FINAL JEE–MAIN EXAMINATION – JANUARY, 2024
(Held On Thursday 01
st
February, 2024) TIME : 3 : 00 PM to 06 : 00 PM
MATHEMATICS
TEST PAPER WITH SOLUTION
SECTION-A
1. Let f(x) = |2x2+5|x|–3|,xR. If m and n denote the
number of points where f is not continuous and not
differentiable respectively, then m + n is equal to :
(1) 5 (2) 2
(3) 0 (4) 3
Ans. (4)
Sol. f(x) = |2x2+5|x|–3|
Graph of y = |2x2 + 5x – 3|
Graph of f(x)
0
Number of points of discontinuity = 0 = m
Number of points of non-differentiability = 3 = n
2. Let and be the roots of the equation px2 + qx –
r = 0, where p 0. If p, q and r be the consecutive
terms of a non-constant G.P and , then
the value of (– is :
(1) (2) 9
(3) (4) 8
Ans. (1)
Sol. px2 + qx – r = 0
p = A, q = AR, r = AR2
Ax2 + ARx – AR2 = 0
x2 + Rx – R2 = 0
()2 = ()2 – 4 = R2 – 4(–R2) = 5
= 80/9
3. The number of solutions of the equation 4 sin2x – 4
cos3x + 9 – 4 cosx = 0; x [ –2, 2] is :
(1) 1
(2) 3
(3) 2
(4) 0
Ans. (4)
Sol. 4sin2x – 4cos3x + 9 – 4 cosx = 0 ; x [ – 2, 2]
4 – 4cos2x – 4cos3x + 9 – 4 cosx = 0
4cos3x + 4cos2x + 4 cosx – 13 = 0
4cos3x + 4cos2x + 4cosx = 13
L.H.S. 12 can’t be equal to 13.
4. The value of dx is equal to:
(1) 0
(2) 1
(3) 2
(4) –1
Ans. (1)
Sol.
Using where f(2a–x) = –f(x)
Here f(1–x) = f(x)
I = 0
–3 1/2
–5/4
1 1 3
4
80
9
20
3
1 1 3
4
2
3 R 4
R
4 R 4 3
16
9
1
132 3
0(2x 3x x 1)
1
13
32
0
I 2x 3x x 1 dx
2a
0
f x dx
5. Let P be a point on the ellipse . Let the
line passing through P and parallel to y-axis meet
the circle x2 + y2 = 9 at point Q such that P and Q
are on the same side of the x-axis. Then, the
eccentricity of the locus of the point R on PQ such
that PR : RQ = 4 : 3 as P moves on the ellipse, is :
(1) (2)
(3) (4)
Ans. (4)
Sol.
h = 3cos;
locus =
6. Let m and n be the coefficients of seventh and
thirteenth terms respectively in the expansion of
. Then is :
(1) (2)
(3) (4)
Ans. (4)
Sol.
:
7. Let be a non-zero real number. Suppose f : R
R is a differentiable function such that f (0) = 2 and
x
lim f x 1
. If f '(x) = f(x) +3, for all x R,
then f (–loge2) is equal to____.
(1) 3 (2) 5
(3) 9 (4) 7
Ans. (3 OR BONUS)
Sol. f(0) = 2,
x
lim f x 1
f’(x) – x.f(x) = 3
I.F = e–x
y(e–x) =
f(x). (e–x) =
x = 0 (1)
f(x) =
x – 1 =
= –3 c = 1
= 1 + e3ln2 = 9
(But should be greater than 0 for finite
value of c)
22
xy
1
94
11
19
13
21
139
23
13
7
Q
P(3cos , 2sin )
Q(3cos , 3sin )
P
2 2
x y 1
9 4
4
P
(3C, 2S) R
(h, k) Q
(3C, 3S)
3
18
k sin
7
22
x 49y 1
9 324
324 117 13
e1
49 9 21 7
18
1
3
2
3
11
x
32x
1
3
n
m
4
9
1
9
1
4
9
4
18
12
33
xx
3
12 6
12
33
18 18
7 6 6 12 6
x x 1 1
t c c .
32 2
3
6 12
12
33
18 18 –6
13 12 12 6 12
x x 1 1
t c c . .x
32 2
3
18 –12 –6
6
m c .3 .2
18 –12 –6
12
n c .2 .3
1
12
–12 –6 3
3
12 6
n 2 .3 3 9
m 2 4
3 .2
x
3.e dx
x
3e c
3
2c
3c2
x
3c.e
3c(0)
x
3
f ( ln 2) c.e
8. Let P and Q be the points on the line
which are at a distance of 6
units from the point R (1,2,3). If the centroid of the
triangle PQR is (,,), then 2+2 +2 is:
(1) 26
(2) 36
(3) 18
(4) 24
Ans. (3)
Sol.
P(8 – 3, 2 + 4, 2 – 1)
PR = 6
(8 – 4)2 + ( 2 + 2)2 + (2 – 4)2 = 36
= 0, 1
Hence P(–3, 4, –1) & Q(5, 6, 1)
Centroid of PQR = (1, 4, 1) (,, )
+ + = 18
9. Consider a ABC where A(1,2,3,), B(–2,8,0) and
C(3,6,7). If the angle bisector of BAC meets the
line BC at D, then the length of the projection of
the vector on the vector is:
(1)
(2)
(3)
(4)
Ans. (1)
Sol.
A(1, 3, 2); B(–2, 8, 0); C(3, 6, 7);
AB =
AC =
Length of projection of on
=
10. Let Sn denote the sum of the first n terms of an
arithmetic progression. If S10 = 390 and the ratio of
the tenth and the fifth terms is 15 : 7, then S15 –S5
is equal to:
(1) 800
(2) 890
(3) 790
(4) 690
Ans. (3)
Sol. S10 = 390
2a + 9d = 78 (1)
(2)
From (1) & (2) a = 3 & d = 8
=790
x 3 y 4 z 1
8 2 2
P Q
R(1,2,3)
AD
AC
37
2 38
38
2
39
2 38
19
A (1,3,2)
C (3, 6, 7)B
(–2, 8, 0)
D
1 : 1
1 7
,7,
2 2
ˆ ˆ ˆ
AC 2i 3j 5k
9 25 4 38
4 9 25 38
1 3 1
ˆ ˆ ˆ ˆ ˆ ˆ
AD i 4j k (i 8j 3k)
2 2 2
AD
AC
AD.AC 37
| AC| 2 38
10 2a 10 1 d 390
2
10
5
t15 a 9d 15 8a 3d
t 7 a 4d 7
15 5
15 5
S – S 6 14 8 6 4 8
22
15 118 5 38
2
11. If , where a and b are
rational numbers, then 9a + 8b is equal to :
(1) 2 (2) 1
(3) 3 (4)
Ans. (1)
Sol.
