A tuning fork A of frequency 200 Hz is sounded with fork B, the number of beats per second is 5. By putting some wax on A, the number of beats increases to 8. The frequency of fork B is
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a
200 Hz
b
195 Hz
c
192 Hz
d
205 Hz
answer is D.
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Detailed Solution
nA↓ – nB = x↑ ... (i) → WrongnB – nA↓ = x↑ ... (ii) → Correct ⇒ nB = nA + x = 200 + 5 = 205 Hz.