FINAL JEE–MAIN EXAMINATION – APRIL, 2024
(Held On Friday 05th April, 2024) TIME : 3 : 00 PM to 6 : 00 PM
MATHEMATICS
TEST PAPER WITH SOLUTION
SECTION-A
1. Let ƒ: [–1, 2] R be given by
ƒ(x) = 2x2 + x + [x2] – [x], where [t] denotes the
greatest integer less than or equal to t. The number
of points, where ƒ is not continuous, is :
(1) 6 (2) 3
(3) 4 (4)5
Ans. (3)
Sol. Doubtful points : –1, 0, 1,
2
,
3
, 2
at x =
2
,
3
22
Cont.
Cont.
f(x) 2x x [x] x Discount
at x = –1 :
RHL f(x) (2 1 ( 1)) 0 2 Dis.
f( 1) 2 1 ( 1) 1 3
at x = 2 :
LHL f(x) 8 2 1 3 12 Cont.
f(2) 8 2 2 4 12
at x = 0 :
LHL 0 0 ( 1) 0 1 Dis.
f(0) 0
at x = 1
LHL 2 1 0 0 3
f(1) 3 1 1 3 Cont.
RHL 2 1 1 1 3
2. The differential equation of the family of circles
passing the origin and having center at the line
y = x is :
(1) (x2 – y2 + 2xy)dx = (x2 – y2 + 2xy)dy
(2) (x2 + y2 + 2xy)dx = (x2 + y2 – 2xy)dy
(3) (x2 – y2 + 2xy)dx = (x2 – y2 – 2xy)dy
(4) (x2 + y2 – 2xy)dx = (x2 + y2 + 2xy)dy
Ans. (3)
Sol. C x2 + y2 + gx + gy = 0 .....(1)
2x + 2yy' + g + gy' = 0
2x 2yy '
g1 y'
Put in (1)
222x 2yy'
x y (x y) 0
1 y'
(x2 – y2 – 2xy)y' = x2 – y2 + 2xy
3. Let S1 = {z C : |z| 5},
2
z 1 3 i
S z C : Im 0
1 3 i
and
S3 = {z C : Re (z) }. Then
(1)
125
6
(2)
125
24
(3)
125
4
(4)
125
12
Ans. (4)
Sol. S1 : x2 + y2 25 .....(1)
S2 : lm of
z 1 3 i
0
1 3 i
lm of
x iy 10
1 3 i
lm of
x iy 1 3 i
0
4
3 x y 0
......(2)
S3 : x 0 ......(3)
Area =
2
5(5)
12
60°
4. The area enclosed between the curves y = x|x| and
y = x – |x| is :
(1)
8
3
(2)
2
3
(3) 1 (4)
4
3
Ans. (4)
Sol.
y = 2x
y = x2
y = 0
(0,0)
(–2,–4)
y = –x2
0
2
2
A x 2x
4
3
5. 60 words can be made using all the letters of the
word BHBJO, with or without meaning. If these
words are written as in a dictionary, then the 50th
word is :
(1) OBBHJ (2) HBBJO
(3) OBBJH (4) JBBOH
Ans. (3)
Sol. B B H J O
B _____ 4! 24
4!
H _____ 12
2!
4!
J _____ 12
2!
O B B H J
O B B J H 50th rank
6. Let
ˆˆ
ˆ
a 2i 5j k
,
ˆˆˆ
b 2i 2j 2k
and
c
be three vectors such that
ˆ ˆ ˆ
c i a b i a c i
.
a.c 29
,
then
ˆˆˆ
c. 2i j k
is equal to :
(1) 10 (2) 5
(3) 15 (4) 12
Ans. (2)
Sol. Let's assume
ˆ
v a b i
ˆˆ
ˆ
5i 3j k
and
ˆ
c i p
So,
p v a p
p v p a 0
p v a 0
p v a
ˆˆ
c i 7i 8j
ˆ ˆ ˆ
a.c a.i a. 7i 8j
29 2 14 40
1
2
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ
c. 2i j k i. 2i j k 7i 8j . 2i j k
114 8 2 5
2
7. Consider three vectors
a, b,c
. Let
a 2, b 3
and
a b c
. If
0, 3
is the angle between
the vectors
b
and
c
, then the minimum value of
2
27 c a
is equal to :
(1) 110 (2) 105
(3) 124 (4) 121
Ans. (3)
Sol.
22
c a c a 2a.c
2
c 4 0
a b c
a b c
2 3 c sin
2
c cosec
3
0, 3
min
22
c33
2
cosec ,
3
2
min
16
27 c a 27 4 124
27
8. Let A(–1, 1) and B(2, 3) be two points and P be a
variable point above the line AB such that the area
of PAB is 10. If the locus of P is ax + by = 15,
then 5a + 2b is :
(1)
12
5
(2)
6
5
(3) 4 (4) 6
Ans. (1)
Sol.
P(h,k)
B(2,3)
A(–1,1)
h k 1
11 1 1 10
22 3 1
–2x + 3y = 25
69
x y 15
55
a =
6
5
, b =
9
5
5a = –6, 2b =
18
5
9. Let (, , ) be the point (8, 5, 7) in the line
x 1 y 1 z 2
2 3 5
. Then + + is equal to
(1) 16 (2) 18
(3) 14 (4) 20
Ans. (3)
Sol.
(2 +1, 3 – 1, 5 + 2)
M
A'
A
(8, 5, 7)
ˆˆˆ
AM. 2i 3j 5k 0
(2 – 7)(2) + (3 – 6)(3) + (5 – 5)(5) = 0
38 = 57
3
2
7 19
M 4, ,
22
A'(0,2,12)
10. If the constant term in the expansion of
12
5
3
3 2x
x5
, x 0, is ×28×
53
, then 25 is
equal to :
(1) 639 (2) 724
(3) 693 (4) 742
Ans. (3)
Sol.
12 r r
1/5
12
r 1 r 1/3
3 2x
TC
x5
12 r r 2r 12
12 5
r
r1 r /3
C 3 2 x
T
5
r = 6
6/5 6
12
68 1/5
72
C 3 2 9 11 7
T 2 .3
25
5
25 = 693
11. Let ƒ, g : R R be defined as : ƒ(x) = |x – 1| and
x
e , x 0
g(x) x 1, x 0
. Then the function ƒ(g(x)) is
(1) neither one-one nor onto.
(2) one-one but not onto.
(3) both one-one and onto.
