The pressure inside two soap bubbles is 1.01 and 1.02 atmosphere. The ration of their respective volumes is

Options

A. `101:102`
B. `102:101`
C. `8:1`
D. `2:1`

Answer

C

Explanation

Let `P_(1) and P_(2) ` be the excess pressures in the bubbles A and B .
Then `P_(1) = 1.01 -1 = 0.01 ` atmosphere
and `P_(2) =1.02 -1 =0.02 ` atmosphere
` :. P = (4T)/R " " :. (P_(1))/(P_(2)) = (R_(2))/(R_(1))`
` :. (R_(2))/(R_(1)) = (0.01)/(0.02) =1/2 " " :. (R_(1))/(R_(2)) = 2 `
` and (V_(A))/(V_(B)) = (4/3 piR_(1)^(3))/(4/3piR_(2)^(3))=((R_(1))/(R_(2)))^(3) = (2/1)^(3) = 8 `