FINAL JEE–MAIN EXAMINATION – APRIL, 2024
(Held On Thursday 04th April, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS
TEST PAPER WITH SOLUTION
SECTION-A
1. Let ƒ : R R be a function given by
2
1 cos2x , x 0
x
ƒ x , x 0
1 cosx , x 0
x
, where , R. If
ƒ is continuous at x = 0, then 2 + 2 is equal to :
(1) 48 (2) 12
(3) 3 (4)6
Ans. (2)
Sol. ƒ(0–) =
2
2
x0
2sin x
lim x
= 2 =
ƒ(0+) =
x0
x
sin 2
lim 2 2
x2
22
=
22
2+2 = 4 + 8 = 12
2. Three urns A, B and C contain 7 red, 5 black;
5 red, 7 black and 6 red, 6 black balls, respectively.
One of the urn is selected at random and a ball is
drawn from it. If the ball drawn is black, then the
probability that it is drawn from urn A is :
(1)
4
17
(2)
5
18
(3)
7
18
(4)
5
16
Ans. (2)
Sol.
A
7R, 5B
B
5R, 7B
C
6R, 6B
P(B) =
1 5 1 7 1 6
...
3 12 3 12 3 12
required probability =
15
.5
3 12
1 5 7 6 18
.
3 12 12 12
3. The vertices of a triangle are A(–1, 3), B(–2, 2) and
C(3, –1). A new triangle is formed by shifting the sides
of the triangle by one unit inwards. Then the equation
of the side of the new triangle nearest to origin is :
(1) x – y –
22
= 0
(2) –x + y –
22
= 0
(3) x + y –
22
= 0
(4) x + y +
22
= 0
Ans. (3)
Sol.
C
(–2,2)
m = –1
(3, –1)
B
A
(–1,3)
equation of AC x + y = 2
equation of line parallel to AC x + y = d
d2 1
2
d 2 2
eqnof new required line
x y 2 2
4. If the solution y = y(x) of the differential equation
(x4 + 2x3 + 3x2 + 2x + 2)dy – (2x2 + 2x + 3)dx = 0
satisfies y(–1) =
4
, then y(0) is equal to :
(1)
12
(2) 0
(3)
4
(4)
2
Ans. (3)
Sol.
2
4 3 2
2x 2x 3
dy dx
x 2x 3x 2x 2
2
22
2x 2x 3
y dx
x 1 x 2x 2
22
dx dx
yx 2x 2 x 1
y = tan–1(x + 1) + tan–1x + C
y(–1) =
4
0C
44
C = 0
y = tan–1(x + 1) + tan–1x
y(0) = tan–11 =
4
5. Let the sum of the maximum and the minimum
values of the function f(x) =
2
2
2x 3x 8
2x 3x 8
be
m
n
,
where gcd(m, n) = 1. Then m + n is equal to :
(1) 182 (2) 217
(3) 195 (4) 201
Ans. (4)
Sol.
2
2
2x 3x 8
y2x 3x 8
x2(2y – 2) + x(3y + 3) + 8y – 8 = 0
use D 0
(3y + 3)2 – 4(2y – 2) (8y – 8) 0
(11y – 5) (5y – 11) 0
5 11
y,
11 5
y = 1 is also included
6. One of the points of intersection of the curves
y = 1 + 3x – 2x2 and y =
1
x
is
1,2
2
. Let the area
of the region enclosed by these curves be
15m
24
– nloge
15
, where , m, n
N. Then + m + n is equal to
(1) 32 (2) 30
(3) 29 (4) 31
Ans. (2)
Sol.
15
2
2
1
2
1
A 1 3x 2x dx
x
15
23 2
1
2
3x 2x
A x n x
23
23
1 5 3 1 5 2 1 5 1 5
An
2 2 2 3 2 2
1 3 1 2 1 1
n
2 2 4 3 8 2
1 5 3 3 15 4 2
A 5 5
2 2 8 4 8 3 3
1 3 1 n 1 5
2 8 12
1 3 2 15 4 1
5 n 1 5
2 4 3 8 3 12
14 15
5 n 1 5
24 24
7. If the system of equations
x 2 sin y 2 cos z 0
x + (cos )y + (sin )z = 0
x + (sin )y – (cos )z = 0
has a non-trivial solution, then
0, 2
is equal to :
(1)
3
4
(2)
7
24
(3)
5
24
(4)
11
24
Ans. (3)
Sol.
1 2 sin 2 cos
1 sin cos 0
1 cos sin
1 2 sin sin cos 2 cos cos sin 0
1 2 cos2 2 sin2 0
1
cos2 sin 2 2
1
cos 2 42
2
2 2n
43
n
83
n = 0,
5
x3 8 24
8. There are 5 points P1, P2, P3, P4, P5 on the side AB,
excluding A and B, of a triangle ABC. Similarly
there are 6 points P6, P7, …, P11 on the side BC and 7
points P12, P13, …, P18 on the side CA of the triangle.
The number of triangles, that can be formed using the
points P1, P2, …, P18 as vertices, is :
(1) 776 (2) 751
(3) 796 (4) 771
Ans. (2)
Sol. 18C3 – 5C3 – 6C3 – 7C3
= 751
9. Let ƒ(x) =
2, 2 x 0
x 2, 0 x 2
and h(x) = f(|x|) + |f(x)|.
Then
2
2
h x dx
is equal to :
(1) 2 (2) 4
(3) 1 (4) 6
Ans. (1)
Sol.
y=f(x
x
2
x–2
–2
–2
f(|x|) |f(x)|
x
(2,0)
x–2
(0,–2)
(–2,0)
0
–x–2
x
2
–2
0
2
x 2 2 x 0, 0 x 2
h(x) x 2 2 x 2 x 0
2
0
–2
2
0
h(x)dx 0
and
0
2
h x dx 2
10. The sum of all rational terms in the expansion of
15
11
53
25
is equal to :
(1) 3133 (2) 633
(3) 931 (4) 6131
Ans. (1)
Sol. Tr + 1 = 15Cr
15 r
r
11
35
52
r 15 r
15 35
r
C 5 . 2
R = 3, 15µ
r = 0, 15
2 rational terms
5
15 3 15
0 15
C 2 C 5
= 8 + 3125 = 3133
11. Let a unit vector which makes an angle of 60° with
ˆˆ
ˆ
2i 2j k
and an angle of 45° with
ˆˆ
ik
be
C
.
