A) \[\sin \theta =\frac{1}{2}\sqrt{{{\rho }_{0}}/\rho }\]
B) \[\sin \theta =\frac{1}{2}\,.\,\frac{{{\rho }_{0}}}{\rho }\]
C) \[\sin \theta =\sqrt{\rho /{{\rho }_{0}}}\]
D) \[\sin \theta ={{\rho }_{0}}/\rho \]
Correct Answer: A
Solution :
Let L = PQ = length of rod \ \[SP=SQ=\frac{L}{2}\] Weight of rod, \[W=Al\rho g\], acting At point S And force of buoyancy, \[{{F}_{B}}=Al{{\rho }_{0}}g\], [l = PR] which acts at mid-point of PR. For rotational equilibrium, \[Al{{\rho }_{0}}g\times \frac{l}{2}\cos \theta =AL\rho g\times \frac{L}{2}\cos \theta \] Þ \[\frac{{{l}^{2}}}{{{L}^{2}}}=\frac{\rho }{{{\rho }_{0}}}\] Þ \[\frac{l}{L}=\sqrt{\frac{\rho }{{{\rho }_{0}}}}\] From figure, \[\sin \theta =\frac{h}{l}=\frac{L}{2l}=\frac{1}{2}\sqrt{\frac{{{\rho }_{0}}}{\rho }}\]You need to login to perform this action.
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