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A stone hangs from the free end of a sonometer wire whose vibrating length, when tuned to a tuning fork is 40 cm. When the stone hangs wholly immersed in water, the resonant length is reduced to 30 cm in same mode of oscillation and to have the same frequency. The relative density of the stone is

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$\frac{16}{7}$

$\frac{16}{5}$

$\frac{16}{9}$

$\frac{16}{3}$

answer is B.

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Detailed Solution

$\begin{array}{l} f = \frac{\sqrt{T}}{l} \\ \sqrt{\frac{T_{1}}{T_{2}}} = \frac{l_{1}}{l_{2}} = \frac{4}{3} \\ \sqrt{\frac{M g}{M g (1 - \frac{1}{R . D})}} = \frac{4}{3} \\ 1 - \frac{1}{R . D} = \frac{9}{16} \\ R . D = \frac{16}{7} \end{array}$

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