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The radii of two soap bubbles are r(i) a...

The radii of two soap bubbles are `r_(i)` and `r_(2)`. In isothermal conditions, two meet together in vacuum. Then the radius kof the resultant bubble is given by

A

`R = (r_1+r_2)//2`

B

`R=r_1(r_1r_2 +r_2)`

C

`R^2 =r_1^2 +r_2^2`

D

`R = r_1+r_2`

Text Solution

Verified by Experts

The correct Answer is:
C
Since, the bubble coalesce in vacuum and there is no change in temp, hence its surface energy does not change, this means that the surface area remains unchanged.
`4pir_(1)^(2) +4pir_(2)^(2) =4piR^2 implies R^2 =r_1^2 +r_2^2`
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