PER unit time. In other words, Mass flow rate is defined as the rate of movement of liquid mass through a unit area. The mass flow is directly depended on the density, velocity of the liquid and area of cross section. It is the movement of mass per unit time.GIVEN,Density of the liquid = ρRadius of the hose pipe = aVelocity of the liquid = vPercentage of liquid passes through the mesh = 50%Percentage of liquid which loses its momentum = 25%Percentage of liquid RETREATED (repelled) = 25%Mass per unit time of a liquid flow is given by\(\frac{{dm}}{{dt}} = \rho Av\) Where, ρ is density of liquid,A is area through which it is flowingand v is velocity.Calculation:Rate of change in momentum of the 25% of liquid which loses all momentum is\({F_1} = \frac{{d{p_1}}}{{dt}} = \frac{{25}}{{100}}\left( {\frac{{dm}}{{dt}}} \right)v\) \(\frac{{d{p_1}}}{{dt}} = \frac{1}{4}\left( {\frac{{dm}}{{dt}}} \right)v\) \({F_1} = \frac{{d{p_1}}}{{dt}} = \frac{1}{4}\rho A{v^2}\) ----(1)Rate of change in momentum of the 25% of the liquid which comes back with same speed.\(\frac{{d{p_2}}}{{dt}} = \frac{{25}}{{100}}\left( {\frac{{dm}}{{dt}}} \right) \times 2v\) \({F_2} = \frac{{d{p_2}}}{{dt}} = \frac{1}{4}\left( {\frac{{dm}}{{dt}}} \right) \times 2v\) \({F_2} = \frac{{d{p_2}}}{{dt}} = \frac{1}{2}\rho A{v^2}\) ----(2)[∵ Net change in velocity is = 2v]∴ Net pressure on the mesh is\(p = \frac{{{F_{net}}}}{A} = \frac{{\left( {{F_1} + {F_2}} \right)}}{A}\;\left[ {\; \because F = \frac{{dp}}{{dt}}} \right]\) ∴ From equations (i) and (ii), we get\(p = \frac{{\frac{1}{4}\rho A{v^2} + \frac{1}{2}\rho A{v^2}}}{A}\) \(p = \frac{{\frac{{\rho A{v^2} + 2\rho A{v^2}}}{4}}}{A}\) \(p = \frac{{3\rho A{v^2}}}{{4A}}\) \(p = \frac{3}{4}\rho {v^2}\) Therefore, the resultant pressure on the mesh will be \(\frac{3}{4}\rho {v^2}\).