Question

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

• 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. $\frac{L^2-l^2}{2L^2}$

Answer 1

The correct option is C

$\frac{L^2}{L^2-l^2}$

Frequence of vibrations in stretched string, n = $\frac{1}{2\left(Length\right)}$ $\sqrt{\frac{T}{\mu }}$

When the stone is completely immersed in water , length changes but frequence doesn't ( $\because$ unison is reestablished )

Hence length $\alpha$ $\sqrt{T}$ $\Rightarrow$ $\frac{L}{l}$ = $\sqrt{\frac{T_{air}}{T_{water}}}$ = $\sqrt{\frac{V_{\rho g}}{v(\rho -1)g}}$

(Specific gravity of stone = $\rho$ and of water = 1)

$\Rightarrow$ $\frac{L}{l}$ = $\sqrt{\frac{\rho }{\rho -1}}$ $\Rightarrow$ $\rho$ = $\frac{L^2}{L^2-l^2}$