A gardener waters the plants by a pipe of diameter 1mm. The water comes out at the rate or 10 $\frac{c{m}^3}{s}$.The reactionary force exerted on the hand of the gardener is

Rate of flow of water $\frac{\mathrm{V}}{\mathrm{t}}=\frac{10{\mathrm{cm}}^3}{\sec }=10\times {10}^{-6}\frac{{\mathrm{m}}^3}{\sec }$
Density of water $\mathrm{\rho }=\frac{{10}^3\mathrm{kg}}{{\mathrm{m}}^3}$
Cross-sectional area of pipe $\mathrm{A}=\mathrm{\pi }{(0.5\times {10}^{-3})}^2$
Force $=\mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{mv}}{\mathrm{t}}=\frac{\mathrm{V\rho v}}{\mathrm{t}}=\frac{\mathrm{\rho V}}{\mathrm{t}}\times \frac{\mathrm{V}}{\mathrm{At}}={\left(\frac{\mathrm{V}}{\mathrm{t}}\right)}^2\frac{\mathrm{\rho }}{\mathrm{A}}$
                                                            $\left(\because \mathrm{v}=\frac{\mathrm{V}}{\mathrm{At}}\right)$
By substituting the value in the above formula we get $\mathrm{F}=0.127\mathrm{N}$