wave_mechanics critical_thinking Question 43
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
Options:
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}} $
Show Answer
Answer:
Correct Answer: C
Solution:
Frequency of vib.
is stretched string $ n=\frac{1}{2(\text{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} $
Therefore $ \frac{L}{l}=\sqrt{\frac{T _{air}}{T _{water}}}=\sqrt{\frac{V\rho g}{V(\rho -1)g}} $ (Density of stone = r and density of water =1)
Therefore $ \frac{L}{l}=\sqrt{\frac{\rho }{\rho -1}} $
Therefore $ \rho =\frac{L^{2}}{L^{2}-l^{2}} $
BETA
