A stretched sonometer wire is in unison with a tuning fork. When the length is increased by 2% the number of beats per second is 5. Find the frequency of the fork in Hz

Let f be the initial frequency of sonometer wire as well as of tuning fork.

Let f' be the final frequency of sonometer wire.

$\mathrm{f}\alpha \frac{1}{\ell };\therefore \frac{f\text{'}}{\mathrm{f}}=\frac{\ell }{1.02\ell }$

$f\text{'}=\frac{f}{1.02}$

Given, $f-f\text{'}=5$

$$\begin{gathered} \Rightarrow \mathrm{f}-\frac{\mathrm{f}}{1.02}=5 \\ \Rightarrow \mathrm{f}=255\mathrm{Hz} \end{gathered}$$