Question

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

• A. Zero
• B. $1.27 \times 10^{-2}\,N$
• C. $1.27 \times 10^{-4}\,N$
• D. $0.127 \,N$

Answer 1

Answer: (D)

Rate of flow of water $\frac{V}{t}=\frac{10\,cm^{2}}{sec}$
$=10 \times 10^{-6} \frac{m^{3}}{sec}$
Density of water $\rho =\frac{10^{3}\,kg}{m^{3}}$
Cross-sectional area of pipe $A=\pi (0.5 \times 10^{-3})^{2}$
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
$F=0.127\,N$