Then
1 1 2
ˆˆ
ˆ
C i j k
23
32
is :
(1)
2 2 1 2 2
ˆˆ ˆ
i j k
3 3 2 3
(2)
2 1 1
ˆˆ
ˆ
i j k
32
32
(3)
1 1 1 1 1 2
ˆˆ
ˆ
i j k
23
3 3 3 2 3
(4)
21
ˆˆ
ik
32
Ans. (4)
Sol.
1 2 3
ˆˆˆ
C C i C j C k
C1
2 + C2
2 + C3
2 = 1
ˆˆ
ˆ
C. 2i 2 j k C 9 cos60
2C1 + 2C2 – C3 =
3
2
C1 – C3 = 1
C1 + 2C2 =
1
2
C1 =
21
32
C2 =
1
32
C3 =
21
32
12. Let the first three terms 2, p and q, with q 2, of a
G.P. be respectively the 7th, 8th and 13th terms of an
A.P. If the 5th term of the G.P. is the nth term of the
A.P., then n is equal to
(1) 151 (2) 169
(3) 177 (4) 163
Ans. (4)
Sol. p2 = 2q
2 = a + 6d ...(i)
p = a + 7d ...(ii)
q = a + 12d ...(iii)
p – 2 = d ((ii) – (i))
q – p = 5d ((iii) – (ii))
q – p = 5(p – 2)
q = 6p – 10
p2 = 2(6p – 10)
p2 – 12p + 20 = 0
p = 10, 2
p = 10 ; q = 50
d = 8
a = –46
2, 10, 50, 250, 1250
ar4 = a + (n – 1)d
1250 = –46 + (n – 1)8
n = 163
13. Let a, b R. Let the mean and the variance of 6
observations –3, 4, 7, –6, a, b be 2 and 23,
respectively. The mean deviation about the mean
of these 6 observations is :
(1)
13
3
(2)
16
3
(3)
11
3
(4)
14
3
Ans. (1)
Sol.
i
x2
6
and
2
2
i
x23
N
+ = 10
2 + 2 = 52
solving we get = 4, = 6
i
xx5 2 5 8 2 4 13
6 6 3
14. If 2 and 6 are the roots of the equation ax2 + bx + 1 = 0,
then the quadratic equation, whose roots are
1
2a b
and
1
6a b
, is :
(1) 2x2 + 11x + 12 = 0 (2) 4x2 + 14x + 12 = 0
(3) x2 + 10x + 16 = 0 (4) x2 + 8x + 12 = 0
Ans. (4)
Sol. Sum = 8 =
b
a
Product = 12 =
1
a
a =
1
12
b =
2
3
2a + b =
2 2 1
12 3 2
6a + b =
6 2 1
12 3 6
sum = –8
P = 12
x2 + 8x + 12 = 0
15. Let and be the sum and the product of all the
non-zero solutions of the equation
2
z z 0
, z C.
Then 4(2 + 2) is equal to :
(1) 6 (2) 4
(3) 8 (4) 2
Ans. (2)
Sol. z = x + iy
z x iy
2 2 2
z x y 2ixy
2 2 2 2
x y 2ixy x y 0
x = 0 or y = 0
–y2 + |y| = 0 x2 + |x| = 0
|y| = |y|2 x = 0
y = 0, ±1
i, –i = i – i = 0
are roots = i(–i) = 1
4(0 + 1) = 4
16. Let the point, on the line passing through the points
P(1, –2, 3) and Q(5, –4, 7), farther from the origin
and at a distance of 9 units from the point P, be
(, , ). Then 2 + 2 + 2 is equal to :
(1) 155 (2) 150
(3) 160 (4) 165
Ans. (1)
Sol. PQ line
x 1 y 2 z 3
4 2 4
pt (4t + 1, –2t – 2, 4t + 3)
distance2 = 16t2 + 4t2 + 16t2 = 81
3
t2
pt (7, –5, 9)
2 + 2 + 2 = 155
option (1)
17. A square is inscribed in the circle
x2 + y2 – 10x – 6y + 30 = 0. One side of this square
is parallel to y = x + 3. If (xi, yi) are the vertices of
the square, then
22
ii
xy
is equal to :
(1) 148 (2) 156
(3) 160 (4) 152
Ans. (4)
Sol.
(5,3)
2
y = x + c & x + y + d = 0
5 3 c 2
2
8d 2
2
|c + 2| = 2 8 + d = ±2
c = 0, – 4 d = –10, –6
pts (5, 5), (3, 3), (7, 3), (5, 1)
22
i1
x y 25 25 9 9 49 9 25 1
= 152
Option (4)
18. If the domain of the function
2
1
e2
3x 22 3x 8x 5
sin log
2x 19 x 3x 10
is (, ],
then 3 + 10 is equal to :
(1) 97 (2) 100
(3) 95 (4) 98
Ans. (1)
Sol.
3x 22
11
2x 19
2
2
3x 8x 5 0
x 3x 10
41
x 5, 5
3 + 10 = 97
Option (1)
19. Let ƒ(x) = x5 + 2ex/4 for all x R. Consider a
function g(x) such that (gof) (x) = x for all x R.
Then the value of 8g(2) is :
(1) 16 (2) 4
(3) 8 (4) 2
Ans. (1)
Sol. f(x) = 2
when x = 0
g(f(x)) f(x) = 1
1
g2 f0
f(x) = 5x4 +
2
4
ex/4
g(2) = 2
Ans = 16
Option (1)
20. Let (0, ) and A =
12
1 0 1
0 1 2
.
If det(adj(2A – AT).adj(A – 2AT)) = 28, then
(det(A))2 is equal to :
(1) 1 (2) 49
(3) 16 (4) 36
Ans. (3)
Sol. |adj(A – 2AT) (2A – AT)| = 28
|(A – 2AT) (2A – AT)| = 24
|A – 2AT| |2A – AT| = ±16
(A – 2AT)T = AT – 2A
|A – 2AT| = |AT – 2A|
|A – 2AT|2 = 16
|A – 2AT| = ±4
1 2 2 2 0
1 0 1 4 0 2
0 1 2 2 2 4
10
3 0 1
2 1 2
1 + 3 = 4
3 = 3
= 1
1 2 1
A 1 0 1 1 3 4
012
|A|2 = 16
SECTION-B
21. If
1/3 1/3
1/2 2/3
1/2
x1
5x 1 x 5 m 5
lim
2x 3 x 4 n 2n
, where
gcd(m, n) = 1, then 8m + 12n is equal to _____
Ans. (100)
Sol.
