JEE Main 2020 · 4 September · evening · Single correct · +4 / −1

Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density d. The area of the base of both vessels is S but the height of liquid in one vessel is x₁ and in the other, x₂ . When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is:

  1. (a) gdS(x₂ + x₁)²
  2. (b) gdS( x₂² + x₁² )
  3. (c) 1/4gdS( x₂ - x₁ )²
  4. (d) 3/4gdS( x₂ - x₁ )²
Answer and solution

Answer (c) 1/4gdS( x₂ - x₁ )²

Solution

Physics sir · 3 steps

Find the potential energy before and after, using each column's centre of mass at half its height.

  1. Energy of one column. A liquid column of height h has mass dSh, and its centre of mass sits at h/2. So its potential energy is dSh × g × h/2 = 1/2gdSh².
  2. Before and after. At the start the energy is 1/2gdS(x₁² + x₂²). The total volume is conserved and the bases are equal, so both columns end at h = (x₁ + x₂)/2. The final energy is 2 × 1/2gdSh² = 1/4gdS(x₁ + x₂)².

    Δ U = gdS/4[2x₁² + 2x₂² - (x₁ + x₂)²] = 1/4gdS(x₂ - x₁)²

  3. So. The system loses energy equal to 1/4gdS(x₂ - x₁)², which is option (c). This energy goes into the flow and is finally lost as heat by viscosity.

The idea to keep. For gravitational energy of a liquid, put all the mass at its centre of mass. Then check the limit: if nothing flows, the energy change must be zero.

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

Study Fluid Mechanics

Asked about 90 times in JEE Main since 2020. This question is from 2020.