FINAL JEE–MAIN EXAMINATION – APRIL, 2024
(Held On Saturday 06th April, 2024) TIME : 3 : 00 PM to 6 : 00 PM
NOON
MATHEMATICS
TEST PAPER WITH SOLUTION
SECTION-A
1. Let ABC be an equilateral triangle. A new triangle
is formed by joining the middle points of all sides
of the triangle ABC and the same process is
repeated infinitely many times. If P is the sum of
perimeters and Q is be the sum of areas of all the
triangles formed in this process, then:
(1)
2
P 36 3Q
(2)
2
P 6 3Q
(3)
2
P 36 3Q
(4)
2
P 72 3Q
Ans. (1)
Sol.
a/2
a
a
a
Area of first
2
3a
4
Area of second
2 2 2
3a a 3a
4 4 16
Area of third
2
a
64
sum of area =
2
3a 1 1
1 ....
4 4 16
22
3a 1 a
Q3
43
4
perimeter of 1st = 3a
perimeter of 2nd =
3a
2
perimeter of 3rd =
3a
4
11
P 3a 1 ...
24
P = 3a.2 = 6a
P
a6
2
1P
Q36
3
2
P 36 3Q
2. Let A = {1, 2, 3, 4, 5}. Let R be a relation on A
defined by xRy if and only if 4x 5y. Let m be the
number of elements in R and n be the minimum
number of elements from A × A that are required
to be added to R to make it a symmetric relation.
Then m + n is equal to:
(1) 24 (2) 23
(3) 25 (4) 26
Ans. (3)
Sol. Given : 4x 5y
then
R 1,1 ,(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4)
(2,5),(3,3),(3, 4),(3,5),(4,4),(4,5),(5,4),(5,5)}
i.e. 16 elements.
i.e. m = 16
Now to make R a symmetric relation add
{(2,1)(3,2)(4,3)(3,1)(4,2)(5,3)(4,1)(5,2)(5,1)}
i.e. n = 9
So m + n = 25
3. If three letters can be posted to any one of the 5
different addresses, then the probability that the
three letters are posted to exactly two addresses is:
(1)
12
25
(2)
18
25
(3)
4
25
(4)
6
25
Ans. (1)
Sol. Total method = 53
faverable =
53
2
C 2 – 2 60
probability =
60 12
125 25
4. Suppose the solution of the differential equation
dy (2 )x y 2
dx x 2 y ( 4 )
represents a circle
passing through origin. Then the radius of this
circle is :
(1)
17
(2)
1
2
(3)
17
2
(4) 2
Ans. (3)
Sol.
dy 2 x – y 2
dx x – y 2 4
xdy – (2 )ydy 4 dy 2 xdx – ydx 2dx
(xdy ydx) – (2 )ydy 4 dy 2 xdx 2dx
22
2y 2x
xy – 4 y
22
0
for this to be circle
2
2
x
2 y 2x – 4 y 0
2
2 + a = 2a
coeff. of
x2 = y2
2
i.e. 2x2 + 2y2 + 2x – 8y = 0
x2 + y2 + x – 4y = 0
1 17
rd 4
42
5. If the locus of the point, whose distances from the
point (2, 1) and (1, 3) are in the ratio 5 : 4, is
ax2 + by2 + cxy + dx + ey + 170 = 0, then the value
of a2 + 2b + 3c + 4d + e is equal to:
(1) 5 (2) –27
(3) 37 (4) 437
Ans. (3)
Sol. let P(x, y)
22
22
x – 2 (y – 1) 25
16
x – 1 y – 3
9x2 + 9y2 + 14x – 118y + 170 = 0
a2 + 2b + 3c + 4d + e
= 81 + 18 + 0 + 56 – 118
= 155 – 118
= 37
6.
22 2
3 3 3 2 2 2
n
(1 1)(n 1) (2 2)(n 2) ..... ·1
(n 1) (n 1)
lim (1 2 ..... n ) (1 2 ..... n )
is equal to:
(1)
2
3
(2)
1
3
(3)
3
4
(4)
1
2
Ans. (2)
Sol.
n–1 2
r1
nn
n32
r 1 r 1
r – r n – r
lim
r – r
n–1 32
r1
2
n
–r r n 1 – nr
lim
n n 1 n n 1 2n 1
–
26
22
n
n – 1 n n 1 n – 1 n(2n – 1) n (n – 1)
–
2 6 2
lim n(n 1) n n 1 2n 1
–
2 2 3
2
n
n n – 1 –n n – 1 n 1 (2n – 1) –n
2 2 3
lim n(n 1) 3n 3n – 4n – 2
26
22
2
n
n – 1 –3n 3n 2 2n n – 1 – 6
lim
n 1 3n – n – 2
2
2
n
n – 1 n 5n – 8 1
lim 3
n 1 3n – n – 2
7. Let 0 r n. If n+1Cr+1 : nCr : n–1Cr–1 = 55 : 35 : 21,
then 2n + 5r is equal to:
(1) 60 (2) 62
(3) 50 (4) 55
Ans. (3)
Ans.
n1
r
n
r
C55
35
C
n 1 ! r! n – r ! 11
!
r 1 !(n – r) n! 7
n 1 11
r 1 7
7n = 4 + 11r
n
r
n–1
r–1
C35
21
C
n! (r – 1)!(n – r)! 5
r! n – r ! (n – 1)! 3
n5
r3
3n = 5r
By solving r = 6 n = 10
2n + 5r = 50
8. A software company sets up m number of
computer systems to finish an assignment in
17 days. If 4 computer systems crashed on the start
of the second day, 4 more computer systems
crashed on the start of the third day and so on, then
it took 8 more days to finish the assignment. The
value of m is equal to :
(1) 125 (2) 150
(3) 180 (4) 160
Ans. (2)
Sol. 17m = m + (m – 4) + (m – 4 × 2)...+...(m – 4 × 24)
17 m = 25m – 4 (1 + 2 ...24)
4 24 25
8m 150
2
9. If z1, z2 are two distinct complex number such that
12
12
z 2z
2
1zz
2
, then
(1) either z1 lies on a circle of radius 1 or z2 lies on a
circle of radius
1
2
(2) either z1 lies on a circle of radius
1
2
or z2 lies on
a circle of radius 1.
(3) z1 lies on a circle of radius
1
2
and z2 lies on a
circle of radius 1.
(4) both z1 and z2 lie on the same circle.
Ans. (1)
Sol.
