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Question

(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 bo

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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 of weight $W$ is in air,the tension $T_1 = W$. Thus,$n_1 = \frac{1}{2L} \sqrt{\frac{W}{\mu}}$.When the bob is immersed in water,the buoyant force $F_B$ acts on it,reducing the tension to $T_2 = W - F_B$. Thus,$n_2 = \frac{1}{2L} \sqrt{\frac{W - F_B}{\mu}}$.Taking the ratio: $\frac{n_1}{n_2} = \sqrt{\frac{W}{W - F_B}} \implies \frac{n_1^2}{n_2^2} = \frac{W}{W - F_B}$.Relative density $\sigma = \frac{W}{F_B}$.Rearranging the ratio: $\frac{n_1^2}{n_2^2} = \frac{W}{W - (W/\sigma)} = \frac{\sigma}{\sigma - 1}$.Solving for $\sigma$: $n_1^2(\sigma - 1) = n_2^2 \sigma \implies \sigma(n_1^2 - n_2^2) = n_1^2$.Therefore,$\sigma = \frac{n_1^2}{n_1^2 - n_2^2}$.

$A$ sonometer wire is stretched by hanging a metal bob,the fundamental frequency of the wire is $n_1$. When 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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