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?

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

Tardigrade Admin · Accepted Answer

Excess of pressure inside the soap bubble p=(4 S/R) ∴ (p2/p1)=(R1/R2)=(R/2 R)=(1/2) p2=(1/2) p1=(1/2) × 10 P a=5 Pa

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

20 Pa

40 Pa

2.5 Pa

5 Pa

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

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