FINAL JEE–MAIN EXAMINATION – JANUARY, 2024
(Held On Thursday 01st February, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS
TEST PAPER WITH SOLUTION
SECTION-A
1. A bag contains 8 balls, whose colours are either
white or black. 4 balls are drawn at random
without replacement and it was found that 2 balls
are white and other 2 balls are black. The
probability that the bag contains equal number of
white and black balls is:
(1)
2
5
(2)
2
7
(3)
1
7
(4)
1
5
Ans. (2)
Sol.
P(4W4B/2W2B) =
(4 4 ) (2 2 / 4 4 )
(2 6 ) (2 2 / 2 6 ) (3 5 ) (2 2 / 3 5 )
............. (6 2 ) (2 2 / 6 2 )
P W B P W B W B
P W B P W B W B P W B P W B W B
P W B P W B W B
=
44
22
8
4
2 6 3 5 6 2
2 2 2 2 2 2
8 8 8
4 4 4
CC
1
5C
C C C C C C
1 1 1
...
5 C 5 C 5 C
=
2
7
2. The value of the integral
4
44
0
:
sin (2 ) cos (2 )
xdx equals
xx
(1)
2
2
8
(2)
2
2
16
(3)
2
2
32
(4)
2
2
64
Ans. (3)
Sol.
4
44
0sin (2 ) cos (2 )
xdx
xx
Let
2xt
then
1
2
dx dt
2
44
0
2
44
0
2
44
0
2
44
0
4
2
4
0
1
4 sin cos
12
4sin cos
22
12
4 sin cos
28 sin cos
sec
28 tan 1
tdt
Itt
t dt
I
tt
dt
II
tt
dt
Itt
tdt
It
Let tant = y then sec2t dt = dy
2
4
0
(1 )
281
y dy
Iy
2
2
02
1
1ydy
1
16 yy
Put
1
yp
y
2
2
dp
I16 p2
1p
tan
16 2 2
2
16 2
I
3. If A =
21
12
, B =
10
11
, C = ABAT and X
= ATC2A, then det X is equal to :
(1) 243
(2) 729
(3) 27
(4) 891
Ans. (2)
Sol.
21 det( ) 3
12
10 det( ) 1
11
AA
BB
Now C = ABAT
det(C) = (dct (A))2 x det(B)
9C
Now |X| = |ATC2A|
= |AT| |C|2 |A|
= |A|2 |C|2
= 9 x 81
= 729
4. If tanA =
22
1,tan
( 1) 1
x
B
x x x x x
and
1
3 2 1 2
tan ,0 , , ,
2
C x x x A B C then
A + B is equal to :
(1) C
(2)
C
(3)
2C
(4)
2C
Ans. (1)
Sol.
Finding tan (A + B) we get
tan (A + B) =
22
2
1
( 1) 1
tan tan 1
1 tan tan 11
x
x x x x x
AB
AB xx
tan (A + B) =
2
2
11x x x
x x x
2
2
2
11
1
tan( ) tan
x x x
x x x
xx
A B C
xx
A B C
5. If n is the number of ways five different employees
can sit into four indistinguishable offices where
any office may have any number of persons
including zero, then n is equal to:
(1) 47
(2) 53
(3) 51
(4) 43
Ans. (3)
Sol.
Total ways to partition 5 into 4 parts are :
5, 0, 0, 0
1 way
4, 1, 0, 0
5!
4!
5 ways
3, 2, 0, 0,
5! 10
3!2!
ways
5!
2,2,0,1 15
2!2!2!
ways
3
5!
2,1,1,1 10
2!(1!) 3!
ways
5!
3,1,1,0 10
3!2!
ways
Total
1+5+10+15+10+10 = 51 ways
6. LetS={
: 1 1z C z
and
2 1 2 2z z i z z
}. Let z1, z2
S
be such that
12
max min
zs
zs
z z and z z
.
Then
2
12
2zz
equals :
(1) 1 (2) 4
(3) 3 (4) 2
Ans. (4)
Sol. Let Z = x + iy
Then (x - 1)2 + y2 = 1
(1)
&
2 1 2 (2 ) 2 2
( 2 1) 2 (2)
x i iy
xy
Solving (1) & (2) we get
Either x = 1 or
1(3)
22
x
On solving (3) with (2) we get
For x = 1
y = 1
Z2 = 1 + i
& for
1
1 1 1
21
2 2 2 2 2
i
x y Z
Now
2
12
2
2
2
11 2 (1 )
2
2
2
zz
ii
7. Let the median and the mean deviation about the
median of 7 observation 170, 125, 230, 190, 210, a, b
be 170 and
205
7
respectively. Then the mean
deviation about the mean of these 7 observations is :
(1) 31
(2) 28
(3) 30
(4) 32
Ans. (3)
Sol. Median = 170
125, a, b, 170, 190, 210, 230
Mean deviation about
Median =
0 45 60 20 40 170 170 205
77
ab
a + b = 300
Mean =
170 125 230 190 210 175
7
ab
Mean deviation
About mean =
50 175 175 5 15 35 55
7
ab
= 30
8. Let
ˆˆ
ˆ ˆ ˆ ˆ
5 3 , 2 4a i j k b i j k
and
ˆˆˆ
.c a b i i i
Then
ˆ
ˆˆ
c i j k
is
equal to
(1) –12 (2) –10
(3) –13 (4) –15
Ans. (1)
Sol.
ˆ
ˆ
53 a i j k
ˆ
ˆˆ
24 b i j k
ˆ ˆ ˆ
() a b i a i b b i a
5 ba
ˆˆ
5 b a i i
ˆ
ˆ ˆ ˆ
11 23 j k i i
ˆˆˆ
11 23 k j i
ˆ
ˆ
11 23jk
ˆ
ˆˆ
. 11 23 12 c i j k
9. Let S =
{ :( 3 2) ( 3 2) 10}.
xx
xR
Then the number of elements in S is :
(1) 4 (2) 0
(3) 2 (4) 1
Ans. (3)
Sol.
xx
3 2 3 2 10
Let
x
3 2 t
1
t 10
t
t2 – 10t + 1 = 0
10 100 4
t 5 2 6
2
x2
3 2 3 2
x = 2 or x = –2
Number of solutions = 2
10. The area enclosed by the curves xy + 4y = 16 and
x + y = 6 is equal to :
(1) 28 – 30
log 2
e
(2) 30 – 28
log 2
e
(3) 30 – 32
log 2
e
(4) 32 – 30
log 2
e
Ans. (3)
Sol. xy + 4y = 16 , x + y = 6
y(x + 4) = 16 ____(1) , x + y = 6___(2)
on solving, (1) & (2)
we get x = 4, x = –2
–4 –2 4(6,0)
(0,6)
Area =
4
2
16
64
30 32ln 2
x dx
x
11. Let f : R R and g : R R be defined as
f(x) =
e
x
log x , x 0
e , x 0
and
g(x) =
x
x , x 0
e , x 0
. Then, gof : R R is :
(1) one-one but not onto
(2) neither one-one nor onto
(3) onto but not one-one
(4) both one-one and onto
Ans. (2)
Sol.
