A tuning fork of frequency 392 Hz, resonates with 50 cm length of a string under tension (T). If length of the string is decreased by 2%, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is

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answer is C.

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Detailed Solution

$n \propto \frac{1}{l} \Rightarrow \frac{Δn}{n} = - \frac{Δl}{l}$

If length is decreased by 2% then frequency increases by 2% i.e., $\frac{n_{2} - n_{1}}{n_{1}} = \frac{2}{100}$

$\Rightarrow n_{2} - n_{1} = \frac{2}{100} \times n_{1} = \frac{2}{100} \times 392 = 7 .8 \approx 8 .$

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