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Two tuning forks P and Q when set vibrat...

Two tuning forks `P` and `Q` when set vibrating , give `4` beats per second. If a prong of the fork `P` is filed, the beats are reduced to `2` per second . If a prong of the fork `P` is filed, the beats are reduced to `2` per second . Determine the original frequency of `P` , if that of `Q` is `250 H_(Z)`

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The correct Answer is:
B
There are four beats between `P` and `Q` , therefore the possible frequencies of `P` are `246` or `254` (that is `250 +- 4) H_(Z)` .
When the prong of `P` is filed, its frequency become greater then the original frequency.
If we assume that the original frequency of `P` is `254` , then on filing its frequency will be greater than `254` . The beats between `P` and `Q` will be more than `4` . But it is given that the beats are reduced to `2` , therfore, `254` not possible.
therefore, the required frequency must be `246 H_(Z)` .
(This is true, because on filling the frequency may increase to `248` , giving `2` beats with `Q` of frequency `250 H_(Z)`)
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