$\mathrm{n}=\frac{1}{2 \mathrm{L}} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$
The tuning fork produces $5$ beats per second with lengths $20 \mathrm{\,cm}$ and $21 \mathrm{\,cm}$. If $\mathrm{n}$ be frequency of fork, then
$\mathrm{n}+5=\frac{1}{2 \times 0.20} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$ .........$(i)$
$\mathrm{n}-5=\frac{1}{2 \times 0.21} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$ .........$(ii)$
Dividing eqn. $(i)$ by $(ii)$ we get:
hence, $\quad \frac{n+5}{n-5}=\frac{0.21}{0.20}$
or $0.2 n+1=0.21 n-1.05$
or $1+1.05=0.21 n-0.2 n^{\prime}$
or $0.05=0.01\, n$
or $\mathrm{n}=205 \mathrm{\,Hz}$
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