1 2 1 2
1 2 1 2
z – 2z z – 2z 4
11
– z z – z z
22
22
1 1 2 1 2 2
z 2z z – 2z z 4 z
22
1 2 1 2
12
z z z z
1
4 – – z z
4 2 2
1 1 2 2 1 1 2 2
z z 2z 2z – z z 2z 2z – 1 0
1 2 2
z, z – 1 1– 2z 2z 0
22
12
z –1 2z – 1 0
10. If the function
2x
1
f(x) x
; x > 0 attains the
maximum value at
1
xe
then :
(1)
e
e
(2)
2e
e (2 )
(3)
e
e
(4)
(2e)
(2e)
Ans. (3)
Sol. Let
2x
1
yx
ny =
1
2x n x
ny –2x nx
1 dy –2 1 nx
y dx
for
1
xe
fn is decreasing
so, e <
2e 2
11
e
e > e
11. Let
ˆˆˆ
a 6i j k
and
ˆˆ
b i j
. If
c
is a is vector
such that
6, a.c 6 , 2 2
c c c a
and the
angle between
ab
and
c
is 60°, then
c
ab
is equal to:
(1)
966
2
(2)
33
2
(3)
36
2
(4)
966
2
Ans. (4)
Sol.
3
a b c a b c 2
c – a 2 2
22
c a – 2c a 8
2
z 38 –12 z 8
2
z – 12 z 30 0
12 144 – 120
z2
12 2 6
2
z 6 6
ˆˆ ˆ
jk
a b 6 1 –1
1 1 0
ˆˆ ˆ
– j 5k
a b 27
3
a b z 27 6 6 2
966
2
12. If all the words with or without meaning made
using all the letters of the word "NAGPUR" are
arranged as in a dictionary, then the word at 315th
position in this arrangement is :
(1) NRAGUP (2) NRAGPU
(3) NRAPGU (4) NRAPUG
Ans. (3)
Sol. NAGPUR
A 5! = 120
G ® 5! = 120 240
NA ® 4! = 24 264
NG ® 4! = 24 288
NP ® 4! = 24 312
NRAGPU = 1 313
NRAGUP 314
NRAPGU 315
13. Suppose for a differentiable function h, h(0) = 0,
h(1) = 1 and h'(0) = h'(1) = 2. If g(x) = h(ex) eh(x),
then g'(0) is equal to:
(1) 5 (2) 3
(3) 8 (4) 4
Ans. (4)
Sol.
x h(x)
g x h e e
x h(x) h(x) x x
g' x h(e ) e h '(x) e h ' e e
h(0) h(0)
g'(0) h(1)e h'(0) e h '(1)
= 2 + 2 = 4
14. Let P () be the image of the point Q(3, –3, 1)
in the line
x 0 y 3 z 1
1 1 1
and R be the point
(2, 5, –1). If the area of the triangle PQR is and
2 = 14K, then K is equal to:
(1) 36 (2) 72
(3) 18 (4) 81
Ans. (4)
Sol.
R(2,5,–1)
S
P(,,)
Q(3,–3,1)
RQ 1 64 4 69
ˆˆˆ
RQ – 8j 2k
ˆˆˆ
RS j – k
RQ RS 1 – 8 – 2 9
cos 69 3 3 23
RQ RS
3 RS RS
cos RQ
23 69
RS 3 3
14 QS
sin 23 69
QS 42
area =
12QS RS 42 3 3
2
9 14
281.14 14k
k = 81
15. If P(6, 1) be the orthocentre of the triangle whose
vertices are A(5, –2), B(8, 3) and C(h, k), then the
point C lies on the circle.
(1) x2 + y2 – 65 = 0 (2) x2 + y2 – 74 = 0
(3) x2 + y2 – 61 = 0 (4) x2 + y2 – 52 = 0
Ans. (1)
Sol.
B(8,3)
D
C(h, k)
(6,1)P
F
E
A(5,–2)
Slope of AD = 3
Slope of BC
1
–3
equation of BC = 3y + x – 17 = 0
slope of BE = 1
Slope of AC = –1
equation of AC is x + y – 3 = 0
point C is (–4, 7)
16. Let
1
f(x) 7 sin 5x
be a function defined on R.
Then the range of the function f(x) is equal to:
(1)
11
,
85
(2)
11
,
76
(3)
11
,
75
(4)
11
,
86
Ans. (4)
Sol. sin5x [–1,1]
–sin5x [–1, 1]
7 – sin5x [6, 8]
1 1 1
,
7 – sin 5x 8 6
17. Let
ˆˆˆ
a 2i j k
,
ˆˆ
ˆˆ
b a i i
ij
.
Then the square of the projection of
a
on
b
is :
(1)
1
5
(2) 2
(3)
1
3
(4)
2
3
Ans. (2)
Sol.
ˆˆ ˆ
i j k
ˆˆ
a i j 2 1 –1
1 1 0
ˆˆˆ
i – j k
ˆ ˆ ˆ ˆ
ˆ
a i j i k j
ˆ ˆ ˆ ˆ ˆ
a i j i i j – k
projection of
a
on
ˆ
b
ab
b
11 2
2
18. If the area of the region
2
a1
(x,y) : y , 1 x 2, 0 a 1
x
x
is
(loge2) –
1
7
then the value of 7a – 3 is equal to:
(1) 2 (2) 0
(3) –1 (4) 1
Ans. (3)
Sol.
a
1
2
area
2
2
1
1a
– dx
xx
2
1
a
nx x
e
a1
n2 – a log 2 –
27
–a 1
–
27
2
a7
7a = 2
7a – 3 = –1
19. If
1
2 2 2 2
11
dx tan (3tan x)
12
a sin x b cos x
+
constant, then the maximum value of
asinx + bcosx, is :
(1)
40
(2)
39
(3)
42
(4)
41
Ans. (1)
Sol.
2
2 2 2
sec xdx
a tan x b
let tanx = t
sec2dx = dt
2 2 2
dt
a t b
22
2
1 dt
ab
ta
–1
2
1 1 t
tan a c
bb
a
a
–1
1tan tan x c
ab b
on comparing
a3
b
ab = 12
a = 6, b = 2
maximum value of
6 sinx + 2cosx is
40
20. If A is a square matrix of order 3 such that
det(A) = 3 and
det(adj(–4 adj(–3 adj(3 adj((2A)–1))))) = 2m3n,
then m +| 2n is equal to:
(1) 3 (2) 2
(3) 4 (4) 6
Ans. (3)
Sol.
