Question

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

• A. 30
• B. 80
• C. 98
• D. 90

Answer 1

Answer: (B)

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

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