# Work done by gravity on Water

Now according to me we would see change in potential energy of system and equate it to the work done by gravity.

But when we see this the first column lowers by $H/2$ and right one rises by $H/2$ and thus there should be no net work done. But it's not so. Work done by gravity is +ve.

Where is it wrong?

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The argument is wrong by implicitly assuming that the two columns are equally large. The work is proportional not only to the change of height but also to the "amount of water", $$\Delta E = mg\cdot \Delta H,\quad m=\rho\cdot A\cdot h$$ and the left column is larger at the beginning. So the left column's drop is numerically more important than the right column's rise.
One may use the fact that as far as the potential energy goes, it doesn't matter which water molecule is which. The process of bringing them at the same level may be visualized as cutting the excess piece of the left column at the top, $H$, to two equal parts and moving the upper half to the right column. The result is that the height will be the same in both columns.
We have lowered the body of water whose volume is $(H/2)\cdot A$ by $(H/2)$ so the total change of the potential energy – work done by gravity – is $$|\Delta E| = \rho\cdot A\cdot (H/2)\cdot g\cdot (H/2) =\frac{\rho A g H^2}{4}$$ If it's homework and you will use the ideas above, it is a matter of honor to indicate this fact in your solution.
Yes the answer is $\dfrac{\rho AgH^2}{4}$. (+1) for simplified approach. –  ABC Apr 4 '13 at 10:02