Question

consider an ice cube of edge 1.0 cm kept in a gravity -free hall. Find the surface area of the water when the ice melts. Neglect the difference in densities of ice and water.

Answer 1

In the absence of gravity the water which form due to melting of ice tends to acquire spherical surface.

So, $6a^3=\left(\frac{4}{3}\right)\pi r^3$

$\Rightarrow r={\left[\frac{18a^3}{4\pi }\right]}^{\frac{1}{3}}$

So, surface area = $4\pi r^2$

= $4\pi {\left[\frac{18a^3}{4\pi }\right]}^{\frac{2}{3}}$

= $(36\pi {)}^{\frac{1}{3}}c{m}^2$