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)`)

Options

A. 30
B. 80
C. 98
D. 90

Answer

B

Explanation

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
`dm =rho dV = rho pi r^(2) dx`
So, the potential energy of the mass dm,
`dU = dmg x rArr dU = g rho pi r^(2) xdx`
Integrating on the both sides, we get
`U= g rho pir^(2) int _(6)^(20) x dx`
`10 m//s^(2)` , density of water `, rho = 10^(3) kg// m ^(2) `
`U = 10 xx 10^(3) xx 3.14 xx (2.5) ^(2) [20^(2)/2 - 6^(2)/ 2]`
` = 10^(4) x 3.14 xx ( 2.5)^(2) xx 182`
`U = 35.71 xx 10^(6) J`
Now, the time taken by pump of power 10 HP
`"time " = ("work done")/("power"),`
`t= (35.71 xx 10^(6))/(10xx 746 xx 60) approx 80 min [ because "1 HP = 746 W"]`
Hence, the correct option is (b).