33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Waves
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125
42. (b) : Pressure change will be minimum at both ends.
43. (d) : Let L(= 100 cm) be the length of the wire and
L 1 , L 2 and L 3 are the lengths of the segments as shown in
the figure.
47. (a)
48. (c) : Distance between a node and adjoining
antinode = l/4
Fundamental frequency, υ∝ 1 L
As the fundamental frequencies are in the ratio of 1 : 3 : 5,
∴ L1 : L2 : 1 1 1
L3
= =
1
: 3
: 5 15: 5:
3
Let x be the common factor. Then
15x + 5x + 3x = 100
100
23x= 100 or x=
23
∴ L 1 = 15 × 100 1500
=
23 23 cm
L2 5 100 500
L3
3 100 300
= × = cm ; = × = cm
23 23
23 23
\ The bridges should be placed from A at
1500
23
cm and 2000 cm respectively.
23
44. (c) : Let l be the length of the string.
Fundamental frequency is given by
υ = 1 2l
T
µ
or υ ∝ 1 ( Q T andµ
are constants)
l
k k k k
Here, l1
= , l 2 = , l 3 = and l =
υ υ υ υ
1
But l = l 1 + l 2 + l 3
2
3
1 1 1 1
∴ = + +
υ υ1 υ2 υ3
. V
45. (b) : Reverberationtime, T = 061
aS
where V is the volume of room in cubic metres, a is the
average absorption coefficient of the room, S is the total
surface area of room in square metres.
or, T
V T
∝
1 V1
S2
V 4S
1
or, = ⎛ ⎞ ⎛ ⎞
S T2
⎝ ⎜ V2
⎠
⎟
⎝
⎜ S1 ⎠
⎟ = ⎛ ⎝ ⎜
⎞
8V
⎠
⎟ ⎛ ⎝ ⎜
⎞
S ⎠
⎟ = 2
or, T 2 = 2T 1 = 2 × 1 = 2 s (Q T 1 = 1 s)
T
46. (c) : n = 1 2l
2
πr
ρ
ρ
ρ′ 1 = ; T′ = 2T and D′ = 2D or r′ = 2r
2
n′= 1 2l
2T
1 T
= = n
2 ρ
π( 2r)
2l
2
πr
ρ
2
From figure, distance between two atoms
λ
= 4 × = 121 . Å or, λ = 121 . Å
4
49. (a) : The frequency of standing wave,
n
υ= = × =
2l v 5 20
5Hz.
2×
10
50. (d) : For the cylindrical tube open at both ends,
f = v/2l
When half of the tube is in water, it behaves as a closed
pipe of length l/2.
v v
∴ f ′ = = ⇒ f′ = f
4(/ l 2)
2l
51. (a) : Length of sonometer wire (l) = 110 cm
and ratio of frequencies = 1 : 2 : 3.
Frequency( υ)∝ 1 or l ∝
1
l υ
1 1 1
Therefore AC : CD : DB = : : = 6: 3:
2
1 2 3
ThereforeAC = 6 × 110 = 60 cmand
11
CD = 3 × 110 = 30 cm
11
Thus AD = 60 + 30 = 90 cm.
52. (a) : Frequency (u) = 100 Hz and distance from
fixed end = 10 cm = 0.1 m. When a stationary wave is
produced, the fixed end behaves as a node.
Thus wavelength (l) = 2 × 0.1 = 0.2 m.
Therefore velocity v = ul = 100 × 0.2 = 20 m/s.
53. (a) : y = Asin(100t)cos(0.01x).
Comparing it with standard equation
y = ⎛ 2π
A
⎝
⎜
T t ⎞ ⎛ 2π
⎠
⎟
⎝
⎜ x ⎞
sin cos
⎠
⎟,
λ
π
we getT = and λ=
200 π
50
Thereforevelocity,( v)
= λ
T
= 200π
π /
= 200
50
× 50
= 10000 = 10 4 m/s
1
1 ⎡T
⎤2
54. (c) : f =
2l
⎢ ⎥ when f is halved, the length will be
⎣ µ ⎦
doubled.
Hence required length of string is 1 m.
55. (d) : Third overtone has a frequency 4n, 4 th harmonic
= three full loops + one half loop, which would make four
nodes and four antinodes.