91Ó°ÊÓ

In this problem, the volume rate of flow of water from faucet will be calculated by using the product of area of the faucet and velocity of the faucet.

The diameter of the faucet is \(d = 1.{\rm{85}}\;{\rm{cm}}\).

The pressure head is \(h = 12.0\;{\rm{m}}\).

The area of faucet is calculated below:

\(\begin{array}{c}A = \frac{\pi }{4}{d^2}\\ = \frac{\pi }{4}{\left( {1.85\;{\rm{cm}} \times \frac{{{{10}^{ - 2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}} \right)^2}\\ = 2.69 \times {10^{ - 4}}\;{{\rm{m}}^2}\end{array}\)

The velocity at faucet is calculated below:

\(\begin{array}{c}v = \sqrt {2gh} \\ = \sqrt {2\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {12.0\;{\rm{m}}} \right)} \\ = 15.34\;{\rm{m/s}}\end{array}\)

Here, g is the gravitational acceleration.

The volume rate of flow water from faucet is calculated below:

\(\begin{array}{c}\dot V = Av\\ = \left( {2.69 \times {{10}^{ - 4}}\;{{\rm{m}}^2}} \right)\left( {15.34\;{\rm{m/s}}} \right)\\ = 4.12 \times {10^{ - 3}}\;{{\rm{m}}^3}{\rm{/s}}\end{array}\)

Hence, the volume rate of flow water from faucet is \(4.12 \times {10^{ - 3}}\;{{\rm{m}}^3}{\rm{/s}}\).