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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

292 cps

c

294 cps

d

288 cps

answer is B.

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Detailed Solution

nA = Known frequency = 288 cps, nB = ?x = 4 bps, which is decreasing (from 4 to 2) after loading i.e. x↓Unknown fork is loaded so nB↓Hence nA – nB↓ = x↓       →        Wrong           nB↓ – nA↓ = x↓      →       Correct⇒ nB = nA + x = 288 + 4 = 292 Hz.
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