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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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The correct answer is:
$n: 1$
Power of motor initially $=P_0$
Let, rate of flow of motor $=(x)$
$\text { Since, power, } \begin{aligned}
P_0 & =\frac{\text { work }}{\text { time }}=\frac{m g y}{t} \\
& =m g\left(\frac{y}{t}\right), \\
\frac{y}{t} & =x=\text { rate of flow of water }
\end{aligned}$
If rate of flow of water is increased by $n$ times, $i e$, (nx).
Increased power
$\begin{aligned}
P_1 & =\frac{m g y^{\prime}}{t} \\
& =m g\left(\frac{y^{\prime}}{t}\right)=m g n \cdot x
\end{aligned}$
The ratio of power
$\begin{gathered}
\frac{P_1}{P_0}=\frac{n m g x}{m g x} \\
\frac{P_1}{P_0}=\frac{n}{1} \Rightarrow P_1: P_0=n: 1
\end{gathered}$
Let, rate of flow of motor $=(x)$
$\text { Since, power, } \begin{aligned}
P_0 & =\frac{\text { work }}{\text { time }}=\frac{m g y}{t} \\
& =m g\left(\frac{y}{t}\right), \\
\frac{y}{t} & =x=\text { rate of flow of water }
\end{aligned}$
If rate of flow of water is increased by $n$ times, $i e$, (nx).
Increased power
$\begin{aligned}
P_1 & =\frac{m g y^{\prime}}{t} \\
& =m g\left(\frac{y^{\prime}}{t}\right)=m g n \cdot x
\end{aligned}$
The ratio of power
$\begin{gathered}
\frac{P_1}{P_0}=\frac{n m g x}{m g x} \\
\frac{P_1}{P_0}=\frac{n}{1} \Rightarrow P_1: P_0=n: 1
\end{gathered}$
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