Two spherical soap bubbles of radii a and b in vacuum coalesce under isothermal conditions. The resulting bubble has a radius given by

Options

A. `((a+b))/(2)`
B. `(ab)/(a+b)`
C. `sqrt(a^(2)+b^(2))`
D. a+b

Answer

C

Explanation

Since, the bubbles coalesce in vacuum and there is no change in temperature, hence its surface energy does not change. This means that the surface area remains unchanged. Hence,
`4pi a^(2) + 4pi b^(2) = 4pi R^(2) rArr`
The radius of resulting bubble `R = sqrt(a^(2)+b^(2))`