Two tuning forks when sounded together produced $4$ beats/sec. The frequency of one fork is $256.$ The number of beats heard increases when the fork of frequency $256$ is loaded with wax. The frequency of the other fork is
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(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
$n_B = n_A + x = 256 + 4 = 260 \,Hz.$
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