
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?
Answer
500.7k+ views
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}$
Complete answer:
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.
$ F = \dfrac{{{{(Rate)}^2} \times d}}{A}$
Complete answer:
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.
Recently Updated Pages
Why is there a time difference of about 5 hours between class 10 social science CBSE

In cricket, what is a "pink ball" primarily used for?

In cricket, what is the "new ball" phase?

In cricket, what is a "death over"?

What is the "Powerplay" in T20 cricket?

In cricket, what is a "super over"?

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

