91Ó°ÊÓ

i) Volume of wooden block in freshwater is$\frac{2}{3}\mathrm{V}$

ii) Volume of wooden block in oil is$0.9\mathrm{V}$

iii) Density of fresh water,role="math" ${\mathrm{ÒÏ}}_{\mathrm{l}}=1000\mathrm{kg}/{\mathrm{m}}^3$

We can write the force of buoyancy in terms of density of the liquid, gravitational constant, and volume of the liquid displaced. The weight of the wooden block is balanced by this force of buoyancy. We can find the density of the liquid and the oil by equating the force of buoyancy to the weight of the wooden block.

Formula:

The buoyant force on a completely submerged object,${\mathrm{F}}_{\mathrm{b}}={\mathrm{ÒÏ}}_1{\mathrm{gV}}_1$(i)

The buoyant force is balanced by the weight of the wooden block.

${\mathrm{F}}_{\mathrm{b}}={\mathrm{F}}_{\mathrm{d}}$

Force of Buoyancy for freshwater using equation (i) is written as:

$F_{bHW}={\mathrm{ÒÏ}}_{w}g{V}_{\mathrm{w}}$

${\mathrm{P}}_{\mathrm{w}}$is the density of the liquid, and ${\mathrm{V}}_3$is the volume of liquid displaced.

Since$2/3$ volume of the wood is submerged in the water, we can write

${\mathrm{V}}_{\mathrm{w}}=\frac{2}{3}\mathrm{V}$

$\mathrm{V}$is the volume of wood.

Weight of the wood is balanced by the buoyance force, so we can write it as

$\mathrm{ÒÏgV}={\mathrm{ÒÏ}}_{\mathrm{w}}{\mathrm{gV}}_{\mathrm{v}}$

$\mathrm{ÒÏgV}={\mathrm{ÒÏ}}_{\mathrm{w}}\mathrm{g}\frac{2}{3}\mathrm{V}$

Canceling $\mathrm{V}$on both sides and substituting the value of density of water, we have

$\mathrm{ÒÏ}=666.66\mathrm{kg}/{\mathrm{m}}^3$

$≈667\mathrm{kg}/{\mathrm{m}}^3$

Hence, the density of wood is$667\mathrm{kg}/{\mathrm{m}}^3$

Buoyance force by oil should be equal to the weight of the wood to make wood float. Therefore, we can write

${\mathrm{ÒÏ}}_{o}g0.9\mathrm{V}=\mathrm{ÒÏ}g\mathrm{V}$ (density of wood,$\mathrm{ÒÏ}=667\mathrm{kg}/{\mathrm{m}}^3$

${\mathrm{ÒÏ}}_{\mathrm{o}}=741.11\mathrm{kg}/{\mathrm{m}}^3$

$≈741\mathrm{kg}/{\mathrm{m}}^3$

Hence, the density of oil is$741\mathrm{kg}/{\mathrm{m}}^3$