g(f(x)) =
()
( ), ( ) 0
, ( ) 0
fx
f x f x
e f x
g(f(x)) =
ln
, ,0
,(0,1)
ln , 1,
x
x
e
e
x
(1,0)
(0,1)
Graph of g(f(x))
g(f(x))
Many one into
12. If the system of equations
2x + 3y – z = 5
x + y + 3z = –4
3x – y + z = 7
has infinitely many solutions, then 13 is equal
to
(1) 1110 (2) 1120
(3) 1210 (4) 1220
Ans. (2)
Sol. Using family of planes
2x + 3y –z – 5 = k1 (x +
y + 3z + 4) + k2 (3x – y
+
z - 7)
2 = k1 + 3k2, 3 = k1
- k2, -1 = 3k1 +
k2, -5 =
4k1 – 7k2
On solving we get
21
13 1 16
, , 70,
19 19 13
kk
13
= 13 (-70)
16
13
1120
13. For 0 < < /2, if the eccentricity of the hyperbola
x2 – y2cosec2 = 5 is
7
times eccentricity of the
ellipse x2 cosec2 + y2 = 5, then the value of is :
(1)
6
(2)
5
12
(3)
3
(4)
4
Ans. (3)
Sol.
2
2
1 sin
1 sin
h
c
e
e
7
hc
ee
22
2
1 sin 7(1 sin )
63
sin 84
3
sin 2
3
14. Let y = y(x) be the solution of the differential
equation
dy
dx
= 2x (x + y)3 – x (x + y) – 1, y(0) = 1.
Then,
2
11
y
22
equals :
(1)
4
4e
(2)
3
3e
(3)
2
1e
(4)
1
2e
Ans. (4)
Sol.
3
2 ( ) ( ) 1
dy x x y x x y
dx
xyt
3
1 2 1
dt xt xt
dx
3
2
dt xdx
tt
42
tdt xdx
2t t
Let
2tz
2
dz xdx
2 2z z
1
42
dz xdx
zz
2
1
2
ln
zxk
z
1
2
ze
15. Let f : R R be defined as
f(x) =
2
2
a bcos2x ; x 0
x
x cx 2 ; 0 x 1
2x 1 ; x 1
If f is continuous everywhere in R and m is the
number of points where f is NOT differential then
m + a + b + c equals :
(1) 1 (2) 4
(3) 3 (4) 2
Ans. (4)
Sol. At x = 1, f(x) is continuous therefore,
f(1–) = f(1) = f(1+)
f(1) = 3 + c ….(1)
f(1+) =
0
lim
h
2(1 + h) + 1
f(1+) =
0
lim
h
3 + 2h = 3 ….(2)
from (1) & (2)
c = 0
at x = 0, f(x) is continuous therefore,
f(0–) = f(0) = f(0+) ….(3)
f(0) = f(0+) = 2 ….(4)
f(0–) has to be equal to 2
2
0
cos 2
lim
h
a b h
h
24
2
0
4 16
1 ...
2! 4!
lim
h
hh
ab
h
24
2
0
2
2 ...
3
lim
h
a b b h h
h
for limit to exist a – b = 0 and limit is 2b ….(5)
from (3), (4) & (5)
a = b = 1
checking differentiability at x = 0
LHD :
2
0
1 cos2 2
lim
h
h
hh
24 2
3
0
4 16
1 1 ... 2
2! 4!
lim 0
h
hh h
h
RHD :
2
0
0 2 2
lim 0
h
h
h
Function is differentiable at every point in its domain
m = 0
m + a + b + c = 0 + 1 + 1 + 0 = 2
16. Let
22
22
xy
ab
= 1, a > b be an ellipse, whose
eccentricity is
1
2
and the length of the latus
rectum is
14
. Then the square of the eccentricity
of
22
22
xy
ab
= 1 is :
(1) 3 (2) 7/2
(3) 3/2 (4) 5/2
Ans. (3)
Sol.
22
22
2
2
2
2
11
11
2
2
214
13
11
22
3
2
H
H
bb
eaa
b
a
b
ea
e
17. Let 3, a, b, c be in A.P. and 3, a – 1, b + l, c + 9 be
in G.P. Then, the arithmetic mean of a, b and c is :
(1) –4 (2) –1
(3) 13 (4) 11
Ans. (4)
Sol.
3, a, b, c
A.P
3, 3+d, 3+2d, 3+3d
3, a–1,b +1, c + 9
G.P
3, 2+d, 4+2d, 12+3d
a = 3 + d
2
2 3 4 2dd
b = 3 + 2d d = 4, -2
c = 3 + 3d
If d = 4 G.P
3, 6, 12, 24
a = 7
b = 11
c = 15
11
3
abc
18. Let C : x2 + y2 = 4 and C’ : x2 + y2 - 4x + 9 = 0 be
two circles. If the set of all values of so that the
circles C and C’ intersect at two distinct points, is
R– [a, b], then the point (8a + 12, 16b – 20) lies on
the curve :
(1) x2 + 2y2 – 5x + 6y = 3
(2) 5x2 – y = – 11
(3) x2 – 4y2 = 7
(4) 6x2 + y2 = 42
Ans. (4)
Sol. x2 + y2 = 4
C (0, 0) r1 = 2
C' (2, 0) r2 =
2
49
|r1 – r2| < CC' < |r1 + r2|
22
2 4 9 2 2 4 9
4 + 42 – 9 – 4
2
49
< 42
True R…. (1)
42 < 4 + 42 – 9 + 4
2
49
5 < 4
2
49
and 2
9
4
25
16
< 42 – 9
33
,,
22
2
169
64
13 13
,,
88
…(2)
from (1) and (2)
13 13
,,
88
R –
13 13
,
88
as per question
13
a8
and
13
b8
required point is (–1, 6) with satisfies option (4)
19. If 5f(x) + 4f
1
x
= x2 – 2, x 0 and y = 9x2f(x),
then y is strictly increasing in :
(1)
11
0, ,
55
(2)
11
,0 ,
55
(3)
11
,0 0,
55
(4)
11
, 0,
55
Ans. (2)
Sol. 5 f(x) + 4 f
1
x
= x2 – 2,
0x
…(1)
Substitute
1
xx
2
11
5 4 2f f x
xx
…(2)
On solving (1) and (2)
42
2
5x 2x 4
fx 9x
y = 9x2f(x)
y = 5x4 – 2x2 – 4 …(3)
3
dy 20x 4x
dx
for strictly increasing
dy 0
dx
4x(5x2 – 1) > 0
x
11
,0 ,
55
20. If the shortest distance between the lines
x y 2 z 1
2 1 1
and
3 1 2
1 2 1
x y z
is 1, then the sum of all possible values of is :
(1) 0 (2)
23
(3)
33
(4)
23
Ans. (2)
Sol. Passing points of lines L1 & L2 are
,2,1 & 3,1,2
S.D =
3 1 1
2 1 1
1 2 1
ˆ
ˆˆ
2 1 1
1 2 1
i j k
1 =
3
3
0, 2 3
SECTION-B
21. If x = x(t) is the solution of the differential
equation (t + 1)dx = (2x + (t + 1)4) dt, x(0) = 2,
then, x(1) equals ___________.