2/3 2/3
x1 1/2 1/2
11
(5x 1) 5 (x 5)
33
lim 11
(2x 3) .2 (x 4)
22
2/3
85
m8
n3
36
8m + 12n = 100
22. In a survey of 220 students of a higher secondary
school, it was found that at least 125 and at most
130 students studied Mathematics; at least 85 and
at most 95 studied Physics; at least 75 and at most
90 studied Chemistry; 30 studied both Physics and
Chemistry; 50 studied both Chemistry and
Mathematics; 40 studied both Mathematics and
Physics and 10 studied none of these subjects. Let
m and n respectively be the least and the most
number of students who studied all the three
subjects. Then m + n is equal to _____
Ans. (45)
Sol.
M
P
P
m
40–x
x
50–x
30–x
C
C
125 m + 90 – x 130
85 P + 70 – x 95
75 C + 80 – x 90
m + P + C + 120 – 2x = 210
15 x 45 & 30 – x 0
15 x 30
30 + 15 = 45
23. Let the solution y = y(x) of the differential
equation
dy
dx
– y = 1 + 4sinx satisfy y() = 1. Then
y2
+ 10 is equal to _____
Ans. (7)
Sol.
x x x
ye e 4e sin x dx
ye–x = –e–x – 2(e–xsinx e–xcosx) + C
y = – 1 – 2(sinx + cosx) + cex
y() = 1 c = 0
y( /2) = –1 – 2 = –3
Ans = 10 – 3 = 7
24. If the shortest distance between the lines
x 2 y 3 z 5
2 3 4
and
x 3 y 2 z 4
1 3 2
is
38 k
35
and
k
2
0
x dx
, where [x]
denotes the greatest integer function, then 63
is equal to _____
Ans. (48)
Sol.
ˆˆˆ
i j k
ˆˆˆ
38 (5i 5j 9k)
ˆ
k . 2 3 4
3 5 5 1 3 2
38 19
ˆ
k
3 5 5
19
k5
3
k2
3/2 1 2 3/2
2
0 0 1 2
x 0 1 2
3
2 1 2 2
2
22
= 2
63 = 48
25. Let A be a square matrix of order 2 such that
|A| = 2 and the sum of its diagonal elements is –3.
If the points (x, y) satisfying A2 + xA + yI = 0 lie
on a hyperbola, whose transverse axis is parallel to
the x-axis, eccentricity is e and the length of the
latus rectum is , then e4 + 4 is equal to ______
Ans. (Bouns)
NTA Ans. (25)
Sol. Given |A| = 2
trace A = –3
and A2 + xA + yI = 0
x = 3, y = 2
so, information is incomplete to determine
eccentricity of hyperbola (e) and length of latus
rectum of hyperbola ()
26. Let
234
2 2 2
C C C
a 1 ...,
3! 4! 5!
1 1 2 2 2 3 3 3 3
0 1 0 1 2 0 1 2 3
C C C C C C C C C
b 1 ...
1! 2! 3!
Then
2
2b
a
is equal to ______
Ans. (8)
Sol. f(x) = 1 +
23
(1 x) (1 x) (1 x) .....
1! 2! 3!
(1 x) 2 2
e 1 (1 x) (1 x) (1 x)
1
1 x 1 x 2! 3! 4!
coef x2 in RHS : 1 +
23
22
CC
.... a
34
coeff. x2 in L.H.S.
22
xx
e 1 x .... 1 x ......
2! 2!
is
e
e e a
2!
b =
23
2
2 2 2
1 ...... e
1! 2! 3!
2
2b 8
a
27. Let A be a 3 × 3 matrix of non-negative real
elements such that
11
A 1 3 1
11
. Then the
maximum value of det(A) is _____
Ans. (27)
Sol. Let A =
1 2 3
1 2 3
1 2 3
a a a
b b b
c c c
11
A 1 3 1
11
a1 + a2 + a3 = 3 …..(1)
b1 + b2 + b3 = 3 …..(2)
c1 +ca2 + c3 = 3 …..(3)
Now,
|A| = (a1b2c3 + a2b3c1 + a3b1c2)
– (a3b2c1 + a2b1c3 + a1b3c2)
From above in formation, clearly |A|max = 27,
when a1 = 3, b2 = 3, c3 = 3
28. Let the length of the focal chord PQ of the
parabola y2 = 12x be 15 units. If the distance of PQ
from the origin is p, then 10p2 is equal to ____
Ans. (72)
Sol.
y2=12x
15
(3,0)
Q
P
P
length of focal chord = 4a cosec2 = 15
12cosec2 = 15
sin2 =
4
5
tan2 = 4
tan = 2
equation
y0 2
x3
y = 2x – 6
2x – y –6 = 0
P =
6
5
10p2 =
36
10. 72
5
29. Let ABC be a triangle of area
15 2
and the
vectors
ˆˆˆ
AB i 2j 7k
,
ˆˆˆ
BC ai bj ck
and
ˆˆ ˆ
AC 6i dj 2k
, d > 0. Then the square of the
length of the largest side of the triangle ABC is
Ans. (54)
Sol.
A
C
B
Area =
ˆˆ ˆ
i j k
11 2 7 15 2
26 d 2
(–4 + 7d)
ˆ
i
–
ˆ
j
(–2 + 42) +
ˆ
k
(d – 12)
(7d – 4)2 + (40)2 + (d – 12)2 = 1800
50d2 – 80d – 40 = 0
5d2 – 8d – 4 = 0
5d2 – 10d – 2d – 4
5d(d – 2) + 2(d – 2) = 0
d = 2 or d =
2
5
d > 0, d = 2
(a + 1)
ˆ
i
+ (b + 2)
ˆ
j
+ (c – 7)
ˆ
k
= 6
ˆ
i
+ 2
ˆ
j
– 2
ˆ
k
a + 1 = 6 & b + 2 = 2, c – 7 = –2
a = 5 b = 0 c = 5
|AB| =
1 4 49 54
|BC| =
25 25 50
|AC| =
86 4 4 44
Ans. 54
30. If
2
4
e
0
sin x 1 a
dx log
1 sin x cosx a 3 b3
, where a,
b N, then a + b is equal to _____
Ans. (8)
Sol.
