Question

A boat floating in fresh water displaces water weighing $\mathbf{35}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{kN}$. (a) What is the weight of the water this boat displaces when floating in salt water of density $\mathbf{1}\mathbf{.10}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$? (b) What is the difference between the volume of fresh water displaced and the volume of salt water displaced?

Short answer

a) Weight of water that gets displaced is $35.6\mathrm{kN}$

b) The difference in the volume of displaced fresh water and salt water is $0.45{\mathrm{m}}^3$

Step-by-step solution

The given data

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

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

Understanding the concept of Archimedes’ Principle

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, $\mathrm{W}=\mathrm{\rho gV}$(i)

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

a) Calculation of weight of displaced water

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$35.6\mathrm{kN}$.

b) Calculation of the difference in the volume of displaced fresh water and saltwater

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

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

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

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

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

${\mathrm{V}}_{\text{fresh}}=\frac{\mathrm{W}}{{\mathrm{\rho }}_{\mathrm{fresh}}\times \mathrm{g}}$

$=\frac{35.6\times {10}^3}{1000(9.8)}({\mathrm{P}}_{\mathrm{fresh}}=1000\raisebox{1ex}{$\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 $0.45{\mathrm{m}}^3$