Question

A ship, carrying fresh water to a desert island in the Carib-bean, has a horizontal cross-sectional area of \(2240\;{{\rm{m}}^2}\) at the waterline. When unloaded, the ship rises 8.25 m higher in the sea. How much water \(\left( {{m^3}} \right)\) was delivered?

Short answer

The volume of water delivered is \(18900\;{{\rm{m}}^3}\).

Step-by-step solution

Given Data

The cross-sectional area is \(A = 2240\;{{\rm{m}}^2}\).

The distance above the sea is \(d = 8.25\;{\rm{m}}\).

Understanding the buoyancy force

In this problem, the loaded fresh water creates the additional buoyant force on the ship, that is equal to the weight of the sweater displaced.

Calculate the mass of the fresh water and volume of the water delivered

The relation ofbuoyant forceis given by,

\(\begin{array}{c}F = {\rho _s}Vg\\mg = {\rho _s}Vg\\m = {\rho _s}\left( {A \times d} \right)\end{array}\)

Here, \({\rho _s}\) is the density of seawater, V is the volume, \(m\) is the mass of the fresh water and \(g\) is the gravitational acceleration.

On plugging the values in the above relation.

\(\begin{array}{l}m = \left( {1025\;{\rm{kg/}}{{\rm{m}}^3}} \right)\left( {2240\;{{\rm{m}}^2} \times 8.25\;{\rm{m}}} \right)\\m = 1.89 \times {10^7}\;{\rm{kg}}\end{array}\)

The volume of the water delivered will be calculated as:

\(\begin{array}{l}{V_0} = \frac{m}{{{\rho _w}}}\\{V_0} = \left( {\frac{{1.89 \times {{10}^7}\;{\rm{kg}}}}{{1000\;{\rm{kg/}}{{\rm{m}}^3}}}} \right)\\{V_0} = 18900\;{{\rm{m}}^3}\end{array}\)

Thus, the volume of water delivered is \(18900\;{{\rm{m}}^3}\).