A cube of ice floats partly in water and partly in kerosene oil. The ratio of volume of ice immersed in water to that in kerosene oil (specific gravity of Kerosene oil = 0.8, specific gravity of ice = 0.9):

Solution

Physics sir · 4 steps

The cube is fully under the surface, so its weight is balanced by two buoyant forces, one from each liquid.

1. Read the figure. The top of the cube is below the kerosene surface, so the whole cube is submerged. A volume V_w sits in water and the rest, Vₖ, sits in kerosene. The total volume is V_w + Vₖ.
2. Balance the forces. The cube floats, so its weight equals the total upthrust. Each liquid pushes up with the weight of the liquid it displaces. Write densities as specific gravity times ρ_(water). Then ρ_(water) and g cancel from every term.

   0.9 (V_w + Vₖ) = 1 × V_w + 0.8 Vₖ

3. Solve. Expanding gives 0.9V_w + 0.9Vₖ = V_w + 0.8Vₖ, so 0.1Vₖ = 0.1V_w. That means V_w = Vₖ.
4. So. The ratio of the volume in water to the volume in kerosene is 1 : 1. The answer is (d). This makes sense because 0.9 lies exactly midway between 0.8 and 1.

The idea to keep. For a body floating across two liquids, equate its weight to the sum of the two upthrusts. If its density lies exactly midway between the two liquids, it splits equally.

Answer from the NTA final answer key. Solution by PrepWiser.