=
9a + 8b =
12. If z is a complex number such that |z|1, then the
minimum value of is:
(1) (2) 2
(3) 3 (4)
Ans. (Bonus)
Sol. |z| 1
Min. value of is actually zero.
13. If the domain of the function f(x) =
+log10 (x2 + 2x – 15) is (– , ) U [,), then
2 + 3 is equal to :
(1) 140 (2) 175
(3) 150 (4) 125
Ans. (3)
Sol. ƒ(x) = + log10(x2 + 2x - 15)
Domain : x2 – 25 0 x (–, -5] [5, )
4 – x2 0 x {–2, 2}
x2 + 2x – 15 > 0 (x + 5) (x – 3) > 0
x (–, –5) (3, )
x (–, –5) [5,)
= –5; = 5
3= 150
14. Consider the relations R1 and R2 defined as aR1b
a2 + b2 = 1 for all a , b, R and (a, b) R2(c, d)
a + d = b + c for all (a,b), (c,d) N × N. Then
(1) Only R1 is an equivalence relation
(2) Only R2 is an equivalence relation
(3) R1 and R2 both are equivalence relations
(4) Neither R1 nor R2 is an equivalence relation
Ans. (2)
Sol. aR1 b a2 + b2 = 1; a, b R
(a, b) R2 (c, d) a + d = b + c; (a, b), (c, d) N
for R1 : Not reflexive symmetric not transitive
for R2 : R2 is reflexive, symmetric and transitive
Hence only R2 is equivalence relation.
34
0
cos x dx a b 3
3
2
/3 4
0
cos xdx
2
/3
0
1 cos2x dx
2
/3 2
0
1(1 2cos2x cos 2x)dx
4
/3 /3 /3
0 0 0
1 1 cos4x
dx 2 cos2xdx dx
42
/3 /3
00
1 1 1
(sin2x) (sin4x)
4 3 2 3 8
/3 /3
00
1 1 1
(sin2x) (sin4x)
4 3 2 3 8
1 3 1 3
4 2 2 8 2
73
2 64
17
a ; b
8 64
972
88
1
z (3 4
2
i)
5
2
3
2
O
P
3, 2
2
3
z 2i
2
2
2
x 25
(4 x )
2
2
x 25
4x
15. If the mirror image of the point P(3,4,9) in the line
is (,,), then 14 (+ + )
is :
(1) 102 (2) 138
(3) 108 (4) 132
Ans. (3)
Sol.
?
3(3 – 2) + 2 (2 – 5) + ( – 7) = 0
14 = 23
Ans. 14 (r) = 108
16. Let f(x) = x N. If for some
a N, f(f(f(a))) = 21, then
where [t] denotes the greatest integer less than or
equal to t, is equal to :
(1) 121
(2) 144
(3) 169
(4) 225
Ans. (2)
Sol. ƒ(x) =
ƒ(ƒ(ƒ(a))) = 21
C–1: If a = even
ƒ(a) = a – 1 = odd
f(f(a)) = 2(a – 1) = even
ƒ(ƒ(ƒ(a))) = 2a – 3 = 21 a = 12
C–2: If a = odd
ƒ(a) = 2a = even
ƒ(ƒ(a)) = 2a – 1 = odd
ƒ(ƒ(ƒ(a))) = 4a – 2 = 21 (Not possible)
Hence a = 12
Now
= 144 – 0 = 144.
17. Let the system of equations x + 2y +3z = 5, 2x +
3y + z = 9, 4x + 3y + z = have infinite number
of solutions. Then + 2is equal to :
(1) 28 (2) 17
(3) 22 (4) 15
Ans. (2)
Sol. x + 2y + 3z = 5
2x + 3y + z = 9
4x + 3y + z = µ
for infinite following = 1 = 2 = 3 = 0
= 0 –13
= 0 = 15
= 0
x 1 y 1 z 2
3 2 1
P(3, 4, 9)
A( , )
N
(3 + 1, 2 –1, + 2)
PN.b 0
23
14
83 32 51
N , ,
14 14 14
3 83 62
2 14 7
4 32 4
2 14 7
9 51 12
2 14 7
x 1, x is even,
2x, x is odd,
xa
lim
3
| x | x ,
aa
x 1; x even
2x; x odd
3
x 12
| x | x
lim 2 12
3
x 12 x 12
| x | x
lim lim
12 12
1 2 3
2 3 1
43
5 2 3
9 3 1
µ 3 13
1 5 3
2 9 1
4 15 13
= 0
for = –13, µ=15 system of equation has infinite
solution hence + 2µ = 17
18. Consider 10 observation x1, x2,…., x10. such that
(xi – ) = 2 and (xi – )2
= 40, where ,
are positive integers. Let the mean and the variance
of the observations be and respectively. The
is equal to :
(1) 2 (2)
(3) (4) 1
Ans. (1)
Sol. x1, x2…….x10
Mean
xi
Now Let yi = xi –
6 – 5 = ± 4 (not possible) or = 2
Hence
19. Let Ajay will not appear in JEE exam with
probability p = , while both Ajay and Vijay will
appear in the exam with probability q = . Then
the probability, that Ajay will appear in the exam
and Vijay will not appear is :
(1)
(2)
(3)
(4)
Ans. (2)
Sol.
Ans.
