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(D) The mass flow rate of water is given by $\frac{dm}{dt} = Av\rho$,where $A$ is the cross-sectional area,$v$ is the velocity of flow,and $\rho$ is t

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$n^3$

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(D) The mass flow rate of water is given by $\frac{dm}{dt} = Av\rho$,where $A$ is the cross-sectional area,$v$ is the velocity of flow,and $\rho$ is the density of water.To obtain $n$ times the water in the same time,the new mass flow rate must be $\left(\frac{dm}{dt}\right)' = n \frac{dm}{dt}$.Since $A$ and $\rho$ are constant,$v' = nv$.The power required to pump the water is $P = \frac{1}{2} \left(\frac{dm}{dt}\right) v^2$.Therefore,the ratio of the new power $P'$ to the original power $P$ is:$\frac{P'}{P} = \frac{\frac{1}{2} (n \frac{dm}{dt}) (v')^2}{\frac{1}{2} (\frac{dm}{dt}) v^2} = n \cdot \frac{(nv)^2}{v^2} = n \cdot n^2 = n^3$.Thus,the power must be increased by a factor of $n^3$.

$A$ pump is used to deliver water at a certain rate from a given pipe. To obtain $n$ times the water in the same time from the same pipe,by what factor should the power of the motor be increased?

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