FINAL JEE–MAIN EXAMINATION – JANUARY, 2024
(Held On Wednesday 31
st
January, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS
TEST PAPER WITH SOLUTION
SECTION-A
1. For 0 < c < b < a, let (a + b – 2c)x2 + (b + c – 2a)x
+ (c + a – 2b) = 0 and
1
be one of its root.
Then, among the two statements
(I) If
1,0
, then b cannot be the geometric
mean of a and c
(II) If
0,1
, then b may be the geometric
mean of a and c
(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. (1)
Sol. f(x) = (a + b – 2c) x2 + (b + c – 2a) x + (c + a – 2b)
f(x) = a + b – 2c + b + c – 2a + c + a – 2b = 0
f(1) = 0
c a 2b
1a b 2c
c a 2b
a b 2c
If, –1 < < 0
c a 2b
10
a b 2c
b + c < 2a and
ac
b2
therefore, b cannot be G.M. between a and c.
If,
01
c a 2b
01
a b 2c
b > c and
ac
b2
Therefore, b may be the G.M. between a and c.
2. Let a be the sum of all coefficients in the
expansion of (1 – 2x + 2x2)2023 (3 – 4x2+2x3)2024
and
x
2024
0
2
x0
log 1 t dt
t1
b lim .
x
If the equations
cx2 + dx + e = 0 and 2bx2 + ax + 4 = 0 have a
common root, where c, d, e R, then d : c : e
equals
(1) 2 : 1 : 4 (2) 4 : 1 : 4
(3) 1 : 2 : 4 (4) 1 : 1 : 4
Ans. (4)
Sol. Put x = 1
a1
x
2024
0
2
x0
ln 1 t dt
1t
b lim x
Using L’ HOPITAL Rule
2024
x0
ln 1 x 11
b lim 2x 2
1x
Now, cx2 + dx + e = 0, x2 + x + 4 = 0
(D < 0)
c d e
1 1 4
3. If the foci of a hyperbola are same as that of the
ellipse
22
xy
1
9 25
and the eccentricity of the
hyperbola is
15
8
times the eccentricity of the
ellipse, then the smaller focal distance of the point
14 2
2, 35
on the hyperbola, is equal to
(1)
28
753
(2)
24
14 53
(3)
2 16
14 53
(4)
28
753
Ans. (1)
Sol.
22
xy
1
9 25
a = 3, b = 5
94
e1
25 5
foci 0, be
= (0, ± 4)
H
4 15 3
e5 8 2
Let equation hyperbola
22
22
xy 1
AB
H
B e 4
8
B3
2 2 2
H
64 9
A B e 1 1
94
280
A9
22
xy 1
80 64
99
Directrix :
H
B 16
ye9
PS =
3 14 2 16
e PM 2 3 5 9
28
753
4. If one of the diameters of the circle x2 + y2 – 10x +
4y + 13 = 0 is a chord of another circle C, whose
center is the point of intersection of the lines 2x +
3y = 12 and 3x – 2y = 5, then the radius of the
circle C is
(1)
20
(2) 4
(3) 6 (4)
32
Ans. (3)
Sol.
C
M
(3, 2)
4
P
(5, –2)
2x + 3y = 12
3x – 2y = 5
13 x = 39
x = 3, y = 2
Center of given circle is (5, –2)
Radius
25 4 13 4
CM 4 16 5 2
CP 16 20 6
5. The area of the region
2xy x 1 x 2
x, y : y 4x, x 4, 0, x 3
x 3 x 4
is
(1)
16
3
(2)
64
3
(3)
8
3
(4)
32
3
Ans. (4)
Sol.
2
y 4x,x 4
xy x 1 x 2 0
x 3 x 4
Case – I :
y0
x x 1 x 2 0
x 3 x 4
x 0,1 2,3
Case – II : y < 0
x x 1 x 2 0, x 1,2 3,4
x 3 x 4
4
0
Area 2 x dx
4
3/2
0
2 32
2x
33
6. If
4x 3 2
f x , x
6x 4 3
and (fof) (x) = g(x), where
22
g : ,
33
then (gogog) (4) is equal
to
(1)
19
20
(2)
19
20
(3) – 4 (4) 4
Ans. (4)
Sol.
4x 3
fx 6x 4
4x 3
43
34x
6x 4
g x x
4x 3 34
64
6x 4
g x x g g g 4 4
7.
2 sin x
2
x0
e 2 sin x 1
lim x
(1) is equal to – 1 (2) does not exist
(3) is equal to 1 (4) is equal to 2
Ans. (4)
Sol.
2 sin x
2
x0
e 2 sin x 1
lim x
2 sin x 2
22
x0
e 2 sin x 1 sin x
lim x
sin x
Let |sinx| = t
2t 2
22
t 0 x 0
e 2t 1 sin x
lim lim
tx
2t
t0
2e 2
lim 1 2 1 2
2t
8. If the system of linear equations
x 2y z 4
2x y 3z 5
3x y z 3
has infinitely many solutions, then 12 + 13 is
equal to
(1) 60 (2) 64
(3) 54 (4) 58
Ans. (4)
Sol.
1 2 1
D 2 3
31
= 1( + 3) + 2(2 – 9) + 1(–2 – 3)
= + 3 + 4 – 18 – 2 – 3
For infinite solutions D = 0, D1 = 0, D2 = 0 and
D3 = 0
D = 0
– 3 + 4 = 17 ….(1)
1
4 2 1
D 5 3 0
31
2
1 4 1
D 2 5 3 0
33
1 5 9 4 2 9 1 6 15 0
13 9 36 9 0
54
13 54, 13
put in (1)
54 54
3 4 17
13 13
54 39 216 221
15 5
1
3
Now,
1 54
12 13 12. 13.
