# 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 .

Fb = 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
Fb = (mass of the water displaced)g
Fb = ρwVg = ρwL2hg    …(2)
Equating (1) and (2), we obtain,
ρwL2hg = Mg

ρwL2h = Vice ρice
ρwL2h = L3ρice
h/L = ; ρicew

putting  ρw = 1 gm/cc and ρice = 0.96 gm/cc
we obtain, h/L  = 0.96
The fraction of its length exposed to air = 1 – h/L = 0.04

## Author: Rajesh Jha

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