Question

Two identical cylindrical vessels, each of base area A, have their bases at the same horizontal level. They contain a liquid of density $\rho$. In one vessel, the height of the liquid is $h_1$ and in the other $h_2>h_1$. When the two vessels are connected, the work done by gravity in equalizing the levels is

• A. $2\rho Ag(h_2-h_1{)}^2$
• B. $\rho Ag(h_2-h_1{)}^2$
• C. $\frac{1}{2}\rho Ag(h_2-h_1{)}^2$
• D. $\frac{1}{4}\rho Ag(h_2-h_1{)}^2$

Answer 1

The correct option is D

$\frac{1}{4}\rho Ag(h_2-h_1{)}^2$

Let the liquid from height $h_2$ come down through a distance $x$. In the other vessel, liquid will rise by the same amount $x$.
Hence, $h_2-x=h_1+x\Rightarrow x=\frac{h_2-h_1}{2}$
Thus, effectively a slab of liquid $\frac{1}{2}(h_2-h_1)$ in thickness falls a vertical distance equal to its thickness under the action of gravity.
Work done by the gravity, $W=mg\times \frac{1}{2}(h_2-h_1)$
where mass of the slab m is given by
$m=\rho \times V=\rho \times \frac{1}{2}(h_2-h_1)\times A$
$\therefore W=\frac{1}{2}\rho (h_2-h_1)\times A\times g\times \frac{1}{2}(h_2-h_1)=\frac{1}{4}\rho Ag(h_2-h_1{)}^2$
Hence, the correct choice is (d).