Q 11-05-029JEE MainEAMCET 2009 (Engineering)Medium
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
Answer: (C) $n^3 : 1$
Through the same pipe, the rate of flow $Av$ increases $n$ times only if the speed $v$ increases $n$ times.
Power = kinetic energy delivered per second $= \dfrac{1}{2}(\rho Av)v^2 \propto v^3$.
So $P_1 : P_0 = n^3 : 1$.
Solution by Sreeraj P, M.Sc Physics