Question

Two identical cylindrical vessels with their bases at the same level, each contain a liquid of density $\rho$. The area of either base is $A$ but in one vessel the liquid height is $h_1$ and in the other liquid height is $h_2(h_2<h_1)$. If the two vessels are connected, the work done by gravity in equalizing the levels is

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

Answer 1

The correct option is D $\frac{1}{4}(h_1-h_2{)}^{2}A\rho g$

Work done by gravity $=-$(change in potential energy)
Assuming final equilibrium is $h$
$W_{Cylinder}=U_i-U_f$
$W_{cylinder1}=$$A\rho g\frac{h_1}{2}(h_1-h)$
$W_{cylinder2}=$$A\rho g\frac{h_2}{2}(h_2-h)$
Total work done by gravity $\left(W_g\right)$ $=W_{cylinder1}+W_{cylinder2}$
$\Rightarrow W_g=A\rho g\frac{h_1}{2}(h_1-h)+A\rho g\frac{h_2}{2}(h_2-h)$
Using volume conservation
$h=\frac{h_1+h_2}{2}$
$W_g=A\rho g\frac{h_1}{2}[h_1-(\frac{h_1+h_2}{2}\left)\right]+A\rho g\frac{h_2}{2}[h_2-(\frac{h_1+h_2}{2}\left)\right]$
$W_g=A\rho g{\left(\frac{h_1-h_2}{2}\right)}^2$