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)

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 = L²h

The volume of water displaced = V = L²h

The buoyant force = weight of the water displaced

F_b = (mass of the water displaced)g

F_b = ρ_wVg = ρ_wL²hg    …(2)

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

ρ_wL²hg = Mg

ρ_wL²h = V_ice ρ_ice

ρ_wL²h = L³ρ_ice

h/L = ; ρ_ice/ρ_w

putting  ρw = 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