MCQ
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)
- ✓$\frac{4 S}{R}$
- B$\frac{4 R}{S}$
- C$\frac{S}{R}$
- D$\frac{2 S}{R}$
✓
Answer
Correct option: A.
$\frac{4 S}{R}$
a
There are two liquid-air surfaces in bubble so
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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