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If the excess pressure inside a soap bub...

If the excess pressure inside a soap bubble of radius `r_(1)` in air is equal to the excess pressure inside air bubble of radius `r_(2)` inside the soap solution, then `r_(1):r_(2)` is

A

`2:1`

B

`1:1`

C

`1:4`

D

`sqrt2:1`

Text Solution

Verified by Experts

The correct Answer is:
A
For soap bubble
`P=(4sigma)/(r_1)`
For air bubble, `P =(2sigma)/(r_2)`
`P = P. implies (4sigma)/(r_1) =(2sigma)/r_2 implies r_1/r_2 =2`
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