Ans. (14)
Sol. (t + 1)dx = (2x + (t + 1)4)dt
4
dx 2x (t 1)
dt t 1
dx 2x
dt t 1
= (t + 1)3
F =
2dt
t1
e
=
2ln(t 1)
e
=
2
1
(t 1)
3
22
x1
(t 1) dt c
(t 1) (t 1)
2
2
x (t 1) c
2
(t 1)
c =
3
2
x =
42
(t 1) 3(t 1)
22
put, t = 1
x = 23 + 6 = 14
22. The number of elements in the set
S = {(x, y, z) : x, y,z Z, x + 2y + 3z = 42, x, y, z
0} equals __________.
Ans. (169)
Sol. x + 2y + 3z = 42, x, y, z 0
z = 0 x + 2y = 42 22
z = 1 x + 2y = 39 20
z = 2 x + 2y = 36 19
z = 3 x + 2y = 33 17
z = 4 x + 2y = 30 16
z = 5 x + 2y = 27 14
z = 6 x + 2y = 24 13
z = 7 x + 2y = 21 11
z = 8 x + 2y = 18 10
z = 9 x + 2y = 15 8
z = 10 x + 2y = 12 7
z = 11 x + 2y = 9 5
z = 12 x + 2y = 6 4
z = 13 x + 2y = 3 2
z = 14 x + 2y = 0 1
Total : 169
23. If the Coefficient of x30 in the expansion of
6
1
1x
(1 + x2)7 (1 – x3)8 ; x 0 is , then ||
equals ___________.
Ans. (678)
Sol. coeff of x30 in
78
623
6
x 1 1 x 1 x
x
coeff. of x36 in
78
623
1 x 1 x 1 x
General term
31 2 3
1 2 3
rr 2r 3r
6 7 8
r r r
C C C 1 x
1 2 3
r 2r 3r 36
Case-I :
1 2 3
12
r r r
0 6 8
r 2r 12
2 5 8
4 4 8
6 3 8
(Taking r3 = 8)
Case-II :
1 2 3
12
r r r
1 7 7 r 2r 15
3 6 7
5 5 7
(Taking r3 = 7)
Case-III :
1 2 3
12
r r r
4 7 6 r 2r 18
6 6 6
(Taking r3 = 6)
Coeff. = 7 + (15 × 21) + (15 × 35) + (35)
–(6 × 8) – (20 × 7 × 8) – (6 × 21 × 8) + (15 × 28)
+ (7 × 28) = –678 =
24. Let 3, 7, 11, 15, ...., 403 and 2, 5, 8, 11, . . ., 404
be two arithmetic progressions. Then the sum, of
the common terms in them, is equal to _________.
Ans. (6699)
Sol. 3, 7, 11, 15, ….., 403
2, 5, 8, 11, …., 404
LCM (4, 3) = 12
11, 23, 35,….. let (403)
403 = 11 + (n - 1) x 12
392 1
12 n
33 66
33
n
n
Sum
33 22 32 12
2
=6699
25. Let {x} denote the fractional part of x and
f(x) =
1 2 1
3
cos (1 {x} )sin (1 {x})
{x} {x}
, x 0. If L
and R respectively denotes the left hand limit and the
right hand limit of f(x) at x = 0, then
2
32
(L2 + R2) is
equal to ____________.
Ans. (18)
Sol. Finding right hand limit
h0
x0
lim f x lim f 0 h
h0
lim f h
1 2 1
2
0
cos (1 )sin (1 )
lim (1 )
h
hh
hh
=
12 1
h0
cos 1 h sin 1
lim h1
Let
1 2 2
cos 1 h cos 1 h
0
lim
21 cos
0
2
1
lim
21 cos
1
21 / 2
R2
Now finding left hand limit
x0
L lim f x
h0
lim f h
2
11
3
h0
cos 1 h sin 1 h
lim
hh
2
11
3
h0
cos 1 h 1 sin 1 h 1
lim
h 1 h 1
1 2 1
2
h0
cos h 2h sin h
lim
1 h 1 1 h
1
2
h0
sin h
lim 21 1 h
1
2
h0
sin h
lim
2 h 2h
1
h0
sin h 1
lim
2 h h 2
L4
22
22
22
32 32
LR 2 16
= 16 + 2
= 18
26. Let the line L :
2
x + y =
pass through the point
of the intersection P (in the first quadrant) of the circle
x2 + y2 = 3 and the parabola x2 = 2y. Let the line L
touch two circles C1 and C2 of equal radius
23
. If
the centres Q1 and Q2 of the circles C1 and C2 lie on the
y-axis, then the square of the area of the triangle
PQ1Q2 is equal to ___________.
Ans. (72)
Sol. x2 + y2 = 3 and x2 = 2y
y2 + 2y – 3 = 0
(y + 3) (y – 1) = 0
y = – 3 or y = 1
y = 1 x =
2
( 2,1)P
p lies on the line
2
2( 2) 1
3
xy
For circle C1
Q1 lies on y axis
Let Q1
0,
coordinates
R1 =
23
(Given
Line L act as tangent
Apply P = r (condition of tangency)
323
3
36
36
or
36
9
3
12
2
12
2 1 1
10 9 1
20 3 1
12(12) 6 2
2
72
PQQ
PQQ
27. Let P = {z : |z + 2 – 3i | 1} and
Q = {z : z (l + i) +
z
(1 – i) –8} . Let in
P Q, |z – 3 + 2i| be maximum and minimum at
z1 and z2 respectively. If |z1|2 + 2|z|2 = +
2
,
where , are integers, then + equals
________.
Ans. (36)
Sol.
