Two tuning forks, A and B, give 4 beats per second when sounded together. The frequency of A is 320 Hz. When some wax is added to B and it is sounded with A, 4 beats per second are again heard. The frequency of B is
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a
312 Hz
b
316 Hz
c
324 Hz
d
328 Hz
answer is C.
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Detailed Solution
nA – nB↓ = x (same) ... (i) → WrongnB↓ – nA = x (same) ... (ii) → Correct⇒ nB = nA + x = 320 + 4 = 324 Hz.