wave_mechanics mock_test_waves_and_acoustics Question 4
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}-t^{2}} $
D) $ \frac{L^{2}-l^{2}}{L^{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Frequency of vib.
is 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 ($ \therefore $ unison reestablished) Hence length $ \propto \sqrt{T}\Rightarrow \frac{L}{l}=\sqrt{\frac{T _{air}}{T _{water}}=\sqrt{\frac{Vpg}{V(p-1)g}}} $ (Density of stone = p and density of water =1) $ \Rightarrow \frac{L}{l}=\sqrt{\frac{\rho }{\rho -1}}\Rightarrow \rho =\frac{L^{2}}{L^{2}-l^{2}} $
BETA