2
24
00
sin x 1 cos2x
dx dx
12 sin 2x
1 sin 2x
2
1 cos2x
2 sin 2x 2 sin 2x
(I1) – (I2)
(I1) =
2
dx
2 tan x
21 tan x
2
4
2
0
sec x dx
2 tan x 2 tan x 2
tanx = t
1
2
0
1 dt
213
t24
=
63
I2 =
/4
0
cos2x
2 sin 2x
dx =
13
n
22
12
1 1 2
I I n
6 2 3
3
a 2,b 6
Ans. 8
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
31. An electron is projected with uniform velocity
along the axis inside a current carrying long
solenoid. Then :
(1) the electron will be accelerated along the axis.
(2) the electron will continue to move with uniform
velocity along the axis of the solenoid.
(3) the electron path will be circular about the axis.
(4) the electron will experience a force at 45° to the
axis and execute a helical path.
Ans. (2)
Sol.
e
B
Since
v || B
so force on electron due to magnetic
field is zero. So it will move along axis with
uniform velocity.
32. The electric field in an electromagnetic wave is
given by
1
z
ˆ
E i40 cos NC
tc
. The
magnetic field induction of this wave is (in SI
unit):
(1)
40 z
ˆ
B i cos t
cc
(2)
z
ˆ
B j40 cos tc
(3)
40 z
ˆ
B k cos t
cc
(4)
40 z
ˆ
B j cos t
cc
Ans. (4)
Sol.
z
ˆ
E i40 cos tc
E
is along +x direction
v
is along +z direction
So direction of
B
will be along +y and magnitude
of B will be
E
c
So answer is
40 zˆ
cos j
t
cc
33. Which of the following nuclear fragments
corresponding to nuclear fission between neutron
1
0n
and uranium isotope
235
92 U
is correct:
(1)
144 89 1
56 36 0
Ba Kr 4 n
(2)
140 94 1
56 38 0
Xe Sr 3 n
(3)
153 99 1
51 41 0
Sb Nb 3 n
(4)
144 89 1
56 36 0
Ba Kr 3 n
Ans. (4)
Sol. Balancing mass number and atomic number
235 1 144 89 1
92 0 56 36 0
U n Ba Kr 3 n
34. In an experiment to measure focal length (f) of
convex lens, the least counts of the measuring
scales for the position of object (u) and for the
position of image (v) are u and v, respectively.
The error in the measurement of the focal length of
the convex lens will be :
(1)
uv
uv
(2)
2
22
uv
fuv
(3)
uv
2f uv
(4)
uv
fuv
Ans. (2)
Sol. f–1 = v–1 – u–1
–f–2 df = – v–2dv – u–2du
2 2 2
df dv du
f v u
2
22
dv du
df f vu
35. Given below are two statements :
Statement I : When speed of liquid is zero
everywhere, pressure difference at any two points
depends on equation P1 – P2 = g (h2 – h1)
Statement II : In ventury tube shown
22
12
2gh
h
1
2
P1, A1
P2, A2
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 incorrect but Statement II is
correct.
(3) Both Statement I and Statement II are incorrect.
(4) Statement I is correct but Statement II is
incorrect.
Ans. (4)
Sol.
h
2
P1
A1
P2, A2
1
Applying Bernoulli's equation
2
1 1 1
1
P gh v
2
=
2
2 2 2
1
P gh v
2
[h1 & h2 are height of point from any reference
level]
Given V1 = V2 = 0 (for statement-1)
P1 – P2 = g(h2 – h2)
For statement-2
22
1 1 2 2
11
P v P v
22
P1 – P2 = gh
22
1 2 2 1
11
P P v v
22
22
21
11
gh v v
22
22
21
2gh v v
Hence answer (4)
36. The resistances of the platinum wire of a platinum
resistance thermometer at the ice point and steam
point are 8 and 10 respectively. After
inserting in a hot bath of temperature 400°C, the
resistance of platinum wire is :
(1) 2 (2) 16 (3) 8 (4) 10
Ans. (2)
Sol. Given R0 = 8, R100 = 10
R100 = R0 (1 + T)
Also, R400 = R0 (1 + T1)
10 = 8 (1 + × 100) 100 =
1
4
R400 = 8 (1 + 400) = 8 (1 + 1) = 16
Hence option (2)
37. A metal wire of uniform mass density having
length L and mass M is bent to form a semicircular
arc and a particle of mass m is placed at the centre
of the arc. The gravitational force on the particle
by the wire is:
(1)
2
GMm
2L
(2) 0
(3)
2
2
GmM
L
(4)
2
2GmM
L
Ans. (4)
Sol.
O
R
M,L
We have
L
R
02
M
2G 2GM
L
gRL
m0 2
2GM m
F mg L
Hence option (4)
38. On celcius scale the temperature of body increases
by 40°C. The increase in temperature on
Fahrenheit scale is:
(1) 70°F
(2) 68°F
(3) 72°F
(4) 75°F
Ans. (3)
Sol. We know that per °C change is equivalent to 1.8°
change in °F.
40° change on celcius scale will corresponds to
72° change on Fahrenheit scale.
Hence option (3)
39. An effective power of a combination of 5 identical
convex lenses which are kept in contact along the
principal axis is 25 D. Focal length of each of the
convex lens is :
(1) 20 cm
(2) 50 cm
(3) 500 cm
(4) 25 cm
Ans. (1)
Sol. We know that Peq = Pi
given all lenses are identical
5P = 25D
P = 5D
11
5 f m 20cm
f5
Hence option (1)
40. Which figure shows the correct variation of
applied potential difference (V) with photoelectric
current (I) at two different intensities of light (I1 <
I2) of same wavelengths :
(1)
I2
I1
V
i
–V0
(2)
I2
I1
V
i
–V0
(3)
I2
I1
V
i
–V0
(4)
I2
I1
V
i
–V0
Ans. (3)
Sol. Given lights are of same wavelength.
Hence stopping potential will remain same.
Since I2 > I1, hence saturation current
corresponding to I2 will be greater than that
corresponding to I1.
Hence option (3)
41. A wooden block, initially at rest on the ground, is
pushed by a force which increases linearly with
time t. Which of the following curve best describes
acceleration of the block with time :
(1)
a
(0,0)
t
(2)
a
(0,0)
t
(3)
a
(0,0)
t
(4)
a
(0,0)
t
Ans. (2)
Sol.
m
F
F kt
F ma a mm
a vs t will be straight line passing through origin.
Since option (2).
