The correct answer Is:
(d) : Here, surface tension, \(S=0.03 \mathrm{Nm}^{-1}\) \(r_{1}=3 \mathrm{~cm}=3 * 10^{-2} \mathrm{~m}, r_{2}=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}\)
Since bubble has two surfaces,
Initial surface area of the bubble
$$
\begin{aligned}
&=2 \times 4 \pi r_{1}{ }^{2}=2 \times 4 \pi \times\left(3 \times 10^{-2}\right)^{2} \\
&=72 \pi \times 10^{-4} \mathrm{~m}^{2}
\end{aligned}
$$
Final surface area of the bubble
$$
=2 \times 4 \pi r^{2}{ }_{2}=2 \times 4 \pi\left(5 \times 10^{-2}\right)^{2}=200 \pi \times 10^{-4} \mathrm{~m}^{2}
$$
Increase in surface energy
$$
=200 \pi \times 10^{-4}-72 \pi \times 10^{-4}=128 \pi \times 10^{-4}
$$
\(\therefore\) Work done \(-S \times\) increase in surface energy
$$
\begin{aligned}
&=0.03 \times 128 \times \pi \times 10^{-4}=3.84 \pi \times 10^{-4} \\
&\left.=4 \pi \times 10^{-4} \mathrm{~J}=0.4 \pi \mathrm{m}\right]
\end{aligned}
$$
