Question

Given the surface tension of a liquid as ${\text{0}}{\text{.2 N}}{{\text{m}}^{ - 1}}$ , what is the increase in the potential energy of the liquid film formed in a loop of $0.05{\text{ }}{{\text{m}}^2}$ ?
A. $5 \times {10^{ - 2}}{\text{ J}}$
B. $2 \times {10^{ - 2}}{\text{ J}}$
C. $3 \times {10^{ - 2}}{\text{ J}}$
D. None of these

Answer 1

Hint: The above question is asking to calculate the increase in the potential energy. When a liquid film is formed in a loop, it will have some surface energy corresponding to the surface tension acting over the surface area of the film. This surface energy gets stored in the form of the potential energy. Thus, use the formula of surface energy to calculate the increase in the potential energy.

Complete answer:
Surface area of the loop of film, $A = 0.05{\text{ }}{{\text{m}}^2}$ … (1)
Surface tension of the liquid film, $T = 0.2{\text{ N}}{{\text{m}}^{ - 1}}$ … (2)

Considering both the inner and outer surfaces of the film, we get:
Total surface area of the film, $dA = 2 \times 0.05 = 0.1{\text{ }}{{\text{m}}^2}$ … (3)

Now, we know that
Surface Energy = Surface Tension $ \times $ Surface Area, that is,
$U = TdA$

Substituting the values from equations (2) and (3) In the above equation,
Surface Energy $ = (0.2)(0.1) = 0.02{\text{ J}}$

Hence, the surface energy of the liquid film is $0.02{\text{ J}}$ . Thus, the correct option is B.

Note: A film has two layers. This means that it has two surfaces. Surface energy is defined as the product of surface tension of the liquid and the area over which it is acting. Therefore, the film will have some surface energy corresponding to its two surfaces. In the above question, you just simply have to find that surface energy. However, keep in mind that for a film, the surface energy, to be calculated, is there because of the two surfaces.