Two soap bubbles of radii `a` and `b` coalesce to form a single bubble of radius `c`. If the external pressure is `P`, find the surface tension of the soap solution.

Options

A. `(P(c^(3) + a^(3) + b^(3)))/(4(a^(2) + b^(2) -c^(2)))`
B. `(P(c^(3) -a^(3) -b^(3)))/(4(a^(2) + b^(2)-c^(2)))`
C. `Pc^(3) -4a^(2)-4b^(2)`
D. `Pc^(2) - 2a^(2)- 3b^(2)`

Answer

B

Explanation

As the total mass of the air inside the bubble and the temperature remains constant, we can say,
`P_(a) V_(a) +P_(b) V_(b) = P_(c ) V_(c)`
As pressure inside the soap bubble is `(4 sigma)/(r )` more than the external pressure, (here `sigma` is the surface tension)
`(P + (4 sigma)/(a)) .((4)/(3) pi a^(3)) + (P (4 sigma)/(b)) .((4)/(3) pi b^(3))= (P + (4sigma)/(c )) .((4)/(3) pi c^(3))`
`rArr P (a^(3) + b^(3) -c^(3)) = 4 sigma (c^(2) -a^(2) -b^(2))`
`or sigma = (P (c^(3) - a^(3) - b^(3)))/(4(a^(2) + b^(2) -c^(2)))`
`rArr` b is correct