Pressure inside a soap bubble is greater than the pressure outside by an amount;(given : R = Radius of bubble, S = Surface tension of bubble)
- $\frac{4S}{R}$
- $\frac{4R}{S}$
- $\frac{S}{R}$
- $\frac{2S}{R}$
The Correct Option is A
Solution and Explanation
For a soap bubble, there are two liquid-air surfaces, so the excess pressure \( \Delta P \) inside the bubble is given by:
\[\Delta P = 2 \left( \frac{2S}{R} \right) = \frac{4S}{R}\]
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