Solveeit Logo
Login

Question

Question: Unit of \(JP{a^{ - 1}}\) is equivalent to: A. \({m^3}\) B. \(c{m^3}\) C. \(d{m^3}\) D. none...

Unit of JPa−1JP{a^{ - 1}} is equivalent to:
A. m3{m^3}
B. cm3c{m^3}
C. dm3d{m^3}
D. none of these.

Explanation

Solution

The work done is defined as the product of the force exerted on a body and the displacement caused in the body due to this force. Pressure is defined as the force exerted per unit area of the surface. The force exerted on a body is defined as the product of its mass and the acceleration produced in it.

Complete step by step answer:
A physical quantity which relates to any other physical quantity must have units and dimensions. In the given question, two different physical quantities, work and pressure are related to each other. We can say this by looking at the units given in the question.
The unit of work is joule whose symbol is (JJ ). The unit of pressure is pascal whose symbol is (PaPa ).
Work done = Force x displacement
Force = mass x acceleration
Pressure = Force/ Area
The units of the inter-related physical quantities are as such:
Mass = kgkg
Acceleration = ms−2m{s^{ - 2}}
Force = kgms−2kgm{s^{ - 2}}
Area = m2{m^2}
Displacement = mm
Work done = kgms−2×m=kgm2s−2=Jkgm{s^{ - 2}} \times m = kg{m^2}{s^{ - 2}} = J
Pressure = kgms−2m2=kgm−1s−2=Pa\dfrac{{kgm{s^{ - 2}}}}{{{m^2}}} = kg{m^{ - 1}}{s^{ - 2}} = Pa
According to the question, the unit of JPa−1JP{a^{ - 1}} is:
JPa−1=kgm2s−2kgm−1s−2=m3JP{a^{ - 1}} = \dfrac{{kg{m^2}{s^{ - 2}}}}{{kg{m^{ - 1}}{s^{ - 2}}}} = {m^3}

Thus, the correct option is A. m3{m^3} .

Note:
A physical quantity can be classified generally into two categories: scalar and vector quantity. Whenever a physical quantity has a magnitude but no direction, it is termed as a scalar quantity. The examples of scalar quantity include mass, length, time, etc. Whenever a physical quantity has a magnitude as well as a direction, it is termed as a vector quantity such as force, pressure, etc.