Question

A motor of power $P_0$ is used to deliver water at a certain rate through a given horizontal pipe. To increase the rate of flow of water through the same pipe $n$ times, the power of the motor is increased to $P_1$. The ratio of $P_1 $ to $P_0$ is

• A. n : 1
• B. $n^2 : 1 $
• C. $n^3 : 1 $
• D. $n^4 : 1 $

Answer 1

The Correct Option is C

Power of motor initially $=P_{0}$
Let, rate of flow of motor $=(x)$
Since, power, $P_{0}=\frac{\text { work }}{\text { time }}=\frac{m g y}{t}$
$=m g\left(\frac{y}{t}\right)$
$\frac{y}{t}=x=$ rate of flow of water
$=m g x\,\,\,...(i)$
If rate of flow of water increased by $n$ times, ie, $(n x)$
Increased power $P_{1}=\frac{m g y^{'}}{t}$
$=m g\left(\frac{y^{'}}{t}\right)$
$=m g n \cdot x$
$=n m g x\,\,\,\,...(ii)$
The ratio of power
$\frac{P_{1}}{P_{0}}=\frac{n m g x}{m g x}$
$\frac{P_{1}}{P_{0}}=\frac{n}{1} $
$\Rightarrow P_{1}: P_{0}=n: 1$