as $L_{2}>L_{1},$ so $n_{2}$
For open pipe,
$n=\frac{v}{2 L}$
$n_{1}-n_{2}=3$ beats $/ \mathrm{s}$
$\frac{v}{2}\left(\frac{1}{L_{1}}-\frac{1}{L_{2}}\right)=3$
$\frac{v}{10^{-2}}\left(\frac{1}{50}-\frac{1}{50.5}\right)=6$
$v=\frac{6 \times 50 \times 50.5 \times 10^{-2}}{0.5}=303 \mathrm{m} / \mathrm{s}$