1
2
Px+y–1 = 0 (L )
1
(L ) x–y+4 = 0
2(3,–2)
(–2,3)
Clearly for the shaded region z1 is the intersection
of the circle and the line passing through P (L1)
and z2 is intersection of line L1 & L2
Circle : (x + 2)2 + (y - 3)2 =1
L1 : x + y – 1 = 0
L2 : x – y + 4 =0
On solving circle & L1 we get
z1 :
11
2 ,3
22
On solving L1 and z2 is intersection of line L1 & L2
we get z2 :
35
,
22
So
22
12
2 14 5 2 17
31 5 2
31
5
36
zz
28. If
2
2sinx 4
8 2 cosxdx
(1 e )(1 sin x)
= + loge (3 + 2
2
), where , are integers, then 2 + 2 equals
___________.
Ans. (8)
Sol.
2
sin 4
2
8 2 cos
1 1 sin
x
x
I dx
ex
Apply king
sin
2
sin 4
2
2
4
2
2
4
0
1
4
0
122
22
022
22
2
8 2 cos ....(2)
1 1 sin
1 & 2
8 2 cos
21 sin
8 2 cos ,
1 sin
sin
82
1
11
11
42 11
11
11
42 12
x
x
xe
I dx
ex
adding
x
I dx
x
x
I dx
x
xt
I dx
t
tt
I dt
tt
tt
tt
I
tt
t
1
2
012
11
&
dt
t
Let t z t k
tt
02
22
2
0
1
42 22
1 1 2
4 2 tan ln
2 2 2 2 2
1 2 2
4 2 ln
2 2 2 2 2 2
2 2ln(3 2 2)
2
2
dz dk
zk
zk
k
29. Let the line of the shortest distance between
the lines
L1 :
ˆ ˆ ˆ ˆ ˆ ˆ
r i 2j 3k i j k
and
L2 :
ˆ ˆ ˆ ˆ ˆ ˆ
r 4i 5j 6k i j k
intersect L1 and L2 at P and Q
respectively. If ( , , ) is the midpoint
of the line segment PQ, then 2( + + )
is equal to ___________.
Ans. (21)
Sol.
ˆˆˆ
b i j k
(DR's of L1)
ˆˆˆ
d i j k
(DR's of L2)
ˆˆ ˆ
i j k
b d 1 1 1
1 1 1
ˆˆˆ
0i 2 j 2k
(DR's of Line perpendicular to
L1and L2)
DR of AB line
=
0,2,2 3 ,3 ,3
3 3 3
0 2 2
Solving above equation we get = –
3
2
and
3
2
point
5 1 9
A , ,
222
5 7 15
B , ,
2 2 2
Point of AB =
5,2,6 , ,
2
2() = 5 + 4 + 12 = 21
30. Let A= {1, 2, 3, . . 20}. Let R1 and R2 two
relation on A such that
R1 = {(a, b) : b is divisible by a}
R2 = {(a, b) : a is an integral multiple of b}.
Then, number of elements in R1 – R2 is equal
to ________.
Ans. (46)
Sol. n(R1) = 20 + 10 + 6 + 5 + 4 + 3 + 2+ 2 +2
10 times
2 1 ... 1
n(R1) = 66
12
R R 1,1 , 2,2 ,... 20,20
12
n R R 20
1 2 1 1 2
n R R n R n R R
1
n R 20
= 66 – 20
R1 – R2 = 46 Pair
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
31. With rise in temperature, the Young's modulus of
elasticity
(1) changes erratically
(2) decreases
(3) increases
(4) remains unchanged
Ans. (2)
Sol. Conceptual questions
32. If R is the radius of the earth and the acceleration
due to gravity on the surface of earth is g = 2 m/s2,
then the length of the second's pendulum at a
height h = 2R from the surface of earth will be,:
(1)
2m
9
(2)
1m
9
(3)
4m
9
(4)
8m
9
Ans. (2)
Sol. g’ =
2
GMe 1g
9
(3R)
T = 2
g'
Since the time period of second pendulum is 2 sec.
T = 2 sec
2 = 2
9
g
1m
9
33. In the given circuit if the power rating of Zener
diode is 10 mW, the value of series resistance Rs to
regulate the input unregulated supply is :
8 V
RS
R = 1k
L
Vz = 5V
+
–
Vs
(1) 5k (2) 10
(3) 1k (4) 10k
Ans. (BONUS)
Sol.
RS
1k
Vz = 5V
V = 8V
s
Is
V1
I
Iz
Pd across Rs
V1 = 8 – 5 = 3V
Current through the load resistor
I =
3
5
1 10
= 5mA
Maximum current through Zener diode
Iz max. =
10
5
= 2mA
And minimum current through Zener diode
Iz min. = 0
Is max. = 5 + 2 = 7mA
And Rs min =
1
smax
V3k
I . 7
Similarly
Is min. = 5mA
And Rs max. =
1
smin.
V
I
=
3
5
k
3k
7
< Rs <
3
5
k
34. The reading in the ideal voltmeter (V) shown in the
given circuit diagram is :
V
5V 0.2
0.2
0.2
0.2
0.25V
0.2
0.2
0.2
5V 5V 5V
5V 5V 5V
(1) 5V (2) 10V
(3) 0 V (4) 3V
Ans. (3)
Sol. i =
eq
eq
E85
r 8 0.2
I = 25A
V = E – ir
= 5 – 0.2 × 25
= 0
35. Two identical capacitors have same capacitance C.
One of them is charged to the potential V and other
to the potential 2V. The negative ends of both are
connected together. When the positive ends are
also joined together, the decrease in energy of the
combined system is :
(1)
2
1CV
4
(2) 2 CV2
(3)
2
1CV
2
(4)
2
3CV
4
Ans. (1)
Sol. VC =
net
net
qCV 2CV
C 2C
VC =
3V
2
Loss of energy
=
2
22
1 1 1 3V
CV C(2V) 2C
2 2 2 2
=
2
CV
4
36. Two moles a monoatomic gas is mixed with six
moles of a diatomic gas. The molar specific heat of
the mixture at constant volume is :
(1)
9R
4
(2)
7R
4
(3)
3R
2
(4)
5R
2
Ans. (1)
Sol. CV =
12
1 v 2 v
12
n C n C
nn
=
5
2 R 6 R
22
26
=
9R
4
37. A ball of mass 0.5 kg is attached to a string of
length 50 cm. The ball is rotated on a horizontal
circular path about its vertical axis. The maximum
tension that the string can bear is 400 N. The
maximum possible value of angular velocity of the
ball in rad/s is,:
(1) 1600 (2) 40
(3) 1000 (4) 20
Ans. (2)
Sol. T =
2
m
400 = 0.52 × 0.5
= 40 rad/s.