A3
–1
adj(–4adj(–3adj (3adj (2A) )))
2
–1
–4adj –3adj(3adj 2A
2
6 –1
4 adj –3adj 3adj(2A)
8
–1
12 12
2 3 3adj 2A
12 8
–1
12 24
2 3 3 adj 2A
12 16
–1
36
2 3 2A
12 36
16
1
23
2A
12 36
16
48
1
23
2A
12 36
48 16
1
23
23
20
–36 20
36
33
2
m = – 36 n = 20
m + 2n = 4
SECTION-B
21. Let [t] denote the greatest integer less than or equal
to t. Let f: [0, ) R be a function defined by
x
f(x) 3x
2
. Let S be the set of all points
in the interval [0, 8] at which f is not continuous.
Then
aS
a
is equal to ________.
Ans. (17)
Sol.
x3
2
is discontinuous at x = 2,4,6,8
x
is discontinuous at x = 1,4
F(x) is discontinuous at x = 1,2,6,8
a 1 2 6 8 17
22. The length of the latus rectum and directrices of a
hyperbola with eccentricity e are 9 and x =
4
3
,
respectively. Let the line y –
3
x +
3
= 0 touch
this hyperbola at (x0, y0). If m is the product of the
focal distances of the point (x0, y0), then 4e2 + m is
equal to _______.
NTA Ans. (61)
Ans. (Bonus)
Sol. Given
2
2b
a
= 9 and
a4
e3
equation of tangent y –
3
x +
3
= 0
by equation of tangent
Let slope = S =
3
Constant = –
3
By condition of tangency
6 = 6a2 – 9a
a = 2, b2 = 9
Equation of Hyperbola is
22
xy
–1
49
and for tangent
Point of contact is (4, 3
3
) = (x0, y0)
Now e =
9
14
=
13
2
Again product of focal distances
m = (x0e + a) (x0e – a)
m + 4e2 = 20e2 – a2
= 20 ×
13
4
– 4 = 61
(There is a printing mistake in the equation of
directrix x =
4
3
.
Corrected equation is x =
4
13
for directrix, as
eccentricity must be greater than one, so question
must be bonus)
23. If S(x) = (1 + x) + 2(1 + x)2 + 3(1 + x)3+.....
+ 60(1 + x)60, x 0, and (60)2 S(60) = a(b)b + b,
where a, b N, then (a + b) equal to ______
Ans. (3660)
Sol.
S(x)=(1+x) + 2(1+x)2 + 3(1+x)3+..+ 60(1 + x)60
(1+x)S = (1+x)2 + ……. 59 (1+x)60+ 60(1+x)61
60
61
1 x 1 x 1
xS 60 1 x
x
Put x = 60
60
61
61 61 1
60S 60 61
60
on solving 3660
24. Let [t] denote the largest integer less than or equal
to t. If
2
32
0
x
[x ] dx a b 2 3 5 c 6 7
2
,
where a, b, c z, then a + b + c is equal to ______
Ans. (23)
Sol.
33
2
2
00
x
x dx dx
2
1 12 3
01 2
0 dx 1dx 2 dx
2 5 6
2
35
3dx 4dx 5dx
7 8 3
6 7 8
6dx 7dx 8dx
22
02
0 dx 1dx
+
68
26
2dx 3dx
+
3
8
4dx 31 6 2 3 5
2 6 7
a = 31 b = –6 c = – 2
a + b + c = 31 – 6 – 2 = 23
25. From a lot of 12 items containing 3 defectives, a
sample of 5 items is drawn at random. Let the
random variable X denote the number of defective
items in the sample. Let items in the sample be
drawn one by one without replacement. If variance
of X is
m
n
, where gcd(m, n) = 1, then n – m is
equal to _______.
Ans. (71)
Sol.
3
5
12
5
C
a1 C
9
4
12
5
C
b 3. C
9
3
12
5
C
c 3. C
9
2
12
5
C
d 1. C
u = 0.a + 1.b + 2.c + 3.d = 1.25
2 = 0.a +1.b + 4.c + 9d – u2
2105
176
Ans. 176 – 105 = 71
26. In a triangle ABC, BC = 7, AC = 8, AB =
and cosA =
2
3
. If 49cos(3C) + 42 =
m
n
, where
gcd(m, n) = 1, then m + n is equal to _______
Ans. (39)
26. In a triangle ABC, BC = 7, AC = 8, AB =
and cosA =
2
3
. If 49cos(3C) + 42 =
m
n
, where
gcd(m, n) = 1, then m + n is equal to _______
Ans. (39)
Sol.
222
b c a
cosA 2bc
2 2 2
2 8 c 7
3 2 8 c
C = 9
2 2 2
7 8 9 2
cosC 2 7 8 7
49 cos3C + 42
49(4 cos3 C – 3 cosC) + 42
3
22
49 4 3 42
77
32
7
m + n = 32 + 7 = 39
27. If the shortest distance between the lines
x y 2 z 1
3 1 1
and
x 2 y 5 z 4
3 2 4
is
44
30
, then the largest possible value of || is equal
to _______.
Ans. (43)
Sol.
1ˆˆ
ˆ
a i 2j k
2ˆˆ ˆ
a 2i 5j 4k
ˆˆˆ
p 3i j k
ˆˆˆ
q 3i 2j 4k
12
ˆˆˆ
2 i 7j 3k a a
ˆˆ
ˆ
p q 6i 15j 3k
22
2
6 12 105 9
44
30 6 15 3
6 126
44
30 3 30
132 6 126
= 1, = –43
|| = 43
28. Let , be roots of
2
x 2x – 8 0
.
If
nn
n
U,
then
10 9
8
U 12U
2U
is equal to _______.
Ans. (4)
Sol.
10 10 9 9
88
2
2
8 2 8 2
88
22
2
88
88
88 4
2
29. If the system of equations
2x + 7y + z = 3
3x + 2y + 5z = 4
x + y + 32z = –1
has infinitely many solutions, then ( – ) is equal
to ________ :
Ans. (38)
Sol. D = D1 = D2 = D3 = 0
D3 =
2 7 3
3 2 4 0 39
11
27
D 3 2 5 0 1
1 39 32
38
30. If the solution y(x) of the given differential
equation (ey + 1) cosx dx + ey sin x dy = 0 passes
through the point
,0
2
, then the value of
y6
e
is equal to ______.
