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

Options

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`

Answer

C

Explanation

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`