Two cylindrical vessels of equal cross sectional area of 2 m² contain water upto heights 10 m and 6 m , respectively. If the vessels are connected at their bottom then the work done by the force of gravity is (Density of water is 10³ kg / m³ and g=10 m / s² )

Solution

Physics sir · 4 steps

Only a 2 m slab of water actually moves. Track how far its centre of mass falls and you are done.

1. Final level. The vessels have equal areas, so the water settles at the average height, (10 + 6)/2 = 8 m in each. The first vessel loses a 2 m layer (from 8 m to 10 m) and the second gains a 2 m layer (from 6 m to 8 m).
2. Mass that moves. That layer has volume 2 × 2 = 4 m³, so its mass is m = 10³ × 4 = 4000 kg. The rest of the water stays where it was.
3. Drop in height. The layer's centre of mass starts at 9 m (middle of 8 to 10 m) and ends at 7 m (middle of 6 to 8 m). It falls Δ h = 2 m.

   W = m g Δ h = 4000 × 10 × 2 = 8 × 10⁴ J

4. So. Gravity does 8 × 10⁴ J of work, which is option (c).

The idea to keep. Work done by gravity equals the loss in potential energy. Find which part of the fluid really moves and how far its centre of mass drops, then use mgΔ h.

Answer from the NTA final answer key. Solution by PrepWiser.