Question

Calculate the work done in blowing a soap bubble from a radius of $2cm$ to $3cm$. The surface tension of the soap solution is $30dynec{m}^{-1}$.

Answer 1

Work done $=$ Surface tension $\left(S\right)\times$ (increase in surface area)
$W=S\times \left(△surfacearea\right)$
now soap bubble has two layers so change in surface area $=2(4\pi r_t^2-4n{r}_i^2)$
$=8\pi (r_t^2-r_i^2)$
$W=S\times 8\pi (r_t^2-r_i^2)$
$r_t=3cm=0.3m{r}_i=2cm=0.2m$
$S=30\frac{dyne}{cm}=30\frac{{10}^{-5}N}{{10}^{-2}m}$
$S=30\times {10}^{-3}N/m$
$W=30\times {10}^{-3}\times 8\pi ({0.3}^2-{0.2}^2)$
$=37.69\times {10}^{-3}J$