Q. 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) $n^{4}\ :1$
Explanation
(a) Power of motor initially $= p_{0}$
Let, rate of flow of motor $= (x)$
Since, power, $p_{0} = \frac{work}{time} = \frac{mgy}{t} = mg\left( \frac{y}{t} \right)$,
$\frac{y}{t} = x =$ rate of flow of water
$= mgx$ ….(i)
If rate of flow of water is increased by $n$ times, $i.e$., $(nx)$
Increased power, $p_{1} = \frac{mgy'}{t} = mg\left( \frac{y^{'}}{t} \right)$,
$= nmgx$ ….(ii)
The ratio of power
$\frac{p_{1}}{p_{0}} = \frac{n\ mgx}{mgx} = \frac{n}{1} \Rightarrow p_{1}:p_{0} \Rightarrow n:1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.