33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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128 NEET-AIPMT Chapterwise Topicwise Solutions Physics
77. (b) : 1st the car is the source and at the cliff, one
observes f ′.
v
∴ f ′ =
v − v f
s
Now cliff is source. It emits frequency f ′ and the observer
is now the driver who observes f ′′.
⎡v+
vo⎤
⎡v+
vo⎤
∴ f ′′ = ⎢ ⎥ f′ or 2 f = ⎢ ⎥ f
⎣ v ⎦
⎣ v−
vs
⎦
⇒ 2v – 2v o = v + v o [as v s = v o ]
v
⇒ vo
=
3
78. (b) : Apparent frequency,
v vo
v v
f ′= + f = + (/) 1 5
f = 12 . f
v
v
Wavelength does not change by motion of observer.
79. (b) : The whistle is revolving in a circle of radius
50 cm. So the source (whistle) is moving and the observer
is fixed.
The minimum frequency will be heard by the observer
when the linear velocity of the whistle (source) will be in
a direction as shown in the figure, i.e. when the source is
receding.
The apparent frequency heard by the observer is then
⎛ v ⎞
given by υ′ = υ
⎝
⎜
+ ⎠
⎟
V v
where V and v are the velocities of sound and source
respectively and u is the actual frequency.
Now, v = r w = 0.5 × 20 = 10 m/s
V = 340 m/s, u = 385 Hz.
340
∴ υ′ = 385 × = 374 Hz.
340 + 10
v u v u
80. (a) : f ′= − f;
f′′ = + f
v
v
Number of beats = f′′− f′=
81. (b) : n ′= nv
n + n1
= v − v s cosθ = nv
v
= n
⎡⎣ Q cos90°=
0⎤ ⎦
\ n 1 = 0
9
82. (a) : Here υ′ = υ
8
Source and observer are moving in opposite direction,
apparent frequency
+
υ′ = υ ×
( v u )
( v−
u )
or, 9 340 + u
υ= υ×
8 340 − u
⇒ 9 × 340 – 9u = 8 × 340 + 8u
340
⇒ 17u= 340 × 1 ⇒ u= = 20 m/s
17
2u
λ
vvv