JEE Physics 2025 Question with Solution

Given: The excess pressure inside soap bubble A is half the excess pressure inside soap bubble B.

Find: The value of $n$ if $V_A = nV_B$.

For a soap bubble, the excess pressure is

$$\Delta P = \frac{4T}{R}$$

where $T$ is the surface tension.

Let the excess pressures be $\Delta P_A$ and $\Delta P_B$, and the radii be $R_A$ and $R_B$. Given,

$$\Delta P_A = \frac{1}{2}\Delta P_B$$

Using $\Delta P = \frac{4T}{R}$,

$$\frac{4T}{R_A} = \frac{1}{2}\left(\frac{4T}{R_B}\right)$$ $$\frac{1}{R_A} = \frac{1}{2R_B}$$ $$R_A = 2R_B$$

The volume of a spherical bubble is

$$V = \frac{4}{3}\pi R^3$$

So,

$$\frac{V_A}{V_B} = \frac{\frac{4}{3}\pi R_A^3}{\frac{4}{3}\pi R_B^3} = \left(\frac{R_A}{R_B}\right)^3$$

Since $\frac{R_A}{R_B} = 2$,

$$n = \frac{V_A}{V_B} = 2^3 = 8$$

Therefore, the value of $n$ is $8$.

Given: Excess pressure in bubble A is half that in bubble B.

Find: The ratio of volumes and hence $n$.

Since for a soap bubble,

$$\Delta P \propto \frac{1}{R}$$

if $\Delta P_A = \frac{1}{2}\Delta P_B$, then radius must double:

$$R_A = 2R_B$$

Now volume of a spherical bubble varies as

$$V \propto R^3$$

Therefore,

$$\frac{V_A}{V_B} = \left(\frac{R_A}{R_B}\right)^3 = 2^3 = 8$$

Hence, $n = 8$.