91Ó°ÊÓ

In this problem, the volume flow rate can be calculated by using the product of area as well as velocity of fluid.

The diameter of garden hose is \(d = \frac{5}{8}\;{\rm{in}}\).

The diameter of the round swimming pool is \(D = 6.1\;{\rm{m}}\).

The depth of the pool is \(h = 1.4\;{\rm{m}}\).

The speed of water flows from the hose is \(v = 0.40\;{\rm{m/s}}\).

The volume of swimming pool is calculated as:

\(V = \frac{\pi }{4}{D^2}h\)

Substitute the values in the above equation.

\(\begin{array}{c} = \frac{\pi }{4}{\left( {6.1\;{\rm{m}}} \right)^2}\left( {1.4\;{\rm{m}}} \right)\\ = 40.91\;{{\rm{m}}^3}\end{array}\)

The area of hose is calculated below:

\(A = \frac{\pi }{4}{d^2}\)

Substitute the values in the above equation.

\(\begin{array}{c} = \frac{\pi }{4}{\left( {\frac{5}{8}\;{\rm{in}} \times \frac{{0.0254\;{\rm{m}}}}{{1\;{\rm{in}}}}} \right)^2}\\ = 2.01 \times {10^{ - 4}}\;{{\rm{m}}^2}\end{array}\)

The time taken by hose to fill the pool is calculated below:

\(\begin{array}{c}\frac{V}{{\Delta t}} = A \cdot v\\\Delta t = \frac{V}{{A \cdot v}}\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c} = \frac{{40.91\;{{\rm{m}}^3}}}{{\left( {2.01 \times {{10}^{ - 4}}\;{{\rm{m}}^2}} \right)\left( {0.40\;{\rm{m/s}}} \right)}}\\ = 508674.1\;{\rm{s}} \times \left( {\frac{{1\;{\rm{day}}}}{{86400\;{\rm{s}}}}} \right)\\ = 5.88\;{\rm{days}}\end{array}\)

Hence, the time taken by hose to fill the pool is \(5.88\;{\rm{days}}\).