Two identical containers X and Y are connected at the bottom by a thin tube of negligible volume. The tube has a valve in it, as shown in the figure. Initially container X has a liquid filled up to height h in it and container Y is empty. When the valve is opened, both containers have equal amount of liquid in equilibrium. If the initial (before the valve is opened) potential energy of the liquid is P₁ and the final potential energy is P₂ then :
- (a)P₁ = P₂
- (b)P₁ = 4P₂
- (c)P₁ = 2P₂
- (d)P₁ = 8P₂
Correct — C, P₁ = 2P₂. Take the potential energy of a liquid column as its weight acting at its centre of mass, which lies at its mid-height. Initially all the liquid, of mass m, sits in container X filled to height h, so its centre of mass is at h/2 and P₁ = mg(h/2). When the valve is opened the liquid divides equally between the two identical containers, so each holds height h/2 with mass m/2, and each half's centre of mass is at h/4. The total final energy is P₂ = 2 × (m/2)g(h/4) = mgh/4. Comparing, P₁ = mgh/2 and P₂ = mgh/4, so P₁ = 2P₂ — the liquid loses half its potential energy as it levels out.
- (a)P₁ = P₂ — The stored potential energy is not conserved here — the liquid falls to a lower average height as it spreads out, so P₂ is smaller than P₁, not equal.
- (b)P₁ = 4P₂ — This overstates the drop. The centre of mass falls from h/2 to h/4 — a factor of 2, not 4 — so P₁ = 2P₂.
- (d)P₁ = 8P₂ — Far too large a ratio; the centre of mass only halves (h/2 → h/4), giving P₁ = 2P₂.
The gravitational potential energy of a body of liquid can be taken as Mgh꜀, where h꜀ is the height of its centre of mass. When liquid in one container flows to fill two identical containers to half the height, the mass stays the same but its centre of mass drops from h/2 to h/4, halving the potential energy. The lost energy is dissipated as heat and small motions as the liquid settles.
Do not track the water molecule by molecule; use the centre of mass. Initially a full column has its centre at h/2. Finally two half-columns each have their centre at h/4. The same total weight acting at half the height means half the potential energy, so P₁ = 2P₂.
- Potential energy of a liquid column = (its weight) × (height of its centre of mass).
- A column of height H has its centre of mass at H/2.
- Initially P₁ = mg(h/2); finally two half-columns give P₂ = mg(h/4).
- So P₁ = 2P₂ — the liquid loses half its potential energy when it levels out.
- Assuming potential energy is conserved as the liquid levels out — some is lost as heat and motion.
- Forgetting the mass also splits between the two containers, not just the height.
A liquid redistributes between connected containers and you compare initial and final potential energy — track the centre of mass of the whole liquid, not individual parcels.
No directly related past PYQ was found.
- practice — not a real PYQ
A uniform column of water of height h in a tank has its centre of mass at a height of:
- (a)h
- (b)h/2
- (c)h/4
- (d)2h/3
Answer(b) h/2 — the centre of mass of a uniform column lies at its mid-height.
- practice — not a real PYQ
Water filled to height h in one container is allowed to level out equally into two identical containers. The potential energy of the water:
- (a)stays the same
- (b)doubles
- (c)halves
- (d)becomes zero
Answer(c) halves — the centre of mass drops from h/2 to h/4, so the potential energy falls to half.