42. If a rubber ball falls from a height h and rebounds
upto the height of h/2. The percentage loss of total
energy of the initial system as well as velocity ball
before it strikes the ground, respectively, are :
(1) 50%,
gh
2
(2) 50%,
gh
(3) 40%,
2gh
(4) 50%,
2gh
Ans. (4)
Sol. Velocity just before collision =
2gh
Velocity just after collision =
h
2g 2
11
KE m mgh
2gh
22
1mgh
2
% loss in energy
=
i
1mgh
KE 2
100 100 50%
1
KE mg2h
2
Hence option (4)
43. The equation of stationary wave is :
2 nt 2 x
y 2asin cos
Which of the following is NOT correct
(1) The dimensions of nt is [L]
(2) The dimensions of n is [LT–1]
(3) The dimensions of n/ is [T]
(4) The dimensions of x is [L]
Ans. (3)
Sol. Comparing the given equation with standard
equation of standing
2n
&
2k
1
nT
[nt] = [] = L
[n] = [] = LT–1
[x] = [] = L
Hence option (3)
44. A body travels 102.5 m in nth second and 115.0 m
in (n + 2)th second. The acceleration is :
(1) 9 m/s2 (2) 6.25 m/s2
(3) 12.5 m/s2 (4) 5 m/s2
Ans. (2)
Sol. Given,
a
102.5 u 2n 1
2
&
a
115 u 2n 3
2
a
102.5 u an 2
&
3a
115 u an 2
12.5 = 2a a = 6.25 m/s2
Hence option (2)
45. To measure the internal resistance of a battery,
potentiometer is used. For R = 10 , the balance
point is observed at = 500 cm and for R = l the
balance point is observed at = 400 cm. The
internal resistance of the battery is approximately :
(1) 0.2 (2) 0.4
(3) 0.1 (4) 0.3
Ans. (4)
Sol. Let potential gradient be .
i × 10 = × 500 = – irs
500 = – 50rs
Also,
i' × 1 = × 400 = – i'rs
400 = – 400 rs
100 = 350 rs rs =
10
35
0.3
Hence option (4)
46. An infinitely long positively charged straight
thread has a linear charge density Cm–1. An
electron revolves along a circular path having axis
along the length of the wire. The graph that
correctly represents the variation of the kinetic
energy of electron as a function of radius of
circular path from the wire is :
(1)
r
KE
O
(2)
r
KE
O
(3)
r
KE
O
(4)
KE
O
r
Ans. (2)
Sol.
r
–e
+
+
+
+
+
+
+
+
+
+
Electric field E at a distance r due to infinite long
wire is
2k
Er
Force of electron F = eE
2k
Fe r
2k e
Fr
This force will provide required centripetal force
2
mv 2k e
Frr
2k e
vm
2
11
2k e
KE mv m
22
m
= ke
This is constant so option (2) is correct.
47. The value of net resistance of the network as
shown in the given figure is :
–8V
15
10
5
–6V
(1)
5
2
(2)
15
4
(3) 6 (4)
30
11
Ans. (3)
Sol.
–8V
15
10
5
–6V
1
2
Diode 2 is in reverse bias
So current will not flow in branch of 2nd diode, So
we can assume it to be broken wire.
Diode 1 is in forward bias
So it will behave like conducting wire. So new
circuit will be
–8V
15
10
–6V
eq
15 10 15 10
R6
15 10 25
Correct answer (3)
48. P-T diagram of an ideal gas having three different
densities 1, 2, 3 (in three different cases) is
shown in the figure. Which of the following is
correct :
P
T
3
2
1
(1) 2 < 3 (2) 1 > 2
(3) 1 < 2 (4) 1 = 2 = 3
Ans. (2)
Sol. For ideal gas
PV = nRT
m
PV RT
M
RT
M
PM
V
RT
PM
(Where m is mass of gas and M is molecular mass
of gas)
P
T
3
2
1
P1
P2
P3
for same temperature P1 > P2 > P3
So 1 > 2 > 3
So correct answer is (2)
49. The co-ordinates of a particle moving in x-y plane
are given by :
x = 2 + 4t, y= 3t + 8t2.
The motion of the particle is :
(1) non-uniformly accelerated.
(2) uniformly accelerated having motion along a
straight line.
(3) uniform motion along a straight line.
(4) uniformly accelerated having motion along a
parabolic path.
Ans. (4)
Sol. x = 2 + 4t
x
dx v4
dt
x
x
dv a0
dt
y = 3t + 8t2
y
dy v 3 16t
dt
y
y
dv a 16
dt
the motion will be uniformly accelerated motion.
For path
x = 2 + 4t
x2 t
4
Put this value of t is equation of y
2
x 2 x 2
y 3 8
44
this is a quadratic equation so path will be
parabola.
Correct answer (4)
50. In an ac circuit, the instantaneous current is zero,
when the instantaneous voltage is maximum. In
this case, the source may be connected to :
A. pure inductor.
B. pure capacitor.
C. pure resistor.
D. combination of an inductor and capacitor.
Choose the correct answer from the options given
below :
(1) A, B and C only (2) B, C and D only
(3) A and B only (4) A, B and D only
Ans. (4)
Sol. This is possible when phase difference is
2
between current and voltage so correct answer will
be (4)
SECTION-B
51. An infinite plane sheet of charge having uniform
surface charge density +s C/m2 is placed on x-y
plane. Another infinitely long line charge having
uniform linear charge density +e C/m is placed at
z = 4m plane and parallel to y-axis. If the
magnitude values |s| = 2 |e| then at point (0, 0, 2),
the ratio of magnitudes of electric field values due
to sheet charge to that of line charge is
n :1
.
The value of n is________.
Ans. (16)
Sol.
y
x
(0, 0, 2)
z = 4
z
e
+s
S0
0
E 2 r
E2
r
2 2 4
1
n = 16
52. A hydrogen atom changes its state from n = 3 to
n = 2. Due to recoil, the percentage change in the
wave length of emitted light is approximately
1 × 10–n. The value of n is__________.
[Given Rhc = 13.6 eV, hc = 1242 eV nm,
h = 6.6 × 10–34 J s, mass of the hydrogen atom
= 1.6 × 10–27 kg]
Ans. (7)
Sol.
