Question

Two identical cylindrical vessels with their bases at the same level, each contains a liquid of density $\rho$ . The height of the liquid in one vessel is $h_1$ and that in the other is $h_2$ . The area of either base is A. The work done by gravity in equalizing the levels when the vessels are interconnected, is:

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

Answer 1

The correct option is B $A\rho g{\left(\frac{h_1-h_2}{2}\right)}^2$

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
$h=\frac{h_1+h_2}{2}$
$W_g=A\rho g\frac{h_1}{2}[h_1-\left(\frac{h_1+h_2}{2}\right)]+A\rho g\frac{h_2}{2}[h_2-\left(\frac{h_1+h_2}{2}\right)]$
$W_g=A\rho g{\left(\frac{h_1-h_2}{2}\right)}^2$