Short Note

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.

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#### Solution

Given:

Edge of the ice cube (a) = 1.0 cm

The water that is formed due to the melting of ice acquires a spherical surface.

In the absence of gravity, let the radius of the spherical surface be r.

Volume of ice cube = volume of spherical surface of water

\[\Rightarrow a^3 = \frac{4}{3}\pi r^3 \]

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

Surface area of spherical water surface = 4πr^{2}

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

\[ = \left( 36\pi \right)^{1/3} {\text{ cm }}^2\]

Concept: Surface Tension

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