91Ó°ÊÓ

The height of the fountain water is \(h = 12\;{\rm{cm}}\).

The diameter of the nozzle is \({d_1} = 0.60\;{\rm{cm}}\).

The distance below the nozzle is \(d = 1.1\;{\rm{m}}\).

The diameter of the pipe is \({d_{\rm{2}}} = 1.2\;{\rm{cm}}\).

In this problem, the Bernoulli’s equation will be utilized for determining the gauge pressure. Initially, the speed of the water at exit of the nozzle will be evaluated.

The relation of velocity is given by,

\(\begin{array}{l}{v_1} = \sqrt {2gh} \\{v_1} = \sqrt {2\left( {9.80\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {0.12\;{\rm{m}}} \right)} \\{v_1} = 1.53\;{\rm{m/s}}\end{array}\)

Here, \(g\) is the gravitational acceleration.

The relation from equation of continuity is given by,

\(\begin{array}{c}{A_1}{v_1} = {A_2}{v_2}\\{v_2} = \left( {\frac{{{A_1}}}{{{A_2}}}} \right){v_1}\\{v_2} = \left( {\frac{{\frac{1}{2}\pi {{\left( {{d_1}} \right)}^2}}}{{\frac{1}{2}\pi {{\left( {{d_2}} \right)}^2}}}} \right){v_1}\\{v_2} = {v_1}{\left( {\frac{{{d_1}}}{{{d_2}}}} \right)^2}\end{array}\)

Here, \({A_1}\) and \({A_2}\) are the area at the inlet and outlet of the nozzle.

On plugging the values in the above relation.

\(\begin{array}{l}{v_2} = \left( {1.53\;{\rm{m/s}}} \right){\left( {\frac{{0.60\;{\rm{cm}}}}{{1.2\;{\rm{cm}}}}} \right)^2}\\{v_2} = 0.38\;{\rm{m/s}}\end{array}\)

The relation from Bernoulli’s equation is given by,

\({P_{\rm{g}}} = \rho \left( {\frac{1}{2}\left( {{v_1}^2 - {v_2}^2} \right) + gd} \right)\)

Here, \(\rho \) is the density of water.

On plugging the values in the above relation.

\(\begin{array}{l}{P_{\rm{g}}} = \left( {1000\;{\rm{kg/}}{{\rm{m}}^3}} \right)\left[ {\frac{1}{2}\left( {{{\left( {1.53\;{\rm{m/s}}} \right)}^2} - {{\left( {0.38\;{\rm{m/s}}} \right)}^2}} \right) + \left( {9.80\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {1.1\;{\rm{m}}} \right)} \right]\\{P_{\rm{g}}} = 11878.25\;{\rm{N/}}{{\rm{m}}^2}\end{array}\)

Thus, the gauge pressure is \(11878.25\;{\rm{N/}}{{\rm{m}}^2}\).