3 13
= 4 + 54 = 58
9. The solution curve of the differential equation
ee
dx
y x log x log y 1 ,
dy
x > 0, y > 0 passing
through the point (e, 1) is
(1)
e
y
log x
x
(2)
2
e
y
log y
x
(3)
e
x
log y
y
(4)
e
x
2 log y 1
y
Ans. (3)
Sol.
dx x x
ln 1
dy y y
Let
xt x ty
y
dx dt
ty
dy dy
dt
t y t ln t 1
dy
dt dt dy
y t ln t
dy t ln t y
dt dy
t.ln t y
dp dy
py
let ln t = p
1dt dp
t
lnp = lny +c
ln(ln t) = ln y + c
x
ln ln ln y c
y
at x = e, y = 1
e
ln ln ln(1) c c 0
1
x
ln ln ln y
y
ln y
x
ln e
y
x
ln y
y
10. Let Z and let A (), B (1, 0), C ()
and D (1, 2) be the vertices of a parallelogram
ABCD. If AB =
10
and the points A and C lie on
the line 3y = 2x + 1, then 2 () is equal
to
(1) 10 (2) 5
(3) 12 (4) 8
Ans. (4)
Sol.
A( ) B(1, 0)
D(1, 2) C( )
Let E is mid point of diagonals
11
22
&
20
22
2
2
2 2 2 2 8
11. Let y = y(x) be the solution of the differential
equation
tan x y
dy ,
dx sin x secx sin x tan x
x 0, 2
satisfying the condition
y 2.
4
Then,
y3
is
(1)
e
3 2 log 3
(2)
e
32 log 3
2
(3)
e
3 1 2log 3
(4)
e
3 2 log 3
Ans. (1)
Sol.
dy sin x ycos x
1 sin x
dx sin x.cos x sin x.
cos x cos x
2
sin x y cos x
sin x 1 sin x
2
dy sec x y.2 cos ec2x
dx
2
dy 2 cos ec 2x .y sec x
dx
dy p.y Q
dx
pdx 2cosec 2x dx
I.F. e e
Let 2x = t
dx
21
dt
dt
dx 2
cosec(t)dt
e
t
ln tan 2
e
ln tan x 1
etanx
y(IF) Q IF dx c
2
11
y sec x c
tan x tan x
1 dt
y. c
tan x | t |
for tan x = t
1
y. ln | t | c
tan x
y tan x ln | tan x | c
Put
x,
4
y = 2
2 = ln 1 + c
c = 2
y | tan x | ln | tan x | 2
y 3 ln 3 2
3
12. Let
ˆ ˆ ˆ
a 3i j 2k,
ˆ ˆ ˆ
b 4i j 7k
and
ˆ ˆ ˆ
c i 3j 4k
be three vectors. If a vectors
p
satisfies
p b c b
and
p a 0
, then
ˆ ˆ ˆ
p i j k
is equal to
(1) 24
(2) 36
(3) 28
(4) 32
Ans. (4)
Sol.
p b c b 0
p c b 0
p c b p c b
Now,
p.a 0 given
So,
c.a a.b 0
(3 – 3 – 8) + (12 + 1 – 14) = 0
= –8
p c 8b
ˆ ˆ ˆ
p 31i 11j 52k
So,
ˆ ˆ ˆ
p.(i j k)
= –31 + 11 + 52
= 32
13. The sum of the series
24
1
1 3 1 1
24
2
1 3 2 2
24
3
1 3 3 3
….. up to 10 terms
is
(1)
45
109
(2)
45
109
(3)
55
109
(4)
55
109
Ans. (4)
Sol. General term of the sequence,
r24
r
T1 3r r
r4 2 2
r
Tr 2r 1 r
r2
22
r
T
r 1 r
r22
r
Tr r 1 r r 1
22
r22
1r r 1 r r 1
2
Tr r 1 r r 1
22
1 1 1
2 r r 1 r r 1
Sum of 10 terms,
10
r
r1
1 1 1 55
T2 1 109 109
14. The distance of the point Q(0, 2, –2) form the line
passing through the point P(5, –4, 3) and
perpendicular to the lines
ˆˆ
r 3i 2k
+
ˆ ˆ ˆ
2i 3j 5k ,
and
ˆ ˆ ˆ
r i 2j k
+
ˆ ˆ ˆ
µ i 3j 2k
,
µ
is
(1)
86
(2)
20
(3)
54
(4)
74
Ans. (4)
Sol. A vector in the direction of the required line can be
obtained by cross product of
ˆ ˆ ˆ
i j k
2 3 5
1 3 2
ˆ ˆ ˆ
9i 9j 9k
Required line,
ˆ ˆ ˆ ˆ
ˆˆ
r 5i 4 j 3k ' 9i 9 j 9k
ˆ ˆ ˆ ˆ
ˆˆ
r 5i 4 j 3k i j k
Now distance of (0, 2, –2)
P.V. of
ˆ ˆ ˆ
P 5 i 4 j 3 k
ˆ ˆ ˆ
AP 5 i 6 j 5 k
ˆ ˆ ˆ
AP i j k 0
5 6 5 0
2
AP 49 16 9
AP 74
15. For
, , 0.
If sin–1 + sin–1 + sin–1 = and
( + + ) ( – + ) = 3, then equal to
(1)
3
2
(2)
1
2
(3)
31
22
(4)
3
Ans. (1)
Sol. Let
1 1 1
sin A,sin B,sin C
A + B + C =
223
2 2 2
2 2 2 1
22
1
cosC 2
sinC
21
cosC 1 2
3
2
16. Two marbles are drawn in succession from a box
containing 10 red, 30 white, 20 blue and 15 orange
marbles, with replacement being made after each
drawing. Then the probability, that first drawn
marble is red and second drawn marble is white, is
(1)
2
25
(2)
4
25
(3)
2
3
(4)
4
75
Ans. (4)
Sol. Probability of drawing first red and then white
10 30 4
75 75 75
17. Let g(x) be a linear function and
1
x
g x , x 0
f x ,
1x , x 0
2x
is continuous at x = 0.
