Question

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

• A. $\frac{22}{7}\times {10}^{-3}N$
• B. $\frac{7}{22}\times {10}^{-6}N$
• C. $\frac{7}{22}\times {10}^{-3}N$
• D. Zero

Answer 1

The correct option is C $\frac{7}{22}\times {10}^{-3}N$

Let the density of water be ${\rho }_w$
${\rho }_w=1000kg/m^3$
Radius of the pipe, r = 1 cm = ${10}^{–2}m$
Area of the cross-section of the tube, $A=\pi r^2=\frac{22}{7}\times ({10}^{-2}{)}^2{m}^2$
$x=\frac{\Delta V}{t}=10c{m}^3/s={10}^{-5}{m}^3/s$
Rate of change of volume,
Volume, V = area of cross-section $\times$ length of pipe
$\Rightarrow \frac{\Delta V}{t}=\frac{A\times \Delta l}{t}$
Now velocity, $v=\frac{\Delta I}{t}$
$\Rightarrow v=\frac{\Delta V}{t}\times \frac{1}{A}=\frac{x}{A}$
Reactionary force exerted on the hand of the gardener:
F =$\frac{\Delta \left(mv\right)}{t}=\rho \frac{\Delta v}{t}\times v$
$\therefore$ Density $\left(\rho \right)=\frac{Mass\left(m\right)}{Volume\left(V\right)}$
$\Rightarrow F=\rho \times x\times \frac{x}{A}$ $\therefore v=\frac{x}{A}$ and $\frac{\Delta v}{t}=x$
$\Rightarrow F=\rho \frac{x^2}{A}$
$\Rightarrow F=\frac{{10}^3\times ({10}^{-5}{)}^2}{\frac{22}{7}\times {10}^{-4}}=\frac{7}{22}\times {10}^{-3}N$
Hence, the correct answer is option (3).