Q. The surface tension of a liquid is $5\ {Nm}^{- 1}$. If a film is held on a ring of area $0.02\ m^{2}$, its total surface energy is about
A) $2 \times 10^{- 2}\ J$
B) $2.5 \times 10^{- 2}\ J$
C) $2 \times 10^{- 1}\ J$
D) $3 \times 10^{- 1}\ J$
Explanation
(c) Surface energy is related to the surface tension by the relation
$U = Td\ A$
$Given,\ T = 5\ Nm^{- 1}$
$dA = 2A$
$2 \times 0.02 = 0.04\ \ m^{2}$
$\therefore\ U = 5 \times 0.04$
$= 0.20\ J\ \ = 2 \times 10^{- 1}\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.