Question

A motor pump set of efficiency 80%, lifts 800 liters of water in 19.6 seconds over a head of 20 m. Its input power is (Take$g = 9.8\,m/{s^2}$):
(A) 64 kW
(B) 40 kW
(C) 10 kW
(D) 196 kW

Answer 1

Hint: The efficiency of any instrument is the ratio of output power to the input power. Calculate the output power which is work done by the motor pump per unit time. To do so, use the relation between potential energy and work done.

Formula used:

1. Efficiency, $\eta = \dfrac{{{P_O}}}{{{P_I}}}$
   Here, ${P_O}$ is the output power and ${P_I}$ is the input power.
2. $P = \dfrac{W}{t}$
   Here, W is the work done and t is the time.
3. $W = {U_f} - {U_i}$
   Here, ${U_f}$ is the final potential energy and ${U_i}$ is the initial potential energy.

Complete step by step answer:
We have given that the efficiency of the motor pump is 80%.
We have the efficiency of the motor pump is,
$\eta = \dfrac{{{P_O}}}{{{P_I}}}$
Here, ${P_O}$ is the output power and ${P_I}$ is the input power.
Since the efficiency is 80%, substituting we get,
$0.8 = \dfrac{{{P_O}}}{{{P_I}}}$
$\Rightarrow {P_I} = \dfrac{{{P_O}}}{{0.8}}$ …… (1)
We know the output power is given by the relation,
${P_O} = \dfrac{W}{t}$ …… (2)
Here, W is the work done and t is the time.
We have the work done by the motor pump on lifting the water of mass 800 kg is equal to the change in the potential energy of the water. Therefore,
$W = {U_f} - {U_i}$
$\Rightarrow W = mg{h_f} - mg{h_i}$
$\Rightarrow W = mg\left( {{h_f} - {h_i}} \right)$
Here, ${h_f}$ is the final height and ${h_i}$ is the initial height.
Since the water is lifted from the ground, the initial height is 0 m. We have given that the water rose to a height of 20 m.
Substituting 800 kg for m, $9.8\,m/{s^2}$ for g, 20 m for ${h_f}$ and 0 m for ${h_i}$ in the above equation, we get,
$W = \left( {800} \right)\left( {9.8} \right)\left( {20 - 0} \right)$
$\Rightarrow W = 156800\,J$
Substituting 156800 J for W and 19.6 s for t in equation (2), we get,
${P_O} = \dfrac{{156800}}{{19.6}}$
$\Rightarrow {P_O} = 8000\,{\text{W}}$
Substituting 8000 W for ${P_O}$ in equation (1), we get,
${P_I} = \dfrac{{8000}}{{0.8}}$
$\Rightarrow {P_I} = 10000\,{\text{W}}$
$\therefore {P_I} = 10\,{\text{kW}}$

So, the correct answer is option (C).

Note: To cross-check the answer, students can use the fact that if the efficiency is less than 100%, the output power is always less than input power. In the solution, we have taken the mass of water as 800 kg instead of 800 liters because the water has the density of$1000\,kg/{m^3}$ that means 1 liter of water contains exactly 1 kg of its mass. To determine the total work done it should be taken as difference in the final potential energy and initial potential energy. If the sign of the work done is positive, the work is done by the motor pump.