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(A) The frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass densit

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$n \left[ \frac{\varrho-1}{\varrho} \right]^{\frac{1}{2}}$

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(A) The frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density. Since $L$ and $\mu$ are constant,$n \propto \sqrt{T}$.In air,the tension $T_1 = mg = V \varrho g$,where $V$ is the volume of the load and $\varrho$ is its specific gravity (density relative to water).When the load is immersed in water,the buoyant force acts on it. The effective tension becomes $T_2 = V(\varrho - 1)g$ (since the density of water is $1 	ext{ g/cm}^3$).Taking the ratio of the frequencies: $\frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{V(\varrho - 1)g}{V \varrho g}} = \sqrt{\frac{\varrho - 1}{\varrho}}$.Therefore,the new frequency $n_2 = n \sqrt{\frac{\varrho - 1}{\varrho}}$.

$A$ sonometer wire under suitable tension having specific gravity $\varrho$,vibrates with frequency $n$ in air. If the load is completely immersed in water,the frequency of vibration of the wire will become:

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