Question

A motor pump-set efficiency 80% lifts 6000 liters of water per minute from a well 20 meters deep. If $g = 10\dfrac{m}{{{s^2}}}$ then its input power is:
A) 20W
B) 25W
C) 16W
D) 250W

Answer 1

The pump is a mechanical device which pick-ups the water from a low head and delivers the water to a high head utilizing a certain amount of energy in this process.

Formula used: The formula of the power of the pump is given by the,
$P = \dfrac{{\rho Vgh}}{t}$
Where $\rho$ is the density of the fluid $V$ is the volume of the fluid $h$ is the height t is the time taken and g is the acceleration due to gravity.

Complete step-by-step answer:
It is given in the problem that a motor pump with efficiency 80% lifts 6000 liters of water per minute to a height of 20 m in one minute and we need to find the value of the input power.
The power is given by,
$P = \dfrac{{\rho Vgh}}{t}$
Where $\rho$ is the density of the fluid $V$ is the volume of the fluid $h$ is the height t is the time taken and g is the acceleration due to gravity.
Replacing the value of density of the fluid the volume of the fluid lifted and the time taken in the above relation.
$\Rightarrow P = \dfrac{{\rho Vgh}}{t}$
$\Rightarrow P = \dfrac{{\left( {1000} \right) \cdot \left( {6 \times {{10}^{ - 3}}} \right) \cdot \left( {10} \right) \cdot \left( {20} \right)}}{{60}}$
$\Rightarrow P = \dfrac{{6 \cdot \left( {10} \right) \cdot \left( {20} \right)}}{{60}}$
$\Rightarrow P = 20W$.
As the formula of the efficiency is given by,
${\text{Efficiency}} = \dfrac{{{\text{output power}}}}{{{\text{input power}}}}$
$\Rightarrow {\text{Efficiency}} = \dfrac{{{\text{output power}}}}{{{\text{input power}}}}$
Replace the value of the efficiency and output power in the above formula.
$\Rightarrow {\text{Efficiency}} = \dfrac{{{\text{output power}}}}{{{\text{input power}}}}$
$\Rightarrow 0 \cdot 8 = \dfrac{{20}}{{{\text{input power}}}}$
$\Rightarrow {\text{input power}} = \dfrac{{20}}{{0 \cdot 8}}$
${\text{input power}} = 25W$.
The input power on the motor pump is equal to 25W.

The correct answer to this problem is option B.

Note: Basically the pump is increasing the head of the water by providing energy from the source. The density of the fluid which is being pumped has to be taken and not the density of water always in the formula of the work done by the pump.