Question

A cube of ice of edge of 4 cm is placed in an empty cylindrical glass of inner diameter 6 cm. Assume that the ice melts uniformly from each side such that it always retains its cubical shape. Remembering that ice is lighter than water, find the length of the edge of the ice cube at the instant it just leaves the contact with the bottom of the glass.

Answer 1

Let L be the side of the cube when it just floats in the water that is collected by melting of the cube.

Let ρ = density of water

ρ' = density of ice

The area of the one of the surfaces of the cube will be L2

Let up to a height h the ice cube is floating hence the mass of water displaced will be ρhL2

Buoyant force = ρghL2

Weight of the ice cube = ρ’gL3

But

Buoyant force = Weight of the ice cube

ρghL2 = ρ’gL3

=> h = ρ’L/ρ ---1.

Now, to make the cube just float water height must be of the height

Volume of water corresponding to height h in the cylinder will be

volume of the cylinder at height h – volume occupied by ice

V = π(6/2)2h - hL2---2

But this water has come from the melting of the ice cube thus

Mass of water got from melting of ice = ρ’(64 – L3) (remembering density of ice is lesser than that of water)

Volume of this mass of water = ρ’(64 – L3)/ρ ---3.

By eqn. 2 and 3

π(6/2)2h - hL2 = ρ’(64 – L3)/ρ

By 1.

(π(6/2)2 –L2)ρ’L/ρ = ρ’(64 – L3)/ρ

=> 9πL = 64

=> L = 2.26 cm