A motor of power P_0 is used to deliver water | vedclass

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(C) The power required to move a fluid through a pipe is given by the work done per unit time. For a horizontal pipe, the work done is primarily to ov

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$n^3: 1$

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(C) The power required to move a fluid through a pipe is given by the work done per unit time. For a horizontal pipe, the work done is primarily to overcome the kinetic energy of the fluid. The kinetic energy per unit time (power) is given by $P = \frac{1}{2} \dot{m} v^2$, where $\dot{m}$ is the mass flow rate and $v$ is the velocity of the fluid.Since the mass flow rate $\dot{m} = \rho A v$ (where $\rho$ is density and $A$ is the cross-sectional area), we have $\dot{m} \propto v$. Thus, $P \propto \dot{m} v^2 \propto \dot{m} (\frac{\dot{m}}{\rho A})^2 \propto \dot{m}^3$.If the rate of flow $\dot{m}$ is increased by a factor of $n$, the new power $P_1$ will be $P_1 \propto (n \dot{m})^3 = n^3 \dot{m}^3$.Therefore, the ratio $\frac{P_1}{P_0} = \frac{n^3 \dot{m}^3}{\dot{m}^3} = n^3$.Hence, the ratio $P_1: P_0 = n^3: 1$.

$A$ 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

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