Q. The coefficient of surface tension of a liquid is $5 \, N m^{- 1}$ . If a film is held on a ring of area $0.02 \, m^{2}$ , its total surface energy is about

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A

$2\times 10^{- 2} \, J$

B

$2.5\times 10^{- 2} \, J$

C

$2\times 10^{- 1} \, J$

D

$3\times 10^{- 1} \, J$

Solution:

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

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