(4) onto but not one-one.
Ans. (1)
Sol. f(g(x)) = |g(x) – 1|
x
e 1 x 0
fog
x 1 1 x 0
x
e 1 x 0
fog x x 0
12. Let the circle C1 : x2 + y2 – 2(x + y) + 1 = 0 and C2
be a circle having centre at (–1, 0) and radius 2. If
the line of the common chord of C1 and C2
intersects the y-axis at the point P, then the square
of the distance of P from the centre of C1 is :
(1) 2 (2) 1
(3) 6 (4) 4
Ans. (1)
Sol. S1 : x2 + y2 – 2x – 2y + 1 = 0
S2 : x2 + y2 + 2x – 3 = 0
Common chord = S1 – S2 = 0
–4x – 2y + 4 = 0
2x + y = 2 P(0, 2)
2
(c,p)
d
= (1 – 0)2 + (2 – 1)2 = 2
13. Let the set S = {2, 4, 8, 16, ....., 512} be partitioned
into 3 sets A, B, C with equal number of elements such
that A B C = S and A B = B C = A C = .
The maximum number of such possible partitions of S
is equal to :
(1) 1680 (2) 1520
(3) 1710 (4) 1640
Ans. (1)
Sol.
A
B
C
9! 3!
3!3!3! 3!
14. The values of m, n, for which the system of
equations
x + y + z = 4,
2x + 5y + 5z = 17,
x + 2y + mz = n
has infinitely many solutions, satisfy the equation :
(1) m2 + n2 – m – n = 46
(2) m2 + n2 + m + n = 64
(3) m2 + n2 + mn = 68
(4) m2 + n2 – mn = 39
Ans. (4)
Sol.
1 1 1
D 2 5 5 0
1 2 m
m = 2
3
1 1 4
D 2 5 17 0
1 2 n
n = 7
15. The coefficients a, b, c in the quadratic equation
ax2 + bx + c = 0 are from the set {1, 2, 3, 4, 5, 6}.
If the probability of this equation having one real
root bigger than the other is p, then 216p equals :
(1) 57 (2) 38
(3) 19 (4) 76
Ans. (2)
Sol. D > 0
b2 > 4ac
b = 3 : (a, c) = (1, 1)(1, 2)(2,1)
b = 4 : (a, c) = (1, 1)(1, 2)(2,1)(1,3)(3,1)
b = 5 : (a, c) = (1,1)(1,2)(2,1) (1,3)(3,1)(1,4)(4,1)
(1,5)(5,1)(1,6)(6,1)(2,3)(3,2)(2,2)
b = 6 : (a, c) = (1,1)(1,2)(2,1) (1,3)(3,1)(1,4)(4,1)
(1,5)(5,1)(1,6)(6,1)(2,3)(3,2)(2,4)(4,2)(2,2)
fav. cases = 38
Prob. :
38
666
16. Let ABCD and AEFG be squares of side 4 and
2 units, respectively. The point E is on the line
segment AB and the point F is on the diagonal AC.
Then the radius r of the circle passing through the
point F and touching the line segments BC and CD
satisfies :
(1) r = 1 (2) r2 – 8r + 8 = 0
(3) 2r2 – 4r + 1 = 0 (4) 2r2 – 8r + 7 = 0
Ans. (2)
Sol.
A
B
E
C
D
(0, 4)
(4, 4)
(4, 0)
(2, 0)
(0, 0)
F(2,2)
r
r
r
o
x = 4
F
(2,2)
y = 4
(4–r, 4–r )
OF2 = r2
(2 – r)2 + (2 – r)2 = r2
r2 – 8r + 8 = 0
17. Let (m, n) =
1
m 1 n 1
0
x (1 x) dx
, m, n > 0. If
1
10 20
0
(1 x ) dx a (b,c)
, then 100(a + b + x)
equals ______.
(1) 1021 (2) 1120
(3) 2012 (4) 2120
Ans. (4)
Sol.
120
10
0
I 1. 1 x dx
x10 = t
x = t1/10
9/10
1
dx t dt
10
19/10
20
0
1
I (1 t) t dt
10
120
9/10
0
1
I t 1 t dt
10
a =
1
10
b =
1
10
c = 21
18. Let 0 and
3
A
2
.
If
3 9 3
B 7 2
2 5 2
is the matrix of cofactors
of the elements of A, then det(AB) is equal to :
(1) 343 (2) 125
(3) 64 (4) 216
Ans. (4)
Sol. Equating co-factor fo A21
(22 – 3) =
= 0, 2 (accept)
Now, 22 – = 3
= 2 = 1
|AB| = |A cof (A)| = |A|3
1 2 3
A 2 2 1 6 2(9) 3(6) 6
1 2 4
19. If
2cos cos2
y( ) cos3 4 cos2 5cos 2
,
then at
2
, y" + y' + y is equal to:
(1)
2
(2) 1
(3)
1
2
(4) 2
Ans. (4)
Sol.
2
32
2cos 2 cos 1
y4cos 3cos 8cos 4 5cos 2
2
2
2cos 2cos 1
y
2cos 2cos 1 2cos 2
11
y2 1 cos
2
y =
1
2
2
11
y' sin
21 cos
2
y =
1
2
2
4
cos 1 cos sin (2)(1 cos )( sin )
1
y" 2(1 cos )
2
y = 1
20. For x 0, the least value of K, for which 41+x + 41–x,
K
2
, 16x + 16–x are three consecutive terms of an
A.P. is equal to :
(1) 10 (2) 4
(3) 8 (4) 16
Ans. (1)
Sol.
x 2x
x 2x
11
4 4 4
k44
22
k 10
SECTION-B
21. Let the mean and the standard deviation of the
probability distribution
X
1
0
–3
P(X)
1
3
K
1
6
1
4
be and , respectively. If – = 2, then + is
equal to _____.
Ans. (5)
Sol.
1 1 1
k1
3 6 4
1
k4
13
–
3 4 4
1
–
32
2
21 1 1 1
9 – –
3 4 4 3 2
2
29
9 3 4
= + 2
22
2
22 9 9
29 3 4 9 4
22
–0
93
= 0, (reject) or = 6
(
x0
is already given)
+ = 2 + 2
= 5
22. Let y = y(x) be the solution of the differential
equation
2
1
1x
2
2
dy 2x y xe
dx 1x
; y(0) = 0.
Then the area enclosed by the curve
2
1
1x
ƒ(x) y(x)e
and the line y – x = 4 is ____.