38. A parallel plate capacitor has a capacitance
C = 200 pF. It is connected to 230 V ac supply
with an angular frequency 300 rad/s. The rms
value of conduction current in the circuit and
displacement current in the capacitor respectively
are :
(1) 1.38 A and 1.38 A
(2) 14.3 A and 143 A
(3) 13.8 A and 138 A
(4) 13.8 A and 13.8 A
Ans. (4)
Sol. I =
C
V
X
= 230 × 300 × 200 × 10–12 = 13.8 A
39. The pressure and volume of an ideal gas are related
as PV3/2 = K (Constant). The work done when the
gas is taken from state A (P1, V1, T1) to state
B (P2, V2, T2) is :
(1) 2(P1V1 – P2V2)
(2) 2(P2V2 – P1V1)
(3)
1 1 2 2
2( P V P V )
(4)
2 2 1 1
2(P V P V )
Ans. (1 or 2)
Sol. For PVx = constant
If work done by gas is asked then
W =
nR T
1x
Here x =
3
2
W =
2 2 1 1
P V P V
1
2
= 2(P1V1 – P2V2) ….. Option (1) is correct
If work done by external is asked then
W = –2(P1V1 – P2V2) ….. Option (2) is correct
40. A galvanometer has a resistance of 50 and it
allows maximum current of 5 mA. It can be
converted into voltmeter to measure upto 100 V by
connecting in series a resistor of resistance
(1) 5975
(2) 20050
(3) 19950
(4) 19500
Ans. (3)
Sol.
Rg
Ig
G
R
R =
g3
g
V 100
R 50
I5 10
= 20000 – 50
= 19950
41. The de Broglie wavelengths of a proton and an
particle are and 2 respectively. The ratio of the
velocities of proton and particle will be :
(1) 1 : 8
(2) 1 : 2
(3) 4 : 1
(4) 8 : 1
Ans. (4)
Sol. =
hh
p mv
v =
h
m
p
pp
vm
vm
= 4 × 2 = 8
42. 10 divisions on the main scale of a Vernier calliper
coincide with 11 divisions on the Vernier scale. If
each division on the main scale is of 5 units, the
least count of the instrument is :
(1)
1
2
(2)
10
11
(3)
50
11
(4)
5
11
Ans. (4)
Sol. 10 MSD = 11 VSD
1 VSD =
10
11
MSD
LC = 1MSD – 1VSD
= 1 MSD
10
11
MSD
=
1MSD
11
=
5
11
units
43. In series LCR circuit, the capacitance is changed
from C to 4C. To keep the resonance frequency
unchanged, the new inductance should be :
(1) reduced by
1L
4
(2) increased by 2L
(3) reduced by
3L
4
(4) increased to 4L
Ans. (3)
Sol. ’ =
11
L'C' LC
L’C’ = LC
L’(4C) = LC
L’ =
L
4
Inductance must be decreased by
3L
4
44. The radius (r), length (l) and resistance (R) of a
metal wire was measured in the laboratory as
r = (0.35 ± 0.05) cm
R = (100 ± 10) ohm
l = (15 ± 0.2) cm
The percentage error in resistivity of the material
of the wire is :
(1) 25.6% (2) 39.9%
(3) 37.3% (4) 35.6%
Ans. (2)
Sol. = R
Rr
2
Rr
=
10 0.05 0.2
2
100 0.35 15
=
1 2 1
10 7 75
= 39.9%
45. The dimensional formula of angular impulse is :
(1) [M L–2 T–1] (2) [M L2 T–2]
(3) [M L T–1] (4) [M L2 T–1]
Ans. (4)
Sol. Angular impulse = change in angular momentum.
[Angular impulse] = [Angular momentum] = [mvr]
= [M L2 T–1]
46. A simple pendulum of length 1 m has a wooden
bob of mass 1 kg. It is struck by a bullet of mass
10–2 kg moving with a speed of 2 × 102 ms–1.
The bullet gets embedded into the bob. The height
to which the bob rises before swinging back is.
(use g = 10 m/s2)
(1) 0.30 m (2) 0.20 m
(3) 0.35 m (4) 0.40 m
Ans. (2)
Sol.
mu = (M + m)V
10–2 × 2 × 102 1 × V
V 2m/s
h =
2
V
2g
= 0.2 m
47. A particle moving in a circle of radius R with
uniform speed takes time T to complete one
revolution. If this particle is projected with the
same speed at an angle to the horizontal, the
maximum height attained by it is equal to 4R. The
angle of projection is then given by :
(1)
1
22
12
2gT
sin R
(2)
1
22
12
R
sin 2gT
(3)
1
22
12
2gT
cos R
(4)
1
2
12
R
cos 2gT
Ans. (1)
M
1m
u
m
Sol.
2R
T
= V
Maximum height H =
22
v sin
2g
4R =
22 2
2
4R
sin
T 2g
sin =
2
2
2gT
R
=
1
22
12
2gT
sin R
48. Consider a block and trolley system as shown in
figure. If the coefficient of kinetic friction between
the trolley and the surface is 0.04, the acceleration
of the system in ms–2 is :
(Consider that the string is massless and
unstretchable and the pulley is also massless and
frictionless) :
(1) 3 (2) 4
(3) 2 (4) 1.2
Ans. (3)
Sol. fk = N = 0.04 × 20g = 8 Newton
a =
60 8
26
= 2m/s2
49. The minimum energy required by a hydrogen atom
in ground state to emit radiation in Balmer series is
nearly :
(1) 1.5 eV (2) 13.6 eV
(3) 1.9 eV (4) 12.1 eV
Ans. (4)
Sol. Transition from n = 1 to n = 3
E = 12.1eV
50. A monochromatic light of wavelength 6000Å is
incident on the single slit of width 0.01 mm. If the
diffraction pattern is formed at the focus of the
convex lens of focal length 20 cm, the linear width
of the central maximum is :
(1) 60 mm
(2) 24 mm
(3) 120 mm
(4) 12 mm
Ans. (2)
Sol. Linear width
W =
7
5
2 d 2 6 10 0.2
a1 10
= 2.4 × 10–2 = 24 mm
SECTION-B
51. A regular polygon of 6 sides is formed by bending
a wire of length 4 meter. If an electric current of
43
A is flowing through the sides of the
polygon, the magnetic field at the centre of the
polygon would be x × l0–7 T. The value of x is
____.
Ans. (72)
Sol.
30º30º
4
6m
B =
0I
64r
(sin 30º + sin 30º)
= 6
7
10 4 3
34
26
= 72 × 10–7T
fk
20 kg
60 N
6 kg
52. A rectangular loop of sides 12 cm and 5 cm, with
its sides parallel to the x-axis and y-axis
respectively moves with a velocity of 5 cm/s in the
positive x axis direction, in a space containing a
variable magnetic field in the positive z direction.
The field has a gradient of 10–3T/cm along the
negative x direction and it is decreasing with time
at the rate of 10–3 T/s. If the resistance of the loop
is 6 m, the power dissipated by the loop as heat is
_______ × 10–9 W.
Ans. (216)
Sol.