If
f 1 f 1 ,
then the value of g(3) is
(1)
e1/3
14
log
3 9e
(2)
e
14
log 1
39
(3)
e
4
log 1
9
(4)
e1/3
4
log 9e
Ans. (4)
Sol. Let g(x) = ax + b
Now function f(x) in continuous at x = 0
x0
lim f x f 0
1
x
x0
1x
lim b
2x
0b
g x ax
Now, for x > 0
11
x
2
1 1 x 1
fx x 2 x 2x
1
x
2
1 x 1 x 1
ln
2 x 2 x x
1 2 2
f 1 ln
9 3 3
And f(–1) = g(–1) = –a
2 2 1
a ln
3 3 9
21
g 3 2ln 33
1/3
4
ln 9e
18. If
32
23
32
x 2x 1 1 3x
f (x) 3x 2 2x x 6
x x 4 x 2
for all
x
, then 2f(0) +
f0
is equal to
(1) 48 (2) 24
(3) 42 (4) 18
Ans. (3)
Sol.
0 1 1
f 0 2 0 6 12
0 4 2
+
32
2
32
x 2x 1 1 3x
6x 2 3x
x x 4 x 2
+
32
23
2
x 2x 1 1 3x
3x 2 2x x 6
3x 1 0 2x
0 0 3 0 1 1 0 1 1
f 0 2 0 6 0 2 0 2 0 6
0 4 2 0 4 2 1 0 0
= 24 – 6 = 18
2f 0 f 0 42
19. Three rotten apples are accidently mixed with
fifteen good apples. Assuming the random variable
x to be the number of rotten apples in a draw of
two apples, the variance of x is
(1)
37
153
(2)
57
153
(3)
47
153
(4)
40
153
Ans. (4)
2
23
32
3x 4x 3
f x 3x 2 2x x 6
x x 4 x 2
Sol. 3 bad apples, 15 good apples.
Let X be no of bad apples
Then P(X = 0)
15
2
18
2
C105
C 153
3 15
11
18
2
CC
45
P X 1 C 153
3
2
18
2
C3
P X 2 C 153
105 45 3 51
E X 0 1 2
153 153 153 153
1
3
2
2
Var X E X E X
2
105 45 3 1
0 1 4
153 153 153 3
57 1 40
153 9 153
20. Let S be the set of positive integral values of a for
which
2
2
ax 2 a 1 x 9a 4 0, x
x 8x 32
.
Then, the number of elements in S is :
(1) 1
(2) 0
(3)
(4) 3
Ans. (2)
Sol. ax2 + 2 (a + 1) x + 9a + 4 < 0
xR
a0
SECTION-B
21. If the integral
1
11 5
22
22
0
525 sin 2x cos x 1 cos x dx
is equal to
n 2 64
, then n is equal to ______
Ans. (176)
Sol.
1
211 5 2
22
0
I sin 2x cos x 1 cos x dx
Put cosx = t2 sinx dx = – 2 t dt
1
2 11 5
0
I 4 t t 1 t t dt
1
14 5
0
I 4 t 1 t dt
Put 1 + t5 = = k2
5t4dt = 2k dk
22
2
1
2k
I 4 k 1 k dk
5
2
6 4 2
1
8
I k 2k k dk
5
2
7 5 3
1
8 k 2k k
I5 7 5 3
8 8 2 8 2 2 2 1 2 1
I5 7 5 3 7 5 3
8 22 2 8
I5 105 105
525 I 176 2 64
22. Let
S 1,
and
f :S
be defined as
x11 7 12 61
t5
1
f (x) e 1 (2t 1) t 2 t 3 2t 10 dt
Let p = Sum of square of the values of x, where
f(x) attains local maxima on S. and q = Sum of the
values of x, where f(x) attains local minima on S.
Then, the value of p2 + 2q is ________
Ans. (27)
Sol.
11 5 7 12 61
x
f x e 1 2x 1 x 2 x 3 2x 10
Local minima at
1
x,
2
x = 5
Local maxima at x = 0, x = 2
1 11
p 0 4 4, q 5
22
Then p2 + 2q = 16 + 11 = 27
23. The total number of words (with or without
meaning) that can be formed out of the letters of
the word ‘DISTRIBUTION’ taken four at a time,
is equal to _____
Ans. (3734)
Sol. We have III, TT, D, S, R, B, U, O, N
Number of words with selection (a, a, a, b)
8
1
4!
C 32
3!
Number of words with selection (a, a, b, b)
4! 6
2!2!
Number of words with selection (a, a, b, c)
28
12
4!
C C 672
2!
Number of words with selection (a, b, c, d)
9
4
C 4! 3024
total = 3024 + 672 + 6 + 32
= 3734
24. Let Q and R be the feet of perpendiculars from the
point P(a, a, a) on the lines x = y, z = 1 and x = –y,
z = –1 respectively. If QPR is a right angle, then
12a2 is equal to _____
Ans. (12)
Sol.
x y z 1 r Q r, r,1
1 1 0
x y z 1 k R k, k, 1
1 1 0
ˆ ˆ ˆ
PQ a r i a r j a 1 k
a = r + a – r = 0.
2a = 2r a = r
ˆˆ
PR a k i a k j a 1 k
a – k – a – k = 0 k = 0
As,
PQ PR
a r a k a r a k a 1 a 1 0
a = 1 or –1
12a2 = 12
25. In the expansion of
5
2
23
3 3 1
1 x 1 x 1 x x x
,
x0
, the
sum of the coefficient of x3 and x–13 is equal to ___
Ans. (118)
Sol.
5
2
23
3 3 1
1 x 1 x 1 x x x
5
3
21
1 x 1 x 1 x
2 15
15
1 x 1 x 1 x
x
17 17
15
1 x x 1 x
x
3
coeff x
in the expansion
18
coeff x
in
17 17
1 x x 1 x
= 0 – 1
= –1
13
coeff x
in the expansion
2
coeff x
in
(1 + x)17 – x(1 + x)17
17 17
21
= 17 × 8 – 17
= 17 × 7
= 119
Hence Answer = 119 – 1 = 118
26. If denotes the number of solutions of |1 - i|x = 2x
and
| z |
arg(z)
, where
41 i i
z 1 i
4i 1 i
,
i1
, then
the distance of the point () from the line
4x – 3y = 7 is _____
Ans. (3)
Sol.