22
11
E 13.6 23
= 1.9 eV
hc
E
hc
E
Pi = Pf
h
0 mv '
h
vm'
2
1 hc
E mv
2'
2
1 hc
h
m
2'
m'
Now
2
2
h hc
E'
2m '
2
2h
' E hc ' 0
2m
2
22 4 Eh
hc h c 2m
'2E
2
2E
hc hc 1 mc
'2E
1
2
22
2E E
1111
'mc mc
22
2
'E
12mc
19
2 27 16
' E 1.9 1.6 10
2mc 2 1.67 10 9 10
= 10–9
% change 10–7
Correct answer 7
53. The magnetic field existing in a region is given by
ˆ
B 0.2 kT
1 2x
. A square loop of edge 50 cm
carrying 0.5 A current is placed in x-y plane with
its edges parallel to the x-y axes, as shown in
figure. The magnitude of the net magnetic force
experienced by the loop is______ mN.
y
(0,0)
2m
2m
x
Ans. (50)
Sol. Force on segment parallel to x-axis will cancel
each other. Hence Fnet will be due to portion
parallel to y-axis.
F = 0.5 × 0.5 × 6 × 0.2 – 0.5 × 0.5 × 0.2 × 5
= 0.5 × 0.5 × 0.2
= 0.25 × 0.2
= 50 × 10–3 N
= 50 mN
54. A alternating current at any instant is given by
i 6 56 sin A
100 t 3
. The rms value
of the current is___________ A.
Ans. (8)
Sol.
2
rms
i dt
Idt
2
2
rms
56
I62
36 28
64
= 8A
55. Twelve wires each having resistance 2 are joined
to form a cube. A battery of 6 V emf is joined
across point a and c. The voltage difference
between e and f is_____ V.
6V
f
a
b
c
d
e
g
h
Ans. (1)
Sol.
6V
f
a
b
c
d
e
g
h
i1
i1
i2
i2/2
i2/2
From symmetry, current through e-b & g-d = 0
eq
33
RR
42
Current through battery
62 4A
3
2
4
i 2 1A
8
V across e-f =
2
i1
R 2 1V
22
56. A soap bubble is blown to a diameter of 7 cm.
36960 erg of work is done in blowing it further. If
surface tension of soap solution is 40 dyne/cm then
the new radius is ______ cm. Take :
22
7
.
Ans. (7)
Sol. = U = SA
36960 erg =
2
2
2
40dyne 7
8 cm
r
cm 2
r = 7 cm
57. Two wavelengths 1 and 2 are used in Young's
double slit experiment 1 = 450 nm and
2 = 650 nm. The minimum order of fringe
produced by 2 which overlaps with the fringe
produced by 1 is n. The value of n is ____.
Ans. (9)
Sol. n22 = n11
21
12
n450 9
n 650 13
n2 = 9
58. An elastic spring under tension of 3 N has a length
a. Its length is b under tension 2 N. For its length
(3a – 2b), the value of tension will be_____ N.
Ans. (5)
Sol. 3 = K (a – )
2 = K (b – )
T = K (3a – 2b – )
T = K (3(a – ) – 2 (b – )
32
K 3 2
KK
= 9 – 4
= 5 N
59. Two forces
1
F
and
2
F
are acting on a body. One
force has magnitude thrice that of the other force
and the resultant of the two forces is equal to the
force of larger magnitude. The angle between
1
F
and
2
F
is
11
cos n
. The value of |n| is ____.
Ans. (6)
Sol.
1F
F
R2
3F
FF
2 2 2
R 1 2 1 2
F F F 2F F cos
9F2 = F2 + 9F2 + 6F2cos
1
cos 6
11
cos 6
n = –6
|n| = 6
60. A solid sphere and a hollow cylinder roll up
without slipping on same inclined plane with same
initial speed v. The sphere and the cylinder reaches
upto maximum heights h1 and h2, respectively,
above the initial level. The ratio h1 : h2 is
n
10
. The
value of n is_____.
Ans. (7)
Sol Gain in P.E. = Loss in K.E.
2
2
2
1K
mgh mv 1
2R
2
2
K
h1
R
1
2
2
1
h77
5
h 1 1 5 2 10
n = 7
CHEMISTRY
TEST PAPER WITH SOLUTION
SECTION-A
61. What pressure (bar) of H2 would be required to
make emf of hydrogen electrode zero in pure water
at 25ºC ?
(1) 10–14 (2) 10–7 (3) 1 (4) 0.5
Sol. 2e– + 2H+(aq) H2(g)
2
H
o
2
P
0.059
E E log
n [H ]
2
H
72
P
0.059
0 0 log
2 (10 )
2
H
72
P
log 0
(10 )
2
H
14
P1
10
2
14
H
P 10 bar
62. The correct sequence of ligands in the order of
decreasing field strength is :
(1)
2
2
CO H O F S
(2)
3
OH F NH CN
(3) NCS– > EDTA4– > CN– > CO
(4) S2– > –OH > EDTA4– > CO
Ans. (1)
Sol. According to spectrochemical series ligand field
strength is CO > H2O > F– > S2–
63. Match List -I with List II:
List - I
Mechanism steps
List - II
Effect
(A)
(I)
– E effect
(B)
(II)
– R effect
(C)
(III)
+ E effect
(D)
(IV)
+ R effect
Choose the correct answer from the options given
below :
(1) (A) – (IV), (B) – (III), (C) – (I), (D) – (II)
(2) (A) – (III), (B) – (I), (C) – (II), (D) – (IV)
(3) (A) – (II), (B) – (IV), (C) – (III), (D) – (I)
(4) (A) – (I), (B) – (II), (C) – (IV), (D) – (III)
Ans. (1)
+
O N = O
O
N O
–
–
+
CN
CN
+H+
+
+
NH2
NH2
H
NTA Ans. (3)
Sol.
NH2
. .
NH2
+
+ R effect
(NH2 electron
donating)
. .
N
– R effect
O
+
O
N
O
O–
(NO
2
electron
withdrawing)
+
+E Effect
+ H+
H
H+ electron
deficient
species)
– E Effect
+ CN
CN
(CN electron
deficient
species)
–
64. What will be the decreasing order of basic strength
of the following conjugate bases ?
3
OH, RO, CH COO, C l
(1)
Cl
> OH > R
O
> CH3CO
O
(2) R
O
>–OH > CH3CO
O
>
Cl
(3) –OH > R
O
> CH3 CO
O
>
Cl
(4)
Cl
> R
O
> –OH > CH3CO
O
Ans. (2)
Sol. Strong acid have weak conjugate base
Acidic strength :
H–Cl > CH3COOH > H2O > R–OH
Conjugate base strength :
Cl– < CH3COO– <
OH
< RO–
65. In the precipitation of the iron group (III) in
qualitative analysis, ammonium chloride is added
before adding ammonium hydroxide to :
(1) prevent interference by phosphate ions
(2) decrease concentration of –OH ions
(3) increase concentration of Cl– ions
(4) increase concentration of NH4
+ ions
Ans. (2)
Sol.
