Question

A cube of wood floats in water, with $42$% of its volume is submerged, then the density of the wood is

• A. $42gc{m}^{-3}$
• B. $0.42gc{m}^{-3}$
• C. $0.58kgc{m}^{-3}$
• D. $600gc{m}^{-3}$

Answer 1

Answer: (B)

$0.42gc{m}^{-3}$

By Archimedes Principle, the buoyant force on a body partially or fully immersed in a fluid is given by the weight of the fluid displaced.

Let the volume of wood be $V$

Thus, volume of wood submerged is $0.42V$

Thus, the buoyant force acting on the wood is $B=0.42\rho gV$

Weight of wood is ${\rho }_{\text{wood}}gV$

Thus, in equilibrium, $0.42\rho gV={\rho }_{\text{wood}}gV\Rightarrow {\rho }_{\text{wood}}=0.42\rho$

As Density of water is $\rho =1{\text{g cm}}^{-3}$, we have ${\rho }_{\text{wood}}=0.42{\text{g cm}}^{-3}$