The surface tension of a soap bubble is 0.03 N/m. The work done in increasing the diameter of bubble from 2 cm to 6 cm is 10-4 J. The value of is . (Take =3.14)
- 0.86
- 0.64
- 1.92
- 7.68
Answer
(3) 1.92
Solution
A soap bubble has two surfaces (inner and outer).
W = T A = T 2 4 (r22 - r12)
r1 = 1 cm = 10-2 m, r2 = 3 cm = 3 10-2 m
W = 0.03 8 (9 - 1) 10-4
⇒ W = 1.92 10-4 J
∴ = 1.92
Answer: (3)