D
(0, 0)
C
(012, 0)
A
1
2
V = 5cm/s
y
x
B
B
B0 is the magnetic field at origin
3
2
dB 10
dx 10
0
Bx
1
B0
dB 10 dx
B – B0 = –10–1x
B =
0x
B10
Motional emf in AB = 0
Motional emf in CD = 0
Motional emf in AD = 1 =
0
Bv
Magnetic field on rod BC B
=
2
0( 12 10 )
B10
Motional emf in BC = 2 =
2
012 10
Bv
10
eq = 2 – 1 = 300 × 10–7 V
For time variation
(eq)’ = A
dB
dt
= 60 × 10–7 V
(eq)net = eq + (eq)’ = 360 × 10–7 V
Power =
2
eq net
R
= 216 × 10–9 W
53. The distance between object and its 3 times
magnified virtual image as produced by a convex
lens is 20 cm. The focal length of the lens used is
__________ cm.
Ans. (15)
Sol.
I O
Vu
v = 3u
v – u = 20 cm
2u = 20 cm
u = 10 cm
1 1 1
( 30) ( 10) f
f = 15 cm
54. Two identical charged spheres are suspended by
strings of equal lengths. The strings make an angle
with each other. When suspended in water the
angle remains the same. If density of the material
of the sphere is 1.5 g/cc, the dielectric constant of
water will be ________
(Take density of water = 1 g/cc)
Ans. (3)
Sol.
r
mg
F
/2
In air
2
2
0
Fq
tan 2 mg 4 r mg
In water
tan 2
=
2
2
0 r eff
F' q
mg' 4 r mg
Equate both equations
0g = 0r g
1
11.5
r = 3
55. The radius of a nucleus of mass number 64 is
4.8 fermi. Then the mass number of another
nucleus having radius of 4 fermi is
1000
x
, where
x is _____.
Ans. (27)
Sol. R = R0A1/3
R3 A
3
4.8 64
4A
=
3
64 1.2
A
64 1.44 1.2
A
64 1000
A1.44 1.2 x
144 12
x64
= 27
56. The identical spheres each of mass 2M are placed
at the corners of a right angled triangle with
mutually perpendicular sides equal to 4 m each.
Taking point of intersection of these two sides as
origin, the magnitude of position vector of the
centre of mass of the system is
42
x
, where the
value of x is ___________
Ans. (3)
Sol.
4m
4m
2M
2M
2M
Position vector
1 1 2 2 3 3
COM
1 2 3
m r m r m r
rm m m
COM
ˆˆ
2M 0 2M 4i 2M 4j
r6M
44
ˆˆ
rij
33
42
| r | 3
x = 3
57. A tuning fork resonates with a sonometer wire of
length 1 m stretched with a tension of 6 N. When
the tension in the wire is changed to 54 N, the
same tuning fork produces 12 beats per second
with it. The frequency of the tuning fork is
_______ Hz.
Ans. (6)
Sol. f =
1T
2L
f1 =
16
2
f2 =
1 54
2
1
2
f1
f3
f2 – f1 = 12
f1 = 6HZ
58. A plane is in level flight at constant speed and each
of its two wings has an area of 40 m2. If the speed
of the air is 180 km/h over the lower wing surface
and 252 km/h over the upper wing surface, the
mass of the plane is ________kg. (Take air density
to be 1 kg m–3 and g = 10 ms–2)
Ans. (9600)
Sol. A = 80 m2
Using Bernonlli equation
A(P2 – P1) =
22
12
1V V A
2
mg =
1
2
× 1 (702 – 502) × 80
mg = 40 × 2400
m = 9600 kg
59. The current in a conductor is expressed as
I = 3t2 + 4t3, where I is in Ampere and t is in
second. The amount of electric charge that flows
through a section of the conductor during t = 1s to
t = 2s is ____________ C.
Ans. (22)
Sol. q =
22
23
11
i dt (3t 4t )dt
q =
2
34
1
tt
q = 22C
60. A particle is moving in one dimension
(along x axis) under the action of a variable force.
It's initial position was 16 m right of origin. The
variation of its position (x) with time (t) is given as
x = –3t3 + 18t2 + 16t, where x is in m and t is in s.
The velocity of the particle when its acceleration
becomes zero is _________ m/s.
Ans. (52)
Sol. x = 3t3 + 18t2 + 16t
v = –9t2 + 36 + 16
a = – 18t + 36
a = 0 at t = 2s
v = –9(2)2 + 36 × 2 + 16
v = 52 m/s
CHEMISTRY
TEST PAPER WITH SOLUTION
SECTION-A
61. If one strand of a DNA has the sequence
ATGCTTCA, sequence of the bases in
complementary strand is:
(1) CATTAGCT (2) TACGAAGT
(3) GTACTTAC (4) ATGCGACT
Ans. (2)
Sol. Adenine base pairs with thymine with 2 hydrogen
bonds and cytosine base pairs with guanine with 3
hydrogen bonds.
62. Given below are two statements : one is labelled as
Assertion (A) and the other is labelled as Reason
(R).
Assertion (A) : Haloalkanes react with KCN to
form alkyl cyanides as a main product while with
AgCN form isocyanide as the main product.
Reason (R) : KCN and AgCN both are highly
ionic compounds.
In the light of the above statement, choose the most
appropriate answer from the options given below:
(1) (A) is correct but (R) is not correct
(2) Both (A) and (R) are correct but (R) is not the
correct explanation of (A)
(3) (A) is not correct but (R) is correct
(4) Both (A) and (R) are correct and (R) is the
correct explanation of (A)
Ans. (1)
Sol.
AgCN is mainly covalent in nature and nitrogen is
available for attack, so alkyl isocyanide is formed
as main product.
63. In acidic medium, K2Cr2O7 shows oxidising action
as represented in the half reaction
X, Y, Z and A are respectively are:
(1) 8, 6, 4 and Cr2O3 (2) 14, 7, 6 and Cr3+
(3) 8, 4, 6 and Cr2O3 (4) 14, 6, 7 and Cr3+
Ans. (4)
Sol. The balanced reaction is,
X = 14
Y = 6
A = 7
64. Which of the following reactions are
disproportionation reactions?
(A) Cu+ Cu2+ + Cu
(C) 2KMnO4 K2MnO4 + MnO2 + O2
Choose the correct answer from the options given
below:
(1) (A), (B) (2) (B), (C), (D)
(3) (A), (B), (C) (4) (A), (D)
Ans. (1)
Sol. When a particular oxidation state becomes less
stable relative to other oxidation state, one lower,
one higher, it is said to undergo disproportionation.
