Question

A film of a liquid is held on a circular ring of radius r. If the surface tension of the liquid is T, the surface energy of liquid is:
A-$\pi {{r}^{2}}T$
B-$2\pi {{r}^{2}}T$
C-$4\pi {{r}^{2}}T$
D-none of the above

Answer 1

Here we need to evoke the concept of surface tension and surface energy. Surface tension is the property of liquids that is governed by the interactions among the molecules of the liquid. The cause of origin of surface tension are the cohesive forces between molecules in a liquid.

Step by step answer:
According to the rule of thermodynamics, liquids try to attain a state with a maximum amount of favorable interactions. This means that the liquids try to assume such a shape that the number of bulk molecules is at a maximum and the amount of surface molecules is at a minimum.
Now surface energy is given by the formula, $E=2TA$, where T is the surface tension and A is the area.
Here, the liquid is on the ring of radius, r, so the area will be equal to $\pi {{r}^{2}}$. Thus, $E=2\pi {{r}^{2}}T$.

Hence the correct option is (B).

Additional information: The molecules at the surface experience less amount of force as comparison to the molecules present inside the bulk. Therefore, energy is needed to move molecules from the bulk of the liquid to the surface.

Note: Here the shape was the ring so we took the area of the ring. Had it been any other surface say circular then we would have taken the area of the circle. Also, greater is the magnitude of the interaction between the molecules, the more energy is required to increase the surface area of a liquid.