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 calculate the work performed by gravity upon connecting two cylindrical vessels, we begin by establishing the relevant physical principles:

1. Each cylindrical vessel possesses a cross-sectional area of \(A = 2 \, \text{m}^2\).
2. One vessel contains water to a height of \(h_1 = 10 \, \text{m}\), while the other contains water to a height of \(h_2 = 6 \, \text{m}\).
3. The density of water is \(\rho = 10^3 \, \text{kg/m}^3\).
4. The acceleration due to gravity is \(g = 10 \, \text{m/s}^2\).

Upon connecting the vessels at their base, water will redistribute until the water level is uniform in both. Let \(h_f\) represent this final equilibrium height.

The final height \(h_f\) is determined by the principle of volume conservation:

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

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

The change in potential energy, and hence the work done, is calculated as:

• \(W = \Delta U = \rho g (A h_1) \frac{h_f - h_1}{2} + \rho g (A h_2) \frac{h_2 - h_f}{2}\)
• Individual potential energy changes are as follows:
• For water moving 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: \(\Delta U_1 = m_1 g \frac{(h_1 - h_f)}{2} = 20000 \times 10 \times \frac{(10 - 8)}{2} = 20000 \times 10 \times 1 = 20000 \, \text{J}\)
• For water moving 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:

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

Therefore, the work performed by gravity after the vessels are connected is \(8 \times 10^4 \, \text{J}\).