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

Minimum work done is equal to 5 KJ.
Check the attachment for solution.

Answer 2

Initial Potential energy of centre of mass of water is
P₁ = mgh
= Vdgh ………[∵ Mass = Volume × Density]
= 1 m³ × 10³ kg/m³ × 10 m/s² × 0.5 m
= 5 kJ

Final Potential energy of centre of mass of water is
P₂ = mgh’
= Vdgh ………[∵ Mass = Volume × Density]
= 1 m³ × 10³ kg/m³ × 10 m/s² × 1 m
= 10 kJ

Work done = Change in Potential energy
= 10 kJ - 5 kJ
= 5 kJ