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Q.

A tuning fork arrangement (pair) produces 4 beats /sec with one fork of frequency 288 cps. A little wax is placed on the unknown fork and it then produces 2 beats/sec. The frequency of the unknown fork is

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a

286 cps

b

294 cps

c

288 cps

d

292 cps

answer is B.

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Detailed Solution

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Frequency of unknown fork=known frequency  Beat frequency =288+4 cps or 288-4cps i.e. 292 cps or 284 cps. When a little wax is placed on the unknown fork, it produce 2 Beats/sec. When a little wax is placed on the unknown fork, its frequency decreases and Simultaneously the beat frequency decreases confirming that the frequency of the unknown Fork is 292 cps. 

Note: Had the frequency of unknown fork been 284 cps, then on [lacing wax its frequency Would have decreased thereby increasing the gap between its frequency and the frequency of Known fork. This would produce high beat frequency.

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