Ans. (3)
Sol. (ey + 1) cosx dx + ey sinx dy = 0
y
d e 1 sin x
= 0
y
e 1 sin x C
It passes through
,0
2
c2
Now,
x6
y
e3
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
31. The longest wavelength associated with Paschen
series is : (Given RH =1.097 × 107 SI unit)
(1) 1.094 × 10–6m (2) 2.973 × 10–6m
(3) 3.646 × 10–6m (4) 1.876 × 10–6m
Ans. (4)
Sol. For longest wavelength in Paschen's series:
22
12
1 1 1
Rnn
For longest n1 = 3
n2 = 4
22
1 1 1
R(3) (4)
1 1 1
R9 16
1 16 9
R16 9
=
7
16 9 16 9
7R 7 1.097 10
= 1.876 × 10–6 m
32. A total of 48 J heat is given to one mole of helium
kept in a cylinder. The temperature of helium
increases by 2°C. The work done by the gas is :
(Given, R = 8.3 J K–1mol–1.)
(1) 72.9 J (2) 24.9 J
(3) 48 J (4) 23.1 J
Ans. (4)
Sol. 1st law of thermodynamics
Q = U + W
+48 = nCvT + W
48 = (1)
3R
2
(2) + W
W = 48 – 3 × R
W = 48 – 3 × (8.3)
W 23.1 Joule
33. In finding out refractive index of glass slab the
following observations were made through
travelling microscope 50 vernier scale division =
49 MSD; 20 divisions on main scale in each cm
For mark on paper
MSR = 8.45 cm, VC = 26
For mark on paper seen through slab
MSR = 7.12 cm, VC = 41
For powder particle on the top surface of the glass
slab
MSR = 4.05 cm, VC = 1
(MSR = Main Scale Reading, VC = Vernier
Coincidence)
Refractive index of the glass slab is:
(1) 1.42 (2) 1.52
(3) 1.24 (4) 1.35
Ans. (1)
Sol. 1 MSD =
1cm
20
= 0.05 cm
1 VSD =
49
50
MSD =
49
50
×0.05 cm = 0.049 cm
LC = 1MSD – 1VSD = 0.001 cm
For mark on paper, L1 = 8.45 cm + 26 × 0.001 cm
= 84.76 mm
For mark on paper through slab, L2 = 7.12 cm +
41× 0.001 cm = 71.61 mm
For powder particle on top surface, ZE = 4.05 cm
+ 1 × 0.001 cm = 40.51 mm
actual L1 = 84.76 – 40.51 = 44.25 mm
actual L2 = 71.61 – 40.51 = 31.10 mm
1
2
L
L
1
2
L44.25 1.42
L 31.10
34. In the given electromagnetic wave
Ey = 600 sin (t – kx) Vm–1, intensity of the
associated light beam is (in W/m2); (Given 0 =
9×10–12C2N–1m–2)
(1) 486 (2) 243
(3) 729 (4) 972
Ans. (1)
Sol. Intensity =
2
00
1Ec
2
=
1
2
× 9 ×10–12 × (600)2 × 3 × 108
=
9
2
× 36 × 3 = 486 w/m2
35. Assuming the earth to be a sphere of uniform mass
density, a body weighed 300 N on the surface of
earth. How much it would weigh at R/4 depth
under surface of earth ?
(1) 75 N (2) 375 N
(3) 300 N (4) 225 N
Ans. (4)
Sol. At surface: mg = 300 N
m =
s
300
g
At Depth
R
4
: gd =
s
d
g1 R
ds
R
g g 1 4R
s
d
3g
g4
weight at depth = m × gd
=
s
3g
m4
=
3300
4
= 225 N
36. The acceptor level of a p-type semiconductor is
6eV. The maximum wavelength of light which can
create a hole would be : Given hc = 1242 eV nm.
(1) 407 nm (2) 414 nm
(3) 207 nm (4) 103.5 nm
Ans. (3)
Sol. Energy =
hc
;
1240
E eV
(nm)
1240
6(nm)
1240 207nm
6
37. A car of 800 kg is taking turn on a banked road of
radius 300 m and angle of banking 30°. If
coefficient of static friction is 0.2 then the
maximum speed with which car can negotiate the
turn safely : (g = 10 m/s2,
3
=1.73)
(1) 70.4 m/s (2) 51.4 m/s
(3) 264 m/s (4) 102.8 m/s
Ans. (2)
Sol. m = 800 kg
r = 300 m
= 30°
s = 0.2
max
tan
V Rg 1 tan
=
tan30 0.2
300 g 1 0.2 tan30
=
0.57 0.2
300 10 1 0.2 0.57
Vmax = 51.4 m/s
38. Two identical conducting spheres P and S with
charge Q on each, repel each other with a force
16N. A third identical uncharged conducting
sphere R is successively brought in contact with
the two spheres. The new force of repulsion
between P and S is :
(1) 4 N (2) 6 N
(3) 1 N (4) 12 N
Ans. (2)
Sol.
Q
Q
0
P
S
R
FPS Q2
FPS = 16 N
Now If P & R are brought in contact then
Q/2
Q
Q/2
P
S
R
Now If S & R are brought in contact then
Q/2
3Q/4
3Q/4
P
S
R
New force between P & S is :
FPS
Q 3Q
24
FPS
2
3Q
8
=
316
8
6N
39. In a coil, the current changes form –2 A to +2A in
0.2 s and induces an emf of 0.1 V. The self-
inductance of the coil is :
(1) 5 mH (2) 1 mH
(3) 2.5 mH (4) 4 mH
Ans. (1)
Sol. (Emf)induced =
di
–L dt
In magnitude form,
ind
di
Emf (–)L dt
0.1 =
(L)[ 2 ( 2)]
0.2
0.1 0.2
L 5mH
4
40. For the thin convex lens, the radii of curvature are
at 15 cm and 30 cm respectively. The focal length
the lens is 20 cm. The refractive index of the
material is :
(1) 1.2 (2) 1.4
(3) 1.5 (4) 1.8
Ans. (3)
Sol.
lens
air 1 2
1 1 1
1
f R R
1 1 1
1
20 1 15 ( 30)
1
20
= (–1)
3
30
–1 =
1
2
13
1 1 5
22
41. Energy of 10 non rigid diatomic molecules at
temperature T is :
(1)
7
2
RT (2) 70 KBT
(3) 35 RT (4) 35 KBT
Ans. (4)
Sol. Degree of freedom(f) = 5 + 2(3N – 5)
f = 5 + 2(3 × 2 –1) = 7
energy of one molecule =
B
fKT
2
energy of 10 molecules
=
B
f
10 K T
2
=
B
7
10 K T
2
= 35 KBT
42. A body of weight 200 N is suspended form a tree
branch thought a chain of mass 10 kg. The branch
pulls the chain by a force equal to (if g = 10 m/s2):
(1) 150 N (2) 300 N
(3) 200 N (4) 100 N
Ans. (2)
Sol.
total weight
= 200+10×10
=300N
200N
10×10
T
Chain block system is in equilibrium so
T = 200 + 100 = 300 N.
