Correct Answer - B
(b) A tuning fork produces `4 beats//sec` with another tuning fork of frequency `288 cps`. From this information we can conclude that the frequency of unknown fprk is `288 + 4 cps` or `288 - 4 cps` i.e. `292 cps` or `284 cps`. When a little wax is placed on the unknown fork, it produces `2beats//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`.