20. Let the locus of the mid points of the chords of
circle x2+(y–1)2 =1 drawn from the origin intersect
the line x+y = 1 at P and Q. Then, the length of PQ
is :
(1)
(2)
(3)
(4) 1
Ans. (1)
1 2 5
2 3 9
4 3 15
10
1
i
10
1
i
6
5
84
25
3
2
5
2
10
i
i1
(x ) 2
10
i
i1
x 10 2
i
6x
µ5 10
10 2
i
i1
(x ) 40
2
22
yi
1y (y)
10
2
10
i
2
2i1
xi
(x )
1(x )
10 10
2
84 12 10
4
25 10
2
6 5 84 16
4
5 25 25
2
5
2
2
7
1
5
9
35
18
35
24
35
3
35
AV
18
35
1
5
2
P(A) p
7
1
P(A V) q
5
5
P(A) 7
18
P(A V) 35
1
2
2
1
2
Sol.
mOM . mCM = –1
locus is x2 + y(y – 1) = 0
x2 + y2 – y = 0
PQ =
= =
SECTION-B
21. If three successive terms of a G.P. with common
ratio r(r > 1) are the lengths of the sides of a
triangle and [r] denotes the greatest integer less
than or equal to r, then 3[r] + [–r] is equal to :
Ans. (1)
Sol. a, ar, ar2 G.P.
Sum of any two sides > third side
a + ar > ar2, a + ar2 > ar, ar + ar2 > a
r2 – r – 1 < 0
(1)
r2 – r + 1 > 0
always true
r2 + r – 1 > 0
(2)
Taking intersection of (1) , (2)
As r > 1
[r] = 1 [ –r] = – 2
3 [ r] + [ – r] = 1
22. Let A = I2 – MMT, where M is real matrix of order
2 × 1 such that the relation MT M = I1 holds. If is
a real number such that the relation AX = X holds
for some non-zero real matrix X of order 2 × 1,
then the sum of squares of all possible values of
is equal to :
Ans. (2)
Sol. A = I2 –2 MMT
A2 = (I2 – 2MMT) (I2–2MMT)
= I2 – 2MMT – 2MMT + 4MMTMMT
= I2 – 4MMT + 4MMT
=I2
AX = X
A2X = AX
X = (X)
X = X
X (–1) = 0
= 1
+ 1
Sum of square of all possible values = 2
C(0, 1)
(0,0)
Om(h,k)
k k 1
.1
hh
P Q x+y–1=0
1
0, 2
1/ 2
p2
1
p22
22
2 r p
11
248
1
2
1 5 1 5
r,
22
,15
r2
15
,
2
1 5 1 5
r,
22
15
r 1, 2
23. Let f : (0, ) R and . If F(x2) =
x4 + x5, then is equal to :
Ans. (219)
Sol. F(x) =
F1(x) = xf(x)
Given F(x2) = x4 + x5, let x2 = t
F(t) = t2 + t5/2
F’(t) = 2t + 5/2 t3/2
t·f(t) = 2t + 5/2 t3/2
f(t) = 2 + 5/2 r1/2
= 24 + 5/2
=219
24. If ,
then is equal to :
Ans. (105)
Sol. + (3cos2x–5)cos3x
+ cos5x – cos3x
+ cos5x – cos3x
y’ = 1 – cos4x· (sinx) + cos2x (sinx)
= 96 = 105
25. Let , and
be three vectors such that
. If the angle between the vector
and the vector is , then the greatest
integer less than or equal to tan2 is :
Ans. (38)
Sol.
+ 4 = –1 = – 5
+ c2 = –8 c2 = – 3
+ c3 = 2 c3 = 7
[tan2] = 38
x
0
F(x) tf(t)dt
12
2
r1
f(r )
x
0
t f (t)dt
12 12
2
r 1 r 1
5
f (r ) 2 r
2
12(13)
2
2
23
x 1 x x 1
y (3cos x 5)cos x
15
x x x x
96y' 6
2
x 1 x x
yx x x x
1
15
3
2
x 1 x x 1
y
x x x 1
1
5
1
3
y x 1
x1
1
5
1
3
9 1 3 1
y' 1
6 16 2 4 2
32 9 12
32
35
32
y' 6
ˆˆˆ
a i j k
ˆˆ ˆ
b i 8j 2k
23
ˆˆˆ
c 4i c j c k
b a c a
c
ˆˆ
ˆ
3i 4 j k
a i j k
b i 8j 2k
23
c 4i c j c k
b a c a
b c a 0
bc
bc
23
i 8j 2k 4i c j c k i j k
c 4i 3j 7k
12 12 7 7 7
cos 26 74 26 74 2 481
2625 3
tan 49
26. The lines L1, L2,…., I20 are distinct. For
n = 1, 2, 3,…., 10 all the lines L2n–1 are parallel to
each other and all the lines L2n pass through a
given point P. The maximum number of points of
intersection of pairs of lines from the set
{L1, L2,…., L20} is equal to :
Ans. (101)
Sol. L1, L3, L5, - - L19 are Parallel
L2, L4, L6, - - L20 are Concurrent
Total points of intersection = 20C2 – 10C2 – 10C2 + 1
= 101
27. Three points O(0,0), P(a, a2), Q(–b, b2), a > 0, b > 0,
are on the parabola y = x2. Let S1 be the area of the
region bounded by the line PQ and the parabola,
and S2 be the area of the triangle OPQ. If the
minimum value of is , gcd(m, n) = 1, then
m + n is equal to :
Ans. (7)
Sol.
PQ:-
y – a2 = (a – b) x – (a – b)a
y = (a – b) x + ab
m + n = 7
28. The sum of squares of all possible values of k, for
which area of the region bounded by the parabolas
2y2 = kx and ky2 = 2(y – x) is maximum, is equal
to :
Ans. (8)
1
2
S
S
m
n
O
P
(a, a )
2
(–b, b )
2Q
2 2 2
2
2
0 0 l
S 1 / 2 | a a l | 1 / 2 ab a b
b b l
22
2ab
y a x a
ab
a2
1
b
S a b x ab x dx
a
23
b
xx
a b ab x
23
233
a b a b (a b )
ab(a b)
23
222
1
2
ab (a b ab)
ab
S23
ab
S
2
2 2 2
3(a b) 6ab 2(a b ab)
3ab
min 2
1 a b 2
3 b a
4m
3n
Sol. ky2 = 2(y – x) 2y2 = kx
Point of intersection
y = 0
Area is maximum when
k = 2, –2
29. If , x(1) = 1, then 5x(2) is equal to :
Ans. (5)
Sol.
