Explanation

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