JEE Physics 2023 Question with Solution

Given:

• Surface tension $S = 3.5 \times 10^{-2} \, \text{N/m}$
• Initial radius $r_i = 10 \, \text{cm} = 0.1 \, \text{m}$
• Final radius $r_f = 20 \, \text{cm} = 0.2 \, \text{m}$

Find: The work done in increasing the radius of the soap bubble.

A soap bubble has two surfaces, so the change in surface area is

$$\Delta A = 2 \times 4\pi (r_f^2 - r_i^2) = 8\pi (r_f^2 - r_i^2)$$

Substituting the radii,

$$\Delta A = 8\pi \left((0.2)^2 - (0.1)^2\right)$$ $$\Delta A = 8\pi (0.04 - 0.01) = 8\pi \times 0.03$$

The work done is equal to the increase in surface energy:

$$W = S \times \Delta A$$

So,

$$W = (3.5 \times 10^{-2}) \times 8\pi \times 0.03$$

Using $\pi = \frac{22}{7}$,

$$W = (3.5 \times 10^{-2}) \times 8 \times \frac{22}{7} \times 0.03$$ $$W = \frac{3.5 \times 8 \times 22 \times 3}{7} \times 10^{-4}$$ $$W = 264 \times 10^{-4} \, \text{J}$$

Therefore, the required value is 264.

Given: A soap bubble is considered, not a single liquid surface.

Find: Why the factor of $2$ appears in the area calculation.

A soap bubble has an inner surface and an outer surface. Each surface contributes surface energy. Therefore total surface area is

$$A = 2 \times 4\pi r^2$$

Hence, when the radius changes from $r_i$ to $r_f$,

$$\Delta A = 2 \times 4\pi r_f^2 - 2 \times 4\pi r_i^2$$ $$\Delta A = 8\pi (r_f^2 - r_i^2)$$

Using only $4\pi r^2$ would correspond to a single surface and would give half the required work. This is why the correct result becomes $264 \times 10^{-4} \, \text{J}$.