Question

A gardener feels it is taking too long to water a garden with a \(\frac{{\bf{3}}}{{\bf{8}}}\;{\bf{in}}\) diameter hose. By what factor will the time be cut using a \(\frac{{\bf{5}}}{{\bf{8}}}\;{\bf{in}}\) diameter hose instead? Assume nothing else is changed.

Short answer

The time will be saved by a factor \(0.36\)of the initial time.

Step-by-step solution

Step 1: Understanding the Volume

The amount of fluid that flows in a given time is equal to the product of area of cross-section of pipe with the distance travelled by water.

The expression for the volume of the water flowing through a pipe is given as,

\(V = Ad\)

Step 2: Given Data

The initial diameter of the hose is\({d_1} = \frac{3}{8}\;{\rm{in}}\).

The final diameter of the hose is \({d_2} = \frac{5}{8}\;{\rm{in}}\).

Calculation of time taken

The area of cross-section of a pipe is,

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

The distance traveled by water in a given time is,

\(d = vt\)

Substitute the values in the equation of volume,

\(V = \left( {\pi {{\left( {\frac{d}{2}} \right)}^2}} \right)\left( {vt} \right)\)

The equation for initial volume of water is given as,

\(V = \pi {\left( {\frac{{{d_1}}}{2}} \right)^2}v{t_1}\)

The equation for final volume of water is given as,

\(V = \pi {\left( {\frac{{{d_2}}}{2}} \right)^2}v{t_2}\)

Divide both the equations of volume,

\(\begin{array}{c}\frac{V}{V} = \frac{{\pi {{\left( {\frac{{{d_1}}}{2}} \right)}^2}v{t_1}}}{{\pi {{\left( {\frac{{{d_2}}}{2}} \right)}^2}v{t_2}}}\\1 = \frac{{d_1^2{t_1}}}{{d_2^2{t_2}}}\\{t_2} = \frac{{d_1^2{t_1}}}{{d_2^2}}\end{array}\)

Substitute the values in the above equation,

\(\begin{array}{c}{t_2} = \frac{{{{\left( {\frac{3}{8}\;{\rm{in}}} \right)}^2}{t_1}}}{{{{\left( {\frac{5}{8}\;{\rm{in}}} \right)}^2}}}\\ = \frac{9}{{25}}{t_1}\\ = 0.36{t_1}\end{array}\)

Therefore, the time will be saved by a factor of \(0.36\).