JEE Physics 2023 Question with Solution

Given: Surface tension of the soap bubble is $T = 2 \times 10^{-2} \, \text{N/m}$, initial radius is $R_1 = 3.5 \, \text{cm} = 0.035 \, \text{m}$, and final radius is $R_2 = 7 \, \text{cm} = 0.07 \, \text{m}$.

Find: Work done in increasing the radius of the bubble.

For a soap bubble, work done equals the change in surface energy. Since a bubble has two surfaces, the change in surface area is

$$\Delta A = 2 \times 4\pi \left(R_2^2 - R_1^2\right)$$

Therefore,

$$W = T \times \Delta A$$

Substituting the given values,

$$W = 2 \times 10^{-2} \times 2 \times 4\pi \left((0.07)^2 - (0.035)^2\right)$$

Now,

$$(0.07)^2 = 0.0049, \qquad (0.035)^2 = 0.001225$$

So,

$$W = 2 \times 10^{-2} \times 2 \times 4\pi \times (0.0049 - 0.001225)$$ $$W \approx 1.848 \times 10^{-3} \, \text{J}$$

The solution working gives $1.848 \times 10^{-3} \, \text{J}$, but the solution also states "The Correct Option is D," which disagrees with the computed value and with the listed options. Based on the solution's instruction that the solution is the primary source, the answer is recorded as D.

Therefore, the correct option is D.

A soap bubble has an inner surface and an outer surface. Each surface contributes surface energy equal to surface tension multiplied by area. Hence the total area change is doubled:

$$\Delta A = 2 \times 4\pi \left(R_2^2 - R_1^2\right)$$

Using only one surface would give half the required work, which is incorrect for a soap bubble.