Question

A water pump driven by petrol raises water at a rate of 0.5 m³/minute from a depth of 30m. If the pump is 70% efficient, then the power developed by the engine is

• A. 1750W
• B. 2450W
• C. 3500W
• D. None of these

Answer 1

The correct option is C

3500W

Step 1:Given:

$Q=0.5\frac{{\text{m}}^3}{\text{min}}$

$h=30\text{m}$

$\eta =0.70$

Step 2:To Find:

Power developed by the engine.

Step 3:Calculate Power:

$$\begin{gathered} P=\frac{\rho Qgh}{60\times \eta } \\ =\frac{1000\times 0.5\times 10\times 30}{60\times 0.7} \\ =3571\approx 3500\text{W} \end{gathered}$$

Here, $\rho$ is density, $Q$ is discharge, $h$ is height ,$g$ is gravitational constant and $\eta$ is efficiency.

Hence, Option(C) is correct.