Q. A stone is hung in air from a wire, which is stretched over a sonometer. The bridges of the sonometer are 40cm apart when the wire is in unison with a tuning fork of frequency 256. When the stone is completely immersed in water, the length between the bridges is 22 cm for re-establishing unison. The specific gravity of material of stone is
A) $\frac{(40^{2})}{\left( 40^{2} \right) + (22)^{2}}$
B) $\frac{\left( 40^{2} \right)}{\left( 40^{2} \right) - (22)^{2}}$
C) $256 \times \frac{22}{40}$
D) $256 \times \frac{40}{22}$
Explanation
(b) When the stone is suspended in air
$n = \frac{1}{2L}\sqrt{\frac{W_{a}}{m}}$
When the stone is suspended in water,
$n = \frac{1}{2L^{'}}\sqrt{\frac{W_{w}}{m}}$
$\therefore\frac{\sqrt{W_{a}}}{L} = \frac{\sqrt{W_{w}}}{L^{'}}\ or\frac{W_{a}}{W_{w}} = \frac{L^{'^{2}}}{L^{2}}$
Specific gravity of stone
$= \frac{W_{a}}{Wa - W_{w}} = \frac{1}{1 - \frac{W_{w}}{W_{a}}} = \frac{1}{1 - \frac{L^{'^{2}}}{L^{2}}}$
$= \frac{L^{2}}{L^{2} - L^{'^{2}}} = \frac{(40)^{2}}{(40)^{2} - (22)^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.