Question

An installation consisting of an electric motor driving a water pump left $75liters$ of water per second to a height of $4.7m$. If the motor consumes a power of $5kW$, then the efficiency of the installation is
$\left( A \right)39\%$
$\left( B \right)69\%$
$\left( C \right)93\%$
$\left( D \right)96\%$

Answer 1

In the above question power consumed by the motor is given. Apply the formula to determine the power used to lift the water. Here the work done is in the form of potential energy. The ratio of power used to power consumed will give the efficiency of the installation.

Formula used:
$P = \dfrac{{mgh}}{t}$
Here $m$ is the mass, $g$ is the acceleration due to gravity and $h$ is the height.

Complete step by step answer:
Energy that is stored in an object is called the potential energy. The stored energy in a system is based on its position, arrangement or state of the object.
It is the work/time ratio. The SI unit of power is the Watt. Watt is equivalent to a Joule/second.
Power consumed by motor is $5kW = 5000W$
Power used in lifting water is given by the ratio work done with respect to time. Here the work done is written in terms of potential energy.
Power used in lifting the water, $P = \dfrac{{mgh}}{t} = \dfrac{{75 \times 9.8 \times 4.7}}{1} = 3454.5W$
Ratio of total power used to power consumed is defined as efficiency.it is expressed in percentage.
$efficiency = \dfrac{{Power\;used}}{{Power\;consumed}} \times 100$
$efficiency = \dfrac{{3454.5}}{{5000}} \times 100 = 69\%$

Hence option $\left( B \right)$ is the correct option.

Note: Ratio between power output (mechanical) and power input (electrical) is the efficiency. Power input is always greater than the Power output, as energy is lost during conversion in various forms, such as heat and friction. Rate of change of work with respect to time is called Power.