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(A) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $T$ is the tension in the wire.When the b

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$\frac{n_1^2}{n_1^2 - n_2^2}$

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(A) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $T$ is the tension in the wire.When the bob is hanging in air, the tension $T_1 = Mg$, so $n_1 \propto \sqrt{Mg}$.When the bob is immersed in water, the buoyant force acts upwards, so the effective weight (tension) becomes $T_2 = Mg - V\rho_w g = Mg(1 - \frac{V\rho_w}{M}) = Mg(1 - \frac{\rho_w}{\rho_m})$, where $\rho_m$ is the density of the metal and $\rho_w$ is the density of water.Thus, $n_2 \propto \sqrt{Mg(1 - \frac{\rho_w}{\rho_m})}$.Taking the ratio: $\frac{n_1}{n_2} = \sqrt{\frac{1}{1 - \frac{\rho_w}{\rho_m}}} = \sqrt{\frac{\rho_m}{\rho_m - \rho_w}}$.Squaring both sides: $\frac{n_1^2}{n_2^2} = \frac{\rho_m}{\rho_m - \rho_w}$.Let the relative density $\sigma = \frac{\rho_m}{\rho_w}$. Then $\frac{n_1^2}{n_2^2} = \frac{\sigma}{\sigma - 1}$.$n_1^2(\sigma - 1) = n_2^2 \sigma \implies \sigma(n_1^2 - n_2^2) = n_1^2$.Therefore, $\sigma = \frac{n_1^2}{n_1^2 - n_2^2}$.

In a sonometer experiment, the fundamental frequency of vibration of a wire is '$n_1$' when the wire is stretched by hanging a metal bob. If the bob is completely immersed in water, the frequency of vibration of the wire becomes '$n_2$'. The relative density of the metal of the bob is

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