43. When UV light of wavelength 300 nm is incident
on the metal surface having work function 2.13 eV,
electron emission takes place. The stopping
potential is : (Given hc = 1240 eV nm)
(1) 4 V (2) 4.1 V (3) 2 V (4) 1.5 V
Ans. (3)
Sol.
s
hc e.V
1240
300
eV – 2.13 eV = eVs
4.13 eV – 2.13 eV = eVs.
So,
s
V 2volt
44. The number of electrons flowing per second in the
filament of a 110 W bulb operating at 220 V is :
(Given e = 1.6 × 10–19 C)
(1) 31.25 × 1017 (2) 6.25 × 1018
(3) 6.25 × 1017 (4) 1.25 × 1019
Ans. (1)
Sol. Power (P) = V.I
110 = (220) (I)
I = 0.5 A
Now, I =
ne
t
0.5 =
n
t
(1.6 × 10–19)
n
t
=
19
0.5
1.6 10
17
n31.25 10
t
45. When kinetic energy of a body becomes 36 times
of its original value, the percentage increase in the
momentum of the body will be :
(1) 500% (2) 600%
(3) 6% (4) 60%
Ans. (1)
Sol. Kinetic energy (K) =
2
P
2m
P 2mK
If Kf = 36 Ki
So, Pf = 6 Pi
% increase in momentum =
fi
i
PP
100%
P
=
ii
i
6P P 100%
P
= 500%
46. Pressure inside a soap bubble is greater than the
pressure outside by an amount :
(given : R = Radius of bubble, S = Surface tension
of bubble)
(1)
4S
R
(2)
4R
S
(3)
S
R
(4)
2S
R
Ans. (1)
Sol. There are two liquid-air surfaces in bubble so
2S 4S
P2
RR
47. Match List-I with List-II
List-I
(Y vs X)
List-II
(Shape of Graph)
(A)
Y = magnetic
susceptibility
X = magnetising
field
(I)
Y
X
(B)
Y = magnetic
field
X = distance
from centre of a
current carrying
wire for x < a
(where a=radius
of wire)
(II)
Y
X
(C)
Y = magnetic
field
X = distance
from centre of a
current carrying
wire for x > a
(where a =
radius of wire)
(III)
X
Y
(D)
Y= magnetic
field inside
solenoid
X = distance
from center
(IV)
Y
X
Choose the correct answer from the options given
below :
(1) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(2) (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
(3) (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
(4) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Ans. (4)
Sol. (A) Graph between Magnetic susceptibility and
magnetising field is :
X
Y
(B) magnetic field due to a current carrying wire
for x a :
0
2
ir
B2a
Y
X
(C) magnetic field due to a current carrying wire
for x a :
0i
B2a
Y
X
(D) magnetic field inside solenoid varies as:
Y
X
48. In a vernier calliper, when both jaws touch each
other, zero of the vernier scale shifts towards left
and its 4th division coincides exactly with a certain
division on main scale. If 50 vernier scale divisions
equal to 49 main scale divisions and zero error in
the instrument is 0.04 mm then how many main
scale divisions are there in 1 cm ?
Sol. 4th division coincides with 3rd division then
0.004 cm = 4VSD – 3MSD
49MSD = 50 VSD
1MSD =
1
N
cm
0.004 =
49
4 MSD
50
– 3MSD
0.004 =
196 3
50
1
N
N =
46 1000 46 1000 230
50 4 200
49. Given below are two statements :
Statement (I) : Dimensions of specific heat is
[L2T–2K–1]
Statement (II) : Dimensions of gas constant is
[M L2T–1K–1]
(1) Statement (I) is incorrect but statement (II) is
correct
(2) Both statement (I) and statement (II) are
incorrect
(3) Statement (I) is correct but statement (II) is
incorrect
(4) Both statement (I) and statement (II) are correct
Ans. (3)
Sol. Q = mST
s =
Q
mT
[s] =
22
ML T
MK
[s] = [L2 T–2 K–1]
Statement-(I) is correct
PV = nRT R =
PV
nT
[R] =
1 2 3
[ML T ][L ]
[mol][K]
[R] = [ML2T–2 mol–1K–1]
Statement-II is incorrect
50. A body projected vertically upwards with a certain
speed from the top of a tower reaches the ground in
t1. If it is projected vertically downwards from the
same point with the same speed, it reaches the
ground in t2. Time required to reach the ground, if
it is dropped from the top of the tower, is :
(1)
12
tt
(2)
12
tt
(3)
1
2
t
t
(4)
12
tt
Ans. (1)
(1) 40 (2) 5
(3) 20 (4) 10
NTA Ans. (3)
Sol.
2
1
u u 2gh
tg
2
2
u u 2gh
tg
2gh
tg
22
2
12 22
(u 2gh) u 2gh
t t t
gg
12
t t t
SECTION-B
51. In Franck-Hertz experiment, the first dip in the
current-voltage graph for hydrogen is observed at
10.2 V. The wavelength of light emitted by
hydrogen atom when excited to the first excitation
level is ______ nm.
(Given hc = 1245 eV nm, e = 1.6 × 10–19C).
Ans. (122)
Sol. 10.2 eV =
hc
1245 eV nm
10.2eV
= 122.06 nm
52. For a given series LCR circuit it is found that
maximum current is drawn when value of variable
capacitance is 2.5 nF. If resistance of 200 and
100 mH inductor is being used in the given circuit.
The frequency of ac source is _____ × 103 Hz.
(given 2 = 10)
Ans. (10)
Sol. for maximum current, circuit must be in resonance.
0
1
f2 L C
0–3 9
1
f
2 100 10 2.5 10
=
11
1
2 25 10
=
5
110 10
25
Hz
=
3
100 10
10
Hz
f0 = 10 × 103 Hz
53. A particle moves in a straight line so that its
displacement x at any time t is given by x2=1 +t2.
