Question

A motor pump lifts $300\,{\text{kg}}$ of water per minute from a well $20\,{\text{m}}$ deep and delivers to height of $20\,{\text{m}}$. Its power is
A. $3\,{\text{kW}}$
B. $1.96\,{\text{kW}}$
C. $0.98\,{\text{kW}}$
D. $3.92\,{\text{kW}}$

Answer 1

Use the formula for the relation between the work done and change in potential energy of an object. Use the formula for the potential energy of an object. Also use the formula for the power in terms of work done and time. First determine the work done by the motor pump from the initial and final potential energies of the water. Then calculate the power of the motor.

Formulae used:
The work done $W$ is given by
$W = - \Delta U$ …… (1)
Here, $\Delta U$ is the change in potential energy.
The potential energy $U$ of an object is given by
$U = mgh$ …… (2)
Here, $m$ is the mass of the object, $\gamma$ is acceleration due to gravity and $h$ is the height of the object from the ground.
The power $P$ is given by
$P = \dfrac{W}{t}$ …… (3)
Here, $W$ is the work done and $t$ is time.

Complete step by step answer:
We have given that the mass of water lifted by the motor pump in one minute is $300\,{\text{kg}}$.
$m = 300\,{\text{kg}}$
$\Rightarrow t = 1\,{\text{min}}$
The depth of the well is $20\,{\text{m}}$ and the height up to which the water is being delivered is $20\,{\text{m}}$.
$d = 20\,{\text{m}}$
$\Rightarrow h = 20\,{\text{m}}$
We have asked to calculate the power of the pump.Let us first calculate the work done by the pump.The initial potential energy of the water is zero.
${U_i} = 0\,{\text{J}}$
The final potential energy of the water is given by
${U_f} = mg\left( {d + h} \right)$

According to equation (1), the work done by the motor pump is
$W = - \left( {{U_i} - {U_f}} \right)$
Substitute $0\,{\text{J}}$ for ${U_i}$ and $mg\left( {d + h} \right)$ for ${U_f}$ in the above equation.
$W = - \left[ {\left( {0\,{\text{J}}} \right) - \left( {mg\left( {d + h} \right)} \right)} \right]$
$\Rightarrow W = mg\left( {d + h} \right)$
Substitute $mg\left( {d + h} \right)$ for $W$ in equation (3).
$P = \dfrac{{mg\left( {d + h} \right)}}{t}$
Substitute $300\,{\text{kg}}$ for $m$, $9.8\,{\text{m/}}{{\text{s}}^2}$ for $g$, $20\,{\text{m}}$ for $d$, $20\,{\text{m}}$ for $h$ and $1\,{\text{min}}$ for $t$ in the above equation.
$P = \dfrac{{\left( {300\,{\text{kg}}} \right)\left( {9.8\,{\text{m/}}{{\text{s}}^2}} \right)\left( {20\,{\text{m}} + 20\,{\text{m}}} \right)}}{{1\,{\text{min}}}}$
$\Rightarrow P = \dfrac{{\left( {300\,{\text{kg}}} \right)\left( {9.8\,{\text{m/}}{{\text{s}}^2}} \right)\left( {20\,{\text{m}} + 20\,{\text{m}}} \right)}}{{60\,{\text{s}}}}$
$\Rightarrow P = 1960\,{\text{J}}$
$\therefore P = 1.96\,{\text{kJ}}$
Therefore, the power of the pump is $1.96\,{\text{kJ}}$.

Hence, the correct option is B.

Note: The students should not forget to convert the unit of time in the SI system of units as all the units used in the formula are in the SI system of units. Also the students may directly use the formula for potential energy. But the students should keep in mind that the work done is equal to the negative of change in potential energy and not equal to the potential energy.