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Standard XII

Physics

Interference in Sound Waves

Frequency of ...

Question

Frequency of tuning fork A is 250 Hz. It produces 4 beats / sec with another tuning fork B. When A is filed, the beat frequency is 6 beats / sec. The frequency of B is:

264 Hz

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254 Hz

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250 Hz

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246 Hz

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Solution

The correct option is D 246 Hz
As A is filed frequency of A increases which led to increasing in beat frequency.
So, A has frequency greater than B frequency.
$δ = δ_{A} - δ_{B}$
$4 = 250 - δ_{B}$
$δ_{B} = 246 H z$

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Q. A tuning fork of unknown frequency produces 4 beats per second when sounded with another tuning fork of frequency 254 Hz. It gives the same number of beats per second when unknown tuning fork loaded with wax. The unknown frequency before loading with wax is

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Frequency of tuning fork A is 250 Hz. It produces 4 beats / sec with another tuning fork B. When A is filed, the beat frequency is 6 beats / sec. The frequency of B is:

The correct option is D 246 HzAs A is filed frequency of A increases which led to increasing in beat frequency. So, A has frequency greater than B frequency.δ=δA−δB4=250−δBδB=246Hz

The correct option is D 246 Hz
As A is filed frequency of A increases which led to increasing in beat frequency.
So, A has frequency greater than B frequency.
$δ = δ_{A} - δ_{B}$
$4 = 250 - δ_{B}$
$δ_{B} = 246 H z$