91Ó°ÊÓ

a) Water displaced by boat,${\mathrm{W}}_{\text{water}}=35.6\mathrm{kN}$

b) Density of saltwater ${\text{ÒÏ}}_{\mathrm{saR}}=1.10×{10}^3\mathrm{kg}/{\mathrm{m}}^3$,

The weight of fresh water and saltwater displaced by the boat should be the same as per Archimedes' principle. Since the density of fresh water and saltwater is different, we can find the volume of saltwater using the given information. By calculating the volume of freshwater using the same concept as above, we can find the difference in the volumes of fresh and saltwater.

Formula:

Weight of a body in terms of density, $\mathrm{W}=\mathrm{ÒÏgV}$(i)

The buoyant force exerted by the fluid on a body,${\mathrm{F}}_{\mathrm{b}}={\mathrm{ÒÏ}}_{\mathrm{w}}\mathrm{gV}$(ii)

From Archimedes’ principle, we can say that the weight of water displaced is equal to the weight of the object that is submerged in liquid, i.e., the water displaced by a boat submerged in water.

So, the weight of displaced water is$35.6\mathrm{kN}$.

Volume of displaced salt water using equation (i) and the given values can be given as:

${\mathrm{V}}_{\text{solf}}=\frac{\mathrm{W}}{{\mathrm{ÒÏ}}_{\text{sat}}\mathrm{g}}$

role="math" localid="1657629806420" $=\frac{35.6×{10}^3}{\left(1.10×{10}^3\right)(9.8)}$

$=3.302{\mathrm{m}}^3$

Similarly in case of fresh water using equation (i) and the given values, the displaced volume is:

${\mathrm{V}}_{\text{fresh}}=\frac{\mathrm{W}}{{\mathrm{ÒÏ}}_{\mathrm{fresh}}×\mathrm{g}}$

$=\frac{35.6×{10}^3}{1000(9.8)}({\mathrm{P}}_{\mathrm{fresh}}=1000\raisebox{1ex}{$\mathrm{kg}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{m}}^3$}\right.$

$=3.75{\mathrm{m}}^3$

Difference in displaced water suing above values can be given as:

$\mathrm{V}={\mathrm{V}}_{\mathrm{fresh}}-{\mathrm{V}}_{\mathrm{salt}}$

$=3.75{\mathrm{m}}^3-3.302{\mathrm{m}}^3$

$=0.45{\mathrm{m}}^3$

Hence, the difference in displaced volume is $0.45{\mathrm{m}}^3$