The potential energy possessed by a soap bubble, having surface tension equal to 0.04N/m of diameter 1 cm. is

Given,

$$\begin{gathered} \mathrm{Surface} \mathrm{tension} \mathrm{T}=0.04\mathrm{N}/\mathrm{m} \\ \mathrm{Diameter} \mathrm{d}=1\mathrm{cm}=0.01\mathrm{m} \\ \mathrm{radius} \mathrm{r}=0.005\mathrm{m} \end{gathered}$$

we know that,

$\mathrm{potential} \mathrm{energy} = \mathrm{Surface} \mathrm{tension} \times \mathrm{Surface} \mathrm{area}$

As soap bubble has two surfaces,

$$\begin{gathered} \mathrm{Total} \mathrm{area}=2\times 4{\mathrm{\pi r}}^2 \\ \mathrm{potential} \mathrm{energy} =(0.04)\times [2\times 4\mathrm{\pi }\times {(0.005)}^2] \\ \mathrm{potential} \mathrm{energy} =8\mathrm{\pi }\times {10}^{-6}\mathrm{J}. \end{gathered}$$

Hence the correct answer is $8\mathrm{\pi }\times {10}^{-6}\mathrm{J}.$