Question

two soap bubbles with radii r1 and r2 (r1>r2) come in contact. their common surface has a radius of curvature r=

Answer 1

Hey buddy,

◆ Answer-
r = r1.r2/(r1-r2)

◆ Explaination-
Lets consider, two soap bubbles with radius r1 nad r2 such that r1 > r2 .

For bubble 1 -
P1 - P0 = 4T/r1
P1 = P0 + 4T/r1 ...(1)

For bubble 2 -
P2 - P0 = 4T/r2
P2 = P0 + 4T/r2 ...(2)

Let's, consider two bubbles merge so that-
P2 - P1 = 4T/r
P0 + 4T/r2 - P0 + 4T/r1 = 4T/r
4T (1/r2 - 1/r1) = 4T/r
1/r = 1/r2 - 1/r1
r = r1.r2 / (r1-r2)

Radius of the common bubble is r1.r2/(r1-r2).

Hope this helps you...

Answer 2

Answer:

Explanation:

◆ Explaination-

Lets consider, two soap bubbles with radius r1 nad r2 such that r1 > r2 .

For bubble 1 -

P1 - P0 = 4T/r1

P1 = P0 + 4T/r1 ...(1)

For bubble 2 -

P2 - P0 = 4T/r2

P2 = P0 + 4T/r2 ...(2)

Let's, consider two bubbles merge so that-

P2 - P1 = 4T/r

P0 + 4T/r2 - P0 + 4T/r1 = 4T/r

4T (1/r2 - 1/r1) = 4T/r

1/r = 1/r2 - 1/r1

r = r1.r2 / (r1-r2)

Radius of curvature of the common surface of bubble bubble is r1.r2/(r1-r2).

HOPE THIS HELPS.....