’t show (contains POS
& COS)
3
22
–
trans [Cr(ox) Cl ]
can’t show (contains
POS & COS)
82. NO2 required for a reaction is produced by
decomposition of N2O5 in CCl4 as by equation
2 2 2
5
24
(g) (g)
g
N O NO O
The initial concentration of N2O5 is 3 mol L–1 and it
is 2.75 mol L–1 after 30 minutes.
The rate of formation of NO2 is x × 10–3 mol L–1
min–1 , value of x is ________.
Ans. (17)
Sol. Rate of reaction (ROR)
2 5 2 2
11
24

[N O ] [NO ] [O ]
t t t
ROR
––
[N O ] ( . ) mol L min
t
11
25
1 1 2 75 3
2 2 30
ROR
11
1 0 25
2 30
 ––
( . ) mol L min
ROR
11
1
240
––
mol L min
Rate of formation of NO2
24
[NO ] ROR
t
––
. molL min
3 1 1 3
416 66 10 17 10
240
.
83. Two reactions are given below:
2 2 3
3
2 822
2
(g) (s)
s
Fe O Fe O , Hº – kJ / mol
2
1110
2
(g) (g)
s
C O CO , Hº – kJ / mol
Then enthalpy change for following reaction
23
3 2 3
(s) (s) (g)
s
C Fe O Fe CO
Ans. (492)
Sol.
2 2 3
3
2 822
2
(g) (s)
s
Fe O Fe O , Hº – kJ / mol
……(1)
2
1110
2
(g) (g)
s
C O CO , Hº – kJ / mol
……(2)
23
3 2 3
(s) (s) (g)
s
C Fe O Fe CO
,
3
H?
(3) = 3 × (2) – (1)
H3 = 3 ×H2 –H1
= 3(–110) + 822
= 492 kJ/mole
84. The total number of correct statements, regarding
the nucleic acids is_________.
A. RNA is regarded as the reserve of genetic
information.
B. DNA molecule self-duplicates during cell
division
C. DNA synthesizes proteins in the cell.
D. The message for the synthesis of particular
proteins is present in DNA
E. Identical DNA strands are transferred to
daughter cells.
Ans. (3)
Sol. A. RNA is regarded as the reserve of genetic
information. (False)
B. DNA molecule self-duplicates during cell
division. (True)
C. DNA synthesizes proteins in the cell. (False)
D. The message for the synthesis of particular
proteins is present in DNA. (True)
E. Identical DNA strands are transferred to
daughter cells. (True)
85. The pH of an aqueous solution containing 1M
benzoic acid (pKa = 4.20) and 1M sodium benzoate
is 4.5. The volume of benzoic acid solution in 300
mL of this buffer solution is _______ mL.
Ans. (100)
Sol.

as
as
1M Benzoic acid 1M Sodium Benzoate
(V ml) (V ml)
Millimole V 1 V 1
pH = 4.5
pH = pka +
[salt]
log [acid]
4.5 = 4.2 +



s
a
V
log V
s
a
V2
V
……(1)

sa
V V 300
...... (2)
Va = 100 ml
86. Number of geometrical isomers possible for the
given structure is/are ___________.
D
H
D
H
Ans. (4)
Sol. 3 stereocenteres, symmetrical
Total Geometrical isomers 4. EE, ZZ, EZ (two
isomers)
D
H
D
H
*
*
*
87. Total number of species from the following which
can undergo disproportionation reaction _______.
1
2 2 3 4 2 2 2
–
H O ,ClO , P ,Cl ,Ag,Cu , F ,NO ,K
Ans. (6)
Sol. Intermediate oxidation state of element can undergo
disproportionation.
1
2 2 3 4 2 2
–
H O ,ClO ,P ,Cl ,Cu ,NO
88. Number of metal ions characterized by flame test
among the following is _______.
2 2 2 2 2 2 2
Sr ,Ba ,Ca ,Cu ,Zn ,Co , Fe
Ans. (4)
Sol. All the following metal ions will respond to flame
test.
2 2 2 2
Sr ,Ba ,Ca ,Cu
89. 2-chlorobutane + Cl2 C4H8Cl2(isomers)
Total number of optically active isomers shown by
C4H8Cl2, obtained in the above reaction is_______.
Ans. (6)
Sol.
90. Number of spectral lines obtained in He+ spectra,
when an electron makes transition from fifth
excited state to first excited state will be
Ans. (10)
Sol. 5th excited state n1 = 6
1st excited state n2 = 2
n = n1– n2 = 6 – 2 = 4
Maximum number of spectral lines
=
1 4 4 1 10
22

n( n ) ( )