Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# A motor pumps water at the rate of $V{{m}^{3}}/s$ , against a pressure $PN{{m}^{-2}}$ . The power of the motor in watt is:A. $PV$ B. $\dfrac{P}{V}$ C. $\dfrac{V}{P}$ D. $V-P$

Last updated date: 13th Jun 2024
Total views: 403.5k
Views today: 7.03k
Verified
403.5k+ views
Hint: You could find the required relation using dimensional analysis. For that, you could begin from expressing each quantity dimensionally. And then you could express the dimension of power in terms of pressure and pumping rate raised to some powers α and β. After that, you could use the principle of homogeneity of dimensions and solve for α and β and hence find the required relation.

Formula used:
Expression for power,
$p=\dfrac{W}{t}$
Expression for pressure,
$P=\dfrac{F}{A}$

In the question we are given a motor pump that is pumping water at the rate of $V{{m}^{3}}/s$ against a pressure of $PN{{m}^{-2}}$ . And we are asked to find the power of the motor in watt in terms of the pressure (P) and rate of pumping is $V{{m}^{3}}/s$ .
Let us solve this question by using the method of dimensional analysis.
We know that power p is the time rate of doing work and can be expressed as,
$p=\dfrac{W}{t}$ ……………………. (1)
Where, W is the work done and t is the time taken.
But we know that work done is product of force and displacement, that is,
$W=F\times S$
But F=ma by Newton’s second law, so, the dimension of force is given by,
$\left[ F \right]=\left[ M \right]\left[ L{{T}^{-2}} \right]=\left[ ML{{T}^{-2}} \right]$
$\Rightarrow \left[ W \right]=\left[ ML{{T}^{-2}} \right]\left[ L \right]=\left[ M{{L}^{2}}{{T}^{-2}} \right]$
$\Rightarrow \left[ p \right]=\dfrac{\left[ M{{L}^{2}}{{T}^{-2}} \right]}{\left[ T \right]}$
The dimension of power is given by,
$\Rightarrow \left[ p \right]=\left[ M{{L}^{2}}{{T}^{-3}} \right]$ ………………………………… (2)
Now pressure by definition is the force acting per unit area, that is,
$P=\dfrac{F}{A}$
Pressure can be dimensionally expressed as,
$\left[ P \right]=\dfrac{\left[ ML{{T}^{-2}} \right]}{{{\left[ L \right]}^{2}}}$
$\Rightarrow \left[ P \right]=\left[ M{{L}^{-1}}{{T}^{-2}} \right]$ ……………………………. (3)
Rate of pumping is given by,
$V=\dfrac{v}{t}$
$\left[ V \right]=\dfrac{{{\left[ L \right]}^{3}}}{\left[ T \right]}$
$\left[ V \right]=\left[ {{L}^{3}}{{T}^{-1}} \right]$ …………………………………….. (4)
Let us now find the required relation, for that let assume,
$p\propto {{P}^{\alpha }}{{V}^{\beta }}$
$p=k{{P}^{\alpha }}{{V}^{\beta }}$
Now let us express them dimensionally,
$\left[ p \right]={{\left[ P \right]}^{\alpha }}{{\left[ V \right]}^{\beta }}$ ………………………….. (5)
Substituting (2), (3) and (4),
$\left[ M{{L}^{2}}{{T}^{-3}} \right]={{\left[ M{{L}^{-1}}{{T}^{-2}} \right]}^{\alpha }}{{\left[ {{L}^{3}}{{T}^{-1}} \right]}^{\beta }}$
$\left[ M \right]{{\left[ L \right]}^{2}}{{\left[ T \right]}^{-3}}={{\left[ M \right]}^{\alpha }}{{\left[ L \right]}^{-\alpha +3\beta }}{{\left[ T \right]}^{-2\alpha -\beta }}$
By the principle of homogeneity of dimensions, the dimensions on both sides of any valid equation should be the same. So,
$\alpha =1$ ……………….. (6)
$-\alpha +3\beta =2$ …………………. (7)
$-2\alpha -\beta =-3$ ……………………….. (8)
Substituting (6) in (7) gives,
$\Rightarrow -\left( 1 \right)+3\beta =2$
$\Rightarrow \beta =\dfrac{3}{3}=1$ ………………………. (9)
Now we can substitute (6) and (9) in (5),
$\left[ p \right]={{\left[ P \right]}^{1}}{{\left[ V \right]}^{1}}$
Therefore the power of the motor in watt could be expressed in terms of pressure P and rate of pumping V as PV.

So, the correct answer is “Option A”.

Note: You should note that the V given in the question is the rate at which the motor is pumping the water not the volume of the water being pumped. Also, in order to avoid confusion, in the question the unit is given for both the quantities. Hence, you should accordingly assign the dimension otherwise you will end up making a huge mistake.