Question

A water pump rated 400 W, has an efficiency of 75%. If it is employed to raise water through a height of 40 m, the volume of water drawn in 10 min is:
$
  {\text{A}}{\text{. 0}}{\text{.45}}{{\text{m}}^3} \\
  {\text{B}}{\text{. 0}}{\text{.75}}{{\text{m}}^3} \\
  {\text{C}}{\text{. 0}}{\text{.62}}{{\text{m}}^3} \\
  {\text{D}}{\text{. 0}}{\text{.02}}{{\text{m}}^3} \\
$

Answer 1

Hint: The efficiency of a system is defined as the ratio of output power and the input power. Power is equal to the rate of doing work. Here we are doing work against gravity so by using various expressions we can calculate the mass of water raised and hence the volume of water.

Formula used:
Efficiency of a system is given as:
$\eta = \dfrac{{{\text{Output power}}}}{{{\text{Input power}}}}$

Complete step-by-step answer:
The efficiency of a mechanical body signifies the amount of power can be generated from it. Mathematically, the efficiency of the system is equal to the ratio of power output obtained from the machine to the power given to the machine. The formula for efficiency is
$\eta = \dfrac{{{\text{Output power}}}}{{{\text{Input power}}}}{\text{ }}...{\text{(i)}}$
We are given the efficiency of the system to be 75% and the water pump is rated at 400 W which means that the input power is 400 W. Therefore we can calculate the output power from input power and efficiency as follows. Putting the values in equation (i), we get
$
  \dfrac{{75}}{{100}} = \dfrac{{{\text{Output power}}}}{{400}} \\
   \Rightarrow {\text{Output power}} = \dfrac{{75}}{{100}} \times 400 = 300W \\
$
Now power is equal to the rate of doing work, therefore, using the output power, we can write
$P = \dfrac{{dW}}{{dt}}$
where W is the work being done. In our case the work is to raise the water to a height of 40m. Since the work is being done against the gravity here, the work done can be given as
$W = mgh$
where m is the mass of water being raised to a height h. Substituting this expression in expression for power we get
$P = \dfrac{d}{{dt}}\left( {mgh} \right)$
Here the value of g and h are constants which can be taken out of the derivative but mass is not constant so
$
  P = gh\dfrac{{dm}}{{dt}} \\
   \Rightarrow \dfrac{{dm}}{{dt}} = \dfrac{P}{{gh}} \\
$
Substituting the various known values, we get
$\dfrac{{dm}}{{dt}} = \dfrac{{300}}{{10 \times 40}} = \dfrac{3}{4}$
We have taken $g = 10m{s^{ - 2}}$. We can rearrange this equation and then integrate it to get the mass of water being lifted in the following way:
$dm = \dfrac{3}{4}dt$
Integrating both sides, we get
$
  \int {dm} = \dfrac{3}{4}\int {dt} \\
   \Rightarrow m = \dfrac{3}{4}t \\
$
We are given $t = 10\min = 10 \times 60s = 600s$. Using this value we get
$m = \dfrac{3}{4} \times 600 = 450kg$
Taking density of water as $\rho = 1000kg/{m^3}$, we get volume to be
$V = \dfrac{m}{\rho } = \dfrac{{450}}{{1000}} = 0.45{m^3}$
Hence, option A is the correct answer.

Note: Output power of the device decides the performance of the device which is why we have used the output power to calculate the work done and not the input power. Also no machine is 100% efficient that is why we have output power less than the input power.