Question

A pump delivers water at the rate of $2400{\text{L}}$ in $6{\text{min}}$over a head of $12{\text{m}}$ . If $g = 10{\text{m}}{{\text{s}}^{ - 2}}$ then find its power.
A. $200{\text{W}}$
B. $800{\text{W}}$
C. $240{\text{W}}$
D. $2{\text{kW}}$

Answer 1

Here the pump delivers the given amount of water by raising it to the height of $12{\text{m}}$ . So a work of some amount is done by the force of the pump. This force will be equal to the weight of the water that is being raised. Power of the pump refers to the amount of work done in a given time.

Formulas used:
The work done by an applied force is given by, $W = F \cdot d$ where $F$ is the applied force and $d$ is the displacement of the body.
The power of an object is given by, $P = \dfrac{W}{t}$ where $W$ is the work done by the object in the time $t$ .

Complete step-by-step solution:
Step 1: List the parameters mentioned in the question.
The amount of water delivered is given to be $m = 2400{\text{L}} = 2400{\text{kg}}$ .
The time taken to deliver the given amount of water is $t = 6{\text{min}}$ .
The height at which it is raised is given to be $h = 12{\text{m}}$ .
Also, the acceleration due to gravity is given to be $g = 10{\text{m}}{{\text{s}}^{ - 2}}$ .
Step 2: Express the work done by the pump in delivering the given amount of water.
The work done by the pump is given by, $W = F \cdot h = Fh$ ------- (1)
where $F$ is the weight of the water and $h$ is the height to which it is raised.
The weight of the water will be $F = mg = 2400 \times 10 = 24000{\text{N}}$
Then substituting for $F = 24000{\text{N}}$ and $h = 12{\text{m}}$ in equation (1) we get, $W = 24000 \times 12 = 288000{\text{J}}$
Thus the work done by the pump is $W = 288000{\text{J}}$ .
Step 3: Express the relation for the power of the pump.
The power of the pump is given by, $P = \dfrac{W}{t}$ --------- (2)
Substituting for $W = 288000{\text{J}}$ and $t = 6{\text{min}}$ in equation (2) we get, $P = \dfrac{{288000}}{{360}} = 800{\text{W}}$
Thus the power of the pump is $P = 800{\text{W}}$ .
So the correct option is B.

Note:- Here we made an approximation in which one litre of water becomes nearly equal to one kilogram of water. The height to which the water had to be raised is considered as its displacement. The force due to gravity and the height are both in the same direction. Also when $t = 6{\text{min}}$ was substituted in equation (2) it was converted into its S.I. unit as $t = 360{\text{s}}$ to get the correct value for the power of the pump.