Question

A block of wood floats in a liquid of density $0.8gc{m}^{-3}$ with one fourth of its volume submerged. In oil the block floats with $60$% of its volume submerged. Find the density of (a) wood and (b) oil

Answer 1

According to Archimedes' principle.

Mass of volume displaced is by object is buoyancy force.

Let the volume of wood is V.

then the volume submerged in liquid is $V/4$

let the density of wood be $\rho$.

let the density of liquid be ${\rho }_l=0.8c{m}^{-3}$.

weight of wood $\rho Vg$.

buoyancy force =$\frac{{\rho }_{l}V}{4}g$

now weight of block is equal to buoyancy force.

$\rho Vg=\frac{{\rho }_{l}V}{4}g$

So $\rho =\frac{0.8}{4}=.2gc{m}^{-3}$

then the volume submerged in oil is 60% of is $3V/5$

weight of wood $\rho Vg$

buoyancy force =$\frac{{\rho }_{oil}3V}{5}g$

now weight of block is equal to buoyancy force.

$\rho Vg=\frac{{\rho }_{oil}3V}{5}g$

So ${\rho }_{oil}=\frac{\rho \times 5}{3}=\frac{0.2\times 5}{3}=0.33gc{m}^{-3}$