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MCQEasyJEE 2025Fluid Pressure & Pascal's Law

JEE Physics 2025 Question with Solution

A 400 g400 \, \text{g} solid cube having an edge of length 10 cm10 \, \text{cm} floats in water. How much volume of the cube is outside the water? (Given: density of water = 1000 kg/m31000 \, \text{kg/m}^3)

  • A

    600 cm3600 \, \text{cm}^3

  • B

    4000 cm34000 \, \text{cm}^3

  • C

    1400 cm31400 \, \text{cm}^3

  • D

    400 cm3400 \, \text{cm}^3

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Mass of the cube is 400 g=0.4 kg400 \, \text{g} = 0.4 \, \text{kg}, edge length is 10 cm10 \, \text{cm}, and density of water is 1000 kg/m31000 \, \text{kg/m}^3.

Find: Volume of the cube outside water.

Using Archimedes’ principle, for a floating body the buoyant force equals the weight of the displaced liquid.

The total volume of the cube is

Vtotal=(10 cm)3=1000 cm3V_{total} = (10 \, \text{cm})^3 = 1000 \, \text{cm}^3

The density of the cube is

ρcube=mVtotal=0.41000×10−6=400 kg/m3\rho_{cube} = \frac{m}{V_{total}} = \frac{0.4}{1000 \times 10^{-6}} = 400 \, \text{kg/m}^3

Since the cube floats, the submerged volume is

Vsubmerged=Vtotal×ρcubeρwaterV_{submerged} = V_{total} \times \frac{\rho_{cube}}{\rho_{water}} Vsubmerged=1000×4001000=400 cm3V_{submerged} = 1000 \times \frac{400}{1000} = 400 \, \text{cm}^3

Hence the volume outside water is

Voutside=Vtotal−VsubmergedV_{outside} = V_{total} - V_{submerged} Voutside=1000−400=600 cm3V_{outside} = 1000 - 400 = 600 \, \text{cm}^3

the solution marks the correct option as D, but the working gives 600 cm3600 \, \text{cm}^3, which matches option A. Therefore, based on the extracted working, the defensible answer should be A.

Common mistakes

  • Using the full cube volume as the displaced volume. For a floating body, only the submerged part displaces water. First find the submerged volume from floating equilibrium, then subtract from the total volume.

  • Confusing volume inside water with volume outside water. After calculating VsubmergedV_{submerged}, do not stop there; the question asks for the part outside water, so use Voutside=Vtotal−VsubmergedV_{outside} = V_{total} - V_{submerged}.

  • Making a unit-conversion error between g\text{g}, kg\text{kg}, cm3\text{cm}^3, and m3\text{m}^3. Convert mass and volume consistently before using density relations.

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