A sonometer wire ' $A$ ' of diameter ' $\mathrm{d}$ ' under tension ' $T$ ' having density ' $\rho_1$ '…
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
$\rho_2=2 \rho_1$
$\rho_2=\rho_1$
$\rho_2=\frac{\rho_1}{2}$
$\rho_2=\frac{\rho_1}{4}$
Solution
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
$\begin{aligned}
& \frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\pi \rho \mathrm{D}^2}}=\frac{1}{2 l} \sqrt{\frac{2 \mathrm{~T}}{\pi \rho_2(2 \mathrm{D})^2}} \\
& \sqrt{\frac{1}{\rho_1}}=\sqrt{\frac{1}{2 \rho_2}} \\
& \frac{1}{\rho_1}=\frac{1}{2 \rho_2} \\
& \therefore \quad \rho_2=\frac{\rho_1}{2}
\end{aligned}$