A sonometer wire is stretched by hanging a me | vedclass

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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 is in air,the tension $T_1 = W = mg$,where $W$ is the weight of the bob.So,$n_1 = \frac{1}{2L} \sqrt{\frac{W}{\mu}}$.When the bob is immersed in water,the effective weight (tension) becomes $T_2 = W - F_B$,where $F_B$ is the buoyant force.$T_2 = W - \frac{W}{\sigma} = W(1 - \frac{1}{\sigma})$,where $\sigma$ is the relative density of the metal.So,$n_2 = \frac{1}{2L} \sqrt{\frac{W(1 - 1/\sigma)}{\mu}}$.Taking the ratio: $\frac{n_1}{n_2} = \sqrt{\frac{1}{1 - 1/\sigma}} = \sqrt{\frac{\sigma}{\sigma - 1}}$.Squaring both sides: $\frac{n_1^2}{n_2^2} = \frac{\sigma}{\sigma - 1}$.$n_1^2(\sigma - 1) = n_2^2 \sigma$.$\sigma(n_1^2 - n_2^2) = n_1^2$.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 vibration 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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