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Q.
A gardner watering the plants by a pipe of diameter 2 mm. The water comes out at the rate of $10\, cm^3/sec$. The reactionary force exerted on the hand of the gardner is
Laws of Motion
Solution:
(V) Rate of flow $=\frac{V}{t}=\frac{\text { volume }}{\text { time }}$
$=\frac{10 \times 10^{-6}}{1}=10^{-5}$
Force exerted on the hand of gardener
$=\frac{P}{t}=\frac{m v}{t}$
Area of cross section $=\frac{\pi D^{2}}{4}=\frac{3.14 \times\left(2 \times 10^{-3}\right)^{2}}{4} $
$=\frac{3.14 \times 4 \times 10^{-6}}{4} $
$=3.14 \times 10^{-6} $
Area $\times$ velocity $=$ rate of flow
$v=\frac{V}{t A}$
Force $=\frac{m v}{t}=\frac{\rho V v}{t}=\frac{\rho}{A}\left(\frac{V}{t}\right)^{2}$
$F=\frac{10^{3}}{3.14 \times 10^{-6}} \times\left(10^{-5}\right)^{2}=0.031$
$F=0.031 N$