
Distinguish between thrust and pressure. Write their units. What is the relation between them? Explain an application where we reduce pressure acting on a body.
Answer
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Hint: First of all, draw a table to distinguish between the terms thrust and pressure. In the next part of question S.I. unit of thrust and pressure is asked. Write the formula of the given quantities and then derive S.I. units. Then, find the relation between thrust and pressure and then write one application from where we can reduce pressure acting on a body.
Complete solution:
Difference between thrust and pressure is as follows:
The S.I. unit of thrust is Newton (represented by ${{N}}$ ) and the S.I. unit of pressure is Pascal (represented by ${{Pa}}$ ).
The relation between thrust and pressure is given as:
${{Thrust = pressure \times \text{Area on which it acts}}}$
An application where we can reduce pressure is given below:
If a thrust of magnitude ${{200 N}}$ is applied on a surface of an object having area ${{20 }}{{{m}}^{{2}}}$, it creates the pressure of ${{10 pa}}$ . We can reduce this pressure, by decreasing the amount of applied force. Now if the applied thrust is reduced to ${{10 N}}$ and the thrust is applied on the same area then the value of pressure becomes ${{0}}{{.5 N}}$.
Note: When a body is submerged in water then it appears to be less in weight than its actual weight. Reason for this fact is that any fluid applies an upward force on a body submerged in it. This force is known as “thrust”. And this force i.e. thrust is in the opposite direction to that of the force of gravity.
Complete solution:
Difference between thrust and pressure is as follows:
| Thrust | Pressure |
| (1) Force acting perpendicular to the surface of an object is known as Thrust. | (1) Thrust acting per unit area is known as pressure. |
| (2) Formula of ${{thrust = pressure \times \text{area on which it acts}}}$ | (2) Formula of ${{pressure = }}\dfrac{{{{Thrust}}}}{{{{\text{Area on which it acts}}}}}$ |
The S.I. unit of thrust is Newton (represented by ${{N}}$ ) and the S.I. unit of pressure is Pascal (represented by ${{Pa}}$ ).
The relation between thrust and pressure is given as:
${{Thrust = pressure \times \text{Area on which it acts}}}$
An application where we can reduce pressure is given below:
If a thrust of magnitude ${{200 N}}$ is applied on a surface of an object having area ${{20 }}{{{m}}^{{2}}}$, it creates the pressure of ${{10 pa}}$ . We can reduce this pressure, by decreasing the amount of applied force. Now if the applied thrust is reduced to ${{10 N}}$ and the thrust is applied on the same area then the value of pressure becomes ${{0}}{{.5 N}}$.
Note: When a body is submerged in water then it appears to be less in weight than its actual weight. Reason for this fact is that any fluid applies an upward force on a body submerged in it. This force is known as “thrust”. And this force i.e. thrust is in the opposite direction to that of the force of gravity.
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