Explanation

(d) $Given,\ f_{0} - f_{c} = 2\ \ \ \ \ \ldots(i)$
Frequency of fundamental mode for a closed organ pipe,
$f_{c} = \frac{v}{4L_{c}}$
Similarly frequency of fundamental mode an open orgen pipe,
$f_{0} = \frac{v}{2L_{0}}$
Given $L_{c} = L_{0}$
$\Longrightarrow f_{0} = 2f_{c}\ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
From Eqs. (i) and (ii), we get
$f_{0} = 4Hz$
And $f_{c} = 2Hz$
When the length of the open pipe is halved, its frequency of fundamental mode is
$f_{0}^{'} = \frac{v}{2\left\lbrack \frac{L_{0}}{2} \right\rbrack}$
$= 2f_{0} = 2 \times 4Hz = 8Hz$
When the length of the closed pipe is doubled, its frequency of fundamental mode is
$f_{0}^{'} = \frac{v}{4\left( 2L_{c} \right)}$
$= \frac{1}{2}f_{c} = \frac{1}{2} \times 2 = 1Hz$
Hence, number of beats produced per second is
$f_{0}^{'} = f^{'} = 8 - 1 = 7$