Q. 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) $\frac{L^{2} + l^{2}}{L^{2}}$
Explanation
(c) Frequency of vib. in stretched string $n = \frac{1}{2(Length)}\sqrt{\frac{T}{m}}$
When the stone is completely immersed in water, length changes but frequency doesn’t ($\because$ unison re-established)
Hence length $\propto \sqrt{T} \Rightarrow \frac{L}{l} = \sqrt{\frac{T_{air}}{T_{water}}} = \sqrt{\frac{V\rho g}{V(\rho - 1)g}}$
[Density of stone $= \rho$ and density of water $= 1$]
$\Rightarrow \frac{L}{l} = \sqrt{\frac{\rho}{\rho - 1}} \Rightarrow \rho = \frac{L^{2}}{L^{2} - l^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.