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a water tank is in the form of a right circular cylinder with 20 ft and radius 6 ft. if the tank is half full of water, find the work required to pump all of its water over the rim?
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\[R=6 ft , H=20 ft\] \[A=mg(H-h_{cm})\] m is whole water mass \[m=\rho\frac{\pi r^2H}{2}\] \[h_{cm}\] is center of mass position \[h_{cm}=\frac{H}{4}\] Now you will get \[A=\rho\frac{\pi r^2H}{2}g(H-H/4)=\frac{3\pi\rho gr^2H^2}{8}\]
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