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JEE Advanced Model Paper 9 with solutions for online practice
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© examsnet.com
Question : 49
Total: 57
Two sound sources are moving away from a stationary observer in oppositedirections with velocity
V
1
and
V
2
(
V
1
>
V
2
) . The frequency of both the series is900Hz.
V
1
and
V
2
are both quite less than speed of sound, V = 300 m/ sec . Find thevalue of (
V
1
−
V
2
) , in m/sec, so that beat frequency observed by observer is 6 Hz.
Your Answer:
Validate
Solution:
f
1
= 900
[
300
300
+
V
1
]
= 900
[
1
+
V
1
300
]
−
1
= 900
[
1
−
V
1
300
]
= 900 -
V
1
Like wise ,
f
2
= 900 -
3
V
2
Given ,
f
2
−
f
1
= 6
3
(
V
1
−
V
2
)
= 6 ⇒
V
1
−
V
2
= 2 m/sec
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