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BITSAT 2016 Solved Question Paper
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© examsnet.com
Question : 37
Total: 150
A source of sound S emitting waves of frequency 100 Hz and an observor O are located at some distance from each other. The source is moving with a speed of 19.4 ms–1 at an angle of 60° with the source observer line as shown in the figure. The observor is at rest. The apparent frequency observed by the observer is (velocity of sound in air
330
m
s
–
1
)
103 Hz
106 Hz
97 Hz
100 Hz
Validate
Solution:
Here, original frequency of sound
f
0
=
100
H
z
Speed of source
V
s
=
19.4
cos
60
°
=
9.7
From Doppler's formula
f
1
=
f
0
(
V
−
V
0
V
−
V
s
)
f
1
=
100
(
V
−
0
V
−
(
+
9.7
)
)
f
1
=
100
V
V
(
1
−
9.7
V
)
f
1
=
100
(
1
+
9.7
330
)
=
103
H
z
Apparent frequency
f
1
=
1.3
H
z
© examsnet.com
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