Integrating factor =
x = –1 – y2 + cy
x(1) = 1
1 = – 1 – 1 + c c = 3
x = – 1 – y2 + 3y
5x (2) = 5(– 1 – 4 + 6)
= 5
30. Let ABC be an isosceles triangle in which A is at
(–1, 0), , AB = AC and B is on the
positive x-axis. If and the line BC
intersects the line y = x + 3 at (, ), then is :
Ans. (36)
Sol.
[By sine rule]
2c = 8 c = 4
2
2y2y
ky k
12y
ky 2 k
4y
ky 2
k
2
2 2k
y4k4
kk
2
2k
22
k4
0
ky 2y
A y .dy
2k
2
2k
23
k4
0
y k 2 y
2 2 k 3
22
22
2k 1 k 4 1 2k
2 2k 3
k 4 k 4
2
11
44
6kk
A M G M
4
kk2
2
4
k4
k
4
kk
2
dx 1 x y
dy y
2
dx x 1 y
dy y y
1dy
y1
y
e
2
2
1 1 y
x dy
yy
x1
yc
yy
2
A3
BC 4 3
4
2
120° 30° B(b,0)
A(–1,0)
30°
C
c 4 3
sin 30 sin120
b = 3, mAB = 0
BC:-
Point of intersection : y = x + 3,
AB b 1 4
BC
1
m3
1
y (x 3)
3
3y x 3
3y x 3
3 1 y 6
6
y31
6
x3
31
6 3 3 3
31
2
13 6
3
13 13
4
236
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
31. In an ammeter, 5% of the main current passes
through the galvanometer. If resistance of the
galvanometer is G, the resistance of ammeter
will be :
(1)
(2)
(3) 199 G
(4) 200 G
Ans. (Bonus)
Sol.
IS S = Ig G
32. To measure the temperature coefficient of resistivity
of a semiconductor, an electrical arrangement
shown in the figure is prepared. The arm BC is
made up of the semiconductor. The experiment is
being conducted at 25°C and resistance of the
semiconductor arm is 3 m. Arm BC is cooled at a
constant rate of 2°C/s. If the galvanometer G shows
no deflection after 10s, then is :
(1) – 2 × 10–2 °C–1
(2) – 1.5 × 10–2 °C–1
(3) – 1 × 10–2 °C–1
(4) – 2.5 × 10–2 °C–1
Ans. (3)
Sol. For no deflection
R = 2.4m
Temperature fall in 10s = 20°C
R = R t
= – 10–2C–1
R t 3 20
R 0.6
13
0.8 R
V = 5mV
B
C
A
3 m1 m
D
G
0.8 m
20
G
R
A
SG
19
R20G
19
G
SG
A
2
S19
G
IS G
100 100
95 5I
I
I
Ig
S
IS
G
199
G
200
G
33. From the statements given below :
(A) The angular momentum of an electron in nth
orbit is an integral multiple of h.
(B) Nuclear forces do not obey inverse square law.
(C) Nuclear forces are spin dependent.
(D) Nuclear forces are central and charge
independent.
(E) Stability of nucleus is inversely proportional to
the value of packing fraction.
Choose the correct answer from the options given
below :
(1) (A), (B), (C), (D) only
(2) (A), (C), (D), (E) only
(3) (A), (B), (C), (E) only
(4) (B), (C), (D), (E) only
Ans. (3)
Sol. Part of theory
34. A diatomic gas ( = 1.4) does 200 J of work when
it is expanded isobarically. The heat given to the
gas in the process is :
(1) 850 J (2) 800 J
(3) 600 J (4) 700 J
Ans. (4)
Sol. = 1 +
2
f
= 1.4
20.4
f
f = 5
W = n R T = 200J
f2
Q nR T
2
7200
2
= 700 J
35. A disc of radius R and mass M is rolling
horizontally without slipping with speed . It then
moves up an inclined smooth surface as shown in
figure. The maximum height that the disc can go
up the incline is :
h
(1)
2
g
(2)
2
3
4g
(3)
2
1
2g
(4)
2
2
3g
Ans. (3)
Sol. Only the translational kinetic energy of disc changes
into gravitational potential energy. And rotational
KE remains unchanged as there is no friction.
2
1mv mgh
2
2
v
h2g
36. Conductivity of a photodiode starts changing only
if the wavelength of incident light is less than
660 nm. The band gap of photodiode is found to be
XeV
8
. The value of X is :
(Given, h = 6.6 × 10–34 Js, e = 1.6 × 10–19C)
(1) 15 (2) 11
(3) 13 (4) 21
Ans. (1)
Sol.
34 8
g9
hc 6.6 10 3 10
E660 10
J
34 8
9 19
6.6 10 3 10 eV
660 10 1.6 10
15 eV
8
So x = 15
37. A big drop is formed by coalescing 1000 small
droplets of water. The surface energy will become :
(1) 100 times (2) 10 times
(3)
1th
100
(4)
1th
10
Ans. (4)
Sol. Lets say radius of small droplets is r and that of big
drop is R
33
44
R 1000 r
33
R = 10r
Ui = 1000 (4r2S)
Uf = 4R2S
= 100 (4r2S)
Uf =
i
1U
10
38. If frequency of electromagnetic wave is 60 MHz
and it travels in air along z direction then the
corresponding electric and magnetic field vectors
will be mutually perpendicular to each other and
the wavelength of the wave (in m) is :
(1) 2.5 (2) 10
(3) 5 (4) 2
Ans. (3)
Sol.
8
6
c 3 10 5m
f60 10
39. A cricket player catches a ball of mass 120 g
moving with 25 m/s speed. If the catching process
is completed in 0.1 s then the magnitude of force
exerted by the ball on the hand of player will be
(in SI unit):
(1) 24 (2) 12
(3) 25 (4) 30
Ans. (4)
Sol.
av
p
Ft
0.12 25 30N
0.1
40. Monochromatic light of frequency 6 × 1014 Hz
is produced by a laser. The power emitted is
2 × 10–3 W. How many photons per second on an
average, are emitted by the source ?
