Question

A pressure of \[10Pa\] acts on an area of \[3.0{m^2}\]. What is the force acting on the area? What force will be exerted by the application of the same pressure if the area is made one-third?

Answer 1

Hint: The answer to the given question can be found using Pascal’s law. Pascal’s law states that when we apply an external static pressure over a confined liquid the pressure in the liquid is distributed evenly throughout the liquid in all directions. We need to use pascal’s formula and substitute all the given values from the question in the formula we will arrive at the answer:
Formula used:
\[ \Rightarrow F = PA\]
Here, $F$ is force applied
$P$ is the pressure transmitted
$A$ is the area

Complete answer:
Given that \[10Pa\] acts on the area of \[3.0{m^2}\]. Therefore,
\[P = 10Pa\]
\[A = 3.0{m^2}\]
Now substituting this in the above-given formula,
\[ \Rightarrow F = 10 \times 3.0\]
\[ \Rightarrow F = 30N\]
Now the second part of the question is to find the force that will be exerted when applying the same pressure but the area becomes one-third of the previously mentioned area.
So we need to make the area to one-third.
Given area \[A = 3.0{m^2}\]. One-third of this will be equal to \[A = 1.0{m^2}\]
Therefore substituting all the known values in the above formula we get,
\[ \Rightarrow F = 10 \times 1.0\]
\[ \Rightarrow F = 10N\]
Correct Answer:
Therefore, the force when the applied pressure is equal to \[10Pa\] that acts on the area of \[3.0{m^2}\] is \[30N\]. Also, the force when we apply same pressure acting on the area that will be equal to the one-third of the previously given area is \[10N\].

Note:
A very useful application of Pascal’s law is a hydraulic lift. It is a device that is used to lift objects. When we apply a small amount of force to an incompressible liquid on the lift side this will create a large amount of force on right. This force helps to lift the heavy object.