The surface tension of a soap solution is $$3.5 \times 10^{-2}$$ N/m. The work required to increase the radius of a soap bubble from 1 cm to 2 cm is $$\alpha \times 10^{-6}$$ J. The value of $$\alpha$$ is _____. $$(\pi = 22/7)$$

A soap bubble possesses two liquid-air surfaces (inner + outer).
Surface energy of a bubble of radius $$r$$ is therefore

$$E = (\text{surface tension}) \times (\text{total surface area})$$
$$E = T \times 2(4\pi r^{2}) = 8\pi T r^{2}$$

Given:
$$T = 3.5 \times 10^{-2}\,\text{N\,m}^{-1}$$
Initial radius $$r_{1}=1\,\text{cm}=0.01\,\text{m}$$
Final radius $$r_{2}=2\,\text{cm}=0.02\,\text{m}$$

The work required to expand the bubble equals the increase in surface energy:

$$W = E_{2}-E_{1}=8\pi T\,(r_{2}^{2}-r_{1}^{2})$$

Compute the change in $$r^{2}$$:
$$r_{2}^{2}-r_{1}^{2}= (0.02)^{2}-(0.01)^{2}=0.0004-0.0001=0.0003\,\text{m}^{2}=3\times10^{-4}\,\text{m}^{2}$$

Insert the numerical values (use $$\pi = \dfrac{22}{7}$$):

$$\begin{aligned} W &= 8 \times \frac{22}{7} \times 3.5 \times 10^{-2} \times 3 \times 10^{-4}\\[4pt] &= \frac{176}{7}\times 3.5 \times 10^{-2} \times 3 \times 10^{-4}\\[4pt] &= 25.142857 \times 0.035 \times 3 \times 10^{-4}\\[4pt] &= 0.88 \times 3 \times 10^{-4}\\[4pt] &= 2.64 \times 10^{-4}\,\text{J} \end{aligned}$$

Express $$W$$ in the form $$\alpha \times 10^{-6}\,\text{J}$$:
$$2.64 \times 10^{-4}\,\text{J}=264 \times 10^{-6}\,\text{J}$$

Hence $$\alpha = 264$$.