Consider a solid sphere of Radius R and mass density \rho (r)=\rho _{0}\left ( 1-\frac{r^{2}}{R^{2}} \right ),\: \: 0<r\leq R. The minimum density of a liquid in which it will float is :   
Option: 1 \frac{2\rho _{0}}{5}

Option: 2 \frac{\rho _{0}}{5}

Option: 3 \frac{2\rho _{0}}{3}  

Option: 4 \frac{\rho _{0}}{3}
 

Answers (1)

 

 

As

Buoyant force-

  • The buoyant force is given as 

F_B=\rho Vg

Where FB=Buoyant force

         \rho= density of the fluid

            V= Volume of the solid body immersed in the liquid or Volume of the fluid displaced

 So

Applying force balance

\rho_l\times\frac{4}{3}\pi R^3=\rho_o(4\pi)\int r^2\left ( 1-\frac{r^2}{R^2} \right )dr

\Rightarrow \rho_l\times\frac{4}{3}\pi R^3=\rho_o(4\pi)\left [ \frac{r^3}{3}-\frac{r^5}{5R^2} \right ]^R_0

\Rightarrow \rho_l\times\frac{4}{3}\pi R^3=\rho_o(4\pi)\left [ \frac{R^3}{3}-\frac{R^3}{5} \right ]

\Rightarrow \rho_l\times\frac{4}{3}\pi R^3=\rho_o(4\pi)\left [\frac{2R^3}{15} \right ]

\Rightarrow \rho_l=\frac{2}{5}\rho_o

So the correct option is 1.

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