JEE MAIN - Physics (2024 - 6th April Evening Shift - No. 17)

Pressure inside a soap bubble is greater than the pressure outside by an amount : (given : $$\mathrm{R}=$$ Radius of bubble, $$\mathrm{S}=$$ Surface tension of bubble)
$$\frac{S}{R}$$
$$\frac{4 \mathrm{~S}}{\mathrm{R}}$$
$$\frac{4 \mathrm{R}}{\mathrm{S}}$$
$$\frac{2 S}{R}$$

Explanation

The difference in pressure inside a soap bubble as compared to the outside is due to the surface tension created by the soap film on the bubble. This difference in pressure can be calculated using the formula that relates the surface tension of the soap bubble to the radius of the bubble. The correct formula for the pressure difference ($$\Delta P$$) across a soap bubble is given by:

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

Here, $$S$$ is the surface tension of the bubble and $$R$$ is the radius of the bubble. The factor of 4 comes from the fact that a soap bubble has two surfaces (an inner and an outer surface), and for each surface, the Laplace pressure (which contributes to the pressure difference due to surface tension) is given by $$\frac{2S}{R}$$. Thus, for two surfaces, you double this amount, resulting in the $$\frac{4 S}{R}$$ term.

Therefore, the correct answer is:

Option B

$$\frac{4 \mathrm{~S}}{\mathrm{R}}$$

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