JEE Mains · Physics · STD 11 - 9.2 surface tension
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)
- A \(\frac{4 S}{R}\)
- B \(\frac{4 R}{S}\)
- C \(\frac{S}{R}\)
- D \(\frac{2 S}{R}\)
Answer & Solution
Correct Answer
(A) \(\frac{4 S}{R}\)
Step-by-step Solution
Detailed explanation
There are two liquid-air surfaces in bubble so \(\Delta \mathrm{P}=2\left(\frac{2 \mathrm{~S}}{\mathrm{R}}\right)=\frac{4 \mathrm{~S}}{\mathrm{R}}\)
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