Its acceleration at any time t is x–n where n =
_____.
Ans. (3)
Sol. x2 = 1 + t2
dx
2x 2t
dt
xv = t
dv dx
x v 1
dt dt
x.a+v2 =1
a =
2 2 2
1 v 1 t / x
xx
a =
3
3
1x
x
54. Three balls of masses 2kg, 4kg and 6kg
respectively are arranged at centre of the edges of
an equilateral triangle of side 2 m. The moment of
inertia of the system about an axis through the
centroid and perpendicular to the plane of triangle,
will be ______ kg m2.
Ans. (4)
Sol.
2kg
4kg
6kg
r
r
c
r
Moment of inertia about C and perpendicular to the
plane is :
I = r2 [2 + 4 + 6]
=
112
3
I = 4 kg-m2
55. A coil having 100 turns, area of 5 × 10–3m2,
carrying current of 1 mA is placed in uniform
magnetic field of 0.20 T such a way that plane of
coil is perpendicular to the magnetic field. The
work done in turning the coil through 90° is _____
J.
Ans. (100)
Sol. W = U = Uf – Ui
W =
fi
(– .B) (– .B)
= 0 +
i
( .B)
= (100 × 5 × 10–3 × 1 × 10–3) × 0.2 J
= 1 × 10–4 J = 100 J
56. In the given figure an ammeter A consists of a
240 coil connected in parallel to a 10 shunt.
The reading of the ammeter is ______ mA.
A
24V
140.4
Ans. (160)
Sol.
24
140.4
I
240
Req
=
240 10
140.4 240 10
Req = 140.4 +
2400
250
Req. = 150
Current in ammeter =
24
150
= 160 mA
57. A wire of cross sectional area A, modulus of
elasticity 2 × 1011 Nm–2 and length 2 m is stretched
between two vertical rigid supports. When a mass
of 2 kg is suspended at the middle it sags lower
from its original position making angle
1
100
radian on the points of support. The value of A is
_____ × 10–4 m2 (consider x<<L).
(given : g=10 m/s2)
2L
x
2kg
Ans. (1)
Sol.
2L
x
20
T
T
In vertical derection
2T sin = 20
using small angle approximation sin =
=
1
100
T =
10
T = 1000N
Change in length L =
22
2 x L 2L
=
2
2
x
2L 1 1
2L
L =
2
x
L
Modulus of elasticity =
stress
strain
2 × 1011 =
3
2
10 2L
x
AL
A = 1 × 10–4 m2
58. Two coherent monochromatic light beams of
intensities I and 4I are superimposed. The
difference between maximum and minimum
possible intensities in the resulting beam is x I. The
value of x is______.
Ans. (8)
Sol. Imax =
2
I 4I 9I
Imin =
2
4I I I
Imax – Imin = 8I
59. Two open organ pipes of length 60 cm and 90 cm
resonate at 6th and 5th harmonics respectively. The
difference of frequencies for the given modes is
_____ Hz.
(Velocity of sound in air = 333 m/s)
Ans. (740)
Sol. The difference in frequency in open organ pipe =
nv
f2L
6v 5v
f2 0.6 2 0.9
v = 333 m/s
f = 740 Hz
60. A capacitor of 10 F capacitance whose plates are
separated by 10 mm through air and each plate has
area 4 cm2 is now filled equally with two dielectric
media of K1 = 2, K2 = 3 respectively as shown in
figure. If new force between the plates is 8 N. The
supply voltage is _____ V.
K1=2
d
K2=3
NTA Ans. (80)
V
C1
C2
Ceq = C1 + C2
0
1
2A
C 10 F
2d
0
2
3A
C 15 F
2d
Ceq = 25 F
Now the charge on C1 = 10V c
C2 = 1.5 V C.
Now force between the plates
2
0
Q
F2A
2 12 2 12
44
00
100V 10 225V 10 8
2 2 10 2 2 10
325 V2 = 8 × 4 × 10–4 × 8.85
V2 =
4
32 8.85 10
325
4
283.2 10
V325
2
V 0.93 10
Sol.
CHEMISTRY
TEST PAPER WITH SOLUTION
SECTION-A
61.
CH3
(I)
(II)
OCH3
(III)
CF3
(IV)
The correct arrangement for decreasing order of
electrophilic substitution for above compounds
(1) (IV) > (I) > (II) > (III)
(2) (III) > (I) > (II) > (IV)
(3) (II) > (IV) > (III) > (I)
(4) (III) > (IV) > (II) > (I)
Ans. (2)
Sol.
OMe(+M)
>
CH3(+H/+I)
>
>
CF3(–I)
62. Molality (m) of 3 M aqueous solution of NaCl is:
(Given : Density of solution = 1.25 g mL–1, Molar
mass in g mol–1 : Na-23, Cl-35.5)
(1) 2.90 m (2) 2.79 m
(3) 1.90 m (4) 3.85 m
Ans. (2)
Sol. 3 moles are present in 1 litre solution
molality =
3 1000
1.25 1000 [3 58.5]
= 2.79 m
63. The incorrect statements regarding enzymes are:
(A) Enzymes are biocatalysts.
(B) Enzymes are non-specific and can catalyse
different kinds of reactions.
(C) Most Enzymes are globular proteins.
(D) Enzyme - oxidase catalyses the hydrolysis of
maltose into glucose.
Choose the correct answer from the option given
below:
(1) (B) and (C) (2) (B), (C) and (D)
(3) (B) and (D) (4) (A), (B) and (C)
Ans. (3)
Sol. Direct NCERT Based
64.
+ NaOH
Cl
CH3
Consider the above chemical reaction. Product "A" is:
(1)
OH
CH3
(2)
OH
CH3
(3)
OH
CH3
(4)
OH
CH3
Ans. (2)
Sol.
Cl
CH3
2
1
N
NaOH/H O
S
+
CH3
1,2–(H–shift)
CH3
CH3
OH
(Major Product )
H
65. During the detection of acidic radical present in a
salt, a student gets a pale yellow precipitate soluble
with difficulty in NH4OH solution when sodium
carbonate extract was first acidified with dil. HNO3
and then AgNO3 solution was added. This
indicates presence of:
(1) Br– (2)
2
3
CO
(3) I– (4) Cl–
Ans. (1)
Sol. Ag+ + I– AgI Yellow ppt.
