Question

A gardener waters plants by a pipe of cross section $1\,{\text{m}}{{\text{m}}^2}$. The water comes out at the rate of 100 cc/s. The reactionary force exerted on the hand of the gardener is
A. 10 N
B. 100 N
C. 1000 N
D. 500 N

Answer 1

Convert the rate from cc/s to ${{\text{m}}^{\text{3}}}{\text{/s}}$. Also, express the rate in terms of flow of mass of water per unit time using the relation between density, mass and volume. The rate is the product of area of cross section and the velocity of the flow. Use Newton’s second law to determine the force exerted by the water flow. Recall Newton’s third law to determine the reactionary force exerted on the hand of the gardener.

Formula used:
Rate of flow, $\dfrac{{dV}}{{dt}} = Av$
Here, A is the cross sectional area and v is the velocity.
Newton’s second law of motion, $F = \dfrac{{dp}}{{dt}}$
Here, p is the linear momentum.

Complete step by step answer:
We have given the rate of water coming out from the pipe,
$\dfrac{{dV}}{{dt}} = 100\,{\text{c}}{{\text{m}}^3}{\text{/s}} = 100 \times {10^{ - 6}}\,{{\text{m}}^3}{\text{/s}}$
$\Rightarrow \dfrac{{dV}}{{dt}} = {10^{ - 4}}\,{{\text{m}}^3}{\text{/s}}$
Let’s express the rate in terms of mass flow as follows,
$\dfrac{{dm}}{{dt}} = \dfrac{{d\left( {{\rho _w}V} \right)}}{{dt}} = {\rho _w}\dfrac{{dV}}{{dt}}$
Here, ${\rho _w}$ is the density of the water.
Substituting $\dfrac{{dV}}{{dt}} = {10^{ - 4}}\,{{\text{m}}^3}{\text{/s}}$ and ${\rho _w} = {10^3}\,{\text{kg/}}{{\text{m}}^3}$ in the above equation, we get,
$\dfrac{{dm}}{{dt}} = \left( {{{10}^3}} \right)\left( {{{10}^{ - 4}}} \right)$
$\Rightarrow \dfrac{{dm}}{{dt}} = {10^{ - 1}}\,{\text{kg/s}}$
We know that the rate is expressed as,
$\dfrac{{dV}}{{dt}} = Av$ …… (1)
Here, A is the cross sectional area of the pipe and v is the velocity of the water coming out of the pipe.
We have given the area of cross section of the pipe is,
$A = 1\,{\text{m}}{{\text{m}}^2} = {\left( {\left( {1\,{\text{m}}} \right)\left( {\dfrac{{{{10}^{ - 3}}}}{{1\,{\text{mm}}}}} \right)} \right)^2}$
$\Rightarrow A = {10^{ - 6}}\,{{\text{m}}^2}$
Substituting $A = {10^{ - 6}}\,{{\text{m}}^2}$ and $\dfrac{{dV}}{{dt}} = {10^{ - 4}}\,{{\text{m}}^3}{\text{/s}}$ in equation (1), we get,
${10^{ - 4}} = \left( {{{10}^{ - 6}}} \right)v$
$\Rightarrow v = \dfrac{{{{10}^{ - 4}}}}{{{{10}^{ - 6}}}}$
$\Rightarrow v = {10^2}\,{\text{m/s}}$
We have from Newton’s second law of motion,
$F = \dfrac{{dp}}{{dt}} = \dfrac{{d\left( {mv} \right)}}{{dt}}$
$\Rightarrow F = v\dfrac{{dm}}{{dt}}$
Here, the factor $mv$ is the linear momentum p of the water.
Substituting $v = {10^2}\,{\text{m/s}}$ and $\dfrac{{dm}}{{dt}} = {10^{ - 1}}\,{\text{kg/s}}$in the above equation, we get,
$F = \left( {{{10}^2}} \right)\left( {{{10}^{ - 1}}} \right)$
$\Rightarrow F = 10\,{\text{N}}$
This is the force exerted by the water. We know that from Newton’s third law, the reactionary force will be equal to the force exerted by the water but in the opposite direction. Therefore, the reactionary force exerted on the hands of the gardener is,
$F = 10\,{\text{N}}$

So, the correct answer is “Option A”.

Note:
Do not get confuse while converting the area of cross section from ${\text{m}}{{\text{m}}^2}$ to ${{\text{m}}^2}$. Remember, $1\,{\text{m}} = {\text{1}}{{\text{0}}^3}\,{\text{mm}}$ and use this to convert the area of cross section into ${{\text{m}}^2}$. Students should always express Newton’s second law as, $F = \dfrac{{dp}}{{dt}}$, where, p is the linear momentum. In this question, the velocity of the water is constant, so the force is proportional to the rate of change of mass of the water.