Two cylindrical vessels of equal cross-sectional area $2\ \text{m}^2$ contain water up to heights $10\ \text{m}$ and $6\ \text{m}$, 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 $g = 10\ \text{m/s}^2$)
Answer: (D) $8\times10^4\ \text{J}$
Water flows until both levels are equal: $h = \dfrac{10 + 6}{2} = 8\ \text{m}$.
A column of height $h$ has potential energy $(\rho Ah)g\cdot\dfrac h2 = \tfrac12\rho Agh^2$.
Initial: $\tfrac12\rho Ag(10^2 + 6^2) = 68\rho Ag$. Final: $\tfrac12\rho Ag(8^2 + 8^2) = 64\rho Ag$.
Work done by gravity = loss in potential energy:
$$W = 4\rho Ag = 4\times1000\times2\times10 = 8\times10^4\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics