Question

A spherical soap bubble of radius 2 cm is attached to a spherical soap bubble of radius 4 cm. Find the radius of the curvature of the common surface?

Answer 1

Answer: 4 cm

Explanation: we know that

Δp =4 T/ r

Where, T =surface tension

And r is radius of curvature

The pressure difference in the merged bubble will be Δp= p1-p2

The above equation is solved in the given picture

Answer 2

☞ QUESTION :-

A spherical soap bubble of radius 2 cm is attached to a spherical soap bubble of radius 4 cm. Find the radius of the curvature of the common surface?

☞ ANSWER :-

⟹$\tt r = {4 \times 10}^{ - 2} m$

☞ SOLUTION :-

[ NOTE : SEE THE ATTACHMENT ]

pressure exerted by 1

$\tt P1 = \dfrac{4T}{r1}$

Pressure exerted by 2

$\tt P2 = \dfrac{4T}{r2}$

Let r is the radius of curvature of common surface.

Net pressure on surface = P2 - P1

⟹$\tt \dfrac{4T}{r} = \dfrac{4T}{r2} - \dfrac{4T}{r1}$

⟹$\tt \dfrac{1}{r} = \dfrac{1}{r2} - \dfrac{1}{r1}$

⟹$\tt \dfrac{1}{r} = \dfrac{1}{2} - \dfrac{1}{4} = \dfrac{1}{4}$

r = 4cm

⟹$\tt r = {4 \times 10}^{ - 2} m$