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A closed organ pipe and an open organ pipe of same length produce 2 beats/second when they are set into vibrations together in fundamental mode. The length of open pipe is now halved and that of closed pipe is doubled. The number of beats produced will be

A
7
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B
4
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C
8
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D
2
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Solution

The correct option is A 7
For a closed organ pipe, the frequency of fundamental mode is
νc=v4Lc

where v is the velocity of sound in air and Lc is the length of the close pipe.

For an open organ pipe, the frequency of fundamental mode is
νo=v2Lo

where Lo is the length of the open pipe.

Given, Lc=Lo

νo=2νc
and νoνc=2

Solving the above equations we get:-
νo=4 Hz
νc=2 Hz

When the length of the open pipe is halved, its frequency of fundamental mode is

νo=v2(Lo2)=2νo=2×4 Hz=8 Hz

When the length of the closed pipe is doubled, its frequency of fundamental mode is

νc=v4(2Lc)=12vc=12×2 Hz=1 Hz

Hence, Number of beats produced per second =νoνc=81=7

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