Given, $\quad \mathrm{f}_{\mathrm{o}}-\mathrm{f}_{\mathrm{c}}=2 \quad \text{...(i)}$
Frequency of fundamental mode for a closed Frequency of fundar organ pipe, $f_{c}=\frac{v}{4 L_{c}}$
Similarly frequency of fundamental mode for an open organ pipe, $\mathrm{f}_{\mathrm{o}}=\frac{\mathrm{V}}{2 \mathrm{~L}_{\mathrm{o}}}$
Given $\quad \mathrm{L}_{\mathrm{c}}=\mathrm{L}_{\mathrm{o}}$
$\Rightarrow$ $f_{0}=2 f_{c} \quad \text{...(ii)}$
From Eqs. (i) and (ii), we get
$\mathrm{f}_{\mathrm{o}}=2 \mathrm{f}_{\mathrm{c}}$ $\mathrm{f}_{\mathrm{o}}=4 \mathrm{~Hz}$ and $\mathrm{f}_{\mathrm{c}}=2 \mathrm{~Hz}$
When the length of the open pipe is halved, frequency of fundamental mode is $$ \mathrm{f}_{\mathrm{o}}^{\prime}=\frac{\mathrm{v}}{2\left[\frac{\mathrm{L}_{\mathrm{o}}}{2}\right]}=2 \mathrm{f}_{\mathrm{o}}=2 \times 4 \mathrm{~Hz}=8 \mathrm{~Hz} $$
When the length of the closed pipe is doubled, its frequency of fundamental mode is
$$
\mathrm{f}_{\mathrm{o}}^{\prime}=\frac{\mathrm{v}}{4\left(2 \mathrm{~L}_{\mathrm{c}}\right)}=\frac{1}{2} \mathrm{f}_{\mathrm{c}}=\frac{1}{2} \times 2=1 \mathrm{~Hz}
$$
Hence, number of beats produced per second is
$$
\mathrm{f}_{\mathrm{o}}^{\prime}-\mathrm{f}_{\mathrm{c}}^{\prime}=8-1=7
$$