Question

A block of wood floats on water with ${\frac{1}{4}}^{\text{th}}$ of its volume above the water surface. The density of the wooden block will be:
(Take, density of water = $1{\text{g cm}}^{-3}$.)

• A. $0.50{\text{g cm}}^{-3}$
• B. $0.25{\text{g cm}}^{-3}$
• C. $1.00{\text{g cm}}^{-3}$
• D. $0.75{\text{g cm}}^{-3}$

Answer 1

The correct option is D $0.75{\text{g cm}}^{-3}$

let the volume of the block be V and density of wood be ρ.

The volume of the block above the surface= $\frac{1}{4}V$

The volume of the immersed block (v) =$V-\frac{1}{4}V$= $\frac{3}{4}V$

We know, Weight of the block = weight of water displaced by immersed part of the block.

i.e. V x ρ x g = $\frac{3}{4}V$ x 1 x g

ρ = $\frac{3}{4}$ = $0.75{\text{g cm}}^{-3}$