Cu+ Cu2+ + Cu
65. In case of isoelectronic species the size of F–, Ne
and Na+ is affected by:
(1) Principal quantum number (n)
(2) None of the factors because their size is the
same
(3) Electron-electron interaction in the outer
orbitals
(4) Nuclear charge (z)
Ans. (4)
Sol. In F–, Ne, Na+ all have 1s2, 2s2, 2p6 configuration.
They have different size due to the difference in
nuclear charge.
Hydrogen bonds
4 4 2 2
3MnO 4H 2MnO MnO 2H O
2
4 2 2
2MnO 3Mn 2H O 5MnO 4H
(D)
2
4 4 2 2
3MnO 4H 2MnO MnO 2H O
(B)
2
2 7 2
Cr O 14H 6e 2Cr 7H O
23
2 7 2
Cr O XH Ye 2A ZH O
2
(ii)
(i)
. .
(Major product)
R—NC
AgCN + R — X
(Covalent)
(Major product)
R—CN
KCN + R — X
(Ionic)
T A C G A A G T
A T G C T T C A
Complementary strand
DNA strand
66. According to the wave-particle duality of matter by
de-Broglie, which of the following graph plot
presents most appropriate relationship between
wavelength of electron () and momentum of
electron (p)?
(1)
p
(2)
p
(3)
1/p
(4)
p
Ans. (1)
Sol.
h1
pp
p = h (constant)
So, the plot is a rectangular hyperbola.
p
67. Given below are two statements:
Statement (I): A solution of [Ni(H2O)6]2+ is green
in colour.
Statement (II): A solution of [Ni(CN)4]2– is
colourless.
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 incorrect
(2) Both Statement I and Statement II are correct
(3) Statement I is incorrect but Statement II is
correct
(4) Statement I is correct but Statement II is
incorrect
Ans. (2)
Sol. [Ni(H2O)6]+2 Green colour solution due to d-d
transition.
[Ni(CN)4]–2 is diamagnetic and it is colourless.
68. Given below are two statements: one is labelled
as Assertion (A) and the other is labelled as
Reason (R).
Assertion (A) : PH3 has lower boiling point than NH3.
Reason (R) : In liquid state NH3 molecules are
associated through vander waal’s forces, but PH3
molecules are associated through hydrogen bonding.
In the light of the above statements, choose the
most appropriate answer from the options given
below:
(1) Both (A) and (R) are correct and (R) is not the
correct explanation of (A)
(2) (A) is not correct but (R) is correct
(3) Both (A) and (R) are correct but (R) is the
correct explanation of (A)
(4) (A) is correct but (R) is not correct
Ans. (4)
Sol. Unlike NH3, PH3 molecules are not associated
through hydrogen bonding in liquid state. That is
why the boiling point of PH3 is lower than NH3.
69. Identify A and B in the following sequence of
reaction
CH3
2
Cl /h
A
2
HO
373K
B
(1)
(A) =
COCl
(B) =
CHO
(2)
(A) =
CHCl2
(B) =
CHO
(3)
(A) =
CH Cl
2
(B) =
CHO
(4)
(A) =
CHCl2
(B) =
COOH
Ans. (2)
Sol.
CH3
Toluene
2
Cl /h
CHCl2
Benzal chloride
2
H O
373 K
CHO
Benzaldehyde
70. Given below are two statements:
Statement (I) : Aminobenzene and aniline are
same organic compounds.
Statement (II) : Aminobenzene and aniline are
different organic compounds.
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 correct
(2) Statement I is correct but Statement II is incorrect
(3) Statement I is incorrect but Statement II is correct
(4) Both Statement I and Statement II are incorrect
Ans. (2)
Sol. Aniline is also known as amino benzene.
71. Which of the following complex is homoleptic?
(1) [Ni(CN)4]2–
(2) [Ni(NH3)2Cl2]
(3) [Fe(NH3)4Cl2]+
(4) [Co(NH3)4Cl2]+
Ans. (1)
Sol. In Homoleptic complex all the ligand attached with
the central atom should be the same. Hence
[Ni(CN)4]2– is a homoleptic complex.
72. Which of the following compound will most easily
be attacked by an electrophile?
(1) (2)
CH3
(3)
Cl
(4)
OH
Ans. (4)
Sol. Higher the electron density in the benzene ring
more easily it will be attacked by an electrophile.
Phenol has the highest electron density amongst all
the given compound.
73. Ionic reactions with organic compounds proceed
through:
(A) Homolytic bond cleavage
(B) Heterolytic bond cleavage
(C) Free radical formation
(D) Primary free radical
(E) Secondary free radical
Choose the correct answer from the options given
below:
(1) (A) only
(2) (C) only
(3) (B) only
(4) (D) and (E) only
Ans. (3)
Sol. Heterolytic cleavage of Bond lead to formation of
ions.
74. Arrange the bonds in order of increasing ionic
character in the molecules. LiF, K2O, N2, SO2 and
CIF3.
(1) CIF3 < N2 < SO2 < K2O < LiF
(2) LiF < K2O < CIF3 < SO2 < N2
(3) N2 < SO2 < CIF3 < K2O < LiF
(4) N2 < CIF3 < SO2 < K2O < LiF
Ans. (3)
Sol. Increasing order of ionic character
N2 < SO2 < ClF3 < K2O < LiF
Ionic character depends upon difference of
electronegativity (bond polarity).
75. We have three aqueous solutions of NaCl labelled
as ‘A’, ‘B’ and ‘C’ with concentration 0.1 M,
0.01M & 0.001 M, respectively. The value of van
t’ Haft factor (i) for these solutions will be in the
order.
(1) iA < iB < iC
(2) iA < iC < iB
(3) iA = iB = iC
(4) iA > iB > iC
Ans. (1)
Sol.
Salt
Values of i (for different conc. of a Salt)
0.1 M
0.01 M
0.001 M
NaCl
1.87
1.94
1.94
i approach 2 as the solution become very dilute.
76. In Kjeldahl’s method for estimation of nitrogen,
CuSO4 acts as :
(1) Reducing agent (2) Catalytic agent
(3) Hydrolysis agent (4) Oxidising agent
Ans. (2)
Sol. Kjeldahl’s method is used for estimation of
Nitrogen where CuSO4 acts as a catalyst.
77. Given below are two statements :
Statement (I) : Potassium hydrogen phthalate is a
primary standard for standardisation of sodium
hydroxide solution.
Statement (II) : In this titration phenolphthalein
can be used as indicator.
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 correct
(2) Statement I is correct but Statement II is incorrect
(3) Statement I is incorrect but Statement II is correct
(4) Both Statement I and Statement II are incorrect
Ans. (1)
Sol. Statement (I) : Potassium hydrogen phthalate is a
primary standard for standardisation of sodium
hydroxide solution as it is economical and its
concentration does not changes with time.
