Question

A sphere of gold of radius $1\, cm$ and density $19.2\, g / cm ^{3}$ has a concentric spherical cavity. If it floats with its whole volume just immersed in molten aluminium of density $2.4\, g / cm ^{3}$ then volume of cavity in cubic centimetre would be_________ . (Take $\pi=\frac{22}{7}$ )

Answer 1

Sphere will float in liquid if its weight is less than or equal to weight of liquid displaced by it,
i.e., $\frac{4}{3} \pi\left( R ^{3}- r ^{3}\right) \sigma g \leq \frac{4}{3} \pi R ^{3} \rho g$
Where, $R$ is radius of sphere, $r$ is radius of cavity, $\sigma$ density of gold and $\rho$ is density of molten aluminium.
As the sphere is floating with its whole volume just immersed in liquid,
$\left( R ^{3}- r ^{3}\right) \sigma= R ^{3} \rho$
$\therefore \frac{ R ^{3}- r ^{3}}{ R ^{3}}=\frac{\rho}{\sigma}=\frac{2.4}{19.2}$
$\therefore 1-\frac{r^{3}}{R^{3}}=\frac{1}{8}$
$\therefore \frac{r^{3}}{R^{3}}=1-\frac{1}{8}=\frac{7}{8}$
$\therefore r^{3}=\frac{7}{8} \times 1^{3}$
$\therefore $ Volume of cavity $=\frac{4}{3} \pi r^{3}$
$=\frac{4}{3} \times \frac{22}{7} \times \frac{7}{8} \times 1^{3}$
$\therefore V =\frac{11}{3} cm ^{3}=3.67\, cm ^{3}$