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# A gardener is watering plants at the rate 0.1 liters/sec using a pipe of cross section $1c{m^2}$ . What additional force is required if he desires to increase the rate two times? Verified
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Hint: It is given that the gardener is watering the plants at some rate. This rate will be dependent on the cross section from e=where the water is coming out, velocity of the water. The force is different entity which can be found out using the formula given below:
$F = \dfrac{{{{(Rate)}^2} \times d}}{A}$

It is given that rate of pouring is $0.1\,liters/\sec$
Area of cross section is $1c{m^2}$
And at last we are given increased rate as twice the initial
$F = \dfrac{{{{(Rate)}^2} \times d}}{A}$
Here rare is the product of cross section area and the velocity
As the rate is doubled the vessel is unchanged so the velocity is doubled and cross section kept constant
If we put all the values and compare
${F_{initial}} = \dfrac{{{{(Rat{e_{initial}})}^2} \times d}}{A}$
After,

${F_{final}} = \dfrac{{{{(Rat{e_{final}})}^2} \times d}}{A} \\ \Rightarrow \;{F_{final}} = \dfrac{{{{(2Rat{e_{initial}})}^2} \times d}}{A} \\ \Rightarrow {F_{final}} = 4{F_{initial}} \\$
In the question we are asked to find out the excess force needed to be applied
${F_{excess}} = {F_{final}} - {F_{initial}} \\ \Rightarrow {F_{excess}} = 4{F_{initial}} - {F_{initial}} \\ \Rightarrow {F_{excess}} = 3{F_{initial}} \\$
Finding initial force by putting all the given values in the equation.
The equation is obtained by changing values in SI units
${F_{initial}} = \dfrac{{{{(0.1)}^2} \times {{10}^3}}}{{{{10}^{ - 4}}}} \\ \Rightarrow {F_{initial}} = {10^5}N \\$
The excess force is
$3{F_{initial}} = 3 \times {10^5}N$

Note: Flow rate Q is said to be the volume of fluid passing by some location through an area during a period of time. Flow rate and velocity are related, but are quite different, physical quantities.
To make the distinction we need to think about the flow rate of a river. The greater the velocity of the water, the greater the flow rate of the river. But flow rate also depends on the size of the river.