xx
2 2 x 0 1
4i i i i
z 1 i
4 1 1
23
i1 4i 6i 4i 1
2
2i
24
2
Distance from (1, 4) to
4x 3y 7
Will be
15 3
5
27. Let the foci and length of the latus rectum of an
ellipse
22
22
xy
1,
ab
a > b be
5,0
and
50
,
respectively. Then, the square of the eccentricity of
the hyperbola
22
2 2 2
xy
1
b a b
equals
Ans. (51)
Sol.
focii 5,0
;
2
2b 50
a
ae = 5
25 2a
b2
2 2 2 5 2a
b a 1 e 2
252
a 1 e 2
2
5 5 2
1e
e2
2
2 2e e
2
2e e 2 0
2
2e 2e e 2 0
2e e 2 1 1 2 0
e 2 2e 1 0
e2
;
1
e2
22
2 2 2
xy
1
b a b
a 5 2
b = 5
2 2 2 2 2
11
a b b e 1 e 51
28. Let
a
and
b
be two vectors such that
| a | 1,| b | 4
and
.
a b 2
. If
c 2a b 3b
and the angle between
b
and
c
is , then
192sin2 is equal to______
Ans. (48)
Sol.
2
b c 2a b b 3 b
|b||c| cos = – 3|b|2
|c| cos = –12, as |b| = 4
a b 2
1
cos 23
2
2
| c | 2a b 3b
3
64 144 192
4
22
| c | cos 144
2
192cos 144
2
192sin 48
29. Let A = {1, 2, 3, 4} and R = {(1, 2), (2, 3), (1, 4)}
be a relation on A. Let S be the equivalence
relation on A such that
RS
and the number of
elements in S is n. Then, the minimum value of n
is _______
Ans. (16)
Sol. All elements are included
Answer is 16
30. Let
f:
be a function defined by
x
x
4
f (x) 42
and
f (1 a )
4
f (a)
M x sin x 1 x dx,
f (1 a )
4
f (a)
1
N sin x 1 x dx;a 2
. If
M N, ,
, then the least value of
22
is equal to ______
Ans. (5)
Sol. f(a) + f(1 – a) = 1.
f (1 a )
4
f (a)
M 1 x .sin x 1 x dx
M = N – M 2M = N
= 2; = 1;
Ans. 5
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
31. The parameter that remains the same for molecules
of all gases at a given temperature is :
(1) kinetic energy (2) momentum
(3) mass (4) speed
Ans. (1)
Sol. KE =
fkT
2
Conceptual
32. Identify the logic operation performed by the given
circuit.
(1) NAND (2) NOR
(3) OR (4) AND
Ans. (3)
Sol.
Y A B A B A B
(De-Morgan's law)
33. The relation between time ‘t’ and distance ‘x’ is t =
x2 + x, where and are constants. The relation
between acceleration (a) and velocity (v) is:
(1) a = –2v3 (2) a = –5v5
(3) a = –3v2 (4) a = –4v4
Ans. (1)
Sol.
2
t x x
(differentiating wrt time)
dt 2x
dx
12x
v
(differentiating wrt time)
2
1 dv dx
2
dt dt
v
3
dv 2v
dt
34. The refractive index of a prism with apex angle A
is cot A/2. The angle of minimum deviation is :
(1) m = 180° – A
(2) m = 180° – 3A
(3) m = 180° – 4A
(4) m = 180° – 2A
Ans. (4)
Sol.
Am
sin 2
A
sin 2
Am
Asin
cos 2
2
AA
sin sin
22
m
A
A
sin sin
2 2 2
A A m
2 2 2 2
m2A
35. A rigid wire consists of a semicircular portion of
radius R and two straight sections. The wire is
partially immerged in a perpendicular magnetic
field B = B0
j
as shown in figure. The magnetic
force on the wire if it has a current i is :
(1)
iBR j
(2)
2iBR j
(3)
iBR j
(4)
2iBR j
Ans. (4)
Sol.
i
2R
Note : Direction of magnetic field is in
ˆ
k
F i B
= 2R
F 2iRBj
36. If the wavelength of the first member of Lyman
series of hydrogen is . The wavelength of the
second member will be
(1)
27
32
(2)
32
27
(3)
27
5
(4)
5
27
Ans. (1)
Sol.
2
22
1 13.6z 1 1
hc 12
….. (i)
2
22
1 13.6z 1 1
' hc 13
….. (ii)
On dividing (i) & (ii)
27
'32
37. Four identical particles of mass m are kept at the
four corners of a square. If the gravitational force
exerted on one of the masses by the other masses is
2
2
2 2 1 Gm ,
32 L
the length of the sides of the
square is
(1)
L
2
(2) 4L
(3) 3L (4) 2L
Ans. (2)
Sol.
F
F
F
'
m
m
m
m
a
a
a
a
net
F 2F F '
2
2
Gm
Fa
and
2
2
Gm
F'
2a
22
net 22
Gm Gm
F2
a 2a
2
2
2 2 1 Gm
32 L
2
2
Gm 2 2 1
2
a
a = 4L
38. The given figure represents two isobaric processes
for the same mass of an ideal gas, then
(1)
21
PP
(2)
21
PP
(3)
12
PP
(4)
12
PP
Ans. (4)
Sol. PV = nRT
nR
VT
P
Slope =
nR
P
Slope
1
P
(Slope)2 > (Slope)1
P2 < P1
39. If the percentage errors in measuring the length
and the diameter of a wire are 0.1% each. The
percentage error in measuring its resistance will
be:
(1) 0.2% (2) 0.3%
(3) 0.1% (4) 0.144%
Ans. (2)
Sol.
2
L
Rd
4
R L 2 d
R L d
L0.1%
L
and
d0.1%
d
R
R
= 0.3%
40. In a plane EM wave, the electric field oscillates
sinusoidally at a frequency of 5 × 1010 Hz and an
amplitude of 50 Vm–1. The total average energy
density of the electromagnetic field of the wave is :
[Use
0
= 8.85 × 10–12 C2 / Nm2]
(1) 1.106 × 10–8 Jm–3
(2) 4.425 × 10–8 Jm–3
(3) 2.212 × 10–8 Jm–3
(4) 2.212 × 10–10 Jm–3
Ans. (1)
Sol.
