Q: A cube of ice is floating in water. The fraction of its length that lie outside the water is (Sp. Gravity of ice = 0.96)

(A) 0.04

(B) 0.4

(C) 0.96

(D) None of these.

Solution: Since the ice cube is in equilibrium, the net force acting on it is zero. Hence the buoyant force exerted on it is equal to its weight mg where m is the mass of the ice cube .

F_{b} = Mg …(1)

Let the cube of ice be immersed into water to a depth h.

The volume of the cube immersed = V = L^{2}h

The volume of water displaced = V = L^{2}h

The buoyant force = weight of the water displaced

F_{b} = (mass of the water displaced)g

F_{b} = ρ_{w}Vg = ρ_{w}L^{2}hg …(2)

Equating (1) and (2), we obtain,

ρ_{w}L^{2}hg = Mg

ρ_{w}L^{2}h = V_{ice} ρ_{ice}

ρ_{w}L^{2}h = L^{3}ρ_{ice}

h/L = ; ρ_{ice}/ρ_{w}

putting ρ_{w} = 1 gm/cc and ρ_{ice} = 0.96 gm/cc

we obtain, h/L = 0.96

The fraction of its length exposed to air = 1 – h/L = 0.04