Advertisement Remove all ads

# 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. - Physics

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.

Advertisement Remove all ads

#### 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 = 4πr2

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

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

Is there an error in this question or solution?
Advertisement Remove all ads

#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 14 Some Mechanical Properties of Matter
Q 28 | Page 301
Advertisement Remove all ads

#### Video TutorialsVIEW ALL 

Advertisement Remove all ads
Share
Notifications

View all notifications

Forgot password?