Ag+ + Cl– AgCl White ppt
Ag+ + Br– AgBr Pale yellow ppt
66. How can an electrochemical cell be converted into
an electrolytic cell ?
(1) Applying an external opposite potential greater
than
0
cell
E
(2) Reversing the flow of ions in salt bridge.
(3) Applying an external opposite potential lower
than
0
cell
E
.
(4) Exchanging the electrodes at anode and
cathode.
Ans. (1)
Sol. Applied external potential should be greater than
0
cell
E
in opposite direction.
67. Arrange the following elements in the increasing
order of number of unpaired electrons in it.
(A) Sc (B) Cr
(C) V (D) Ti
(E) Mn
Choose the correct answer from the options given
below:
(1) (C) < (E) < (B) < (A) < (D)
(2) (B) < (C) < (D) < (E) < (A)
(3) (A) < (D) < (C) < (B) < (E)
(4) (A) < (D) < (C) < (E) < (B)
Ans. (4)
Sol. Unpaired electron
Sc[Ar] 4s2 3d1 1
Cr[Ar] 4s1 3d5 6
V[Ar] 4s2 3d3 3
Ti : [Ar] 4s2 3d2 2
Mn : [Ar] 4s2 3d5 5
68. Match List-I with List-II.
List-I List-II
Alkali Metal Emission Wavelength
in nm
(A) Li (I) 589.2
(B) Na (II) 455.5
(C) Rb (III) 670.8
(D) Cs (IV) 780.0
Choose the correct answer from the options given
below:
(1) (A)-(I), (B)-(IV), (C)-(III), (D)-(II)
(2) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(3) (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
(4) (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
Ans. (2)
Sol. Fact Based
69. The major products formed:
OCH3
3 2 4 2
HNO ,H SO Br (excess)
Fe
'A' 'B'
A and B respectively are:
(1)
OCH3
NO2
and
OCH3
NO2
Br
Br
(2)
OCH3
NO2
and
OCH3
Br
Br
NO2
(3)
OCH3
NO2
and
OCH3
NO2
Br
(4)
OCH3
NO2
and
OCH3
Br
NO2
Ans. (2)
Sol.
OMe
HNO3–H2SO4
OMe
NO2
(A)
OMe
Br2 (Excess)
OMe
NO2
(B)
NO2
Fe
Br
Br
70. The incorrect statement regarding the geometrical
isomers of 2-butene is:
(1) cis-2-butene and trans-2-butene are not
interconvertible at room temperature.
(2) cis-2-butene has less dipole moment than
trans-2-butene.
(3) trans-2-butene is more stable than cis-2-butene.
(4) cis-2-butene and trans-2-butene are stereoisomers.
Ans. (2)
Sol.
C = C
H
CH3
CH3
H
Cis–but–2–ene
C = C
H
CH3
H
Trans–but–2–ene
CH3
(Polar)
(Non Polar)
Cis–but–2–ene has higher Dipole moment than
trans–but–2–ene.
71. Given below are two statements:
Statement I: PF5 and BrF5 both exhibit sp3d
hybridisation.
Statement II: Both SF6 and [Co(NH3)6]3+ exhibit
sp3d2 hybridisation.
In the light of the above statements, choose the
correct answer from the options given below:
(1) Statement I is true 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 false but Statement II is true
Ans. (3)
Sol.
Hybridisation
Hybridisation
PF5
sp3d
SF6
sp3d2
BrF5
sp3d2
[Co(NH3)6]+3
d2sp3
Both Statement (1) and (2) are false.
72. The number of ions from the following that are
expected to behave as oxidising agent is:
Sn4+, Sn2+, Pb2+, Tl3+, Pb4+, Tl+
(1) 3 (2) 4
(3) 1 (4) 2
Ans. (4)
Sol. Due to inert pair effect; T+3 and Pb+4 can behave
as oxidising agents.
73. Identify the product
A
in the following reaction.
NH2
A
(1)
NH2
OH
(2)
OH
(3)
NH2
Cl
(4)
Cl
OH
Ans. (2)
Sol.
NH2
NaNO2+HCl
N2Cl
NO2
(A)
Cl
NaOH, 623K, 300 atm
OH
+
Cu2Cl2
–
H+
74. The correct statements among the following, for a
"chromatography" purification method is:
(1) Organic compounds run faster than solvent in
the thin layer chromatographic plate.
(2) Non-polar compounds are retained at top and
polar compounds come down in column
chromatography.
(3) Rf of a polar compound is smaller than that of a
non-polar compound.
(4) Rf is an integral value.
Ans. (3)
Sol. Non polar compounds are having higher value of
Rf than polar compound.
75. Evaluate the following statements related to group
14 elements for their correctness.
(A) Covalent radius decreases down the group
from C to Pb in a regular manner.
(B) Electronegativity decreases from C to Pb down
the group gradually.
(C) Maximum covalence of C is 4 whereas other
elements can expand their covalence due to
presence of d orbitals.
(D) Heavier elements do not form p-p bonds.
(E) Carbon can exhibit negative oxidation states.
Choose the correct answer from the options given
below:
(1) (C), (D) and (E) Only (2) (A) and (B) Only
(3) (A), (B) and (C) Only (4) (C) and (D) Only
Ans. (1)
Sol. (A) Down the group; radius increases
(B) EN does not decrease gradually from C to Pb.
(C) Correct.
(D) Correct.
(E) Range of oxidation state of carbon ; –4 to +4
76. Match List-I with the List-II
List-I List-II
Reaction Type of redox reaction
(A) N2(g) + O2(g) 2NO(g) (I) Decomposition
(B) 2Pb(NO3)2(s) (II) Displacement
2PbO(s) + 4NO2(g) + O2(g)
(C) 2Na(s) + 2H2O(l) (III) Disproportionation
2NaOH(aq.) + H2(g)
(D) 2NO2(g) + 2–OH(aq.) (IV) Combination
2(aq.) 3(aq.) 2 (l)
NO NO H O
Choose the correct answer from the options given
below:
(1) (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
(2) (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
(3) (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
(4) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
Ans. (4)
Sol. A (IV)
B (I)
C (II)
D (III)
77. Consider the given reaction, identify the major
product P.
3
(i) LiAlH (ii) PCC (iii) HCN/OH
4
(iv) H O/OH,
2
CH COOH "P"
(1) CH3 – CH2 – CH2 – OH
(2)
CH3 — CH2 — C — NH2
O
(3)
CH3 — C — CH2CH3
O
(4)
CH3 — CH —COOH
OH
Ans. (4)
Sol.
