Question

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

Answer 1

Answer: 3.67

Sphere will float in liquid if its weight is less than or equal to weight of liquid displaced by it,
i.e., $\frac{4}{3}\pi \left(R^3-r^3\right)\sigma g\le \frac{4}{3}\pi R^3\rho g$
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 $=\frac{4}{3}\pi r^3$
$=\frac{4}{3}\times \frac{22}{7}\times \frac{7}{8}\times 1^3$
$\therefore V=\frac{11}{3}c{m}^3=3.67c{m}^3$