Question

A cubical vessel of height 1 m is full of water . The minimum work done in taking water out from vessel will be?

Answer 1

The vessel is cubical so its dimensions are 1×1
Vol of water
= 1 m³
= 1000 L
Mass of water = 1000 kg
Height = 1 m
g= 9.8 m/s
Thus, work done
= mgh
= 1000 × 9.8 × 1/2
=4900 J

Answer 2

Suppose the cubical vessel is on top of floor. The water flows out and has a potential energy of 0, let us say.

The area of the base of the cubical vessel = 1 m * 1 m = 1 m²
volume = 1 m³

density of water = 1,000 kg/m³

mass of the water inside the vessel = 1,000 kg/m³ * 1 m³ = 1,000 kg

height of the center of gravity of vessel full of water = 1/2 height
  = 1/2 m

To empty water from the vessel, water has to be lifted by 1/2 m from center of mass to the top of the vessel.

Minimum energy = m g h = 1000 * 10 * 1/2 = 5, 000 J