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Asphere of radius $R$ and made of material of relative density $\sigma$ has a concentric cavity of radius $r$. It just floats when placed in a tank full of water. The value of the ratio $R/r$ will be
${\left( {\frac{\sigma }{{\sigma - 1}}} \right)^{1/3}}$
${\left( {\frac{{\sigma - 1}}{\sigma }} \right)^{1/3}}$
${\left( {\frac{{\sigma + 1}}{\sigma }} \right)^{1/3}}$
${\left( {\frac{{\sigma - 1}}{{\sigma + 1}}} \right)^{1/3}}$
Solution
According to law of floatation, a body floats if weight of the body is equal to the weight of water displaced
$\Rightarrow W_{s}=W_{w}$
$\Rightarrow V_{s} \times \rho_{s} \times g=V_{w} \times \rho_{w} \times g$
$\Rightarrow \frac{4}{3} \pi\left(R^{3}-r^{3}\right) \times \sigma \rho_{w}=\frac{4}{3} \pi R^{3} \times \rho_{w}$
$\Rightarrow \sigma\left(R^{3}-r^{3}\right)=R^{3}$
$\Rightarrow R^{3}=\left(\frac{\sigma}{\sigma-1}\right) r^{3}$
$\Rightarrow \frac{R}{r}=\left(\frac{\sigma}{\sigma-1}\right)^{1 / 3}$
