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

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