Question

Given the surface tension of a liquid as $5{\text{ N}}{{\text{m}}^{ - 1}}$ , what is the surface energy of liquid film on a ring of area $0.15{\text{ }}{{\text{m}}^2}$ ?
A. 0.75 J
B. 1.5 J
C. 2.25 J
D. 3.0 J

Answer 1

Hint: The above question is related to surface tension and surface energy of fluids. Surface tension is given by surface energy per unit area. This means that surface energy is given by the product of surface tension and the surface area. Use this relation to solve the above question.

Complete answer:
Given values:
Surface area of the ring $ = 0.15{\text{ }}{{\text{m}}^2}$ … (1)
Surface tension of the liquid film $ = 5{\text{ N}}{{\text{m}}^{ - 1}}$ … (2)

Considering both the inner and outer surfaces of the film, we get:
Total surface area of the film $ = 2 \times 0.15 = 0.3{\text{ }}{{\text{m}}^2}$ … (3)

Now, we know that
Surface Energy = Surface Tension $ \times $ Surface Area
Substituting the values from equations (2) and (3) In the above equation,
Surface Energy $ = (5)(0.3) = 1.5{\text{ J}}$

Hence, the surface energy of the liquid film is 1.5 J. Thus, the correct option is B.

Note: A film has two layers, that is, it has two surfaces. Thus, the surface energy, to be calculated, is there because of the two surfaces. In the above question, it is necessary to consider both the surfaces of the film to calculate the total surface area, and hence, the surface energy.