Question-155293

Question Number 155293 by peter frank last updated on 28/Sep/21

Answered by mr W last updated on 28/Sep/21

Commented by mr W last updated on 28/Sep/21

T_{1} = T_{2} = T_{3} = T = m e m b r a n e t e n s i o n

p_{1}, p_{2}, p_{3} = r e s u l t a n t p r e s s u r e o n m e m b r a n e

p_{1} = \frac{2 T_{1}}{r_{1}}

p_{2} = \frac{2 T_{2}}{r_{2}}

p_{3} = \frac{2 T_{3}}{r_{3}}

p_{3} = p_{1} - p_{2}

\frac{2 T_{3}}{r_{3}} = \frac{2 T_{1}}{r_{1}} - \frac{2 T_{2}}{r_{2}}

\Rightarrow \frac{1}{r_{3}} = \frac{1}{r_{1}} - \frac{1}{r_{2}} = \frac{r_{2} - r_{1}}{r_{1} r_{2}}

\Rightarrow r_{3} = r = \frac{r_{1} r_{2}}{r_{2} - r_{1}}

Commented by peter frank last updated on 28/Sep/21

thank you very much .

but soap bubble has two surface

\frac{4 T}{r}

Commented by mr W last updated on 28/Sep/21

i r e f e r e t o t h e r e s u l t a n t f o r c e a s i f

i t i s a m e m b r a n e l i k e a b a l l o o n .

Commented by peter frank last updated on 28/Sep/21

okay .

Commented by Tawa11 last updated on 28/Sep/21

Great sirs

Answered by peter frank last updated on 28/Sep/21

Commented by peter frank last updated on 28/Sep/21

S = common interface

H = outside pressure

P_{1} & P_{2} = inside pressure for small

and large bubble

γ = surface tension

\underset{―}{F o r s m a l l b u b b l e}

P_{1} - H = \frac{4 γ}{r_{1}} \dots . (i)

\underset{―}{F o r l a r g e b u b b l e}

P_{2} - H = \frac{4 γ}{r_{2}} \dots . (ii)

\underset{―}{F o r t h e i n t e r f a c e}

P_{1} - P_{2} = \frac{4 γ}{r} \dots . (iii)

(i) - (ii)

P_{1} - P_{2} = \frac{4 γ}{r_{1}} - \frac{4 γ}{r_{2}}

but P_{1} - P_{2} = \frac{4 γ}{r}

\frac{4 γ}{r} = \frac{4 γ}{r_{1}} - \frac{4 γ}{r_{2}}

\frac{1}{r} = \frac{1}{r_{1}} - \frac{1}{r_{2}}

r = \frac{r_{1} r_{2}}{r_{2 -} r_{1}}