Q. Surface tension of a soap solution is able of 2.0 cm diameter will be
A) $7.6 \times 10^{- 6}\ \pi$ J
B) $15.2 \times 10^{- 6}\ \pi$ J
C) $1.9 \times 10^{- 6}\ \pi$ J
D) $1 \times 10^{- 4}\ \pi$ J
Explanation
(b) Work done = surface tension $\times$ increase in area
$= 1.9 \times 10^{- 2} \times (4\pi R^{2}) \times 2$
$= 1.9 \times 10^{- 2} \times 4 \times \pi\left( 1 \times 10^{- 2} \right)^{2} \times 2$
$= 15.2 \times 10^{- 6}\pi$J
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.