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The 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 surface energy is about:A) $5 \times {10^{ - 2}}\,J$B) $2.5 \times {10^{ - 2}}\,J$C) $2 \times {10^{ - 1}}\,J$D) $3 \times {10^{ - 1}}\,J$

Last updated date: 13th Jun 2024
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Hint: In the given question, we need to apply the concept and formula of surface energy. Surface energy is the product of surface tension and total surface area. Be careful while calculating the surface area. The film will have two surfaces and hence two surface areas.

Complete step by step solution:
Surface energy is the work done per unit area by a force which produces the new surface. Surface energy is also known as surface free energy. The ring film will have two surfaces, one on the top and other on the bottom. So, while calculating the surface energy we must account for both the surfaces.

We are given that the value of surface tension $T$ is $T = 5\,N{m^{ - 1}}$ and the value of surface area $S$ is $S = 0.02\,{m^2}$ . But as we have discussed that the total surface area will be $2S$ . Now, the surface energy $E$ will be given as:
$E = T \times 2S$
$\Rightarrow E = 5\,N{m^{ - 1}} \times \left( {2 \times 0.02\,{m^2}} \right)$
$\Rightarrow E = 0.2\,J$
This is the surface energy required.

Thus, option C is the correct option.