A gold sphere of radius 1 cm and density `19.2g//cm^3` with a concentric spherical cavity floats with its whole volume just immersed in molten aluminium of density `2. 4g//cm^3`. The volume of cavity in cubic centimetre would be --------(Take `pi = 22/7` )

Answer

`3.67`

Explanation

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