To find the work done by gravity when two identical cylindrical vessels containing liquids of different heights are connected, we need to consider the change in gravitational potential energy as the liquid levels equalize. Let's break this down step by step.
Understanding the Setup
We have two cylindrical vessels with the same base area \( A \) and density \( \rho \) of the liquid. The heights of the liquid in the two vessels are \( h_1 \) and \( h_2 \). When we connect these vessels, liquid will flow from the vessel with the higher liquid level to the one with the lower level until the heights equalize.
Calculating the Initial Potential Energy
The potential energy (PE) of a liquid column in a vessel is given by the formula:
PE = mgh
where \( m \) is the mass of the liquid, \( g \) is the acceleration due to gravity, and \( h \) is the height of the liquid column. The mass \( m \) can be expressed as \( m = \rho V \), where \( V \) is the volume of the liquid.
- For the first vessel with height \( h_1 \):
Volume \( V_1 = A \cdot h_1\)
Mass \( m_1 = \rho \cdot A \cdot h_1\)
Potential Energy \( PE_1 = \rho \cdot A \cdot h_1 \cdot g \cdot \frac{h_1}{2} = \frac{1}{2} \rho g A h_1^2 \) (considering the center of mass of the liquid column).
- For the second vessel with height \( h_2 \):
Volume \( V_2 = A \cdot h_2\)
Mass \( m_2 = \rho \cdot A \cdot h_2\)
Potential Energy \( PE_2 = \rho \cdot A \cdot h_2 \cdot g \cdot \frac{h_2}{2} = \frac{1}{2} \rho g A h_2^2 \).
Equalizing the Liquid Levels
When the vessels are connected, the liquid will adjust to a new height \( h_f \) where:
h_f = \frac{h_1 + h_2}{2}
We can now calculate the potential energy of both vessels after the liquid levels have equalized:
Work Done by Gravity
The work done by gravity in equalizing the levels can be calculated as the difference in potential energy before and after connecting the vessels:
Work = (PE_1 + PE_2) - (2 \cdot PE_f)
Substituting the Values
Substituting the equations we derived earlier:
Work = \left(\frac{1}{2} \rho g A h_1^2 + \frac{1}{2} \rho g A h_2^2\right) - 2 \cdot \left(\frac{1}{2} \rho g A h_f^2\right)
Work = \frac{1}{2} \rho g A (h_1^2 + h_2^2 - 2h_f^2)
Final Result
Since \( h_f = \frac{h_1 + h_2}{2} \), we can simplify \( h_f^2 \) as:
h_f^2 = \frac{(h_1 + h_2)^2}{4} = \frac{h_1^2 + 2h_1h_2 + h_2^2}{4}
So, the work done simplifies to:
Work = \frac{1}{2} \rho g A (h_1^2 + h_2^2 - 2 \cdot \frac{h_1^2 + 2h_1h_2 + h_2^2}{4})
This expression gives us the amount of work done by gravity when the two liquid levels equalize. By following these logical steps, we can better understand how gravity acts on the liquid and the energy transformations involved in this process.
