Q. Surface tension of water is 73 dyne $cm^{- 1}$ at $20℃$. If surface area is increased by $0.10\ m^{2}$, work done is
A) 7.3 erg
B) $7.3 \times 10^{4}\ erg$
C) 73 J
D) 0.73 J
Explanation
(b)
Work done = surface tension $\times$ increase in area
$= 73\ dyne\ cm^{- 1} \times 0.10\ m^{2}$
$= 73\ dyne\ cm^{- 1} \times 0.10 \times 10^{4}\ cm^{2}$
$= 7.3 \times 10^{4}\ ergs$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.