Question

The excess of pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is n times the volume of the second where n is(a) 0.125(b) 0.250(c) 1(d) 2

Answer 1

Answer:

A) 0.125

Explanation:

Let the excess pressure inside the soap bubble = P

Excess pressure inside the first soap bubble = 2P

Let the radius of first bubble = x

Let the radius of second bubble = R

Thus, excess pressure in first bubble = 2P = 4S/x

excess pressure in second bubble = P = 4S/r (1)

Thus,

2(4S/r) = 4S/x

= x = R/2

Volume of first soap bubble (V1) = 4/3πx³

Volume of first soap bubble (V2)= 4/3πR³

ratio of volumes

V1/ V2 = 4/3πx³/4/3πR³

= x³ = nR³

= (R/2)³ = nR³

n = 1/8 = 0.125

Thus, the volume of the first bubble is n times the volume of the second where n is 0.125