Ans. (18)
Sol. IF =
2
22
2x –1
dx
1x 1x
ee
2 2 2
–1 1 –1 dx
1 x 1 x 1 x
y e x e e
2
–1 2
1x x
y e c
2
(0, 0)
C0
2
1
2
1x
x
y(x) e
2
2
x
f(x) 2
(4, 8)
(–2, 2)
y = x2/2
y = x/4
42
–2
x
A (x 4) dx 18
2
23. The number of solutions of
sin2x + (2 + 2x – x2)sinx – 3(x – 1)2 = 0, where
– x , is
Ans. (2)
Sol. sin2x – (x2 – 2x – 2)sinx – 3(x – 1) 2 = 0
sin2x – (x – 1)2)sinx – 3(x–1)2 = 0
roots :
–3
(x–1)2
sinx = –3 (reject) or (x – 1)2
sinx = (x – 1)2
(1,0)
24. Let the point (–1, , ) lie on the line of the
shortest distance between the lines
x 2 y 2 z 5
3 4 2
and
x 2 y 6 z 1
1 2 0
.
Then ( – )2 is equal to _____.
Ans. (25)
Sol.
(–2,–6,1)
Q
(–1,,)
(–2,2,5)
P(–3– 2, 4+ 2, 2+ 5)
Q(– – 2, 2 – 6, 1)
DRS of PQ = (3 – , 2 – 4 – 8, –2 – 4)
DRS of PQ =
ˆˆ
ˆ
i j k
–1 2 0
–3 4 2
ˆˆˆ
4i 2 j 2k
OR
(2, 1, 1)
– 4 – 8 –2 – 4
2 1 1
= + 2 & 7 = – 8
–1
1
Q : (–3, –4, 1)
LPQ =
x 3 y 4 z 1
2 1 1
(–1, , ) 1 =
41
11
= –3, = 2
( – )2 = 25
25. If
3 2 5 2 6 9 3 11 2 49 20 6
1 ....
18 180
2 3 36 3
upto =
e
ba
2 1 log
ab
, where a and b are
integers with gcd(a, b) = 1, then 11a + 18b is equal
to ______.
Ans. (76)
Sol.
2 3 4
x x x x
S 1 ....
18 180
2 3 36 3
Put
xt,
3
where
x 3 – 2
2 3 4
t t t t
S 1 ....
2 6 12 20
2 3 4
1 1 1 1 1 1 1
S 1 t 1 – t – t – t –
2 2 3 3 4 4 5
2 3 3 2 3 4
t t t t t t t
S 1 t ... – ...
2 3 4 2 3 4 5
2 2 3
t 1 t t
S t ... – t ... 2
2 t 2 3
11
S 2 1 – – log 1 – t – 1 log 1 – t 2
tt
3 3 – 2
S 2 – 1 log 1 –
3 – 2 3
22
S 2 loge
3 – 2 3
6 2 2 3 2
S 2 loge 2 1 loge
2 3 2 3
a = 2, b = 3
11a + 18b = 11 × 2 +18 × 3 = 76
26. Let a > 0 be a root of the equation 2x2 + x – 2 = 0.
If
2
2
1
xa
16 1 cos(2 x 2x )
lim 17
(1 ax )
, where
, Z then + is equal to ______.
Ans. (170)
Sol.
2x2 + x – 2 = 0
a
b
2x2 – x – 2 = 0
2
22
1
x2
a
11 1
1 – cos2 x – x – 4 x –
ab b
lim16
11
4 x a x –
ba
2
2
2 1 1
16 –
ab
a
2 2 2
32 17 17.8 17 8 16
4
aa
–1 117
136.16 18 2 7
18.2 7 18 2 7
136 18 2 7 16
256
153 17 17 17
+ = 153 + 17 = 170
27. If
22
0
2xdx
ƒ(t) 1 cos t sin x
, 0 < t < , then the value
of
2
2
0
dt
ƒ(t)
equals _____.
Ans. (1)
Sol.
22
0
2x
f(t) dx
1 – cos t sin x
.....(1)
22
0
( – x)dx
21 – cos t sin x
....(2)
22
0
2f(t) 2 dx
1 – cos t sin x
22
0
f(t) dx
1 – cos t sin x
divide & by cos2x
2
2 2 2
0
sec xdx
f(t) sec x – cos t tan x
/2 2
2 2 2
0
sec xdx
f(t) 2 sec x – cos t tan x
tanx = z
sec2xdx = dz
22
0
dz
f(t) 2 1 sin t z
2
sin t
Then
/2 2
0
dt
f(t)
/2
0
sin t dt
= 1
28. Let the maximum and minimum values of
22
2
8x x 12 4 x 7
, x R be M and m
respectively. Then M2 – m2 is equal to _____.
Ans. (1600)
Sol. (x – 7)2 + (y – 4)2
2
y 8x – x – 12
22
y –(x – 4) 16 –12
22
x – 4 y 4
P(7,4)
A
(2, 0)
B
C(4, 0)
(6, 0)
2
m = 9
M = 41
M2 – m2 = 412 – 92 = 1600
29. Let a line perpendicular to the line 2x – y = 10
touch the parabola y2 = 4(x – 9) at the point P. The
distance of the point P from the centre of the circle
x2 + y2 – 14x – 8y + 56 = 0 is ______.
Ans. (10)
Sol. y2 = 4(x – 9)
slope of tangent =
–1
2
Point of contact
2
1 2 1
P 9 , –1
1
–2
2
P(13, –4)
center of circle C(7, 4)
distance
2
2
CP 13 – 7 –4 – 4
= 10
30. The number of real solutions of the equation
x |x + 5| + 2|x + 7| – 2 = 0 is _____.
Ans. (3)
30. The number of real solutions of the equation
x |x + 5| + 2|x + 7| – 2 = 0 is _____.
Allen Ans. (3)
Sol. Case I : x –5
x2 + 5x + 2x + 12 = 0
x2 + 7x + 12 = 0
x = –3, –4
Case II : –7 < x < –5
–x2 – 5x + 2x + 14 – 2 = 0
–x2 – 3x + 12 = 0
–3 9 48
x2
–3 57
2
–3 – 57 –3 57
x , (rejected)
22
Case III : x –7
–x2 – 5x – 2x – 14 – 2 = 0
x2 + 7x + 16 = 0
D = 49 – 64 < 0
No solutions
No. of solutions = 3
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
31. Given below are two statements :
Statement I : When the white light passed through
a prism, the red light bends lesser than yellow and
violet.
