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

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(C) The power $P$ 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

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$n^3: 1$

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(C) The power $P$ 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 $K$ of a mass $m$ moving with velocity $v$ is $K = \frac{1}{2}mv^2$.The power $P$ is the rate of change of kinetic energy: $P = \frac{dK}{dt} = \frac{1}{2} v^2 \frac{dm}{dt}$.Let the mass flow rate be $\frac{dm}{dt} = \dot{m}$. Since the pipe is the same,the cross-sectional area $A$ is constant. The flow rate (volume per unit time) is $Q = Av$. Thus,the mass flow rate $\dot{m} = \rho A v$,where $\rho$ is the density of water.Substituting $\dot{m}$ into the power equation: $P = \frac{1}{2} v^2 (\rho A v) = \frac{1}{2} \rho A v^3$.Since $Q = Av$,we have $v = Q/A$. Substituting this into the power equation: $P = \frac{1}{2} \rho A (Q/A)^3 = \frac{\rho}{2A^2} Q^3$.This shows that $P \propto Q^3$.Given that the initial flow rate is $Q_0$ and the final flow rate is $Q_1 = n Q_0$,the ratio of powers is:$\frac{P_1}{P_0} = \left( \frac{Q_1}{Q_0} \right)^3 = n^3$.Therefore,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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