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) \[\frac{{{L}^{2}}-{{l}^{2}}}{{{L}^{2}}}\]
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 \[\propto \sqrt{T}\]Þ \[\frac{L}{l}=\sqrt{\frac{{{T}_{air}}}{{{T}_{water}}}}=\sqrt{\frac{V\rho g}{V(\rho -1)g}}\] (Density of stone = r and density of water =1) Þ \[\frac{L}{l}=\sqrt{\frac{\rho }{\rho -1}}\]Þ \[\rho =\frac{{{L}^{2}}}{{{L}^{2}}-{{l}^{2}}}\]
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