Practice portal › Mechanical Properties of Fluids › Archimedes' Principle and Floating
Asked in JEE Main 10th April 2nd Shift 2019 · Floating and submerged bodies
Given: cube of side 0.5 m, so V=0.125 m³, floating with 30% submerged, ρ_w=10³ kg m⁻³.
Floating as it is, the block's own mass equals the water it displaces: m=ρ_w(0.30V)=10³×0.0375=37.5 kg.
Just fully submerged, the displacement is the whole cube: ρ_wV=10³×0.125=125 kg of water.
The extra load is the difference between what it can displace and what it already weighs.
M=125-37.5=87.5 kg.
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