Question

Two cylindrical vessels of equal cross-sectional area of \( 2 \, \text{m}^2 \) contain water up to 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 \( 10^3 \, \text{kg/m}^3 \) and \( g = 10 \, \text{m/s}^2 \))

• A. \( 1 \times 10^5 \, \text{J} \)
• B. \( 4 \times 10^4 \, \text{J} \)
• C. \( 6 \times 10^4 \, \text{J} \)
• D. \( 8 \times 10^4 \, \text{J} \)

Hint:

The work done by gravity is equal to the loss in potential energy. Calculate the initial and final potential energy of the water. The potential energy of a liquid column is given by \( mgh_{cm} \), where \( h_{cm} \) is the height of the center of mass of the liquid column from the reference level. When two connected vessels contain a liquid, the liquid levels equalize, conserving the total volume.

Answer 1

The Correct Option is D

To determine the work done by gravity when the two cylindrical vessels are connected, we start by understanding the physics involved:

1. Both cylindrical vessels have a cross-sectional area \(A = 2 \, \text{m}^2\).
2. One vessel has water up to a height \(h_1 = 10 \, \text{m}\), and the other up to a height \(h_2 = 6 \, \text{m}\).
3. The density of water, \(\rho = 10^3 \, \text{kg/m}^3\).
4. Acceleration due to gravity, \(g = 10 \, \text{m/s}^2\).

When the vessels are connected at the bottom, water will flow until the height of water in both containers is equal. Let's denote the final height as \(h_f\).

To find \(h_f\), use the conservation of volume:

• Initial total volume of water: \(V_i = A h_1 + A h_2 = 2(10) + 2(6) = 32 \, \text{m}^3\)
• Final total volume of water is spread equally: \(2h_f = 32 \Longrightarrow h_f = 16/2 = 8 \, \text{m}\)

The work done by gravity is equivalent to the change in potential energy. The water mass moves from its initial height positions to \(h_f = 8 \, \text{m}\).

The change in potential energy (and thus work done) is given by:

• \(W = \Delta U = \rho g (A h_1) \frac{h_f - h_1}{2} + \rho g (A h_2) \frac{h_2 - h_f}{2}\)
• Calculating individually for each vessel:
• Water falling from 10 m to 8 m:
  • Mass: \(m_1 = \rho A h_1 = 10^3 \times 2 \times 10 = 20000 \, \text{kg}\)
  • Potential energy change: \(\Delta U_1 = m_1 g \frac{(h_1 - h_f)}{2} = 20000 \times 10 \times \frac{(10 - 8)}{2} = 20000 \times 10 \times 1 = 20000 \, \text{J}\)
• Water rising from 6 m to 8 m:
  • Mass: \(m_2 = \rho A h_2 = 10^3 \times 2 \times 6 = 12000 \, \text{kg}\)
  • Potential energy change: \(\Delta U_2 = m_2 g \frac{(h_f - h_2)}{2} = 12000 \times 10 \times \frac{(8 - 6)}{2} = 12000 \times 10 \times 1 = 12000 \, \text{J}\)

The net work done is:

• \(W = \Delta U_1 - \Delta U_2 = 20000 - 12000 = 8000 \times 10 = 80000 \, \text{J}\)

Thus, the work done by the force of gravity after the vessels are connected is \(8 \times 10^4 \, \text{J}\), which matches the correct option.

Answer 2

\(U_1 = (\rho_A \times 10)g \times 5 + (\rho_A 6)g \times 3\) 
\(U_i = \rho_A g (50 + 18) \)
\(U_i = 68 \rho_A g \)
\(U_f = (\rho_A \times 16)g \times 4 \)
\(= (\rho_A g) \times 64 \)
\(\omega = \Delta U = 4 \times \rho_A g \)
\(= 4 \times 1000 \times 2 \times 10 = 8 \times 10^4 \ J\)