(Given h = 6.63 × 10–34 Js)
(1) 9 × 1018 (2) 6 × 1015
(3) 5 × 1015 (4) 7 × 1016
Ans. (3)
Sol. P = nh
3
34 14
P 2 10
nh6.63 10 6 10
= 5 × 1015
41. A microwave of wavelength 2.0 cm falls normally
on a slit of width 4.0 cm. The angular spread of the
central maxima of the diffraction pattern obtained
on a screen 1.5 m away from the slit, will be:
(1) 30° (2) 15°
(3) 60° (4) 45°
Ans. (3)
Sol. For first minima a sin =
1
sin a2
= 30°
Angular spread = 60°
42. C1 and C2 are two hollow concentric cubes
enclosing charges 2Q and 3Q respectively as
shown in figure. The ratio of electric flux passing
through C1 and C2 is :
2Q
3Q
(1) 2 : 5 (2) 5 : 2
(3) 2 : 3 (4) 3 : 2
Ans. (1)
Sol.
smaller cube
0
2Q
bigger cube
0
5Q
smaller cube
bigger cube
2
5
43. If the root mean square velocity of hydrogen
molecule at a given temperature and pressure is
2 km/s, the root mean square velocity of oxygen at
the same condition in km/s is :
(1) 2.0 (2) 0.5
(3) 1.5 (4) 1.0
Ans. (2)
Sol.
rms
3RT
VM
12
2 1 2
VM 2 32
V M V 2
V2 = 0.5 km/s
44. Train A is moving along two parallel rail tracks
towards north with speed 72 km/h and train B is
moving towards south with speed 108 km/h.
Velocity of train B with respect to A and velocity
of ground with respect to B are (in ms–1) :
(1) –30 and 50
(2) –50 and –30
(3) –50 and 30
(4) 50 and –30
Ans. (3)
Sol.
20 m/s
30 m/s
A
B
VA = 20 m/s
VB = –30 m/s
Velocity of B w.r.t. A
VB/A = – 50 m/s
Velocity of ground w.r.t. B
VG/B = 30 m/s
45. A galvanometer (G) of 2 resistance is connected
in the given circuit. The ratio of charge stored in
C1 and C2 is :
6 F
4 F
C1
C2
6
4
G
6 V
(1)
2
3
(2)
3
2
(3) 1
(4)
1
2
Ans. (4)
Sol.
C1
C2
6
4
6 V
2
I
I
In steady state
Req = 12
6
I 0.5A
12
P.D across C1 = 3V
P.D acoross C2 = 4V
q1 = C1V1 = 12 C
q2 = C2V2 = 24 C
1
2
q1
q2
46. In a metre-bridge when a resistance in the left gap
is 2 and unknown resistance in the right gap, the
balance length is found to be 40 cm. On shunting
the unknown resistance with 2, the balance
length changes by :
(1) 22.5 cm (2) 20 cm
(3) 62.5 cm (4) 65 cm
Ans. (1)
Sol.
G
2X
100 –
First case
2X
X3
40 60
In second case
23
X ' 1.2
23
2 1.2
100
200 – 2 = 1.2
200 62.5cm
3.2
Balance length changes by 22.5 cm
47. Match List - I with List - II.
List - I List - II
(Number) (Significant figure)
(A) 1001 (I) 3
(B) 010.1 (II) 4
(C) 100.100 (III) 5
(D) 0.0010010 (IV) 6
Choose the correct answer from the options given
below :
(1) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
(2) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(3) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
(4) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
Ans. (3)
Sol. Theoretical
48. A transformer has an efficiency of 80% and
works at 10 V and 4 kW. If the secondary voltage
is 240 V, then the current in the secondary coil is :
(1) 1.59 A (2) 13.33 A
(3) 1.33 A (4) 15.1 A
Ans. (2)
Sol. Efficiency =
SS
PP
EI
EI
S
240I
0.8 4000
S
3200
I240
= 13.33A
49. A light planet is revolving around a massive star in
a circular orbit of radius R with a period of
revolution T. If the force of attraction between
planet and star is proportional to
32
R
then choose
the correct option :
(1) T2 R5/2 (2) T2 R7/2
(3) T2 R3/2 (4) T2 R3
Ans. (1)
Sol.
2
3/2
GMm
F m R
R
2
5/2
1
R
2
T so
T2 R5/2
50. A body of mass 4 kg experiences two forces
1ˆ
ˆˆ
F 5i 8j 7k
and
2ˆ
ˆˆ
F 3i 4j 3k
. The
acceleration acting on the body is :
(1)
ˆ
ˆˆ
2i j k
(2)
ˆ
ˆˆ
4i 2j 2k
(3)
ˆ
ˆˆ
2i j k
(4)
ˆ
ˆˆ
2i 3j 3k
Ans. (3)
Sol. Net force =
ˆ
ˆˆ
8i 4j 4k
Fˆ
ˆˆ
a 2i j k
m
SECTION-B
51. A mass m is suspended from a spring of negligible
mass and the system oscillates with a frequency f1. The
frequency of oscillations if a mass 9 m is suspended
from the same spring is f2. The value of
1
2
f
f
is ___.
Ans. (3)
Sol.
1
1k
f2m
2
1k
f2 9m
1
2
f93
f 1 1
52. A particle initially at rest starts moving from
reference point. x = 0 along x-axis, with velocity
that varies as
4 xm / s.
The acceleration of
the particle is ____ms–2.
Ans. (8)
Sol.
V 4 x
dv
aV
dx
1/2
1
4 x 4 x
2
= 8 m/s2
53. A moving coil galvanometer has 100 turns and
each turn has an area of 2.0 cm2. The magnetic
field produced by the magnet is 0.01 T and
the deflection in the coil is 0.05 radian when a
current of 10 mA is passed through it.
The torsional constant of the suspension wire is
x × 10–5 N-m/rad. The value of x is___.
Ans. (4)
Sol. = BINAsin
C = BINAsin90°
34
BINA 0.01 10 10 100 2 10
C0.05
= 4 × 10–5 N-m/rad.
x = 4
54. One end of a metal wire is fixed to a ceiling and a
load of 2 kg hangs from the other end. A similar
wire is attached to the bottom of the load and
another load of 1 kg hangs from this lower wire.
Then the ratio of longitudinal strain of upper wire
to that of the lower wire will be___.
[Area of cross section of wire = 0.005 cm2,
Y = 2 × l011 Nm–2 and g = 10 ms–2]
Ans. (3)
Sol.
