Question

A motor pumps water at the rate of volume at the rate of $V{m^3}$ per second, again a pressure $PN{m^{ - 2}}$. The power of the meter in watt is
(A) PV
(B) $\left( {\dfrac{P}{V}} \right)$
(C) $\left( {\dfrac{V}{P}} \right)$
(D) $\left( {\dfrac{{{V^2}}}{P}} \right)$

Answer 1

Hint: We know that with the help of dimensional formulas, we can derive the relation between physical quantities or derive the different expression in physics.
In order to derive the formula of power, firstly we use the dimensional formulas of volume and pressure, then calculate them after then compare the powers of dimensional formulas of power, pressure and volume. Finally we derive the expression of power in terms of pressure and volume.
Formula used
In this problem, we use following dimensional formulas
Power $ = [{M^1}{L^{ - 1}}{T^{ - 3}}]$
Pressure $ = [{M^1}{L^{ - 1}}{T^{ - 2}}]$
Volume $ = [{M^0}{L^3}{T^0}]$

Complete step by step solution:
We know that the power is given by
Power $ = \dfrac{{work}}{{time}}$
& work $ = $ force $ \times $ displacement
So, power $ = \dfrac{{force \times displacement}}{{time}}$
We know that the dimensional formula of force is
$[F] = [mass(m)][acceleration(a)]$
$[F] = [{M^1}][{L^1}{T^{ - 2}}]$
$[F] = [{M^1}{L^1}{T^{ - 2}}]$ …..(1)
So, the dimensions of power is
$[Power] = \dfrac{{[{M^1}{L^1}{T^{ - 2}}][{L^1}]}}{{[{T^1}]}}$
$[Power] = [{M^1}{L^2}{T^{ - 3}}]$ …..(2)
Now, we also know that the pressure is given by
$pressure(P) = \dfrac{{force(F)}}{{area(A)}}$ ……(3)
Dimensions of area $[A] = [{L^2}]$ …..(4)
So, from equation (1), (3) & (4), dimensions of pressure
$[Pressure] = \dfrac{{[{M^1}{L^1}{T^{ - 2}}]}}{{[{L^2}]}}$
$ = [{M^1}{L^{ - 1}}{T^{ - 2}}]$ …..(5)
The dimensional formula of volume is
$[V] = [{L^3}]$ ……(6)
Now, we combine equation (2), (5) & (6)
$[Power] = {[Pressure]^\alpha }{[Volume]^\beta }$ …...(7)
$[{M^1}{L^2}{T^{ - 3}}] = {[{M^1}{L^{ - 1}}{T^{ - 2}}]^\alpha }{[{L^3}]^\beta }$
$[{M^1}{L^2}{T^{ - 3}}] = [{M^\alpha }{L^{ - \alpha + 3\beta }}{T^{ - 2\alpha }}]$
On comparing the power of both sides
$\alpha = 1,$\[\]$ - \alpha + 3\beta = 2$
So, $ - 1 + 3\beta = 2$
$3\beta = 3$ $ \Rightarrow $ $\beta = 1$
Hence, from equation (7)
$Power = PV(Pressure \times Volume)$

So, option A is correct answer PV.

Note: In order to derive the relation between physical quantities we use the dimensional formula.