Two strings X and Y of a sitar produce a beat frequency 4 Hz. When the tension of the string Y is slightly increased the beat frequency is found to be 2 Hz. If the frequency of X is 300 Hz, then the original frequency of Y was
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a
296 Hz
b
298 Hz
c
302 Hz
d
304 Hz
answer is A.
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Detailed Solution
nx=300Hz, ny=?x = beat frequency = 4 Hz, which is decreasing (4→2)after increasing the tension of the string y. Also tension of wire y increasing so ny↑ (∵ n∝T)Hence nx−ny↓=x↓ → Correct ny↑−nx=x↓ → Wrong ⇒ny=nx−x=300−4=296Hz