A sphere of relative density σ and diameter D has concentric cavity of diameter d. The ratio of D/d, if it just floats on water in a tank is :

Solution

Physics sir · 4 steps

Just floats means fully under water: the shell's weight equals the buoyancy on the whole sphere.

1. What just floats means. The sphere is completely submerged, with its top just touching the surface. So it displaces water equal to its full outer volume, π/6D³. The cavity is empty and adds no weight.
2. Weight and buoyancy. Only the material between the two spheres has weight: W = σ ρ_w g π/6(D³ - d³). The buoyant force is B = ρ_w g π/6D³, where ρ_w is the density of water.

   σ(D³ - d³) = D³ ⇒ D³(σ - 1) = σ d³

3. Solve the ratio. Dividing gives D³/d³ = σ/(σ - 1), so D/d = (σ/(σ-1))^(1/3).
4. So. The answer is (d). A quick check: σ > 1 makes this ratio greater than 1, as it must be, since the outer diameter is larger than the cavity.

The idea to keep. For a hollow body that just floats, weigh only the material but count the full outer volume for buoyancy. The cavity removes weight, not displaced water.

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