Two vibrating tuning forks produce progressive waves given by $y_1=4\sin 500\pi t$ and $y_2=2\sin 506\pi t$. Number of beats produced per minute is

$3.\frac{\Delta V}{V}=\frac{P}{B}$ and $n_2=\frac{506\pi }{2\pi }$   

${\text{v}}_{\text{0}}=\frac{\text{v}}{2\left(\frac{{\text{L}}_{\text{0}}}{\text{2}}\right)}=2\left(\frac{\text{v}}{{\text{2L}}_{\text{0}}}\right)={\text{2v}}_{\text{0}}=2\times 4$= 8 Hz

When the length of the closed pipe is doubled, its frequency of fundamental mode is${\text{v}}_c^{\text{'}}=\frac{\text{V}}{{\text{4(2L}}_{\text{0}}\text{)}}=\frac{1}{2}\left(\frac{\text{v}}{{\text{4L}}_{\text{0}}}\right)=\frac{1}{2}\times 2=\text{1 Hz}$

Hence, number of beats produced per second = ${\text{v}}_{\text{0}}-{\text{v}}_{\text{c}}^{\text{'}}=8-1=7\text{ Hz}$