A) \[\frac{2{{\rho }_{0}}}{5}\]
B) \[\frac{{{\rho }_{0}}}{3}\]
C) \[\frac{{{\rho }_{0}}}{5}\]
D) \[\frac{2{{\rho }_{0}}}{3}\]
Correct Answer: A
Solution :
[a] For minimum density of liquid, solid sphere has to float (completely immersed) in the liquid. \[\therefore mg={{F}_{B}}\left( Also\text{ }{{V}_{immersed}}={{V}_{total}} \right)\] or \[\int{\rho dV=\frac{4}{3}\pi {{R}^{3}}\rho \ell }\] \[\Rightarrow \int\limits_{0}^{R}{{{\rho }_{0}}4\pi \left( 1-\frac{{{r}^{2}}}{{{R}^{2}}} \right)\cdot \,{{r}^{2}}dr=\frac{4}{3}\pi {{R}^{3}}\rho \ell }\] \[\Rightarrow 4\pi {{\rho }_{0}}\left[ \frac{{{r}^{3}}}{3}-\frac{{{r}^{5}}}{5R} \right]_{0}^{R}=\frac{4}{3}\pi {{R}^{3}}\rho \ell \] \[\Rightarrow \frac{4\pi {{\rho }_{0}}{{R}^{3}}}{3}\times \frac{2}{5}=\frac{4}{3}\pi {{R}^{3}}\rho \ell \] \[\therefore \rho \ell =\frac{2{{\rho }_{0}}}{5}\]You need to login to perform this action.
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