Question

In order to float a ring of area $0.04{\text{ }}{{\text{m}}^2}$ in a liquid of surface tension $75{\text{ N/m}}$ , what will be the required surface energy?
A. 3 J
B. 6.5 J
C. 1.5 J
D. 4 J

Answer 1

Hint: Use the concept of surface tension and surface energy to solve the above question. Surface tension of a liquid is defined as surface tension per unit area. This means for an area of ‘a’ square meters, the surface energy is given by ‘Ta’, where ‘T’ is the surface tension of the liquid. Use this formula to solve the above question.

Complete answer:
Given values:
Surface area of the ring, $A = 0.04{\text{ }}{{\text{m}}^2}$ … (1)
Surface tension of the liquid, $T = 75{\text{ N}}{{\text{m}}^{ - 1}}$ … (2)

For a ring, only one surface will be considered, hence,
Total surface area of the film, $dA = A = 0.04{\text{ }}{{\text{m}}^2}$ … (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 $ = (0.04)(75) = 3{\text{ J}}$

Hence, the required surface energy of the ring to float is ${\text{3 J}}$ . Thus, the correct option is A.

Note: A film is very thin and has two air-liquid interfaces, one on the outside of the bubble and the other on the inside. Thus, the surface energy, to be calculated, is there because of the two surfaces. In the above question, a ring is to be considered. It has only one interface. In case of films, it is necessary to consider both the surfaces of the film to calculate the total surface area, and hence, the surface energy.