c
(c) Suppose two tuning forks are named \(A\) and \(B\) with frequencies \({n_A} = 256\,Hz\) (known), \(n_B = ?\) (unknown), and beat frequency \(x = 4 \,bps.\)

Frequency of unknown tuning fork may be \({n_B} = 256 + 4 = 260\,Hz\)

or \( = 256 - 4 = 252\,Hz\)

It is given that on sounding waxed fork \(A\) (fork of frequency \(256 Hz\)) and fork \(B\), number of beats (beat frequency) increases.

It means that with decrease in frequency of \(A,\) the difference in new frequency of \(A\) and the frequency of \(B\) has increased.

This is possible only when the frequency of \(A\) while decreasing is moving away from the frequency of \(B.\)

This is possible only if \(n_B = 260 Hz.\)

Alternate method : It is given \({n_A} = 256\,Hz,\,{n_B} = ?\) and \(x = 4 \,bps\)

Also after loading \(A\) (i.e. \(n_A \downarrow\)), beat frequency (i.e. \(x\)) increases (\(\uparrow\)).

Apply these informations in two possibilities to known the frequency of unknown tuning fork.

\(n_A \downarrow -n_B = x\uparrow\) ... \((i)\)

\(n_B -n_A \downarrow = x\uparrow\) ... \((ii)\)

It is obvious that equation \((i)\) is wrong \((ii)\) is correct so

\(n_B = n_A + x = 256 + 4 = 260 \,Hz.\)