Question

The surface tension of a soap solution is $0.03N{m}^1$. How much work is done to produce a soap bubble of radius $0.05m$?

Answer 1

$T=0.03N{m}^{-1}$

$r=0.05m$

$W=2\times T\times 4\pi r^2$

$=2\times 0.03\times 4\pi \left(0.05\times 0.05\right)$

$=6\pi \times {10}^{-4}$

We get,

$=6\pi \times {10}^{-4}.$