Question

The surface tension of a liquid is $5N/m$. If a film of this liquid is held on a ring of area $0.02m^2$, its surface energy is about :

• A. $0.2J$
• B. $2.5\times {10}^{-2}J$
• C. $2.5\times {10}^{-1}J$
• D. $3\times {10}^{-1}J$

Answer 1

Answer: (A)

$0.2J$

Surface Energy $U=A\times T$

where $S$ is surface energy, $T$ is surface tension and $A$ is the surface area.

Two films are formed because of the ring, hence area will be double of the given area.

$U=5\times 2\times 0.02=0.2J$