Thank you for reporting, we will resolve it shortly
Q.
$5$ beats/second are heard when a tuning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either $0.95\, m$ or $1\, m$. The frequency of the fork will be :
Length of wire is $L_{1}=0.95\, m ; L_{2}=1\, m$.
Number of beats per second heard $=5$.
Let frequency of fork be $f$. Therefore,
$\frac{v}{2 L_{1}}-f=5 $
$\Rightarrow \frac{v}{2 L_{1}}=5+f$
and $f-\frac{v}{2 L_{2}}=5 $
$\Rightarrow \frac{v}{2 L_{2}}=f-5$
Dividing Eq. (1) by Eq. (2), we get
$\frac{\frac{v}{2 L_{1}}}{\frac{\nu}{2 L_{2}}}=\frac{5+f}{f-5} $
$\Rightarrow \frac{L_{2}}{L_{1}}=\frac{f+5}{f-5}$
$ \Rightarrow \frac{1}{0.95}=\frac{f+5}{f-5} $
$\Rightarrow f-5=0.95\, f+4.75$
$\Rightarrow f-0.95\, f=5+4.75 $
$\Rightarrow 0.05\, f=9.75$
$ \Rightarrow f=\frac{9.75}{0.05}$
$ \Rightarrow f=195\, Hz$