JEE Main 2026 · 24 January · evening · Single correct · +4 / −1

A cubical block of density ρ_b=600 kg/m³ floats in a liquid of density ρₑ=900kg/m³. If the height of block is H=8.0 cm then height of the submerged part is
____ cm .

  1. (a) 6.3
  2. (b) 4.3
  3. (c) 7.3
  4. (d) 5.3
Answer and solution

Answer (d) 5.3

Solution

Physics sir · 3 steps

Weight equals buoyant force, and since the cube has one cross-section, only heights are left to compare.

  1. Floating condition. A floating block is in equilibrium, so its weight equals the weight of liquid it displaces. Let the face area be A and the submerged height be h.

    ρ_b A H g = ρ_ℓ A h g ⇒ h = ρ_b/ρ_ℓ H

  2. Put in numbers. ρ_b/ρ_ℓ = 600/900 = 2/3, so h = 2/3 × 8.0 cm = 5.33 cm.
  3. So. The submerged height is about 5.3 cm, which is option (d).

The idea to keep. For a floating block, the fraction submerged equals block density divided by liquid density. For a cube or any prism, that fraction applies directly to the height.

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

Study Fluid Mechanics

Asked about 90 times in JEE Main since 2020. This question is from 2026.