Statement II : The refractive indices are different
for different wavelengths in dispersive medium. In
the light of the above statements, choose the
correct answer from the options given below :
(1) Both Statement I and Statement II are true.
(2) Statement I is true but Statement II is false.
(3) Both Statement I and Statement II are false.
(4) Statement I is false but Statement II is true.
Ans. (1)
Sol. As red > yellow > violet
Light ray with longer wavelength bends less.
32. Which of the following statement is not true about
stopping potential (V0)?
(1) It depends on the nature of emitter material.
(2) It depends upon frequency of the incident light.
(3) It increases with increase in intensity of the
incident light.
(4) It is 1/e times the maximum kinetic energy of
electrons emitted.
Ans. (3)
Sol. KEmax = h – 0 = eV
33. The angular momentum of an electron in a
hydrogen atom is proportional to : (Where r is the
radius of orbit of electron)
(1)
r
(2)
1
r
(3) r (4)
1
r
Ans. (1)
Sol.
2
Cmv
Fr
2
12
2
Kq q mv
r
r
mv2r2 = Kq1q2r
2
12
LKq q r
m
Lr
34. A galvanometer of resistance 100 when
connected in series with 400 measures a voltage
of upto 10 V. The value of resistance required to
convert the galvanometer into ammeter to read
upto 10 A is x × 10–2 . The value of x is :
(1) 2 (2) 800
(3) 20 (4) 200
Ans. (3)
Sol.
3
g10
i 20 10 A
400 100
For ammeter
Let shunt resistance = S
igR = (i – ig)S
20 × 10–3 × 100 = 10 S
S = 20 × 10–2
35. The vehicles carrying inflammable fluids usually
have metallic chains touching the ground :
(1) To conduct excess charge due to air friction to
ground and prevent sparking.
(2) To alert other vehicles.
(3) To protect tyres from catching dirt from
ground.
(4) It is a custom.
Ans. (1)
Sol. Static charge is developed due to air friction. This
can result in combustion. So, metallic chains is
used to discharge excess charge.
36. If n is the number density and d is the diameter of
the molecule, then the average distance covered by
a molecule between two successive collisions (i.e.
mean free path) is represented by :
(1)
2
1
2n d
(2)
2
2n d
(3)
2
1
2n d
(4)
2 2 2
1
2n d
Ans. (3)
Sol. n = number of molecule per unit volume
d = diameter of the molecule
2
1
2 d n
(By Theory)
37. A particle moves in x-y plane under the influence
of a force
F
such that its linear momentum is
ˆˆ
P(t) i cos(kt) jsin(kt)
. If k is constant, the
angle between
F
and
P
will be :
(1)
2
(2)
6
(3)
4
(4)
3
Ans. (1)
Sol.
ˆˆ
P cos(kt)i sin(kt)j
;
P1
P mv
ˆˆ
Pv
ˆˆ
ˆ
v cos(kt)i sin(kt)j
ˆˆ
ksin(kt)i kcos(kt)j
ˆ
ak
ˆˆ
ˆ
a sinkti cosktj
ˆˆ
ˆˆ
F a sin kt i coskt j
ˆˆ
F.P sin kt cos t sin kt cos t
cos 0
ˆˆ 11
FP
2
38. The electrostatic force
1
(F )
and magnetic force
2
(F )
acting on a charge q moving with velocity v
can be written :
(1)
1
F qV.E
,
2
F q(B.V)
(2)
1
F qB
,
2
F q(B V)
(3)
1
F qE
,
2
F q(V B)
(4)
1
F qE
,
2
F q(B V)
Ans. (3)
Sol.
1
F qE
(Theory)
2
F q(V B)
39. A man carrying a monkey on his shoulder does
cycling smoothly on a circular track of radius 9m
and completes 120 revolutions in 3 minutes. The
magnitude of centripetal acceleration of monkey is
(in m/s2) :
(1) zero (2) 16 2 ms–2
(3) 42 ms–2 (4) 57600 2 ms–2
Ans. (2)
Sol. Given : R = 9m,
120 revolution in 3 min
120Rev. 120 2 rad 4 rad / s
3min. 3 60 sec 3
acentripetal = 2R =
2
49
3
= 162 m/s2
40. A series LCR circuit is subjected to an AC signal
of 200 V, 50 Hz. If the voltage across the inductor
(L = 10 mH) is 31.4 V, then the current in this
circuit is ________ :
(1) 68 A (2) 63 A
(3) 10 A (4) 10 mA
Ans. (3)
Sol. Voltage across inductor VL = IXL
31.4 = I[L]
31.4 = I[L(2f)]
31.4 = I[10 × 10–3(2 × 3.14) × 50
I = 10 A
41. What is the dimensional formula of ab–1 in the
equation
2
a
PV
(V – b) = RT, where letters
have their usual meaning.
(1) [M0L3T–2] (2) [ML2T–2]
(3) [M–1L5T3] (4) [M6L7T4]
Ans. (2)
Sol. [V] = [b]
Dimension of b = [L3]
&
2
a
[P] V
[a] = [PV2] = [ML–1T–2][L6]
Dimension of a = [ML5T–2]
52
1 2 2
3
[ML T ]
ab [ML T ]
[L ]
42. The output (Y) of logic circuit given below is 0
only when :
A
B
1
Y
(1) A = 1, B = 0 (2) A = 0, B = 0
(3) A = 1, B = 1 (4) A = 0, B = 1
Ans. (2)
Sol.
A
B
1
Y
0
0
0
0
0
0
0
0
OR gate
OR gate
AND gate
43. A body is moving unidirectionally under the
influence of a constant power source. Its
displacement in time t is proportional to :
(1) t2 (2) t2/3
(3) t3/2 (4) t
Ans. (3)
Sol. P = costant FV = constant
dV
mV
dt
= constant
Vt
00
VdV (C) dt
2
VCt
2
V t1/2
1/2
ds t
dt
St
1/2
00
ds K t dt
3/2
2
S K t
3
S t3/2
displacement is proportional to (t)3/2
44. Match List-I with List-II :-
List-I
EM-Wave
List-II
Wavelength
Range
(A)
Infra-red
(I)
< 10–3 nm
(B)
Ultraviolet
(II)
400 nm to 1 nm
(C)
X-rays
(III)
1 mm to 700 nm
(D)
Gamma rays
(IV)
1 nm to 10–3 nm
Choose the correct answer from the options given
below :
(1) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
(2) (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
(3) (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
(4) (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
Ans. (2)
Sol. Infrared is the least energetic thus having biggest
wavelength () & gamma rays are most energetic
thus having smallest wavelength ().
