Question

A stone is hung in air from a wire which is stretched over a sonometre. The bridges of the sonometre are $L\text{cm}$ 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 $l\text{cm}$ for re-establishing unison, the specific gravity of the material of the stone is

• A. $\frac{L^2}{L^2+l^2}$
• B. $\frac{L^2-l^2}{L^2}$
• C. $\frac{L^2}{L^2-l^2}$
• D. $\frac{L^2+l^2}{L^2}$

Answer 1

The correct option is C $\frac{L^2}{L^2-l^2}$

Let the condition be as shown in the figure
From the diagram we have
$T=mg$
and $T^{{}^{\prime }}+F_B=mg$
$T^{{}^{\prime }}+{\rho }_{\omega }\frac{m}{\sigma }g=mg$
$T^{{}^{\prime }}=mg(1-\frac{{\rho }_{\omega }}{\sigma })$
Where, ${\rho }_{\omega }$ is density of water and $\sigma$ is density of block
Now from the question,
$N=\frac{1}{2L}\sqrt{\frac{mg}{\mu }}=\frac{1}{2l}\sqrt{\frac{mg(1-\frac{{\rho }_{\omega }}{\sigma })}{\mu }}$
$\Rightarrow \frac{l}{L}=\sqrt{1-\frac{1}{{\sigma }^{{}^{\prime }}}}$
where ${\sigma }^{{}^{\prime }}$ is specific gravity of stone
Thus, we have
$\frac{l^2}{L^2}=1-\frac{1}{{\sigma }^{{}^{\prime }}}\Rightarrow \frac{1}{{\sigma }^{{}^{\prime }}}=1-\frac{l^2}{L^2}=\frac{L^2-l^2}{L^2}$
Hence,
${\sigma }^{{}^{\prime }}=\frac{L^2}{L^2-l^2}$