T = 30N
1
T = 10N
2
2 kg
1 kg
FL
LAY
LF
L AY
1
11
22
2
L
LF
30 3
LF 10
L
55. A particular hydrogen - like ion emits the radiation
of frequency 3 × 1015 Hz when it makes transition
from n = 2 to n = l. The frequency of radiation
emitted in transition from n = 3 to n = l is
x
9
× 1015 Hz, when x = ______.
Ans. (32)
Sol.
2
22
if
11
E 13.6z nn
22
fi
11
EC
nn
22
fi
11
hC
nn
22
fi
1 2 1
2
22
fi
31
11
nn
11
nn
11
3 / 4
14
11 8 / 9
19
39
48
1
2
27
32
15
21
32 32 3 10 Hz
27 27
15
32 10 Hz
9
56. In an electrical circuit drawn below the amount of
charge stored in the capacitor is ___C.
10 V
+–
10 F5
R2
R1
R36
C
4
Ans. (60)
Sol.
10 V
+–
C=10 F
R=5
2
R=4
1
R=6
3
I2
I1
I3
In steady state there will be no current in branch of
capacitor, so no voltage drop across R2 = 5
I2 = 0
13
10
I I 1A
46
32
R c R
V V V
2
R
V0
I3R3 = Vc
Vc = 1 × 6 = 6 volt
qc = CVc = 10 × 6 = 60 C
57. A coil of 200 turns and area 0.20 m2 is rotated at half
a revolution per second and is placed in uniform
magnetic field of 0.01 T perpendicular to axis of
rotation of the coil. The maximum voltage generated
in the coil is
2
volt. The value of is__.
Ans. (5)
Sol. = NAB cos(t)
d
dt
= NABsin(t)
max = NAB
= 200 × 0.2 × 0.01 ×
42
volt
10 5
58. In Young's double slit experiment, monochromatic
light of wavelength 5000 Å is used. The slits are
1.0 mm apart and screen is placed at 1.0 m away
from slits. The distance from the centre of the
screen where intensity becomes half of the
maximum intensity for the first time is___×10–6 m.
Ans. (125)
Sol. Let intensity of light on screen due to each slit is I0
So internity at centre of screen is 4I0
Intensity at distance y from centre-
I = I0 + I0 +
00
2 I I cos
Imax = 4I0
max
I
2
= 2I0 = 2I0 + 2I0 cos
cos = 0
2
Kx 2
2d sin 2
2 y 1
dD2
7
3
D 5 10 1
y4d 4 10
= 125 × 10–6
= 125
59. A uniform rod AB of mass 2 kg and Length 30 cm
at rest on a smooth horizontal surface. An impulse
of force 0.2 Ns is applied to end B. The time taken
by the rod to turn through at right angles will be
s
x
, where x = ____.
Ans. (4)
Sol.
A
B
JL = 0.3m
m = 2kg
Impulse J = 0.2 N-S
J Fdt 0.2N s
Angular impuls
(M)
c
M dt
L
F dt
2
LL
Fdt J
22
0.3 0.2
2
= 0.03
2
cm
ML
I12
2
2 (0.3)
12
0.09
6
cm f i
M I ( )
f
0.09
0.03 ( )
6
f = 2 rad/s
= t
t sec.
2 2 4
X = 4
60. Suppose a uniformly charged wall provides a
uniform electric field of 2 × l04 N/C normally. A
charged particle of mass 2 g being suspended
through a silk thread of length 20 cm and remain
stayed at a distance of 10 cm from the wall. Then
the charge on the particle will be
1C
x
where
x =______. [use g = 10 m/s2]
Ans. (3)
Sol.
+
+
+
+
+
+
+
20cm
d=10cm
mg
qE
E
(Uniform)
10 1
sin 20 2
= 30°
qE
tan mg
4
3
q 2 10
tan 30 1 10 10
6
1q 10
3
6
1
q 10 C
3
x = 3
CHEMISTRY TEST PAPER WITH SOLUTION
SECTION-A
61. The transition metal having highest 3rd ionisation
enthalpy is :
(1) Cr (2) Mn
(3) V (4) Fe
Ans. (2)
Sol. 3rd Ionisation energy : [NCERT Data]
V : 2833 KJ/mol
Cr : 2990 KJ/mol
Mn : 3260 KJ/mol
Fe : 2962 KJ/mol
alternative
Mn : 3d5 4s2
Fe : 3d6 4s2
Cr : 3d5 4s1
V : 3d3 4s2
So Mn has highest 3rd IE among all the given
elements due to d5 configuration.
62. Given below are two statements :
Statement (I) : A p bonding MO has lower electron
density above and below the inter-nuclear asix.
Statement (II) : The p* antibonding MO has a
node between the nuclei.
In the light of the above statements, choose the
most appropriate answer from the options given
below:
(1) Both Statement I and Statement II are false
(2) Both Statement I and Statement II are true
(3) Statement I is false but Statement II is true
(4) Statement I is true but Statement II is false
Ans. (3)
Sol. A p bonding molecular orbital has higher electron
density above and below inter nuclear axis
Inter nuclear
axis
pu(p )
y
+
–
(+)
py
py
– –
+
+
Nodal Plane
–
+
pg(p )
y
*
+
–
(–)
py
py
–
+
–
+
63. G
iven below are two statements : one is labelled as
Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : In aqueous solutions Cr2+ is
reducing while Mn3+ is oxidising in nature.
Reason (R) : Extra stability to half filled electronic
configuration is observed than incompletely filled
electronic configuration.
In the light of the above statement, choose the most
appropriate answer from the options given below:
(1) Both (A) and (R) are true and (R) is the
correct explanation of (A)
(2) Both (A) and (R) are true but (R) is not the
correct explanation of (A)
(3) (A) is false but (R) is true
(4) (A) is true but (R) is false
Ans. (1)
Sol. Cr2+ is reducing as it configuration changes from d4
to d3 due to formation of Cr3+, which has half filled
t2g level, on other hand, the change Mn3+ to Mn
2+
result half filled d5 configuration which has extra
stability.