2
E0
1
UE
2
2
12
E
1
U 8.85 10 50
2
= 1.106 × 10–8 J/m3
41. A force is represented by F = ax2 + bt1/2
Where x = distance and t = time. The dimensions
of b2/a are :
(1) [ML3T–3] (2) [MLT–2]
(3) [ML–1T–1] (4) [ML2T–3]
Ans. (1)
Sol. F = ax2 + bt1/2
1 1 2
2
[F]
[a] M L T
[x ]
1 1 5/2
1/2
[F]
[b] M L T
[t ]
2 2 5
21 3 3
1 1 2
M L T
bM L T
aM L T
42. Two charges q and 3q are separated by a distance
‘r’ in air. At a distance x from charge q, the
resultant electric field is zero. The value of x is :
(1)
(1 3)
r
(2)
r
3(1 3)
(3)
r
(1 3)
(4)
r (1 3)
Ans. (3)
Sol.
net P
E0
22
kq k 3q
x (r x)
(r – x)2 = 3x2
r x 3x
r
x31
43. In the given arrangement of a doubly inclined
plane two blocks of masses M and m are placed.
The blocks are connected by a light string passing
over an ideal pulley as shown. The coefficient of
friction between the surface of the plane and the
blocks is 0.25. The value of m, for which M = 10
kg will move down with an acceleration of 2 m/s2,
is : (take g = 10 m/s2 and tan 37° = 3/4)
(1) 9 kg
(2) 4.5 kg
(3) 6.5 kg
(4) 2.25 kg
Ans. (2)
Sol.
a = 2m/s2
a = 2m/s2
Mgsin53°
T
fr
mgsin37
°
fr
For M block
10gsin53° – µ (10g) cos53° – T = 10 × 2
T = 80 – 15 – 20
T = 45 N
For m block
T – mg sin 37° – µmg cos 37° = m × 2
45 = 10 m
m = 4.5 kg
44. A coil is placed perpendicular to a magnetic field
of 5000 T. When the field is changed to 3000 T in
2s, an induced emf of 22 V is produced in the coil.
If the diameter of the coil is 0.02 m, then the
number of turns in the coil is :
(1) 7 (2) 70
(3) 35 (4) 140
Ans. (2)
Sol.
Nt
( B)A
Bi = 5000 T,
Bf = 3000 T
d = 0.02 m
r = 0.01 m
( B)A
= (2000)(0.01)2 = 0.2
Nt
0.2
22 N 2
N = 70
45. The fundamental frequency of a closed organ pipe
is equal to the first overtone frequency of an open
organ pipe. If length of the open pipe is 60 cm, the
length of the closed pipe will be :
(1) 60 cm (2) 45 cm
(3) 30 cm (4) 15 cm
Ans. (4)
Sol.
L1
/4
/2
/4
1
L
4
22
v = f
22
2v
f2L
v = f1(4L1)
22
v
fL
11
v
f4L
f1 = f2
12
vv
4L L
L2 = 4L1
60 = 4 × L1
L1 = 15 cm
46. A small steel ball is dropped into a long cylinder
containing glycerine. Which one of the following
is the correct representation of the velocity time
graph for the transit of the ball?
(1) (2)
(3) (4)
Ans. (2)
Sol.
F
mg
Fv
mg – FB – Fv = ma
33
L
4 4 dv
r g r g 6 rv m
3 3 dt
Let
3
L1
4R g K
3m
and
2
6rK
m
12
dv K K v
dt
vt
12
00
dv dt
K K v
v
12
0
2
1n K K v t
K
12 2
1
K K v
n K t
K
2
Kt
1 2 1
K K v K e
2
Kt
1
2
K
v 1 e
K
47. A coin is placed on a disc. The coefficient of
friction between the coin and the disc is µ. If the
distance of the coin from the center of the disc is r,
the maximum angular velocity which can be given
to the disc, so that the coin does not slip away, is :
(1)
g
r
(2)
r
g
(3)
g
r
(4)
rg
Ans. (3)
Sol.
r
N
Coin
µ
mg
f
N = mg
f = m2r
f = µN
µmg = mr2
g
r
48. Two conductors have the same resistances at 0°C
but their temperature coefficients of resistance are
1 and 2. The respective temperature coefficients
for their series and parallel combinations are :
(1)
12
12
,2
(2)
1 2 1 2
,
22
(3)
12
12
12
,
(4)
12
12
,
2
Ans. (2)
Sol. Series :
Req = R1 + R2
eq 1 2
2R(1 ) R(1 ) R(1 )
eq 1 2
2R(1 ) 2R R
12
eq 2
Parallel :
eq 1 2
1 1 1
R R R
12
eq
1 1 1
RR(1 ) R(1 )
(1 )
2
eq 1 2
2 1 1
1 1 1
21
eq 1 2
11
2
1 1 1
12
2 1 1
=
1 2 eq
21
1 2 1 2
21
=
2
eq 1 2 eq 1 2
22
Neglecting small terms
2 + 2(1 + 2)
eq 1 2
22
1 2 eq
2
12
eq 2
49. An artillery piece of mass M1 fires a shell of mass
M2 horizontally. Instantaneously after the firing,
the ratio of kinetic energy of the artillery and that
of the shell is :
(1) M1 / (M1 + M2) (2)
2
1
M
M
(3) M2 / (M1 + M2) (4)
1
2
M
M
Ans. (2)
Sol.
12
pp
KE =
2
p
2M
; p same
1
KE m
2
1 1 2
2
21
2
KE p / 2M M
KE M
p / 2M
50. When a metal surface is illuminated by light of
wavelength , the stopping potential is 8V. When
the same surface is illuminated by light of
wavelength 3, stopping potential is 2V. The
threshold wavelength for this surface is :
(1) 5
(2) 3
(3) 9
(4) 4.5
Ans. (3)
Sol. E =
max
K
0
hc
max 0
K eV
0
hc hc
8e
......(i)
0
hc hc
2e 3
.......(ii)
on solving (i) & (ii)
09
SECTION-B
51. An electron moves through a uniform magnetic
field
00
B B i 2B j T.
