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Q.

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
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