64. Match List - I with List - II.
List-I List-II
(Reactants) Products
(A) Phenol, Zn/ (I) Salicylaldehyde
(B) Phenol, CHCl3, NaOH, HCl (II) Salicylic acid
(C) Phenol, CO2, NaOH, HCl (III) Benzene
(D) Phenol, Conc. HNO3 (IV) Picric acid
Choose the correct answer from the options given
below.
(1) (A)-(IV), (B), (II), (C)-(I), (D)-(III)
(2) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
(3) (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
(4) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Ans. (3)
Sol.
OH
Zn,
OH
CHCl + NaOH
3
HCl
OH
CHO
Salicylaldehyde
OH
CO + NaOH
2
HCl
OH
COOH
Salicylic acid
OH
Con. HNO3
OH
NO2
Picric Acid
NO2
NO2
65. Given below are two statements :
Statement (I) : Both metal and non-metal exist in
p and d-block elements.
Statement (II) : Non-metals have higher
ionisation enthalpy and higher electronegativity
than the metals.
In the light of the above statements, choose the
most appropriate answer from the option 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. (2)
Sol. I. In p-Block both metals and non metals are
present but in d-Block only metals are present.
II. EN and IE of non metals are greater than that
of metals
I - False, II-True
66. The strongest reducing agent amont the following
is:
(1) NH3 (2) SbH3
(3) BiH3 (4) PH3
Ans. (3)
Sol. Strongest reducing agent : BiH3 explained by its
low bond dissociation energy.
67. Which of the following compounds show colour
due to d-d transition?
(1) CuSO4.5H2O (2) K2Cr2O7
(3) K2CrO4 (4) KMnO4
Ans. (1)
Sol. CuSO4.5H2O
Cu2+ : 3d9 4s0
unpaired electron present so it show colour due to
d-d transition.
68. The set of meta directing functional groups from
the following sets is:
(1) –CN, –NH2, –NHR, –OCH3
(2) –NO2, –NH2, –COOH, –COOR
(3) –NO2, –CHO, –SO3H, –COR
(4) –CN, –CHO, –NHCOCH3, –COOR
Ans. (3)
Sol.
O
N
O,C H
O
,SOH
O
O
,C R
O
All are –M, Hence meta directing groups.
69. Select the compound from the following that will
show intramolecular hydrogen bonding.
(1) H2O
(2) NH3
(3) C2H5OH
(4)
NO2
OH
Ans. (4)
Sol. H2O, NH3, C2H5OH Intermolecular H-Bonding
N
O
O
O
H Intramolecular
H-Bonding
70. Lassaigne's test is used for detection of :
(1) Nitrogen and Sulphur only
(2) Nitrogen, Sulphur and Phosphorous Only
(3) Phosphorous and halogens only
(4) Nitrogen, Sulphur, phosphorous and halogens
Ans. (4)
Sol. Lassaigne's test is used for detection of all element
N, S, P, X.
71. Which among the following has highest boiling
point?
(1) CH3CH2CH2CH3
(2) CH3CH2CH2CH2–OH
(3) CH3CH2CH2CHO
(4) H5C2 – O – C2H5
Ans. (2)
Sol. Due to H-bonding boiling point of alcohol is High.
72. In the given reactions identify A and B.
H + A
2Pd/C CH3
HC C
C H
2 5
H
3 3 2
3
Na/LiquidNH
CH C C CH H "B"
(1) A : 2–Pentyne B : trans – 2 – butene
(2) A : n – Pentane B : trans – 2 – butene
(3) A : 2 – Pentyne B : Cis – 2 – butene
(4) A : n – Pentane B : Cis – 2 – butene
Ans. (1)
Sol.
H + CH – C C – C H
23 2 5
2-pentyne
Pd/C CH3
HC C
C H
2 5
H
CH – C C – CH + H
3 3 2
Na CH3
HC C
H
Liquid NH3CH3
Trans-2-butene
73. The number of radial node/s for 3p orbital is:
(1) 1 (2) 4
(3) 2 (4) 3
Ans. (1)
Sol. For 3p : n = 3, = 1
Number of radial node = n – – 1
= 3 – 1 – 1 = 1
74. Match List - I with List - II.
List - I List - II
Compound Use
(A) Carbon tetrachloride (I) Paint remover
(B) Methylene chloride (II) Refrigerators and air
conditioners
(C) DDT (III) Fire extinguisher
(D) Freons (IV) Non Biodegradable
insecticide
Choose the correct answer from the options given
below :
(1) (A)-(I), (B), (II), (C)-(III), (D)-(IV)
(2) (A)-(III), (B)-(I), (C)-(IV), (D)-( II)
(3) (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
(4) (A)-( II), (B)-(III), (C)-(I), (D)-(IV)
Ans. (2)
Sol. CCl4 used in fire extinguisher. CH2Cl2 used as
paint remover. Freons used in refrigerator and AC.
DDT used as non Biodegradable insecticide.
75. The functional group that shows negative
resonance effect is:
(1) –NH2 (2) –OH
(3) –COOH (4) –OR
Ans. (3)
Sol.
COH
O
shows –R effect, while rest 3 groups
shows +R effect via lone pair.
76. [Co(NH3)6]3+ and [CoF6]3– are respectively known
as:
(1) Spin free Complex, Spin paired Complex
(2) Spin paired Complex, Spin free Complex
(3) Outer orbital Complex, Inner orbital Complex
(4) Inner orbital Complex, Spin paired Complex
Ans. (2)
Sol. [Co(NH3)6]3+
Co3+ (strong field ligand) 3d6
gg
t ,e
60
2
,
Hybridisation : d2sp3
Inner obital complex(spin paired complex)
Pairing will take place.
[CoF6]3–
Co3+ (weak field ligand) 3d6
gg
t ,e
42
2
Hybridisation : sp3d2
Outer orbital complex (spin free complex)
no pairing will take place
77. Given below are two statements :
Statement (I) : SiO2 and GeO2 are acidic while
SnO and PbO are amphoteric in nature.
Statement (II) : Allotropic forms of carbon are
due to property of catenation and p-d bond
formation.
In the light of the above statements, choose the
most appropriate answer from the options given
below:
(1) Both Statement I and Statement II are false
(2) Both Statement I and Statement II are true
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Ans. (3)
Sol. SiO2 and GeO2 are acidic and SnO, PbO are
amphoteric.
