Question

A gardener waters the plants by a pipe of diameter $1\,mm$. The water comes out at the rate of $10\,cm{s^{ - 1}}$. The reaction force exerted on the hand of the gardener is:
(A) $0\,N$
(B) $1.27 \times {10^{ - 2}}\,N$
(C) $1.27 \times {10^{ - 4}}\,N$
(D) $0.127\,N$

Answer 1

By using the expression of the reactionary force that the reactionary force is equal to the rate of change of momentum. In which the mass rate and velocity rate can be replaced by the known parameter and by substituting the known values in it we can find out the reaction force exerted on the hand of the garden.

Useful formula
The reactionary force is equal to the rate of change of momentum is expressed as,
$F = m \times \dfrac{{dv}}{{dt}}$
On further,
$F = \dfrac{{mv}}{t}$
Where, $m$ is the mass, $v$ is the velocity, $t$ is the time

Complete step by step solution
Given data:
Diameter of the pipe is $1\,mm$.
Volume rate of water flow is $10\,c{m^3}{s^{ - 1}}$
The reactionary force is expressed as,
$F = \dfrac{{mv}}{t}\,..................\left( 1 \right)$
Convert the mass rate and velocity into known parameters.
Where,
$m = \dfrac{{v \times \rho }}{t}$
Where,
$\rho$ is the density
$v = \dfrac{l}{t}$
Where,
$l$ is the length
Length can also be written as,
$l = \dfrac{V}{A}$
Where,
$V$ is the volume
$A$ is the cross sectional area
Then,
Substitute the known parameters in the equation (1),
$F = \left( {\dfrac{{V \times \rho }}{t}} \right) \times \left( {\dfrac{V}{{A \times t}}} \right)$

By arranging the terms,
$F = {\left( {\dfrac{V}{t}} \right)^2} \times \dfrac{\rho }{A}\,..................\left( 2 \right)$
Substitute the known value in equation (2),
$F = {\left( {10} \right)^2} \times \dfrac{{1\,gc{m^{ - 3}}}}{A}$
$A = \pi {r^2}$
Where,
$D = 1\,mm$
Then,
$r = \dfrac{1}{2}\,mm$
Therefore,
$F = {\left( {10} \right)^2} \times \dfrac{{1\,gc{m^{ - 3}}}}{{\pi {{\left( {\dfrac{1}{2} \times {{10}^{ - 1}}\,cm} \right)}^2}}} \\\ F = 100 \times \dfrac{{1\,gc{m^{ - 3}}}}{{\pi {{\left( {\dfrac{1}{2} \times {{10}^{ - 1}}\,cm} \right)}^2}}} \\\$
By simplifying the above equation,
$F = 127 \times {10^2}\,gcm{s^{ - 2}} \\\ F = 127 \times {10^2}\,dyne \\\$
Where,
$1\,dyne = {10^{ - 5}}\,N$
Then,
$F = \left( {127 \times {{10}^2}} \right) \times \left( {{{10}^{ - 5}}\,N} \right)$
$F = 127 \times {10^{ - 3}}\,N$
So, the reactionary force be
$F = 0.127\,N$
Thus, the reactionary force exerted on the hand of the gardener is $0.127\,N$.

Hence, the option (D) is correct.

Additional information
The force which is required to accelerate a mass of one gram at the rate of one centimetre per second square is known as $dyne$, which is a CGS system unit.

Note: The rate of flow of liquid can be expressed in terms of velocity of the liquid and cross-section area. It can be expressed in terms of time and volume. Because the liquid incompressible the rate of flow of the liquid into an area should be equivalent to the rate of flow of liquids out of an area.