45. During an adiabatic process, if the pressure of a
gas is found to be proportional to the cube of its
absolute temperature, then the ratio of
P
V
C
C
for the
gas is :
(1)
5
3
(2)
9
7
(3)
3
2
(4)
7
5
Ans. (3)
Sol. P T3
PT–3 = constant
PV nR
T
= constant from ideal gas equation
(P) (PV)–3 = constant
P–2 V–3 = cosntant ...(1)
Process equation for adiabatic process is
PVy = constant ...(2)
Comparing equation (1) and (2)
P
V
C3
y
C2
46. Match List-I with List-II :
List-I
List-II
(A)
A force that
restores an
elastic body of
unit area to its
original state
(I)
Bulk modulus
(B)
Two equal and
opposite forces
parallel to
opposite faces
(II)
Young's modulus
(C)
Forces
perpendicular
everywhere to
the surface per
unit area same
everywhere
(III)
Stress
(D)
Two equal and
opposite forces
perpendicular to
opposite faces
(IV)
Shear modulus
Choose the correct answer from the options given
below :
(1) (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
(2) (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
(3) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(4) (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
Ans. (3)
Sol. (A)
restoring
F
stress A
If A = 1
Stress = Frestoring
(A)-(III)
(B)
Shear stress
(B)-(IV)
(C)
Volumetric stress
(C)-(I)
(D)
Longitudinal stress
(D)-(II)
47. A vernier callipers has 20 divisions on the vernier
scale, which coincides with 19th division on the
main scale. The least count of the instrument is
0.1 mm. One main scale division is equal to
____mm.
(1) 1 (2) 0.5
(3) 2 (4) 5
Ans. (3)
Sol. 20 VSD = 19 MSD
19
1VSD MSD
20
L.C. = 1 MSD – 1 VSD
0.1 mm =
19
1MSD MSD
20
1
0.1 MSD
20
1 MSD = 2mm
48. A heavy box of mass 50 kg is moving on a
horizontal surface. If co-efficient of kinetic friction
between the box and horizontal surface is 0.3 then
force of kinetic friction is :
(1) 14.7 N
(2) 147 N
(3) 1.47 N
(4) 1470 N
Ans. (2)
Sol.
50kg
v
µk = 0.3
Fk = µkN = 0.3 × 50 × 9.8 = 147 N
49. A satellite revolving around a planet in stationary
orbit has time period 6 hours. The mass of planet is
one-fourth the mass of earth. The radius orbit of
planet is : (Given = Radius of geo-stationary orbit
for earth is 4.2 × 104 km)
(1) 1.4 × 104 km
(2) 8.4 × 104 km
(3) 1.68 × 105 km
(4) 1.05 × 104 km
Ans. (4)
Sol.
3/2
2r
TGM
3/2 1/2
1 1 2
2 2 1
T r M
T r M
1/2
3/2
14 3/2
(r )
6M
24 M / 4
(4.2 10 )
r1 = 1.05 × 104 km
50. The ratio of heat dissipated per second through the
resistance 5 and 10 in the circuit given below
is :
20
5
10
10V
(1) 1 : 2 (2) 2 : 1
(3) 4 : 1 (4) 1 : 1
Ans. (2)
Sol.
20
5
10
10V
i1
i2
1
2
i10 2
i 5 1
2
2
1 1 1
2
222
P i R 2 5 2
P 1 10 1
iR
SECTION-B
51. A solenoid of length 0.5 m has a radius of 1 cm
and is made up of 'm' number of turns. It carries a
current of 5A. If the magnitude of the magnetic
field inside the solenoid is 6.28 × 10–3 T, then the
value of m is :
Ans. (500)
Sol. µ0ni = B n = number of turns per unit length
0m
µ i B
3
7
0
B. 6.28 10 0.5
mµi 12.56 10 5
m = 500
52. The shortest wavelength of the spectral lines in the
Lyman series of hydrogen spectrum is 915 Å. The
longest wavelength of spectral lines in the Balmer
series will be _______ Å.
Ans. (6588)
Sol. Lyman Series
n = L
Shortest,
22
12
hc 1 1
13.6 nn
E ;
0
hc 13.6 1
Balmer Series :
n = 2
n = 3
22
1
hc 1 1
13.6 23
1
hc 1 1
13.6 49
1
hc 5
13.6 36
0
1
13.6 5
13.6 36
0
136
5
=
915 36 6588
5
53. In a single slit experiment, a parallel beam of green
light of wavelength 550 nm passes through a slit of
width 0.20 mm. The transmitted light is collected
on a screen 100 cm away. The distance of first
order minima from the central maximum will be
x × 10–5 m. The value of x is :
Ans. (275)
Sol.
y
100cm
= 550 nm
d = 0.2 mm
92
3
D 550 10 100 10
yd0.2 10
= 275
54. A sonometer wire of resonating length 90 cm has a
fundamental frequency of 400 Hz when kept under
some tension. The resonating length of the wire
with fundamental frequency of 600 Hz under same
tension _______ cm.
Ans. (60)
Sol.
L
/2
f0 = 400 Hz ;
T
vµ
= constant
L
2
; v = f0
0
vL
2f
v = 2Lf0
0
2Lf
v
L' 2f ' 2f '
0
Lf 90 400 60
f ' 600
55. A hollow sphere is rolling on a plane surface about
its axis of symmetry. The ratio of rotational kinetic
energy to its total kinetic energy is
x
5
. The value
of x is ________.
Ans. (2)
Sol.
22
2
22 2 2 2
12
1mR
I23
2
11 1 2 1
I mv mR m(R )
22 2 3 2
22
3
25
1
3
x = 2
56. A hydraulic press containing water has two arms
with diameters as mentioned in the figure. A force
of 10 N is applied on the surface of water in the
thinner arm. The force required to be applied on
the surface of water in the thicker arm to maintain
equilibrium of water is _______ N.
14cm
10N
1.4cm
Ans. (1000 N)
Sol.
12
12
FF
AA
122
F10
(7) (0.7)
F1 = 1000 N
57. The electric field at point p due to an electric
dipole is E. The electric field at point R on
equitorial line will be
E
x
. The value of x :
R
Q
2r
r
–q
O
+q
p
r
Ans. (16)
Sol.
