Question

A cylindrical well of radius $2.5\, m$ has water upto a height of $14 \,m$ from the bottom. If the water level is at a depth of $6\, m$ from the top of the well, then the time taken (in minutes) to empty the well using a motor of $10\, HP$ is approximately,

• A. 30
• B. 80
• C. 98
• D. 90

Answer 1

Answer: (B)

A cylindrical well of radius $2.5 \,m$ is shown in the figure,

Let a small volume element $dx$ at a distance $x$ from the surface, then mass of the element,
$d m=\rho d V=\rho \pi r^{2} d x$
So, the potential energy of the mass $d m$,
$d U=d m g x \Rightarrow d U=g \rho \pi r^{2} x d x$
Integrating on the both sides, we get
$U=g \rho \pi r^{2} \int\limits_{6}^{20} x d x$
Here, $g=10 m / s ^{2}$, density of water, $\rho=10^{3} kg / m ^{3}$
$r=2.5 \,m $
$U= 10 \times 10^{3} \times 3.14 \times(2.5)^{2}\left[\frac{20^{2}}{2}-\frac{6^{2}}{2}\right]$
$= 10^{4} \times 3.14 \times(2.5)^{2} \times 182 $
$U= 35.71 \times 10^{6} J$
Now, the time taken by pump of power $10\, HP$
time $=\frac{\text { work done }}{\text { power }}$
$t=\frac{35.71 \times 10^{6}}{10 \times 746 \times 60} \approx 80 \,min $
$[\because 1\, HP =746 \,W ]$