At a particular instant of
time, the velocity of electron is
u 3i 5j
m/s. If
the magnetic force acting on electron is
F 5ek N
,
where e is the charge of electron, then the value of
B0 is ____ T.
Ans. (5)
Sol.
F q v B
00
ˆˆ ˆ ˆ ˆ
5ek e 3i 5j B i 2B j
00
ˆ ˆ ˆ
5ek e 6B k 5B k
0
B 5T
52. A parallel plate capacitor with plate separation 5
mm is charged up by a battery. It is found that on
introducing a dielectric sheet of thickness 2 mm,
while keeping the battery connections intact, the
capacitor draws 25% more charge from the battery
than before. The dielectric constant of the sheet is
____.
Ans. (2)
Sol. Without dielectric
0
A
QV
d
with dielectric
0
AV
Qt
dtK
given
00
A V A V
1.25
td
dtK
2
1.25 3 5
K
K2
53. Equivalent resistance of the following network is
_____ .
Ans. (1)
Sol.
6 is short circuit
A
B
2
2
2
3
3
A
B
3
3
3
eq 1
R 3 1
3
54. A solid circular disc of mass 50 kg rolls along a
horizontal floor so that its center of mass has a
speed of 0.4 m/s. The absolute value of work done
on the disc to stop it is ______ J.
Ans. (6)
Sol. Using work energy theorem
22
11
W KE 0 mv I
22
W = 0 –
2
2
2
1K
mv 1
2R
=
2
11
50 0.4 1
22
= –6J
Absolute work = +6J
W = –6J
W 6J
55. A body starts falling freely from height H hits an
inclined plane in its path at height h. As a result of
this perfectly elastic impact, the direction of the
velocity of the body becomes horizontal. The value
of
H
h
for which the body will take the maximum
time to reach the ground is _____.
Ans. (2)
Sol.
H – h
u = 0
h
Total time of flight = T
2h 2(H h)
Tgg
For max. time =
dT 0
dh
2 1 1 0
g2 H h 2 h
H h h
H
h2
H2
h
56. Two waves of intensity ratio 1 : 9 cross each other
at a point. The resultant intensities at the point,
when (a) Waves are incoherent is I1(b) Waves are
coherent is I2 and differ in phase by 60°. If
1
2
l10
lx
then x = _____.
Ans. (13)
Sol. For incoherent wave
1 A B
I I I
I1 = I0 + 9I0
I1 = 10I0
For coherent wave
2 A B A B
I I I 2 I I cos60
I2 = I0 + 9I0 +
2
0
1
2 9I . 2
= 13 I0
1
2
I10
I 13
57. A small square loop of wire of side is placed
inside a large square loop of wire of side L
(L =
2
). The loops are coplanar and their centers
coinside. The value of the mutual inductance of the
system is
x
× 10–7 H, where x = _____.
Ans. (128)
Sol.
L
Flux linkage for inner loop.
= Bcenter . 2
2
0i
4 sin45 sin45
L
42
2
0i
22 L
2
00
22
M 2 2
iL
7
4
2 2 10
=
7
8 2 10 H
=
7
128 10 H
x = 128
58. The depth below the surface of sea to which a
rubber ball be taken so as to decrease its volume by
0.02% is _____ m.
(Take density of sea water = 103 kgm–3, Bulk modulus
of rubber = 9 × 108 Nm–2, and g = 10 ms–2)
Ans. (18)
Sol.
P
V
V
V
PV
V
gh V
103 × 10 × h =
80.02
9 10 100
h = 18 m
59. A particle performs simple harmonic motion with
amplitude A. Its speed is increased to three times at
an instant when its displacement is
2A .
3
The new
amplitude of motion is
nA .
3
The value of n is ___.
Ans. (7)
Sol.
22
v A x
at x =
2A
3
2
22A 5A
vA33
New amplitude = A'
2
22A
v' 3v 5A (A') 3
7A
A' 3
60. The mass defect in a particular reaction is 0.4g.
The amount of energy liberated is n × 107 kWh,
where n = _____.
(speed of light = 3 × 108 m/s)
Ans. (1)
Sol. E =
2
mc
= 0.4 × 10–3 × (3 × 108)2
= 3600 × 107 kWs
=
7
7
3600 10 kWh 1 10 kWh
3600
CHEMISTRY
TEST PAPER WITH SOLUTION
SECTION-A
61. Give below are two statements:
Statement-I : Noble gases have very high boiling
points.
Statement-II: Noble gases are monoatomic gases.
They are held together by strong dispersion forces.
Because of this they are liquefied at very low
temperature. Hence, they have very high boiling points.
In the light of the above statements. choose the
correct answer from the options given below:
(1) Statement I is false but Statement II is true.
(2) Both Statement I and Statement II are true.
(3) Statement I is true but Statement II is false.
(4) Both Statement I and Statement II are false.
Ans. (4)
Sol. Statement I and II are False
Noble gases have low boiling points
Noble gases are held together by weak dispersion
forces.
62. For the given reaction, choose the correct
expression of KC from the following :-
2
3
(aq) aq aq
Fe SCN FeSCN
(1)
2
C3
FeSCN
KFe SCN
(2)
3
C2
Fe SCN
KFeSCN
(3)
2
C22
3
FeSCN
KFe SCN
(4)
2
2
C3
FeSCN
KFe SCN
Ans. (1)
Sol.
CProducts ion conc.
KReactants ion conc.
2
C3
FeSCN
KFe SCN
63. Identify the mixture that shows positive deviations
from Raoult's Law
(1) (CH3)2CO + C6H5NH2
(2) CHCl3 + C6H6
(3) CHCl3 + (CH3)2CO
(4) (CH3)2CO + CS2
Ans. (4)
Sol. (CH3)2CO + CS2 Exibits positive deviations from
Raoult’s Law
64. The compound that is white in color is
(1) ammonium sulphide
(2) lead sulphate
(3) lead iodide
(4) ammonium arsinomolybdate
Ans. (2)
Sol. Lead sulphate-white
Ammonium sulphide-soluble
Lead iodide-Bright yellow
Ammonium arsinomolybdate-yellow
65. The metals that are employed in the battery
industries are
A. Fe
B. Mn
C. Ni
D. Cr
E. Cd
Choose the correct answer from the options given
below:
(1) B, C and E only
(2) A, B, C, D and E
(3) A, B, C and D only
(4) B, D and E only
Ans. (1)
Sol. Mn, Ni and Cd metals used in battery industries.