Carbon does not have d-orbitals so can not form
p-d Bond with itself. Due to properties of
catenation and p-p bond formation. carbon is
able to show allotropic forms.
78.
C H Br
2 5 alc. KOH ABr2
CCl4BKCN
Excess CH O
3+
Excess
D
Acid D formed in above reaction is :
(1) Gluconic acid
(2) Succinic acid
(3) Oxalic acid
(4) Malonic acid
Ans. (2)
Sol.
C H Br
2 5 alc. KOH CH2Br2
CCl4
(B)
KCN
Excess
H O
3+
(A)
CH2CH2CH2
Br Br
CH2CH2
CN CN
(C)
CH2CH2
COOH COOH
(D)
Succinic
Acid
79. Solubility of calcium phosphate (molecular mass, M)
in water is Wg per 100 mL at 25° C. Its solubility
product at 25°C will be approximately.
(1)
3
7W
10 M
(2)
5
7W
10 M
(3)
5
3W
10 M
(4)
5
5W
10 M
Ans. (2)
Sol.
W
SM
10
23
3 4 2 4
Ca (PO ) (s) 3Ca (aq.) 2PO (aq.)
3s 2s
W 1000 W 10
SM 100 M
Ksp = (3s)3 (2s)2
= 108 s5
5
5W
108 10 M
7W
1. 8 1 M
5
00
80. Given below are two statements :
Statement (I) : Dimethyl glyoxime forms a six-
membered covalent chelate when treated with
NiCl2 solution in presence of NH4OH.
Statement (II) : Prussian blue precipitate contains
iron both in (+2) and (+3) oxidation states. In the
light of the above statements, choose the most
appropriate 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) Both Statement I and Statement II are false
(4) Statement I is true but Statement II is false
Ans. (1)
Sol. Ni2+ + NH4OH + dmg
CH
3C N
CCH3NNi2+
O H
N
O
CCH3
N CCH3
OHO
2 Five member ring
4 6 3
III II
Fe [Fe(CN) ]
Prussian Blue
SECTION-B
81. Total number of isomeric compounds (including
stereoisomers) formed by monochlorination of
2-methylbutane is________.
Ans. (6)
Sol.
Cl
*
(2)
Cl
(1) Cl
(2)
*Cl
(1)
82. The following data were obtained during the first
order thermal decomposition of a gas A at constant
volume:
A(g) 2B(g) + C(g)
S.No Time/s Total pressure/(atm)
1. 0 0.1
2. 115 0.28
The rate constant of the reaction is______× 10–2s–1
(nearest integer)
Ans. (2)
Sol. A(g) 2B(g) + C(g)
t = 0 0.1
t = 115 sec. 0.1 – x 2x x
0.1 + 2x = 0.28
2x = 0.18
x = 0.09
1 0.1
Kn
115 0.1 0.09
= 0.0200 sec–1
= 2 × 10–2 sec–1
83. The number of tripeptides formed by three
different amino acids using each amino acid once
is ______.
Ans. (6)
Sol. Let 3 different amino acid are A, B, C then
following combination of tripeptides can be
formed-
ABC, ACB, BAC, BCA, CAB, CBA
84. Number of compounds which give reaction with
Hinsberg's reagent is_______.
NCl
2
+–
O
NH
NH2
N NNH
2
NH2
N
H
N
NH
2HN
2C NH2
O
NH
Ans. (5)
Sol.
NH2
NH NH2
H N
2
N
NH
H
NH2
85. Mass of ethylene glycol (antifreeze) to be added to
18.6 kg of water to protect the freezing point at
–24°C is________ kg (Molar mass in g mol–1 for
ethylene glycol 62, Kf of water = 1.86 K kg mol–1)
Ans. (15)
Sol. Tf = iKf × molality
W
24 (1) 1.86 62 18.6
W = 14880 gm
= 14.880 kg
86. Following Kjeldahl's method, 1g of organic
compound released ammonia, that neutralised 10
mL of 2M H2SO4. The percentage of nitrogen in
the compound is _______%.
Ans. (56)
Sol. H2SO4 + 2NH3 (NH4)2 SO4
Millimole of H2SO4 10 × 2
So Millimole of NH3 = 20 × 2 = 40
Organic NH3
Compound 40 Millimole
Mole of N =
40
1000
wt. of N =
40 14
1000
% composition of N in organic compound
40 14 100
1000 1
= 56%
87. The amount of electricity in Coulomb required for
the oxidation of 1 mol of H2O to O2 is ____×105C.
Ans. (2)
Sol. 2H2O O2 + 4H+ + 4e–
WQ
E 96500
mole × n-factor =
Q
96500
1 × 2 =
Q
96500
Q = 2 × 96500 C
= 1.93 × 105 C
88. For a certain reaction at 300K, K = 10, then G°
for the same reaction is –_______×10–1 kJ mol–1.
(Given R = 8.314 JK–1 mol–1)
Ans. (57)
Sol. G° = –RT
n
K
= –8.314 × 300
n
(10)
= 5744.14 J/mole
= 57.44 × 10–1 kJ/mole
89. Consider the following redox reaction :
2
4 2 2 4 2 2
MnO H H C O Mn H O CO
The standard reduction potentials are given as
below
red
E
2
4
MnO /Mn
E 1.51V
2 2 2 4
CO / H C O
E 0.49V
If the equilibrium constant of the above reaction is
given as Keq = 10x, then the value of x = _______
(nearest integer)
Ans. (338 OR 339)
Sol. Cell Rxn ; MnO4
– + H2C2O4 Mn2+ + CO2
cell
E
= E°
op of anode + E°
RP of cathode
= 0.49 + 1.51 = 2.00V
At equilibrium
Ecell = 0,
cell
0.059
E logK
n
(As per NCERT
RT 0.059
F
But
RT 0.0591
F
can also be taken.)
0.059
2 logK
10
logK = 338.98
90. 10 mL of gaseous hydrocarbon on combustion
gives 40 mL of CO2(g) and 50 mL of water
vapour. Total number of carbon and hydrogen
atoms in the hydrocarbon is _____.
Ans. (14)
Sol.
CxHy
10ml
+ O2 CO2 + H2O
2 2 2
yy
CxHy x O xCO H O
42
10x 5y
10x = 40
x = 4
5y = 50
y = 10
C4H10