Phenophthalin can acts as indicator in acid base
titration as it shows colour in pH range 8.3 to 10.1
78. Match List – I with List –II.
List – I (Reactions)
List – II (Reagents)
(A)
CH (CH ) –C–OC HCH (CH ) CHO
3 2 5 2 5 3 2 5
O
(I)
CH3MgBr, H2O
(B)
C6H5COC6H5C6H5CH2C6H5
(II)
Zn(Hg) and conc. HCl
(C)
C6H5CHOC6H5CH(OH)CH3
(III)
NaBH4 , H+
(D)
CH COCHCOOC H CH C(OH)CHCOOC H
3 2 2 5 3 2 2 5
H
(IV)
DIBAL-H, H2O
Choose the correct answer from options given
below:
(1) A-(III), (B)-(IV), (C)-(I), (D)-(II)
(2) A-(IV), (B)-(II), (C)-(I), (D)-(III)
(3) A-(IV), (B)-(II), (C)-(III), (D)-(I)
(4) A-(III), (B)-(IV), (C)-(II), (D)-(I)
Ans. (2)
Sol.
2
DIBAL H, H O
3 2 5 2 5 3 2 5
CH (CH ) COOC H CH (CH ) CHO
Zn(Hg) & conc. HCl
6 5 6 5 6 5 2 6 5
C H COC H C H CH C H
3
2
CH MgBr
6 5 6 5 3
HO
C H CHO C H CH OH CH
4
NaBH , H
3 2 2 5 3 2 2 5
CH COCH COOC H CH CH(OH)CH COOC H
79. Choose the correct option for free expansion of an
ideal gas under adiabatic condition from the
following :
(1) q = 0, T 0, w = 0
(2) q = 0, T < 0, w 0
(3) q 0, T = 0, w =0
(4) q = 0, T = 0, w =0
Ans. (4)
Sol. During free expansion of an ideal gas under
adiabatic condition q = 0, T = 0, w =0.
80. Given below are two statements:
Statement (I) : The NH2 group in Aniline is ortho
and para directing and a powerful activating group.
Statement (II) : Aniline does not undergo Friedel-
Craft’s reaction (alkylation and acylation).
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 correct
(2) Both Statement I and Statement II are incorrect
(3) Statement I is incorrect but Statement II is correct
(4) Statement I is correct but Statement II is incorrect
Ans. (1)
Sol. The NH2 group in Aniline is ortho and para
directing and a powerful activating group as NH2
has strong +M effect.
Aniline does not undergo Friedel-Craft’s reaction
(alkylation and acylation) as Aniline will form
complex with AlCl3 which will deactivate the
benzene ring.
SECTION-B
81. Number of optical isomers possible for
2 – chlorobutane ………
Ans. (2)
Sol.
Cl
*
There is one chiral centre present in given
compound.
So, Total optical isomers = 2
Cl
and
Cl
82. The potential for the given half cell at 298K is
(-)………… × 10-2 V.
2H+
(aq) + 2e- H2 (g)
[H+] = 1M,
2
H
P
= 2 atm
(Given: 2.303 RT/F = 0.06 V, log2 = 0.3)
Ans. (1)
Sol.
2
2
H
o
2
H /H
P
0.06
E = E log
2[H ]
2
0.06 2
E = 0.00 log
2[1]
2
E = 0.03 0.3 0.9 10 V
83. The number of white coloured salts among the
following is …………….
(A) SrSO4 (B) Mg(NH4)PO4 (c) BaCrO4
(D) Mn(OH)2 (E) PbSO4 (F) PbCrO4
(G) AgBr (H) PbI2 (I) CaC2O4
(J) [Fe(OH)2(CH3COO)]
Ans. (5)
Sol. SrSO4 – white
Mg(NH4)PO4 – white
BaCrO4 – yellow
Mn(OH)2 – white
PbSO4 – white
PbCrO4 – yellow
AgBr – pale yellow
PbI2 – yellow
CaC2O4 – white
[Fe(OH)2(CH3COO)] – Brown Red
84. The ratio of
14
12
C
C
in a piece of wood is
1
8
part that
of atmosphere. If half life of 14C is 5730 years, the
age of wood sample is ….. years.
Ans. (17190)
Sol.
14 12
atmosphere
14 12
wood sample
( C / C)
t ln ( C / C)
As per the question,
14 12
wood
14 12
atmosphere
( C / C) 1
8
( C / C)
So,
t ln8
1/2
ln 2 t ln8
t
t = 3 × t1/2 = 17190 years
85. The number of molecules/ion/s having trigonal
bipyramidal shape is …….. .
PF5, BrF5, PCl5, [PtCl4]2–, BF3, Fe(CO)5
Ans. (3)
Sol. PF5, PCl5, Fe(CO)5 ; Trigonal bipyramidal
BrF5 ; square pyramidal
[PtCl4]- 2 ; square planar
BF3 ; Trigonal planar
86. Total number of deactivating groups in aromatic
electrophilic substitution reaction among the
following is
O
OCH ,
3
O
CH ,
3
–N
H
CH ,
3
–N
H
–C N, –OCH3
Ans. (2)
Sol.
O
OCH ,
3
O
CH ,
3
–N
H
CH ,
3
–N
H
–C N, –OCH3
(–M group) (+M group) (+M group) (–M group) (+M group)
87. Lowest Oxidation number of an atom in a
compound A2B is -2. The number of an electron in
its valence shell is
Ans. (6)
Sol. A2B 2A+ + B– 2, B-2 has complete octet in its di-
anionic form, thus in its atomic state it has 6
electrons in its valence shell. As it has negative
charge, it has acquired two electrons to complete
its octet.
88. Among the following oxide of p - block elements,
number of oxides having amphoteric nature is
Cl2O7, CO, PbO2, N2O, NO, Al2O3, SiO2, N2O5,
SnO2
Ans. (3)
Sol. Acidic oxide: Cl2O7, SiO2, N2O5
Neutral oxide: CO, NO, N2O
Amphoteric oxide: Al2O3 , SnO2 , PbO2
89. Consider the following reaction:
3PbCl2 + 2(NH4)3PO4 Pb3(PO4)2 + 6NH4Cl
If 72 mmol of PbCl2 is mixed with 50 mmol of
(NH4)3PO4 , then amount of Pb3(PO4)2 formed is
…… mmol. (nearest integer)
Ans. (24)
Sol. Limiting Reagent is PbCl2
mmol of Pb3(PO4)2 formed
=
2
mmol of PbCl reacted
3
= 24 mmol
90. Ka for CH3COOH is 1.8 × 10– 5 and Kb for NH4OH
is 1.8 × 10–5. The pH of ammonium acetate
solution will be
Ans. (7)
Sol.
w a b
pK pK pK
pH 2
pKa = pKb
w
pK
pH 7
2
