JEE Main 17 March 2021 Shift 1 Solved Paper

When two soap bubbles of radii a and b(b>a) coalesce, it can be shown as followsPressure inside the bubble 1,p1=p0+‌

4T

a

where, p0 is the initial pressure,
T is the surface tension
and a is the radius of the bubble 1.
Similarly, pressure inside the bubble 2,p2=p0+‌

4T

b

where, b is the radius of the bubble 2 .
Excess pressure at common surface is given by
p‌ex ‌=p1−p2
Let, r be the radius of common surface, then
⇒‌‌‌

4T

r

=(p0+‌

4T

a

)−(p0+‌

4T

b

)
⇒‌‌‌

4T

r

=p0+‌

4T

a

−p0−‌

4T

b

⇒r=‌

ab

b−a

, which is the required expression for radius of curvature at common surface.