Question
Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly. (Surface tension of soap solution = 0.03 Nm-1)
-
4π mJ
-
0.2π mJ
-
2π mJ
-
0.4π mJ
Solution
D.
0.4π mJ
Work done = Change in surface energy
⇒ W = 2T x 4π (R22-R12)
= 2 x 0.03 x 4π [ (5)2-(3)2] x 10-4 J
= 0.4 π mJ
