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(C) The frequency of vibration of a stretched string is given by $n = \frac{1}{2L} \sqrt{\frac{T}{m}}$.Since the frequency $N$ remains constant,we hav

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$\frac{L^2}{L^2 - l^2}$

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(C) The frequency of vibration of a stretched string is given by $n = \frac{1}{2L} \sqrt{\frac{T}{m}}$.Since the frequency $N$ remains constant,we have $L \propto \sqrt{T}$,where $T$ is the tension in the wire.In air,the tension $T_{air} = V \rho g$,where $V$ is the volume of the stone and $\rho$ is its density.When immersed in water,the tension becomes $T_{water} = V(\rho - \sigma)g$,where $\sigma$ is the density of water. Taking $\sigma = 1$ (specific gravity unit),$T_{water} = V(\rho - 1)g$.Thus,$\frac{L}{l} = \sqrt{\frac{T_{air}}{T_{water}}} = \sqrt{\frac{V \rho g}{V(\rho - 1)g}} = \sqrt{\frac{\rho}{\rho - 1}}$.Squaring both sides,$\frac{L^2}{l^2} = \frac{\rho}{\rho - 1}$.$\rho(l^2) = L^2(\rho - 1) \implies \rho l^2 = L^2 \rho - L^2$.$L^2 = \rho(L^2 - l^2)$.Therefore,$\rho = \frac{L^2}{L^2 - l^2}$.

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