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 $\frac{R}{r}={\left(\frac{1}{1-\frac{1}{\sigma }}\right)}^{1/3}={\left(\frac{\sigma }{\sigma -1}\right)}^{1/3}$