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A motor is used to deliver water at a certain rate through a given horizontal pipe. To deliver $n$-times the water through the same pipe in the same time the power of the motor must be increased as follows :
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The correct answer is:
$n^3$-times
If the motor pumps water (density $=\rho$ ) continuously through a pipe of area of crosssection $A$ with velocity $v$, then mass flowing out per second,
$m=A v \rho$ ...(i)
Rate of increase of kinetic energy
$=\frac{1}{2} m v^2=\frac{1}{2}(A v \rho) v^2$ ..(ii)
Mass $m$, flowing out per sec, can be increased to $m^{\prime}$ by increasing $v$ to $v^{\prime}$, then power increases from $P$ to $P^{\prime}$.
$\frac{P^{\prime}}{P}=\frac{\frac{1}{2} A \rho v^3}{\frac{1}{2} A \rho v^3}$
or $\quad \frac{P^{\prime}}{P}=\left(\frac{v^{\prime}}{v}\right)^3$
Now $\frac{m^{\prime}}{m}=\frac{A \rho v^{\prime}}{A \rho v}=\frac{v^{\prime}}{v}$
As $\quad m^{\prime}=n m, v^{\prime}=n v$
$\therefore \quad \frac{P^{\prime}}{P}=n^3$
$\Rightarrow \quad P^{\prime}=n^3 P$
$m=A v \rho$ ...(i)
Rate of increase of kinetic energy
$=\frac{1}{2} m v^2=\frac{1}{2}(A v \rho) v^2$ ..(ii)
Mass $m$, flowing out per sec, can be increased to $m^{\prime}$ by increasing $v$ to $v^{\prime}$, then power increases from $P$ to $P^{\prime}$.
$\frac{P^{\prime}}{P}=\frac{\frac{1}{2} A \rho v^3}{\frac{1}{2} A \rho v^3}$
or $\quad \frac{P^{\prime}}{P}=\left(\frac{v^{\prime}}{v}\right)^3$
Now $\frac{m^{\prime}}{m}=\frac{A \rho v^{\prime}}{A \rho v}=\frac{v^{\prime}}{v}$
As $\quad m^{\prime}=n m, v^{\prime}=n v$
$\therefore \quad \frac{P^{\prime}}{P}=n^3$
$\Rightarrow \quad P^{\prime}=n^3 P$
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