A sonometer wire ' \(A\) ' of diameter ' \(\mathrm{d}\) ' under tension ' \(T\) ' having density ' \(\rho_1\) ' vibrates with fundamental frequency ' \(n\) '. If we use another wire 'B' which vibrates with same frequency under tension ' \(2 \mathrm{~T}\) ' and diameter ' \(2 \mathrm{D}\) ' then density ' \(\rho_2\) ' of wire ' \(B\) ' will be

The formula for frequency of a sonometer is \(\mathrm{f}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\pi \rho \mathrm{D}^2}}\)
Here \(l\) is length, \(\mathrm{T}\) is tension, \(\mathrm{D}\) is diameter and \(\rho\) is density.
The frequency of both the wires is same.
The frequency of the wire \(\mathrm{A}\) is \(\mathrm{f}_{\mathrm{A}}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\pi \rho_1 \mathrm{D}^2}}\)
The frequency of the wire \(\mathrm{B}\) is \(f_{\mathrm{B}}=\frac{1}{2 l} \sqrt{\frac{2 \mathrm{~T}}{\pi \rho_2(2 \mathrm{D})^2}}\)
Equating both the frequencies
\(\begin{aligned}
& \frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\pi \rho \mathrm{D}^2}}=\frac{1}{2 l} \sqrt{\frac{2 \mathrm{~T}}{\pi \rho_2(2 \mathrm{D})^2}} \\
& \sqrt{\frac{1}{\rho_1}}=\sqrt{\frac{1}{2 \rho_2}} \\
& \frac{1}{\rho_1}=\frac{1}{2 \rho_2} \\
& \therefore \quad \rho_2=\frac{\rho_1}{2}
\end{aligned}\)