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ExamsJEE AdvancedPhysics

A cube of ice floats partly in water and partly in a liquid X. Given that the specific gravity of liquid X is 0.4 and the specific gravity of ice is 0.9, find the ratio of the volume of ice submerged in water to the volume of ice submerged in liquid X.

  1. 1:1
  2. 2:1
  3. 1:2
  4. 9:4

Correct answer: 1:2

Solution

Equilibrium: rho_w * V_w + rhoₓ * Vₓ = rho_ice * (V_w + Vₓ). With rho_w = 1, rhoₓ = 0.4, rho_ice = 0.9: V_w + 0.4*Vₓ = 0.9*V_w + 0.9*Vₓ. Rearranging: V_w - 0.9*V_w = 0.9*Vₓ - 0.4*Vₓ => 0.1*V_w = 0.5*Vₓ => V_w/Vₓ = 0.5/0.1 = 5/1. Wait — that gives 5:1, not 1:2. Let me redo: V_w + 0.4Vₓ = 0.9(V_w + Vₓ) = 0.9V_w + 0.9Vₓ. So V_w - 0.9V_w = 0.9Vₓ - 0.4Vₓ => 0.1V_w = 0.5Vₓ => V_w = 5Vₓ => V_w:Vₓ = 5:1. But this is not among the options! Check: ice density 0.9, water density 1, liquid density 0.4. Buoyancy = weight: 1*V_w + 0.4*Vₓ = 0.9*(V_w+Vₓ). V_w + 0.4Vₓ = 0.9V_w + 0.9Vₓ. 0.1V_w = 0.5Vₓ. V_w:Vₓ = 5:1. None of the given options (1:1, 2:1, 1:2, 9:4) match. The question may have swapped water and liquid X, or the ice density differs. If we swap: rho_ice = 0.9, rho_water (liquid X at bottom) = 0.4, rhoₓ (liquid on top) = 1 — that does not physically make sense. The problem is likely defective or has a typo in densities. Given available options and the closest algebraic scenario, marking defective.

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