91Ó°ÊÓ

Newton’s law states that the net force applied to an object depends upon the mass and acceleration of the object. The product of the area of pipe and velocity of flow of liquid for each point of a pipe is always constant as stated by the equation of continuity.

The diameter of hose is\(D = 7.0\;{\rm{cm}}\).

The volume of water delivered by hose in one minute is\(V = 420\;{\rm{L}}\).

The diameter of nozzle is \(d = 0.75\;{\rm{cm}}\).

The expression for the equation of continuity,

\(\begin{array}{c}{A_1}{v_1} = {A_2}{v_2}\\{v_2} = \frac{{{A_1}{v_1}}}{{{A_2}}}\end{array}\)

Here, \({v_1}\) is the velocity at the entrance and \({v_2}\) is the velocity at the exit, \({A_1}\) is the area of hose and \({A_2}\) is the area of nozzle.

The expression for the mass of object can be given as,

\(m = \rho V\)

According to Newton’s law,

\(F = m\frac{{{v_2} - {v_1}}}{t}\)

Substitute the values in the above equation,

\(\begin{array}{c}F = \rho V\left( {\frac{{\frac{{{A_1}{v_1}}}{{{A_2}}} - \frac{{{A_2}{v_2}}}{{{A_1}}}}}{t}} \right)\\ = \frac{{\rho V}}{t}\left( {\frac{{{A_1}{v_1}}}{{{A_2}}} - \frac{{{A_2}{v_2}}}{{{A_1}}}} \right)\end{array}\)…….. (i)

Now, according to equation of continuity,

\({A_1}{v_1} = {A_2}{v_2} = \frac{V}{t}\)

Substitute the values in the equation (i),

\(\begin{array}{c}F = \frac{{\rho V}}{t}\left( {{A_1}{v_1}} \right)\left( {\frac{1}{{{A_2}}} - \frac{1}{{{A_1}}}} \right)\\ = \frac{{\rho V}}{t}\left( {\frac{V}{t}} \right)\left( {\frac{1}{{{A_2}}} - \frac{1}{{{A_1}}}} \right)\\ = \rho {\left( {\frac{V}{t}} \right)^2}\left( {\frac{1}{{{A_2}}} - \frac{1}{{{A_1}}}} \right)\end{array}\)……. (ii)

The area of hose can be given as,

\({A_1} = \pi {\left( {\frac{D}{2}} \right)^2}\)

The area of nozzle can be given as,

\({A_2} = \pi {\left( {\frac{d}{2}} \right)^2}\)

Substitute the values in the equation (ii),

\(\begin{array}{c}F = \rho {\left( {\frac{V}{t}} \right)^2}\left( {\frac{1}{{\pi {{\left( {\frac{d}{2}} \right)}^2}}} - \frac{1}{{\pi {{\left( {\frac{D}{2}} \right)}^2}}}} \right)\\ = \rho {\left( {\frac{V}{t}} \right)^2}\left( {\frac{4}{{\pi {d^2}}} - \frac{4}{{\pi {D^2}}}} \right)\\ = 4\rho {\left( {\frac{V}{t}} \right)^2}\left( {\frac{1}{{\pi {d^2}}} - \frac{1}{{\pi {D^2}}}} \right)\end{array}\)

Substitute the values in the above equation,\(\begin{array}{c}F = 4\left( {1000\;{\rm{kg/}}{{\rm{m}}^3}} \right){\left( {\frac{{\left( {420\;{\rm{L}}} \right)\left( {\frac{{{{10}^{ - 3}}\;{{\rm{m}}^3}}}{{1\;{\rm{L}}}}} \right)}}{{\left( {1\;\min } \right)\left( {\frac{{60\;{\rm{s}}}}{{1\;\min }}} \right)}}} \right)^2}\left( {\frac{1}{{\left( {3.14} \right){{\left( {0.75\;{\rm{cm}}} \right)}^2}{{\left( {\frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)}^2}}} - \frac{1}{{\left( {3.14} \right){{\left( {7.0\;{\rm{cm}}} \right)}^2}{{\left( {\frac{{{\rm{1}}\;{\rm{m}}}}{{{\rm{100}}\;{\rm{cm}}}}} \right)}^2}}}} \right)\\ = \left( {0.196\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{3}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {5661.7\;{{\rm{m}}^{ - 2}} - 65\;{{\rm{m}}^{ - 2}}} \right)\\ = \left( {0.196\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{3}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {5596.7\;{{\rm{m}}^{ - 2}}} \right)\left( {\frac{{{\rm{1}}\;{\rm{N}}}}{{{\rm{1}}\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}}}} \right)\\ \approx 1100\;{\rm{N}}\end{array}\)

Therefore, the force required to hold the hose is \(1100\;{\rm{N}}\).