Question

The air pressure inside a soap bubble of radius $R$ exceeds the outside air pressure by $10 Pa$. By how much will the pressure inside a bubble of radius $2 R$ exceed the outside air pressure?

• A. 20 Pa
• B. 40 Pa
• C. 2.5 Pa
• D. 5 Pa

Answer 1

Answer: (D)

Excess of pressure inside the soap bubble
$p=\frac{4 S}{R}$
$\therefore \frac{p_{2}}{p_{1}}=\frac{R_{1}}{R_{2}}=\frac{R}{2 R}=\frac{1}{2}$
$p_{2}=\frac{1}{2} p_{1}=\frac{1}{2} \times 10\, P a=5\, Pa$