CH3
COOH
LiAlH4
CH3–CH2–OH
PCC
CH3–C–H
OH
HCN/OH
CH3—C—CN
H
H2O/OH
,
OH
CH3—CH—COOH
O
78. The correct IUPAC name of [PtBr2(PMe3)2] is:
(1) bis(trimethylphosphine)dibromoplatinum(II)
(2) bis[bromo(trimethylphosphine)]platinum(II)
(3) dibromobis(trimethylphosphine)platinum(II)
(4) dibromodi(trimethylphosphine)platinum(II)
Ans. (3)
Sol. Dibromo bis(trimethylphosphine) platinum (II)
79. Match List-I with List-II
List-I List-II
Tetrahedral Complex Electronic configuration
(A) TiCl4 (I)
20
2
e ,t
(B) [FeO4]2– (II)
43
2
e ,t
(C) [FeCl4]– (III)
00
2
e ,t
(D) [CoCl4]2– (IV)
23
2
e ,t
Choose the correct answer from the option given
below:
(1) (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
(2) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(3) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
(4) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Ans. (4)
Sol.
TiC4
t2
0
FeO4
2–
t2
0
e0
e2
FeC4
1–
t2
3
e2
CoC4
2–
t2
3
e4
0
VI
III
II
80. The ratio
P
C
K
K
for the reaction:
(g) 2(g) 2(g)
1
CO O CO
2
is:
(1) (RT)1/2 (2) RT
(3) 1 (4)
1
RT
Ans. (4)
Sol. CO(g) +
1
2
O2(g) CO2(g)
ng = 1 –
1
12
=
1
2
g
n
P
C
K1
(RT)
KRT
SECTION-B
81. An amine (X) is prepared by ammonolysis of
benzyl chloride. On adding p-toluenesulphonyl
chloride to it the solution remains clear. Molar
mass of the amine (X) formed is______ g mol–1.
(Given molar mass in gmol–1 C : 12, H : 1, O : 16, N : 14)
Ans. (287)
Sol.
CH2Cl
NH3
(excess)
PhCH2 –N–CH2Ph
CH2Ph
(X) (3° amine)
Molar Mass of (X) is 287 g mol–1
82. Consider the following reactions
NiS + HNO3 + HCl A + NO + S + H2O
A + NH4OH + H3C – C = N – OH
H3C – C = N – OH
B + NH4Cl + H2O
The number of protons that do not involve in
hydrogen bonding in the product B is______.
Ans. (12)
Sol. B
H
O
O
N
N
Ni
C
CH3
N=C
CH3
C
H3C
C
N
O
H
O
H3C
3NiS + 2HNO3 + 6HCl
3NiCl2 + 2NO + 3S + 4H2O
NiCl2 + 2NH4OH +
H3C – C = N – OH
H3C – C = N – OH
NH4Cl + H2O + (B)
83. When 'x' × 10–2 mL methanol (molar mass = 32 g;
density = 0.792 g/cm3) is added to 100 mL
water (density = 1 g/cm3), the following diagram is
obtained.
Vapour pressure
270.65
273.15
Temperature/K
x =………………(nearest integer)
[Given: Molal freezing point depression constant
of water at 273.15 K is 1.86 K kg mol–1]
Ans. (543)
Sol. Tf = 273.15 – 270.65 = 2.5 K
Tf = Kf m 2.5 = 1.86 ×
n
0.1
n = 0.1344 moles
w = 0.1344 × 32 = 4.3 g
Volume =
4.3
0.792
= 5.43 ml = 543 × 10–2 ml
84.
OC2H5
3 2 4 2
HNO ,H SO 2Br ,Fe
major major
product product
PQ
The ratio of number of oxygen atoms to bromine
atoms in the product Q is_____ × 10–1.
Ans. (15)
Sol.
OC2H5
HNO3+H2SO4
OC2H5
NO2
NO2
Br
Br
2Br2–Fe
OC2H5
85. Number of carbocation from the following that are
not stabilized by hyperconjugation is……….. .
+
(tert.-Butyl) (tert.-Butyl)'
+
,
+
,
CH2 – OCH3
+
+
,
N – CH2
+
CH3
Ans. (5)
Sol.
C
+
CH3
+
+
CH2–O–CH3
+
N–CH2
+
86. For the reaction at 298 K, 2A + B C. H
= 400 kJ mol–1 and S = 0.2 kJ mol–1 K–1. The
reaction will become spontaneous above______ K.
Ans. (2000)
Sol. G = 0
T =
H 400
S 0.2
= 2000 K
87. Total number of species from the following with
central atom utilising 2p2 hybrid orbitals for
bonding is…………. .
NH3, SO2, SiO2, BeCl2, C2H2, C2H4, BCl3, HCHO,
C6H6, BF3, C2H4Cl2
Ans. (6)
Sol. Central atom utilising sp2 hybrid orbitals
SO2, C2H4, BCl3, HCHO, C6H6, BF3
88. Consider the two different first order reactions
given below
A + B C (Reaction 1)
P Q (Reaction 2)
The ratio of the half life of Reaction 1 : Reaction 2
is 5 : 2. If t1 and t2 represent the time taken to
complete
rd
23
and
th
45
of Reaction 1 and
Reaction 2, respectively, then the value of the ratio
t1 : t2 is________× 10–1 (nearest integer).
[Given: log10(3) = 0.477 and log10(5) = 0.699]
Ans. (17)
Sol.
1/2 I 2
1/2 II 1
(t ) K 5
(t ) K 2
K1t1 = n
1
2
13
= n3
K2t2 = n
1
4
15
= n5
11
22
Kt0.477
K t 0.699
1
2
t0.477 5
t 0.699 2
= 1.7 = 17 × 10–1
89. For hydrogen atom, energy of an electron in first
excited state is – 3.4 eV, K.E. of the same electron
of hydrogen atom is x eV. Value of x is______
× 10–1 eV. (Nearest integer)
Ans. (34)
90. Among
24
VO ,MnO
and
2
27
Cr O
, the spin-only
magnetic moment value of the species with least
oxidising ability is……………….BM (Nearest
integer).
(Given atomic member V = 23, Mn = 25, Cr = 24)
Ans. (0)
Sol. For 3d transition series;
Oxidising power : V+5 < Cr+6 < Mn+7
V+5 : [Ar] 4s0 3d0
Number of unpaired electron = 0
0
