Q. Work done in increasing the size of soap bubble from radius of 3 cm to 5 cm is nearly (surface tension of soap solution$= 0.03{Nm}^{- 1})$
A) $0.2\ \pi\ mJ$
B) $2\ \pi\ mJ$
C) $0.4\ \pi\ mJ$
D) $4\ \pi\ mJ$
Explanation
(c) Work done = Change in surface energy
$w = 2T \times 4\pi\left( R_{2}^{2} - R_{1}^{2} \right)$
$= 2 \times 0.03 \times 4\pi\left\lbrack (5)^{2} - (3)^{2} \right\rbrack \times 10^{- 4}$
$= 0.4\pi\ mJ$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.