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Asked by Prashant DIGHE | 20 Oct, 2019, 07:23: AM
Expert Answer
Let Pa be the pressure inside soap bubble of radius a . Let Pb be the pressure inside soap bubble of radius b .
Let Pc be the pressure inside soap bubble of radius c .
if P is pressure inside a bubble and Po be the pressure outside, then we have, P = [ Po + (4T/r) ] ...............(1)
where T is surface tension and r is radius.
For isothermal condition, we have, Pc Vc = Pa Va + Pb Vb .......................(2)
where V is volume of the drop whose subscripts are same as that of pressure.
By substituting for pressure from eqn.(1) and using volume in terms of radius of bubble,
eqn.(2) is rewritten as , [ Po + (4T/c) ] [ (4/3)πc3 ] = [ Po + (4T/a) ] [ (4/3)πa3 ] + [ Po + (4T/b) ] [ (4/3)πb3 ] ..................(3)
From eqn.(3), we get, T = Po [c3 -a3 -b3 ] / [ 4 ( a2 + b2 - c2 ) ]
Answered by Thiyagarajan K | 20 Oct, 2019, 10:42: PM
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