A stone is hung in air from a wire which is stretched over a sonometer. The bridges of the sonometer are $L \,cm$ apart when the wire is in unison with a tuning fork of frequency $ N$. When the stone is completely immersed in water, the length between the bridges is $l \,cm$ for re-establishing unison, the specific gravity of the material of the stone is
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(c) Frequency of vib. is stretched string $n = \frac{1}{{2({\rm{Length)}}}}\sqrt {\frac{T}{m}} $
When the stone is completely immersed in water, length changes but frequency doesn’t ( $\because$ unison reestablished)
Hence length $ \propto \sqrt T $==> $\frac{L}{l} = \sqrt {\frac{{{T_{air}}}}{{{T_{water}}}}} = \sqrt {\frac{{V\rho g}}{{V(\rho - 1)g}}} $
(Density of stone $= \rho$ and density of water $ =1$)
==> $\frac{L}{l} = \sqrt {\frac{\rho }{{\rho - 1}}} $==> $\rho = \frac{{{L^2}}}{{{L^2} - {l^2}}}$
When the stone is completely immersed in water, length changes but frequency doesn’t ( $\because$ unison reestablished)
Hence length $ \propto \sqrt T $==> $\frac{L}{l} = \sqrt {\frac{{{T_{air}}}}{{{T_{water}}}}} = \sqrt {\frac{{V\rho g}}{{V(\rho - 1)g}}} $
(Density of stone $= \rho$ and density of water $ =1$)
==> $\frac{L}{l} = \sqrt {\frac{\rho }{{\rho - 1}}} $==> $\rho = \frac{{{L^2}}}{{{L^2} - {l^2}}}$
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