44
NH OH NH OH
44
NH Cl NH Cl
Due to common ion effect of
4
NH
,
[OH–] decreases in such extent that only group-III
cation can be precipitated , due to their very low
Ksp in the range of 10–38.
66.
Br
CH2Br
NaOH(alc)
HBr
Ether
B
C
Identify
B
C
and
and how are
A
and
C
related ?
(B) (C)
(1)
functional
group
isomers
(2)
Derivative
(3)
position
isomers
(4)
chain
isomers
Ans. (3)
Sol.
CH2
H
alc.
Br
Br
B
r
Br
NaOH
(E2)
Br
(A)
(B)
HBr ether
(Electrophilic
Addition
Reaction)
(C)
A and C are position isomer.
Br
Br
Br
Br
Br
Br
Br
OH
Br
OH
OH
OH
Br
OH
67. One of the commonly used electrode is calomel
electrode. Under which of the following categories
calomel electrode comes ?
(1) Metal – Insoluble Salt – Anion electrodes
(2) Oxidation – Reduction electrodes
(3) Gas – Ion electrodes
(4) Metal ion – Metal electrodes
Ans. (1)
Sol. Theory based
68. Number of complexes from the following with
even number of unpaired "d" electrons is____.
[V(H2O)6]3+, [Cr(H2O)6]2+, [Fe(H2O)6]3+,
[Ni(H2O)6]3+, [Cu(H2O)6]2+
[Given atomic numbers : V = 23, Cr = 24, Fe = 26,
Ni = 28, Cu = 29]
(1) 2 (2) 4
(3) 5 (4) 1
Ans. (1)
Sol. [V(H2O)6]3+ d2sp3
23V :- [Ar]3d34s2
V+3 :- [Ar]3d2 , n = 2 (even number of unpaired e–)
[Cr(H2O)6]2+ sp3d2
24Cr :- [Ar]3d54s1
Cr+2 :- [Ar]3d4 , n = 4 (even number of unpaired e–)
eg
t2g
[Fe(H2O)6]3+ sp3d2
Fe3+ [Ar]3d54s0
n = 5 (odd number of unpaired e–)
[Ni(H2O)6] 3+ sp3d2
Ni :- [Ar]3d84s2
Ni+3 :- [Ar]3d7 , n = 3 (odd number of unpaired e–)
[Cu(H2O)6]2+ sp3d2
Cu :- [Ar]3d94s0
n = 1 (odd number of unpaired e–)
69. Which one of the following molecules has
maximum dipole moment ?
(1) NF3 (2) CH4
(3) NH3 (4) PF5
Ans. (3)
Sol. CH4 & PF5 , µnet = 0 (non polar)
33
NH NF
Vector addition of bond Vector subtraction of bond
moment & lone pair moment moment & lone pair moment
µ µ
70. Number of molecules/ions from the following in
which the central atom is involved in sp3
hybridization is ________.
NO3
–, BCl3, ClO2
–, ClO3
(1) 2 (2) 4
(3) 3 (4) 1
Ans. (1)
Sol.
O = N
O
sp2
Cl
Cl
Cl
sp2
Cl
sp3
O
O–
O–
Cl
O
O
O
·
sp3
B
71. Which among the following is incorrect statement?
(1) Electromeric effect dominates over inductive
effect
(2) The electromeric effect is, temporary effect
(3) The organic compound shows electromeric
effect in the presence of the reagent only
(4) Hydrogen ion (H+) shows negative electromeric
effect
Ans. (4)
Sol. Hydrogen ion (H+) shows positive electromeric
effect.
72. Given below are two statements :
Statement I : Acidity of -hydrogens of aldehydes
and ketones is responsible for Aldol reaction.
Statement II : Reaction between benzaldehyde
and ethanal will NOT give Cross – Aldol product.
In the light of 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. (4)
Sol. Aldehyde and ketones having acidic
-hydrogen show aldol reaction
H–C=O
+
CH3–C–H
O
(Acidic H)
Benzaldehyde
Ethanal
Base
H–C=CH–C–H
Cross aldol product
O
73. Which of the following nitrogen containing
compound does not give Lassaigne’s test ?
(1) Phenyl hydrazine (2) Glycene
(3) Urea (4) Hydrazine
Ans. (4)
Sol. Hydrazine (NH2–NH2) have no carbon so does not
show Lassaigne’s test.
74. Which of the following is the correct structure of
L-Glucose ?
(1)
CHO
CH2OH
HO
HO
HO
OH
(2)
CHO
CH2OH
HO
HO
OH
OH
(3)
CHO
CH2OH
HO
HO
OH
OH
(4)
CHO
CH2OH
HO
HO
HO
HO
Ans. (1)
Sol. Structure of L-Glucose is
CHO
CH2OH
HO
HO
HO
OH
75. The element which shows only one oxidation state
other than its elemental form is :
(1) Cobalt (2) Scandium
(3) Titanium (4) Nickel
Ans. (2)
Sol. Co, Ti, Ni can show +2, +3 and +4 oxidation state,
But 'Sc' only shows +3 stable oxidation state.
76. Identify the product in the following reaction :
O
O
Zn – Hg
HCl
Product
H
(1)
OH
OH
(2)
O
(3)
OH
O
(4)
Ans. (4)
Sol.
H
O
O
Zn –Hg
HCl
(Clemmensen
reduction)
77. Number of elements from the following that
CANNOT form compounds with valencies which
match with their respective group valencies is
______.
B, C, N, S, O, F, P, Al, Si
(1) 7 (2) 5 (3) 6 (4) 3
Ans. (4)
Sol. N,O, F can’t extend their valencies upto their
group number due to the non-availability of vacant
2d like orbital.
78. The Molarity (M) of an aqueous solution
containing 5.85 g of NaCl in 500 mL water is :
(Given : Molar Mass Na : 23 and Cl : 35.5 gmol–1)
(1) 20 (2) 0.2
(3) 2 (4) 4
Ans. (2)
Sol.
NaCl
sol
n
MV (in L)
5.85
58.5
M 0.2M
0.5
79. Identify the correct set of reagents or reaction
conditions ‘X’ and ‘Y’ in the following set of
transformation.
