Question

A sphere of radius $R$ has a concentric spherical cavity of radius $r .$ The relative density of the material of the sphere is $\sigma .$ It just floats when placed in tank full of water. The value of $\frac{R}{r }$ is

• A. $\left(\frac{\sigma}{\sigma-1}\right)^{\frac{1}{3}}$
• B. $\left(\frac{\sigma-1}{\sigma}\right)^{\frac{1}{3}}$
• C. $\left(\frac{\sigma}{\sigma-1}\right)^{\frac{1}{2}}$
• D. $\left(\frac{\sigma-1}{\sigma}\right)^{\frac{1}{2}}$

Answer 1

Answer: (A)

Buoyancy force due to water, on a sphere of radius $R$,
$F_{w}=\rho_{w} g V_{s}$
where,$\rho_{w} =$ density of water
$V_{s} =$ volume of sphere
$\therefore F_{w} =\rho_{w} g\left(\frac{4}{3} \pi R^{3}\right) \,\,\,\left(\because V_{s}=\frac{4}{3} \pi R^{3}\right)$
weight of the sphere,
$F_{s}=\rho_{s} g\left(\frac{4}{3} \pi\left(R^{3}-r^{3}\right)\right)$
where $\rho_{s}=$ density of material of sphere
According to Archimedes' principle for floating,
$\therefore F_{w}=F_{s}$
$\Rightarrow \rho_{w} g\left(\frac{4}{3} \pi R^{3}\right)=\rho_{s} g\left(\frac{4}{3} \pi\left(R^{3}-r^{3}\right)\right)$
$\Rightarrow \rho_{w} R^{3}=\rho_{s}\left(R^{3}-r^{3}\right)$
$\Rightarrow R^{3}\left(\rho_{s}-\rho_{w}\right)=r^{3} \rho_{s}$
$\Rightarrow \frac{R}{r}=\left(\frac{\rho_{s}}{\rho_{s}-\rho_{w}}\right)^{1 / 3}$ or $\frac{R}{r}=\left(\frac{1}{1-\frac{\rho_{w}}{\rho_{s}}}\right)^{1 / 3}$
$\because$ Relative density of material of the sphere, $\sigma=\frac{\rho_{s}}{\rho_{w}}$
Or $ \frac{R}{r}=\left(\frac{1}{1-\frac{1}{\sigma}}\right)^{1 / 3}=\left(\frac{\sigma}{\sigma-1}\right)^{1 / 3}$