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A sonometer wire under suitable tension having specific gravity ' $\varrho^{\prime}$, vibrates with frequency 'n' in air. If the load is completely immersed in water the frequency of vibration of wire will become
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The correct answer is:
$n\left[\frac{0-1}{\varrho}\right]^{\frac{1}{2}}$
$(\mathrm{C})$
$\mathrm{T}_{1}=\mathrm{mg}=\mathrm{v} \rho \mathrm{g}$
$\mathrm{T}_{2}=\mathrm{v}(\rho-\sigma) \mathrm{g} \quad \sigma=1$ for water
$\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{\mathrm{vpg}}{\mathrm{v}(\rho-1) \mathrm{g}}=\frac{\rho}{\rho-1}$
$\therefore \frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\sqrt{\frac{\rho}{\rho-1}} \quad \therefore \mathrm{n}_{2}=\mathrm{n}_{1} \sqrt{\frac{\rho-1}{\rho}}$
$\mathrm{T}_{1}=\mathrm{mg}=\mathrm{v} \rho \mathrm{g}$
$\mathrm{T}_{2}=\mathrm{v}(\rho-\sigma) \mathrm{g} \quad \sigma=1$ for water
$\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{\mathrm{vpg}}{\mathrm{v}(\rho-1) \mathrm{g}}=\frac{\rho}{\rho-1}$
$\therefore \frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\sqrt{\frac{\rho}{\rho-1}} \quad \therefore \mathrm{n}_{2}=\mathrm{n}_{1} \sqrt{\frac{\rho-1}{\rho}}$
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