Question

An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity 2m/s. The mass per unit length of water in the pipe is 100 kg/m. What is the power of the engine?
$
  (a){\text{ 400W}} \\
  (b){\text{ 200W}} \\
  (c){\text{ 100W}} \\
  (d){\text{ 800W}} \\
 $

Answer 1

Hint: In this question use the concept that power is the rate of doing work, and so power will be work done per unit time that is $P = \dfrac{W}{t}$. Now the work done will simply be the kinetic energy of the water flowing through the hose pipe that is $W = \dfrac{1}{2}m{v^2}$. Use this concept to approach the problem.

Complete step-by-step answer:

As we know that the power developed by the engine or the power of the engine by which water is pumped is proportional to the product of the cube of velocity at which the water is flowing through the pipe and the mass per unit length of the water in the pipe.
As we know that the power is the ratio of amount of work done to the time interval in which the work is done.

In the case of a pump the work done is equal to the kinetic energy.

As we know that the kinetic energy (K.E) = $\dfrac{1}{2}M{v^2}$, where M = mass of the water and v = velocity of the flow of water through the pipe.
So power (P) = $\dfrac{{K.E}}{t} = \dfrac{1}{2}\dfrac{{M{v^2}}}{t}$.................. (1)
And it is often measure in watts which has symbol (W)

Now it is given that the velocity (v) at which water passes through the pipe and leaves it = 2 m/s.
And it is also given that the mass (M) per unit length (L) of the water in the pipe is = 100 (kg/m).

$ \Rightarrow v = 2$ m/s............. (2)

And

$ \Rightarrow \dfrac{M}{L} = 100$ Kg/m................. (3)
So from equation (1) and (3) we have,
$ \Rightarrow P = \dfrac{1}{2}\dfrac{{\left( {100L} \right){v^2}}}{t}$................... (4)
Now as we know that the length per unit time is nothing but velocity.
$ \Rightarrow v = \dfrac{L}{t}$
So substitute this value in equation (4) we have,
$ \Rightarrow P = \dfrac{1}{2}\left( {100} \right){v^3}$
Now substitute the value from equation (2) in the above equation we have,
$ \Rightarrow P = \dfrac{1}{2}\left( {100} \right){.2^3}$ W
Now simplify this equation we have,
$ \Rightarrow P = 4\left( {100} \right) = 400$ W
So the power of the engine is 400 watts.

So this is the required answer.

Hence option (A) is the correct answer.

Note: The S.I unit of power is Watt, there is a general confusion regarding the physical significance of power and energy. Both are different as the energy is defined as the capacity to do work whereas the power is the rate at which this work is done. In this problem we have used the relation $v = \dfrac{L}{t}$ where L is the length of pipe that is being travelled by the water, this relation depicts a similar concept somewhat related to the relationship between distance, speed and time.