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(C) Let the frequency of the tuning fork be $N$.Since the frequency of a vibrating string is inversely proportional to its length $(f \propto 1/l)$,we

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$205$

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(C) Let the frequency of the tuning fork be $N$.Since the frequency of a vibrating string is inversely proportional to its length $(f \propto 1/l)$,we have $f \cdot l = 	ext{constant}$.For a sonometer wire,the frequency $f$ is given by $f = k/l$.When the length is $20 \ cm$,the frequency is $f_1 = k/20$. Since it produces $5$ beats per second with the tuning fork,$f_1 = N \pm 5$.When the length is $21 \ cm$,the frequency is $f_2 = k/21$. Since the beat frequency remains $5$,$f_2 = N \mp 5$.As the length increases,the frequency decreases. Therefore,$f_1 = N + 5$ and $f_2 = N - 5$.Thus,$(N + 5) \cdot 20 = (N - 5) \cdot 21$.$20N + 100 = 21N - 105$.$N = 205 \ Hz$.

$A$ tuning fork vibrating with a sonometer having $20 \ cm$ wire produces $5$ beats per second. The beat frequency does not change if the length of the wire is changed to $21 \ cm$. The frequency of the tuning fork (in Hertz) must be

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