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


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


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 [1]

Advertisement Remove all ads

View all notifications

      Forgot password?
View in app×