Question

The surface tension of a liquid is $5 \,N/m.$ 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. $3 \times 10^{-1} J$

Answer 1

Answer: (C)

Surface energy is related to the surface tension by the relation
$ U=Td\,A $
Given, $ T=5\,N/m $
$ dA=2A=2\times 0.02=0.04\,{{m}^{2}} $
$ \therefore $ $ U=5\times 0.04 $
$ =0.20\,J $
$ =2\times {{10}^{-1}}\,J $