Question

A motor pump is delivering water at a certain rate. In order to increase the rate of delivery by $100\%$ , the power of the motor is to be increased by
A. $300\%$
B. $200\%$
C. $400\%$
D. $700\%$

Answer 1

The power of the motor is proportional to the velocity of the water delivered. Thus if the rate of delivery is to be increased, the velocity of the water is to be increased. This will gradually increase the power of the motor.

Complete step-by-step solution:
The rate at which the energy is consumed is called power. Hence the expression is given as,
$P = \dfrac{E}{t} = \dfrac{{\dfrac{1}{2}m{v^2}}}{t}$
Where, $E$ is the energy, $m$ is the mass, $v$ is the velocity and $t$ is the time. In order to increase the rate of delivery by $100\%$the velocity will be increased by $100\%$from the same pipe at the same time.
Therefore the velocity increases like,
$\left( {100 + 100} \right)v = 200v$
The new power which is increase by $100\%$is given as,
$P' = \dfrac{{\dfrac{1}{2}m{{\left( {200v} \right)}^2}}}{t} \\\ = \dfrac{{\dfrac{1}{2}m \times 40000{v^2}}}{t} \\\ = 400\dfrac{{\dfrac{1}{2}m \times 100{v^2}}}{t} \\\ = 400P \times 100 \\\$

Hence to increase the rate of delivery by 100%, the power of the motor will be increased by 400%.
It is clear that the high power motors will have high velocity in delivering water. So in order to increase the power of the motor, increase the velocity of water.
The answer is option C.

Note:- We have to note some mathematical errors that can happen while doing percentage problems. The percentage is what multiplied $100$ with the value. It is a fraction of $100$ itself.