Question

Water shoots out a pipe and nozzle as shown in the figure. The cross-sectional area for the tube at point $A$ is four times that of the nozzle. The pressure of water at point $A$ is $41\times {10}^{3}N{m}^{-2}$ (guage). If the height ${}^{\prime }{h}^{\prime }$ above the nozzle to which water jet will shoot is $x/10$ meter than $x$ is. (Neglect all the losses occurred in the above process) $[g=10m/s^2]$.

Answer 1

Cross sectional area at point$4A\left(tube\right)$

Cross sectional area at point $=A\left(nozle\right)$

The pressure at a point A is $=41\times {10}^{3}N/m^2$

The height above the nozzle $=x+\frac{x}{10}=\frac{11x}{10}g=10m/s^{2}P=h\rho g41\times {10}^3=\frac{11x}{10}\times 10\times \frac{m}{1.1}x=\frac{41\times {10}^3}{10}x=41\times {10}^{2}m$