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, (Take, g=10 ~ms ⁻² )

Step-by-step solution

A cylindrical well of radius 2.5 ~m is shown in the figure. Let a small volume element d x at a distance x from the surface, then mass of the element, d m= d V= r^2 d x So, the potential energy of the mass d m , d U=d m g x d U=g r^2 x d x Integrating on the both sides, we get U=g r^2 ₆²⁰ x d x Here, g=10 ~m / s ^2 , density of water, =10^3 ~kg / m ^3 aligned r=2.5 ~m & U & =10 10^3 3.14 (2.5)^2 [ 20^2 2 - 6^2 2 ] & =10^4 3.14 (2.5)^2 182 U & =35.71 10^6 ~J aligned Now, the time taken by pump of power 10 HP aligned