The surface tension of a soap bubble is 0.03 N/m. The diameter of the bubble is increased from 2 cm to 6 cm. The work done in increasing the diameter is:
απ × 10-4 J
Find the value of α.
A soap bubble has two liquid surfaces. Therefore, the work done in expanding a soap bubble is equal to the increase in surface energy.
Surface Energy = 2TΔA
where:
Initial radius:
r1 = 1 cm = 0.01 m
Final radius:
r2 = 3 cm = 0.03 m
Change in area:
ΔA = 4π(r22 − r12)
= 4π[(0.03)2 − (0.01)2]
= 4π(0.0009 − 0.0001)
= 4π(0.0008)
= 0.0032π
Work done:
W = 2TΔA
= 2 × 0.03 × 0.0032π
= 1.92π × 10-4 J
Therefore:
α = 1.92 ≈ 2
α = 2