P3
2KP
EE
r
R3
KP E
E16
(2r)
x = 16
58. The maximum height reached by a projectile is
64 m. If the initial velocity is halved, the new
maximum height of the projectile is ______ m.
Ans. (16)
Sol.
22
max u sin
H2g
2
1max 1
2
2max 2
Hu
Hu
2
2
2max
64 u
H(u / 2)
H2max = 16 m
59. A wire of resistance 20 is divided into 10 equal
parts. A combination of two parts are connected in
parallel and so on. Now resulting pairs of parallel
combination are connected in series. The
equivalent resistance of final combination is
_______ .
Ans. (5)
Sol.
10 equal part
20
Each part has resistance = 2
2 parts are connected in parallel so, R = 1
Now, there will be 5 parts each of resistance 1,
they are connected in series.
Req = 5R, Req = 5
60. The current in an inductor is given by I = (3t + 8)
where t is in second. The magnitude of induced
emf produced in the inductor is 12 mV. The self-
inductance of the inductor _______ mH.
Ans. (4)
Sol. I = 3t + 8
= 12 mV
dI
Ldt
12 = L × 3
L = 4 mH
CHEMISTRY
TEST PAPER WITH SOLUTION
SECTION-A
61. Match List - I with List - II.
List - I List - II
(A) ICI (I) T -Shape
(B) ICI3 (II) Square pyramidal
(C) CIF5 (III) Pentagonal
bipyramidal
(D) IF7 (IV) Linear
Choose the correct answer from the options given
below:
(1) (A)–(I), (B)–(IV), C–(III), D–(II)
(2) (A)–(I), (B)–(III), C–(II), D–(IV)
(3) (A)–(IV), (B)–(I), C–(II), D–(III)
(4) (A)–(IV), (B)–(III), C–(II), D–(I)
Ans. (3)
Sol. A. I – Cl (iv) linear
B.
I – Cl
Cl
Cl
(I) T-shape
C.
Cl
F
F
F
F
F
(II) Square pyramidal
D.
I F
F
F
F
F
F
F
(III) Pentagonal bipyramidal
62. While preparing crystals of Mohr's salt, dil. H2SO4
is added to a mixture of ferrous sulphate and
ammonium sulphate, before dissolving this mixture
in water, dil. H2SO4 is added here to:
(1) prevent the hydrolysis of ferrous sulphate
(2) prevent the hydrolysis of ammonium sulphate
(3) make the medium strongly acidic
(4) increase the rate of formation of crystals
Ans. (1)
Sol. Fe+2 ions undergoes hydrolysis, therefore while
preparing aqueous solution of ferrous sulphate and
ammonium sulphate in water dilute sulphuric acid
is added to prevent hydrolysis of ferrous sulphate.
63. Identify the major product in the following
reaction.
Br
CH
3
OH
C
2
H
5
OH
Major Product
(1)
CH2
(2)
Br
(3)
CH3
(4)
CH3
Ans. (3)
Sol.
CH3
Br
H
CH3
64. The correct nomenclature for the following
compound is:
OH
O
H
OH
O
(1) 2–carboxy–4–hydroxyhept–6–enal
(2) 2–carboxy–4–hydroxyhept–7–enal
(3) 2–formyl–4–hydroxyhept–6–enoic acid
(4) 2–formyl–4–hydroxyhept–7–enoic acid
Ans. (3)
Sol.
6
C
H
O
5
3
O
2
1
OH
7
4
OH
CH2
2-formly-4-hydroxyhept-6-enoic acid
65. Given below are two statements : one is labelled as
Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : NH3 and NF3 molecule have
pyramidal shape with a lone pair of electrons on
nitrogen atom. The resultant dipole moment of
NH3 is greater than that of NF3.
Reason (R) : In NH3, the orbital dipole due to lone
pair is in the same direction as the resultant dipole
moment of the N–H bonds. F is the most
electronegative element.
In the light of the above statements, choose the
correct answer from the options given below:
(1) Both (A) and (R) are true and (R) is the correct
explanation of (A)
(2) (A) is false but (R) is true
(3) (A) is true but (R) is false
(4) Both (A) and (R) are true but (R) is NOT the
correct explanation of (A)
Ans. (1)
Sol.
N
F
F
F
Resultant dipole moment = 0.80 × 10–30 Cm
N
H
H
H
Resultant dipole moment = 4.90 × 10–30 cm
66. Given below are two statements:
Statement I : On passing HCl(g) through a
saturated solution of BaCl2, at room temperature
white turbidity appears.
Statement II : When HCl gas is passed through a
saturated solution of NaCl, sodium chloride is
precipitated due to common ion effect.
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) Statement I is incorrect but Statement II is
correct
(4) Both Statement I and Statement II are correct
Ans. (1)
Sol. BaCl2, NaCl are soluble but on adding HCl(g) to
BaCl2, NaCl solutions, Sodium or Barium
chlorides may precipitate out, as a consequence of
the law of mass action.
67. The metal atom present in the complex MABXL
(where A, B, X and L are unidentate ligands and M
is metal) involves sp3 hybridization. The number of
geometrical isomers exhibited by the complex is:
(1) 4 (2) 0
(3) 2 (4) 3
Ans. (2)
Sol. Tetrahedral complex does not show geometrical
isomerism.
68. Match List - I with List - II.
List - I List - II
(Pair of Compounds) (Isomerism)
(A) n-propanol and (I) Metamerism
Isopropanol
(B) Methoxypropane and (II) Chain Isomerism
ethoxyethane
(C) Propanone and (III) Position
propanal Isomerism
(D) Neopentane and (IV) Functional
Isopentane Isomerism
(1) (A)–(II), (B)–(I), (C)–(IV), (D)–(III)
(2) (A)–(III), (B)–(I), (C)–(II), (D)–(IV)
(3) (A)–(I), (B)–(III), (C)–(IV), (D)–(II)
(4) (A)–(III), (B)–(I), (C)–(IV), (D)–(II)
Ans. (4)
Sol.
OH
&
OH
⇒ Position isomers
OCH3
&
O
⇒ Metamers
&
C — H
⇒ Functional isomers
O
O
&
neopentane
isopentane
⇒ Chain isomers
69. The quantity of silver deposited when one coulomb
charge is passed through AgNO3 solution:
(1) 0.1 g atom of silver
(2) 1 chemical equivalent of silver
(3) 1 g of silver
(4) 1 electrochemical equivalent of silver
Ans. (4)
Sol. W = ZIt
W = ZQ
Q =
W
Z
W = ZQ = (electrochemical equivalent)
70. Which one of the following reactions is NOT
possible?
