Q. Work done forming a liquid drop of radius R is $W_{1}$ and that of radius $3R\ is\ W_{2}$. The ratio of work done is
A) 1:3
B) 1:2
C) 1:4
D) 1:9
Explanation
(d) $W = T \times 4\pi R^{2}$
$\Longrightarrow \ \ \frac{W_{1}}{W_{1}} = \frac{T \times 4\pi R^{2}}{T \times 4\pi(3R)^{2}}$
$= \frac{T \times 4\pi R^{2}}{T \times 36\pi R^{2}} = \frac{1}{9}$
$\therefore\ W_{1}:W_{2} = 1:9$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.