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) \[\frac{{{L}^{2}}-{{l}^{2}}}{{{L}^{2}}}\]
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) \[\Rightarrow \frac{L}{l}=\sqrt{\frac{\rho }{\rho -1}}\Rightarrow \rho =\frac{{{L}^{2}}}{{{L}^{2}}-{{l}^{2}}}\]You need to login to perform this action.
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