Surface tension of a soap bubble is $$2.0 \times 10^{-2}$$ N m$$^{-1}$$. Work done to increase the radius of soap bubble from $$3.5$$ cm to $$7$$ cm will be: [Take $$\pi = \frac{22}{7}$$]

We need to find the work done to increase the radius of a soap bubble from 3.5 cm to 7 cm.

A soap bubble has two surfaces (inner and outer), so:

$$W = 2T \times \Delta A = 2T \times 4\pi(R_2^2 - R_1^2) = 8\pi T(R_2^2 - R_1^2)$$

$$T = 2.0 \times 10^{-2}$$ N/m, $$R_1 = 3.5$$ cm $$= 0.035$$ m, $$R_2 = 7$$ cm $$= 0.07$$ m, $$\pi = \dfrac{22}{7}$$.

$$R_2^2 - R_1^2 = (0.07)^2 - (0.035)^2 = 0.0049 - 0.001225 = 0.003675 \text{ m}^2$$

$$W = 8 \times \frac{22}{7} \times 2.0 \times 10^{-2} \times 0.003675$$

$$W = 8 \times \frac{22}{7} \times 7.35 \times 10^{-5}$$

$$W = 8 \times 22 \times \frac{7.35}{7} \times 10^{-5} = 8 \times 22 \times 1.05 \times 10^{-5}$$

$$W = 184.8 \times 10^{-5} = 18.48 \times 10^{-4} \text{ J}$$

The correct answer is Option C: $$18.48 \times 10^{-4}$$ J.