JEE Physics 2025 Question with Solution

Given: Two soap bubbles have radii $R_1 = 2 \, \text{cm}$ and $R_2 = 4 \, \text{cm}$.

Find: The radius of curvature $r$ of the common surface.

For two soap bubbles in contact, the radius of curvature of the common surface is given by

$$r = \frac{r_1 \cdot r_2}{r_1 - r_2}$$

Taking the magnitudes of the given radii,

$$r = \frac{2 \times 4}{4 - 2}$$ $$r = \frac{8}{2} = 4 \, \text{cm}$$

Therefore, the required radius of curvature is $4 \, \text{cm}$.

Given: Two soap bubbles have radii $R_1$ and $R_2$ with values $2 \, \text{cm}$ and $4 \, \text{cm}$.

Find: The radius of curvature $r$ of the common soap film.

Let $p_1$ and $p_2$ be the pressures inside the bubbles, and let the outer atmospheric pressure be $p_0$.

For a soap bubble,

$$p_1 = p_0 + \frac{4\gamma}{R_1}, \qquad p_2 = p_0 + \frac{4\gamma}{R_2}$$

Hence,

$$p_1 - p_2 = 4\gamma\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

For the common soap film of radius of curvature $r$,

$$p_1 - p_2 = \frac{4\gamma}{r}$$

Equating the two expressions,

$$4\gamma\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{4\gamma}{r}$$

Cancelling $4\gamma$,

$$\frac{1}{r} = \frac{1}{R_1} - \frac{1}{R_2}$$

Substituting $R_1 = 2 \, \text{cm}$ and $R_2 = 4 \, \text{cm}$,

$$\frac{1}{r} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$

Therefore,

$$r = 4 \, \text{cm}$$

Thus, the final answer is $4$.