The surface tension of soap solution is $$3.5 \times 10^{-2}$$ N m$$^{-1}$$. The amount of work done required to increase the radius of soap bubble from 10 cm to 20 cm is _______ $$\times 10^{-4}$$ J. (take $$\pi = \frac{22}{7}$$)

A soap bubble has two surfaces, so the work done to change its radius is:

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

We are given: $$T = 3.5 \times 10^{-2}$$ N/m, $$R_1 = 10$$ cm = 0.1 m, $$R_2 = 20$$ cm = 0.2 m

$$W = 2 \times 3.5 \times 10^{-2} \times 4 \times \frac{22}{7} \times (0.04 - 0.01)$$

$$= 2 \times 3.5 \times 10^{-2} \times \frac{88}{7} \times 0.03$$

$$= 7 \times 10^{-2} \times \frac{88}{7} \times 0.03$$

$$= 88 \times 10^{-2} \times 0.03$$

$$= 2.64 \times 10^{-2} = 264 \times 10^{-4} \text{ J}$$