(1)
OCH3
HBr
OH
(2)
OH
HCl
Cl
(3)
Cl
NaOH
OH
High Temp, H
+
(4)
OCH
3
Cl
2
/AlCl
3
OCH
3
Cl
Ans. (2)
Sol.
OH
sp
2
Cl
Not Possible
71. Given below are two statements :
Statement I : The metallic radius of Na is 1.86 A°
and the ionic radius of Na+ is lesser than 1.86 A°.
Statement II : Ions are always smaller in size than
the corresponding elements.
In the light of the above statements, choose the
correct answer from the options given below :
(1) Statement I is correct but Statement II is
false
(2) Both Statement I and Statement II are true
(3) Both Statement I and Statement II are false
(4) Statement I is incorrect but Statement II is
true
Ans. (1)
Sol.
Na Na
rr
+
>
So, Statement (I) is correct but size of anions are
greater than size of neutral atoms.
So statement (II) is incorrect.
72. CH3CH2–OH
(i) Jone's Reagent
(ii) KMnO4
(iii)NaOH, CaO,∆
P
Consider the above reaction sequence and identify
the major product P.
(1) Methane (2) Methanal
(3) Methoxymethane (4) Methanoic acid
Ans. (1)
Sol.
CH3 — CH2 — OH
Joner reagent (CrO
3
+ H
⊕
)
KMnO
4
CH3 — C — OH
O
Soda
lime
process
NaOH
CaI
∆
CH4 + Na2CO3
73. Consider the given chemical reaction :
KMnO
4
–H
2
SO
4
Heat
Product ''A''
Product ''A'' is :
(1) picric acid (2) oxalic acid
(3) acetic acid (4) adipic acid
Ans. (4)
Sol.
KMnO
4
, H
2
SO
4
C — OH
O
C — OH
O
74. For the electro chemical cell
M|M2+||X|X2–
If
( ) ( )
22
00
M /M X/X
E 0.46Vand E 0.34V.
+−
= =
Which of the following is correct ?
(1) Ecell = –0.80 V
(2) M + X → M2 + X2– is a spontaneous reaction
(3) M2+ + X2– → M + X is a spontaneous reaction
(4) Ecell = 0.80 V
Ans. (3)
Sol. M | M+2 || X / X2–
22
oo o
cell M/M X/X
EE E
+−
= +
= –0.46 + 0.34 = – 0.12V
As
o
cell
E
is negative so anode becomes cathode and
cathode become anode. Spontaneous reaction will be
M+2 + X2– → M + X
75. The number of moles of methane required to produce
11g CO2(g) after complete combustion is:
(Given molar mass of methane in g mol–1 : 16)
(1) 0.75 (2) 0.25
(3) 0.35 (4) 0.5
Ans. (2)
Sol. CnH2n+2 +
3n 1
2
+
O2
→
nCO2 + (n + 1)H2O
CH4 + 2O2
→
CO2 + 2H2O
4gm 11gm
0.25 mole 0.25 mole
0.25 mole CH4 gives 0.25 mole (or 11gm) CO2
76. The number of complexes from the following with
no electrons in the t2 orbital is _______.
TiCl4, [MnO4]–, [FeO4]2–, [FeCl4]–, [CoCl4]2–
(1) 3 (2) 1
(3) 4 (4) 2
Ans. (1)
Sol.
TiCl
4
⇒ Ti
+4
⇒ Mn
+7
⇒ Fe
+6
⇒ Fe
+2
⇒ Co
+2
77. The number of ions from the following that have
the ability to liberate hydrogen from a dilute acid is
_______. Ti2+, Cr2+ and V2+
(1) 0 (2) 2
(3) 3 (4) 1
Ans. (3)
Sol. The ions Ti+2, V+2 Cr+2 are strong reducing agents
and will liberate hydrogen from a dilute acid, eg.
23
(aq.) (aq.) (aq.) 2
2Cr 2H 2Cr H (g)
++ +
+ → +
78. Identify A and B in the given chemical reaction
sequence : -
A
AlC3
O
O
O
Zn–Hg
HCl
B
H
+
O
(1)
CH
2
OH
, B -
A -
COOH
(2)
COOH
, B -
A -
COOH
O
(3)
, B -
A -
O
O
(4)
, B -
A -
O
O
OH
OH
Ans. (2)
Sol.
O
O
O
O
AlCl
3
ESR
C
C
O
HO
Zn, Hg
HCl
Clemmension
Reduction
C
O
HO
H
⊕
ESR
O
(A)
(B)
79. The correct statements from the following are :
(A) The decreasing order of atomic radii of group 13
elements is Tl > In > Ga > Al > B.
(B) Down the group 13 electronegativity
decreases from top to bottom.
(C) Al dissolves in dil. HCl and liberate H2 but conc.
HNO3 renders Al passive by forming a
protective oxide layer on the surface.
(D) All elements of group 13 exhibits highly
stable +1 oxidation state.
(E) Hybridisation of Al in [Al(H2O)6]3+ ion is
sp3d2.
Choose the correct answer from the options given
below :
(1) (C) and (E) only
(2) (A), (C) and (E) only
(3) (A), (B), (C) and (E) only
(4) (A) and (C) only
Ans. (1)
Sol. A. size order T > In > Al > Ga > B
B. Electronegativity order B > Al < Ga < In < T
D. B, Al are more stable in +3 oxidation state
So, only C, E statements are correct.
80. Coagulation of egg, on heating is because of :
(1) Denaturation of protein occurs
(2) The secondary structure of protein remains
unchanged
(3) Breaking of the peptide linkage in the primary
structure of protein occurs
(4) Biological property of protein remains
unchanged
Ans. (1)
Sol. Coagulation of egg give primary structure of
protein, which is known as denaturation of protein
SECTION-B
81. Combustion of 1 mole of benzene is expressed at
C6H6(1) +
( ) ( ) ( )
2 22
15 Og COg 3HO1.
2→+
The standard enthalpy of combustion of 2 mol of
benzene is – 'x' kJ.
x = ______.
(1) standard Enthalpy of formation of 1 mol of
C6H6(1), for the reaction
6C(graphite) + 3H2(g) → C6H6(1) is 48.5 kJ mol–1.
