Under isothermal condition, energy E is supplied to a soap bubble of surface Tension $\mathrm{\sigma }$ and radius r, to double the radius of bubble, then E is

Given,

Soap's surface tension is σ

${\mathrm{r}}_1=\mathrm{r}, {\mathrm{r}}_2=2\mathrm{r}$

A soap bubble has two surfaces because it has two layers (inside and outside).
∴ Surface area of the smaller bubble total is $={\mathrm{A}}_1=2\times 4{{\mathrm{\pi r}}_1}^2=8{\mathrm{\pi r}}^2$

also the larger bubble's overall surface area = ${\mathrm{A}}_2=2\times 4{{\mathrm{\pi r}}_2}^2=2\times 4\mathrm{\pi }{\left(2\mathrm{r}\right)}^2=32{\mathrm{\pi r}}^2$

Hence the net charge is,

$∆\mathrm{A}={\mathrm{A}}_2-{\mathrm{A}}_1=32{\mathrm{\pi r}}^2-8{\mathrm{\pi r}}^2=24{\mathrm{\pi r}}^2$

energy used to increase the bubble's radius

$\mathrm{E}=\mathrm{\sigma }\times \mathrm{\Delta A}=24{\mathrm{\pi r}}^2\mathrm{\sigma }$

Hence the  correct answer is $24{\mathrm{\pi r}}^2\mathrm{\sigma }.$