66. A species having carbon with sextet of electrons
and can act as electrophile is called
(1) carbon free radical
(2) carbanion
(3) carbocation
(4) pentavalent carbon
Ans. (3)
Sol.
Six electron species
67. Identify the factor from the following that does not
affect electrolytic conductance of a solution.
(1) The nature of the electrolyte added.
(2) The nature of the electrode used.
(3) Concentration of the electrolyte.
(4) The nature of solvent used.
Ans. (2)
Sol. Conductivity of electrolytic cell is affected by
concentration of electrolyte, nature of electrolyte
and nature of solvent.
68. The product (C) in the below mentioned reaction
is:
alc
aq
KOH HBr
3 2 2 KOH
CH CH CH Br A B C
(1) Propan-1-ol
(2) Propene
(3) Propyne
(4) Propan-2-ol
Ans. (4)
Sol.
CH –CH –Br
32
KOH( alc)
CH –CH=CH
32
HBr
CH –CH–CH
33
Br
KOH( aq)
CH –CH–CH
33
OH
69. Given below are two statements: One is labelled as
Assertion A and the other is labelled as Reason R:
Assertion A: Alcohols react both as nucleophiles
and electrophiles.
Reason R: Alcohols react with active metals such
as sodium, potassium and aluminum to yield
corresponding alkoxides and liberate hydrogen.
In the light of the above statements, choose the
correct answer from the options given below:
(1) A is false but R is true.
(2) A is true but R is false.
(3) Both A and R are true and R is the correct
explanation of A.
(4) Both A and R are true but R is NOT the
correct explanation of A
Ans. (4)
Sol. As per NCERT, Assertion (A) and Reason (R) is
correct but Reason (R) is not the correct
explanation.
70. The correct sequence of electron gain enthalpy of
the elements listed below is
A. Ar
B. Br
C. F
D. S
Choose the most appropriate from the options
given below:
(1) C > B > D > A
(2) A > D > B > C
(3) A > D > C > B
(4) D > C > B > A
Ans. (2)
Sol. Element egH(kJ/mol)
F –333
S –200
Br –325
Ar +96
C
H
HH
71. Identify correct statements from below:
A. The chromate ion is square planar.
B. Dichromates are generally prepared from
chromates.
C. The green manganate ion is diamagnetic.
D. Dark green coloured K2MnO4 disproportionates in
a neutral or acidic medium to give permanganate.
E. With increasing oxidation number of transition
metal, ionic character of the oxides decreases.
Choose the correct answer from the options given
below:
(1) B, C, D only
(2) A, D, E only
(3) A, B, C only
(4) B, D, E only
Ans. (4)
Sol. A. CrO4
2– is tetrahedral
B. 2Na2CrO4 + 2H+ Na2Cr2O7 + 2Na+ + H2O
C. As per NCERT, green manganate is
paramagnetic with 1 unpaired electron.
D. Statement is correct
E. Statement is correct
72. ‘Adsorption’ principle is used for which of the
following purification method?
(1) Extraction
(2) Chromatography
(3) Distillation
(4) Sublimation
Ans. (2)
Sol. Principle used in chromotography is adsorption.
73. Integrated rate law equation for a first order gas
phase reaction is given by (where Pi is initial
pressure and Pt is total pressure at time t)
(1)
i
it
P
2.303
k log
t 2P P
(2)
i
it
2P
2.303
k log
t 2P P
(3)
it
i
2P P
2.303
k log
tP
(4)
i
it
P
2.303
kt 2P P
Ans. (1)
Sol. A B + C
Pi 0 0
Pi – x x x
Pt = Pi + x
Pi - x = Pi – Pt + Pi
= 2Pi – Pt
i
it
P
2.303
K log
t 2P P
74. Given below are two statements: One is labelled as
Assertion A and the other is labelled as Reason R:
Assertion A: pKa value of phenol is 10.0 while
that of ethanol is 15.9.
Reason R: Ethanol is stronger acid than phenol.
In the light of the above statements, choose the
correct answer from the options given below:
(1) A is true but R is false.
(2) A is false but R is true.
(3) Both A and R are true and R is the correct
explanation of A.
(4) Both A and R are true but R is NOT the
correct explanation of A.
Ans. (1)
Sol. Phenol is more acidic than ethanol because
conjugate base of phenoxide is more stable than
ethoxide.
75. Given below are two statements:
Statement I: IUPAC name of HO–CH2–(CH2)3–
CH2– COCH3 is 7-hydroxyheptan-2-one.
Statement II: 2-oxoheptan-7-ol is the correct
IUPAC name for above compound.
In the light of the above statements. choose the
most appropriate answer from the options given
below:
(1) Statement I is correct but Statement II is incorrect.
(2) Both Statement I and Statement II are incorrect.
(3) Both Statement I and Statement II are correct.
(4) Statement I is incorrect but Statement II is correct.
Ans. (1)
Sol. 7-Hydroxyheptan-2-one is correct IUPAC name
76. The
correct statements from following are:
A. The strength of anionic ligands can be
explained by crystal field theory.
B. Valence bond theory does not give a
quantitative interpretation of kinetic stability of
coordination compounds.
C. The hybridization involved in formation of
[Ni(CN)4]2–complex is dsp2.
D. The number of possible isomer(s) of
cis-[PtC12 (en)2]2+ is one
Choose the correct answer from the options given
below:
(1) A, D only
(2) A, C only
(3) B, D only
(4) B, C only
Ans. (4)
Sol. B. VBT does not explain stability of complex
C. Hybridisation of [Ni(CN)4]–2 is dsp2.
77. The linear combination of atomic orbitals to form
molecular orbitals takes place only when the
combining atomic orbitals
A. have the same energy
B. have the minimum overlap
C. have same symmetry about the molecular axis
D. have different symmetry about the molecular
axis
Choose the most appropriate from the options
given below:
(1) A, B, C only
(2) A and C only
(3) B, C, D only
(4) B and D only
Ans. (2)
Sol. * Molecular orbital should have maximum overlap
* Symmetry about the molecular axis should be
similar
78. Match List I with List II
LIST-I LIST-II
A.
