The surface tension of a soap solution is 3.5 10-2 N/m. The work required to increase the radius of a soap bubble from 1 cm to 2 cm is 10-6 J. The value of is . ( =22/7)
Answer
264
Solution
A soap bubble has two free surfaces: increase in surface area =2 4 (r22-r12)
W=T 8 (r22-r12)
⇒ W=3.5 10-2 8 227 [(2 10-2)2-(1 10-2)2 ]
⇒ W=3.5 10-2 1767 3 10-4=264 10-6 J
∴ =264
Answer: 264