QJEEA-2024-12: JEE Advanced 2024 JEEA-2024-12 Solution: Q.12 Two large, identical water tanks, 1 and 2, kept on the top …

Solution: The ratio t₁/t₂ is 3.

Case 1: Tank 1 (No Pipe)

1. Volume Flow Rate:

• The rate of change of water level (y) is given by:
• dy/dt = - (a/A) √(2gy)
• Where:
• a = area of the hole
• A = cross-sectional area of the tank
• g = acceleration due to gravity
• y = water level in the tank

1. Separation and Integration:

• Separate variables: dy/√y = - (a/A) √(2g) dt
• Integrate from initial to final conditions:
• ∫[h to 0] dy/√y = - (a/A) √(2g) ∫[0 to t₁] dt

1. Evaluate Integrals:

• [2√y] [h to 0] = - (a/A) √(2g) [t] [0 to t₁]
• -2√h = - (a/A) √(2g) t₁
• t₁ = (2A√h) / (a√(2g)) = (A/a) √(2h/g)

Case 2: Tank 2 (With Pipe)

1. Volume Flow Rate:

• The total water height is H + y:
• dy/dt = - (a/A) √(2g(H + y))

1. Separation and Integration:

• Separate variables: dy/√(H + y) = - (a/A) √(2g) dt
• Integrate:
• ∫[h to 0] dy/√(H + y) = - (a/A) √(2g) ∫[0 to t₂] dt

1. Evaluate Integrals:

• [2√(H + y)] [h to 0] = - (a/A) √(2g) [t] [0 to t₂]
• 2√H - 2√(H + h) = - (a/A) √(2g) t₂
• t₂ = (2A/a√(2g)) [√(H + h) - √H]

1. Substitute H = (16/9)h:

• t₂ = (A/a√(g/2)) [√(16h/9 + h) - √(16h/9)]
• t₂ = (A/a√(g/2)) [√(25h/9) - √(16h/9)]
• t₂ = (A/a√(g/2)) [(5√h/3) - (4√h/3)]
• t₂ = (A/a√(g/2)) (√h/3)
• t₂ = (A/a) √(2h/9g) = (A/a) (1/3) √(2h/g)

Ratio t₁/t₂:

• Calculate the ratio:t₁/t₂ = [(A/a) √(2h/g)] / [(A/a) (1/3) √(2h/g)]
• t₁/t₂ = 3

Answer:

The ratio t₁/t₂ is 3.