Question

Pressure inside two soap bubbles is 1.02 atm and 1.05 respectively. The ratio of their surface area is

Answer 1

The answer is 6.25 : 1

Solution

P = 4S/r  , where P=excess pressure inside soap bubble, S -= surface tension, r = radius of bubble. Assuming ambient pressure = 1 atm,  

P1/P2 = 0.02/0.05 = (4S/R1)/4S/R2= R2/R/1  

P1/P2 = 0.4 = R2/R1  

So R1/R2 = 1/0.4 = 2.5 to1  

Surface area =4pir^2  

4piR1^2/4piR2^2 = R1^2/R2^2 = (2.5/1)^2 = 6.25 to 1

Answer 2

Concept:

A solid object's surface area is a measurement of the overall space that the object's surface takes up. It is given as- S = 4πr².

Given:

The pressure inside one soap bubble, P1 = 1.02atm

The pressure inside second soap bubble, P2 = 1.05atm

Find:

We need to determine the ratio of the soap bubbles surface area

Solution:

For one soap bubble,

We have, P1 = P₀ + 4T/r1

For second soap bubble,

We have, P2 = P₀ + 4T/r2

We have P₀ = 1atm

Therefore, 1st equation becomes,

1.02 = 1 + 4t/r1

4T/r1 = 0.02

r1 = 4T/0.02

Second equation becomes,

1.05 = 1 + 4T/r2

1.05 - 1 = 4T/r2

r2  = 4T/0.05

Therefore, r1/r2 = 4T/0.02 × 0.05/4T4

r1/r2 = 0.05/0.02

r1/r2 = 5/2

(r1/r2)² = (5/2)²

(r1/r2)² = 25/4

Therefore, For Si = 4πr1²

S2 = 4πr2²

S1/S2 = r1²/r2²

S1/S2 = 25/4

Thus, the ratio of the surface area of two soap bubbles is 25/4.

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