Glucose/NaHCO3/D I. Gluconic acid
B. Glucose/HNO3 II. No reaction
C. Glucose/HI/ D III. n-hexane
D.
Glucose/Bromine
water]
IV. Saccharic acid
Choose the correct answer from the options given
below:
(1) A-IV, B-I, C-III, D-II
(2) A-II, B-IV, C-III, D-I
(3) A-III, B-II, C-I, D-IV
(4) A-I, B-IV, C-III, D-II
Ans. (2)
Sol. Glucose 3
NaHCO
D
¾ ¾ ¾ ¾®
no reaction
Glucose 3
HNO
D
¾ ¾ ¾®
saccharic acid
Glucose HI
D
¾ ¾®
n-hexane
Glucose 2
Br
D
¾ ¾®
Gluconic acid
79. Consider the oxides of group 14 elements
SiO2, GeO2, SnO2, PbO2, CO and GeO. The
amphoteric oxides are
(1) GeO, GeO2
(2) SiO2, GeO2
(3) SnO2, PbO2
(4) SnO2, CO
Ans. (3)
Sol. SnO2 and PbO2 are amphoteric
80. Match List I with List II
LIST I (Technique) LIST II (Application)
A. Distillation I. Separation of
glycerol from
spent-lye
B. Fractional
distillation
II. Aniline - Water
mixture
C. Steam
distillation
III. Separation of crude
oil fractions
D. Distillation
under reduced
pressure
IV. Chloroform-
Aniline
Choose the correct answer from the options given below:
(1) A-IV, B-I, C-II, D-III
(2) A-IV, B-III, C-II. D-I
(3) A-I. B-II, C-IV, D-III
(4) A-II, B-III. C-I, D-IV
Ans. (2)
Sol. Fact (NCERT)
SECTION-B
81. Molar mass of the salt from NaBr, NaNO3, KI and
CaF2 which does not evolve coloured vapours on
heating with concentrated H2SO4 is _____ g mol–1,
(Molar mass in g mol–1 : Na : 23, N : 14, K : 39,
O : 16, Br : 80, I : 127, F : 19, Ca : 40
Ans. (78)
Sol. CaF2 does not evolve any gas with concentrated
H2SO4.
NaBr evolve Br2
NaNO3 evolve NO2
KI evolve I2
82. The 'Spin only’ Magnetic moment for [Ni(NH3)6]2+
is______× 10–1 BM.
(given = Atomic number of Ni : 28)
Ans. (28)
Sol. NH3 act as WFL with Ni2+
Ni2+ = 3d8
No. of unpaired electron = 2
n n 2 8 2.82
BM
= 28.2 × 10–1 BM
x = 28
83. Number of moles of methane required to produce
22g CO2(g) after combustion is x × 10–2 moles. The
value of x is
Ans. (50)
Sol.
4(g) 2(g) 2(g) 2 ( )
CH 2O CO 2H O
2
CO 22
n 0.5
44
moles
So moles of CH4 required = 0.5 moles
i.e. 50 × 10–2 mole
x = 50
84. The product of the following reaction is P.
CHO
OH
(i) PhMgBr(1 equiv.)
(ii) aq. NH Cl
4
The number of hydroxyl groups present in the
product P is________.
Ans. (0)
Sol. Product benzene has zero hydroxyl group
CHO
OH Product
PhMgBr
(1 eq.) +
CHO
OMgBr
aq. NH Cl
4
CHO
OH
Reactant
85. The number of species from the following in which
the central atom uses sp3 hybrid orbitals in its
bonding is__________.
NH3, SO2, SiO2, BeCl2, CO2, H2O, CH4, BF3
Ans. (4)
Sol. NH3 sp3
SO2 sp2
SiO2 sp3
BeCl2 sp
CO2 sp
H2O sp3
CH4 sp3
BF3 sp2
86.
CH CH Br + NaOH
32
C H OH
25
HO
2
Product A
Product B
The total number of hydrogen atoms in product A
and product B is__________.
Ans. (10)
Sol.
CH CH Br + NaOH
32
C H OH
25
HO
2
CH =CH
22
CH –CH –OH
32
Total number of hydrogen atom in A and B is 10
87. Number of alkanes obtained on electrolysis of a
mixture of CH3COONa and C2H5COONa is_____.
Ans. (3)
Sol.
33
CH COONa CH
2 5 2 5
C H COONa C H
2 5 3 2 2 3
2C H CH CH CH CH
3 3 3
2CH CH CH
2
3 5 3 2 3
CH C H CH CH CH
88. Consider the following reaction at 298 K.
29
2(g) 3(g) P
3O O .K 2.47 10 .
2
rG
for the reaction is _________ kJ. (Given R
= 8.314 JK–1 mol–1)
Ans. (163)
Sol.
29
2(g) 3(g) P
3O O .K 2.47 10 .
2
rG
= – RT ln KP
= – 8.314 × 10–3 × 298 × ln (2.47 × 10–29)
= –8.314 × 10–3 × 298 × (–65.87)
= 163.19 kJ
89. The ionization energy of sodium in kJ mol–1
. If
electromagnetic radiation of wavelength 242 nm is
just sufficient to ionize sodium atom is______.
Ans. (494)
Sol.
1240
E eV
nm
1240 eV
242
= 5.12 eV
= 5.12 × 1.6 × 10–19
= 8.198 × 10–19 J/atom
= 494 kJ/mol
90. One Faraday of electricity liberates x × 10–1 gram
atom of copper from copper sulphate, x is______.
Ans. (5)
Sol. Cu2+ + 2e– Cu
2 Faraday mol Cu
1 Faraday mol Cu deposit
0.5 mol = 0.5 g atom = 5 × 10–1
x = 5
