EAMCET Engineering 2006 Solved Paper

Let radii of two soap bubbles are a and b respectively and radius of single larger bubble is c.
As excess pressure for a soap bubble is ‌

4T

r

and external pressure P.
Pi=P+‌

4T

r

So, ‌‌Pa=P+‌

4T

a

,Pb=P+‌

4T

b

and Pc=P+‌

4T

c

. . . (i)
and ‌‌Va=‌

4

3

πa3,‌‌Vb=‌

4

3

πb3
and Vc=‌

4

3

πc3 . . . (ii)
Now as mass is conserved.
µa+µb=µc
‌ i.e., ‌‌‌‌

PaVa

RTa

+‌

PbVb

RTb

=‌

PcVc

RTc

‌‌(‌ as ‌PV=µRT)
As temperature is constant, i.e., Ta=Tb=Tc
So, ‌‌PaVa+PbVb=PcVc
which in the light of Eqs. (i) and (ii) becomes,
(P+‌

4T

a

)(‌

4

3

πa3)+(P+‌

4T

b

)(‌

4

3

πb3)
‌=(P+‌

4T

c

)(‌

4

3

πc3)
‌‌ i.e., ‌4T(a2+b2−c2)=P(c3−a3−b3) . . . (iii)
‌‌ Now, ‌‌‌V=‌

4

3

π(a3+b3−c3)
‌‌ and ‌‌‌A=4π(a2+b2−c2)
‌∴‌‌‌

TA

π

=−‌

3

4π

VP
‌‌ or ‌‌‌4TA+3PV=0
‌