Question:

Pressure inside a soap bubble is greater than the pressure outside by an amount;(given : R = Radius of bubble, S = Surface tension of bubble)

Updated On: Nov 26, 2024
  • $\frac{4S}{R}$
  • $\frac{4R}{S}$
  • $\frac{S}{R}$
  • $\frac{2S}{R}$
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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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