(2) Standard Enthalpy of formation of 1 mol of
CO2(g), for the reaction
C(graphite) + O2(g) → CO2(g) is –393.5 kJ mol–1.
(3) Standard and Enthalpy of formation of
1 mol of H2O(1), for the reaction
H2(g) +
( )
2
1Og
2
→ H2O(1) is –286 kJ mol–1.
Ans. (6535)
Sol. 6C(graphite)+3H2(g) → C6H6(); ∆H = 48.5 kJ/mol
C(graphite)+O2(g)→ CO2(g); ∆H = –393.5 kJ/mol
( )
( )
g
21
Hg
2
+
→ H2O() ; ∆H = – 286 kJ/mol
equation –(1) × 1 + (2) × 6 + (3) × 3
– 48.5 –6 × 393.5 – 3 × 286
= – 3267.5 kJ for 1 mol
= – 6535 kJ for 2 mol
Ans. 6535 kJ
82. The fusion of chromite ore with sodium carbonate
in the presence of air leads to the formation of
products A and B along with the evolution of CO2.
The sum of spin-only magnetic moment values of
A and B is ___ B.M. (Nearest integer)
(Given atomic number : C : 6, Na : 11, O : 8,
Fe : 26, Cr : 24]
Ans. (6)
Sol. 4FeCr2O4 + 8Na2CO3 + 7O2 →
8Na2CrO4 + 2Fe2O3 + 8CO2
A B
Spin only magnetic moment
For Na2CrO4 µB = 0
For Fe2O3 µB = 5.9
sum = 5.9
83. X of enthanamine was subjected to reaction with
NaNO2/HCl followed by hydrolysis to liberate N2
and HCl. The HCl generated was completely
neutralised by 0.2 moles of NaOH. X is ____ g.
Ans. (9)
Sol.
CH3—CH2—NH2
NaNO2 + HCl
CH3—CH2—N2Cl
⊕
Θ
HOH
CH3—CH2—OH + N2 + HCl
(g)
0.2 mole
0.2 mole
MW of ethanamine = 45
45 × 0.2 = 9 gm
84. In an atom, total number of electrons having
quantum numbers n = 4, |ml| = 1 and ms = –
1
2
is
Ans. (6)
Sol. n = 4
m
0 0
1 –1, 0, +1
2 –2, –1, 0, +1, +2, +3
So number of orbital associated with
n = 4, |m| = 1 are 6
Now each orbital contain one e– with ms =
1
–2
85. Using the given figure, the ratio of Rf values of
sample A and sample C is x × 10–2. Value of x is
__________.
12.5 cm
Solvent front
10.0 cm
Sample C
6.5 cm
Sample B
5.0 cm
Sample A
0.0 cm
Base line
Samples (A,B,C)
Fig : Paper chromatography of Samples
Ans. (50)
Sol. Rf of A =
5
12.5
Rf of C =
10
12.5
Ratio =
f(A)
f(C)
R1
R2
=
= 0.5 or 50 × 10–2
86. In the Claisen-Schmidt reaction to prepare 351 g of
dibenzalacetone using 87 g of acetone, the amount
of benzaldehyde required is _________g. (Nearest
integer)
Ans. (318)
Sol. Claisen Schmidt reaction
H
O
C
2 mole
+
O
CH
3
CH
3
1 mole
NaOH
⊕
Θ
— CH = CH — C — CH = CH —
O
Dibanzal acetone
1 mole
3 mole
1.5 mole
351 gm = 1.5 mole
mw of benzaldehyde = 106
106 × 3 = 318 gm. Benzaldehyde is required to
give 1.5 mole (or 351 gm) product
87. Consider the following single step reaction in gas
phase at constant temperature.
2A(g) + B(g) → C(g)
The initial rate of the reaction is recorded as r1
when the reaction starts with 1.5 atm pressure of A
and 0.7 atm pressure of B. After some time, the
rate r2 is recorded when the pressure of C becomes
0.5 atm. The ratio r1 : r2 is ________ × 10–1.
(Nearest integer)
Ans. (315)
Sol. 2A(g) + B(g) → C(g)
r1 1.5 atm 0.7 atm
r2 0.5 atm 0.2 atm 0.5 atm
r = K [PA]2[PB]
r1 = K [1.5]2[0.7]
r2 = K [0.5]2[0.2]
–1
1
2
r7
9 31.5 315 10
r2
=×= = ×
Ans. 315
88. The product © in the following sequence of
reactions has _________ π bonds.
KMnO4–KOH
∆
A
H3O+
B
Br2
C
FeBr3
Ans. (4)
Sol. A =
C — O K
O
⊕
Θ
B =
C — OH
O
C =
C — OH
O
Br
π bonds = 4
89. Considering acetic acid dissociates in water, its
dissociation constant is 6.25 × 10–5. If 5 mL of
acetic acid is dissolved in 1 litre water, the solution
will freeze at –x × 10–2 °C, provided pure water
freezes at 0 °C.
x = _________. (Nearest integer)
Given : (Kf)water = 1.86 K kg mol–1.
density of acetic acid is 1.2 g mol–1
molar mass of water = 18 g mol–1.
molar mass of acetic acid = 60 g mol–1.
density of water = 1 g cm–3
Acetic acid dissociates as
CH3COOH
3
CH COO H
Θ⊕
+
Ans. (19)
Sol. Mass of CH3COOH = V × d
= 5 ml × 1.2 g/ml
= 6 gm
3
CH COOH 6
n 0.1mol
60
= =
33
CH COOH CH COOH
0.1
m M 0.1M
1
≈==
CH3COOH
CH3COO– + H+
C
C – Cα Cα Cα
2
aC
K1
α
=−α
1 – α ≈ 1 ⇒ Ka = Cα2
–5
Ka 6.25 10
C 0.1
×
α= =
= 25 × 10–3
V.f. (i) = 1 + α(n – 1) = 1 + α(2 – 1) = 1 + α
= 1 + 25 × 10–3 = 1.025
∆Tf = iKfm
= (1.025)(1.86)(0.1)
= 0.19
= 19 × 10–2
90. Number of compounds from the following with
zero dipole moment is ___________.
HF, H2, H2S, CO2, NH3, BF3, CH4, CHCl3, SiF4,
H2O, BeF2
Ans. (6)
Sol. H2, CO2, BF3, CH4, SiF4 , BeF2
are symm. molecule so dipole moment is zero
