A conical container of radius 10 ft and height 40 ft is filled to a height of 36 ft of a liquid weighing 60.6 lb divided by ft cubed. How much work will it take to pump the contents to the rim? How much work will it take to pump the liquid to a level of 4 ft above the cone's rim?

The centre of mass of a right cone of height, H lies on the axis of symmetry at a distance of H/4 from the base

= 3H/4 from top.

D = 5 + 36/4

= 14 ft from the upper rim

Mass of liquid in the cone = volume * density (ρ)

= ⅓.π. r². h.ρ

= ⅓ * 36 * [10 * (36/40) ]² * π * 60.6

= 18505 lbs.

where,

r = radius of the liquid surface.

Workdone required to lift this to the rim

= force * distance

= 18505 * 4.45 * 14

= 1152861.5 N. ft.

B.

4 ft above the rim needs an extra 1152861.5 * 4

= 4611446 N. ft.

Total workdone = 1152861.5 + 4611446

= 5764307.5 J.