CH3 – CH2 – CH2 – Br Product
‘X’
‘Y’
CH3 – CH – CH3
Br
(1) X = conc.alc. NaOH, 80°C, Y = Br2/CHCl3
(2) X = dil.aq. NaOH, 20°C, Y = HBr/acetic acid
(3) X = conc.alc. NaOH, 80°C, Y = HBr/acetic
acid
(4) X = dil.aq. NaOH, 20°C, Y = Br2/CHCl3
Ans. (3)
Sol. CH3–CH2–CH2–Br
o
X conc.alc.NaOH
80 C
CH3–CH=CH2
Y HBr /Aceticacid
CH3–CHBr – CH3
80. The correct order of first ionization enthalpy
values of the following elements is :
(A) O (B) N
(C) Be (D) F
(E) B
Choose the correct answer from the options given
below :
(1) B < D < C < E < A (2) E < C < A < B < D
(3) C < E < A < B < D (4) A < B < D < C < E
Ans. (2)
Sol. Correct order of Ist IE
Li < B < Be < C < O < N < F < Ne
E < C < A < B < D
SECTION-B
81. The enthalpy of formation of ethane (C2H6) from
ethylene by addition of hydrogen where the bond-
energies of C – H, C – C, H – H are 414 kJ, 347 kJ,
615 kJ and 435 kJ respectively is - __________ kJ.
Ans. (125)
Sol. C2H4(g) + H2(g) C2H6(g)
H = BE(C = C) + 4BE (C – H) + BE (H – H)
– BE(C – C) – 6BE (C – H)
H = BE(C = C) + BE(H – H) – BE(C – C)
– 2BE(C – H)
= 615 + 435 – 347 – 2 × 414
= – 125 kJ
82. The number of correct reaction(s) among the
following is ________.
(A)
+
C
O
Cl
Anhyd.AlCl3
CH2
(B)
C
O
Cl
H2
COOH
Pd – BaSO4
(C)
CO, HCl
CHO
Anhyd.AlCl3/CuCl
(D)
H3O+
NH2
CONH2
Ans. (1)
Sol.
(A)
+
C
O
Cl
Anhy. AlCl3
CH2
(Incorrect)
(B)
O
C
Cl
H2
(Incorrect)
Pd –BaSO4
COOH
(C)
CO, HCl
(Correct)
Anhy. AlCl3 / CuCl
CHO
(D)
H3O
(Incorrect)
NH2
CONH2
+
83. X g of ethylamine is subjected to reaction with
NaNO2/HCl followed by water; evolved dinitrogen
gas which occupied 2.24 L volume at STP.
X is ______ × 10–1 g.
Ans. (45)
Sol.
3 2 2
Mol.wt.45g
CH CH NH
2
NaNO HCl
2
HO
CH3CH2–OH +
2
14 g
N
given : N2 evolved is 2.24 L i.e. 0.1 mole.
i.e. CH3CH2NH2 (ethyl amine) will be 4.5 g
(=0.1 mole)
Hence the answer = 45 × 10–1 g
84. The de-Broglie’s wavelength of an electron in the
4th orbit is _______ a0. (a0 = Bohr’s radius)
Ans. (8)
Sol. 2rn = nd
2
0d
n
2 a n
Z
2
0d
4
2 a 4
1
d0
8a
85. Only 2 mL of KMnO4 solution of unknown
molarity is required to reach the end point of a
titration of 20 mL of oxalic acid (2 M) in acidic
medium. The molarity of KMnO4 solution should
be ________ M.
NTA Ans. (50)
Sol. eq.(KMnO4) = eq.(H2C2O4)
M × 2 × 5 = 2 × 20 × 2
M = 8M
86. Consider the following reaction
MnO2 + KOH + O2 A + H2O.
Product ‘A’ in neutral or acidic medium
disproportionate to give products ‘B’ and ‘C’ along
with water. The sum of spin-only magnetic
moment values of B and C is __________ BM.
(nearest integer)
(Given atomic number of Mn is 25)
Ans. (4)
Sol. MnO2 + KOH + O2 K2MnO4 + H2O
(A)
K2MnO4
Neutral/acidic solution
KMnO4 + MnO2
Mn+4 :- [Ar]3d3
n = 3, =
3(3 2)
= 3.87 B.M.
Nearest integer is (4)
87. Consider the following transformation involving
first order elementary reaction in each step at
constant temperature as shown below.
Step 1 Step 2
Step 3
A B C P
Some details of the above reaction are listed
below.
Step
Rate constant
(sec–1)
Activation
energy (kJ mol–1)
1
k1
300
2
k2
200
3
k3
Ea3
If the overall rate constant of the above
transformation (k) is given as
12
3
kk
kk
and the
overall activation energy (Ea) is 400 kJ mol–1, then
the value of Ea3 is ______ kJ mol–1 (nearest
integer)
Ans. (100)
Sol.
12
3
KK
KK
aa
12
a
a3
EE
ERT RT
12
RT E
RT
3
A e A e
Ae
Ae
a a a
a 1 2 3
(E E E )
E12
RT RT
3
AA
Ae e
A
1 2 3
a a a a
E E E E
400 = 300 + 200 –
3
a
E
3
a
E
= 100 kJ/mole
88. 2.5 g of a non-volatile, non-electrolyte is dissolved
in 100 g of water at 25°C. The solution showed a
boiling point elevation by 2°C. Assuming the
solute concentration in negligible with respect to
the solvent concentration, the vapour pressure of
the resulting aqueous solution is _______ mm
of Hg (nearest integer)
[Given : Molal boiling point elevation constant of
water (Kb) = 0.52 K. kg mol–1,
1 atm pressure = 760 mm of Hg, molar mass of
water = 18 g mol–1]
Ans. (707)
Sol. 2 = 0.52 × m
2
m0.52
According to question, solution is much diluted
so
solute
osolvent
n
P
Pn
solvent
o
Pm
M
P 1000
osolvent
m
P P M
1000
2
0.52
760 18
1000
= 52.615
P5 = 760 – 52.615 = 707.385 mm of Hg
89. The number of different chain isomers for C7H16 is
__________.
Ans. (9)
Sol.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
90. Number of molecules/species from the following
having one unpaired electron is ________.
1 1 2
2 2 2
O ,O , NO, CN , O
Ans. (2)
Sol. According to M.O.T.
O2 no. of unpaired electrons = 2
O2
– no. of unpaired electron = 1
NO no. of unpaired electron = 1
CN– no. of unpaired electron = 0
2
2
O
no. of unpaired electron = 0
