Question

A gardener waters the plants by a pipe of diameter 1mm. The water comes out at the rate or 10$c{m^3}/\sec$ . 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. 0.127N

Answer 1

When gardener water flowing through pipe with certain amount of rate can be calculated and then we need to the formula of newton’s $2^{nd}$ law of motion and substituting the density formula hence we find the force exerted on the hand of the gardener.

Complete step by step answer:
From the given data:
D=1mm=$1 \times {10^{ - 3}}m$
The water comes out in the pipe or rate of flow of water is given by
$\ \dfrac{V}{t} = \dfrac{{10}}{1} \\\ \Rightarrow \dfrac{V}{t} = 10 \times {10^{ - 6}}{m^3}\sec \\\ \$
The density of water $\rho = {10^3}kg{m^{ - 3}}$
The area of Cross-section of pipe is given by A=$\pi {d^2}$
From newton’s 2nd law of motion, the formula which is given by
F=ma
$\ \Rightarrow F = m\dfrac{{dv}}{{dt}} \\\ \Rightarrow F = \dfrac{{mv}}{t} \\\ \$
We know that the density formula is given by
$density = \dfrac{{mass}}{{volume}}$
Take substitution for the mass for the above equation
$\ F = \dfrac{{V\rho v}}{t} \\\ \Rightarrow F = \dfrac{{\rho V}}{t} \times \dfrac{V}{{At}} \\\ \Rightarrow F = {\left( {\dfrac{V}{t}} \right)^2}\dfrac{\rho }{A} \\\ \$
$\ F = \dfrac{{400 \times 1 \times 7}}{{22 \times {{10}^{ - 2}}}} \\\ \Rightarrow F = 127.27 \times {10^{ - 2}}dyne \\\ \$
We convert dyne into newton
$\ 1dyne = {10^{ - 5}}N \\\ \Rightarrow F = \dfrac{{127.27}}{{{{10}^{ - 2}}}} \times {10^{ - 5}} \\\ F = 0.127N \\\ \$

So, the correct answer is “Option D”.

Note:
Students usually get confused with this type of numerical.
We need to understand here more important that while taking the substitution for density and formula for area of cross section of the pipe.
Also students should be careful while converting the CGS system to SI system.