Question

A motor of power $P_{\circ }$ is used to deliver water at a certain rate through a given horizontal pipe. To increase the water of flow of water through the same pipe n, times the power of water is increased to $P_1$ The ratio of $P_1$ and $P_0$ is:

• A. $n:1$
• B. $n^2:1$
• C. $n^3:1$
• D. $n^4:1$

Answer 1

The correct option is C $n^3:1$

We know that

$P=7.v$

By Newton's second law of motion

$7=\frac{\Delta mv}{\Delta t}=v\frac{Deltam+\Delta v}{\Delta t}$

As $Deltav=0$ and $Deltam=\rho A\left(vDeltat\right)$

So,

$7=v\frac{Deltam+mDeltav}{Deltat}=v\frac{\left(\rho AvDeltat\right)+o}{Deltat}$

$7=\rho A\frac{v^{2}Deltat}{Deltat}=\rho A{v}^2$

So,

$P=7.v=\left(\rho A{v}^2\right).v=\rho A{v}^3$

$\frac{P}{\rho }=\frac{\rho A{v}^{\prime 3}}{\rho A{v}^3}=\frac{\rho Am(nv{)}^3}{\rho A(v{)}^3}{n}^3:1$

Option $C$ is correct