Explanation

(d) Let radii of two soap bubble are a and b respectively and radius of single larger bubble is c .
As excess pressure for a soap bubble is $\frac{4T}{r}$ and external pressure p
$p_{i} = p + \frac{4T}{r}$
$So,\ p_{a} = p + \frac{4T}{a},\ \ \ \ p_{b} = p + \frac{4T}{b}$
$and\ \ \ p_{c} = p + \frac{4T}{c}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots..(i)$
$and\ \ \ V_{a} = \frac{4}{3}\pi a^{3},\ \ \ \ V_{b} = \frac{4}{3}\pi b^{3}$
$and\ \ \ V_{c} = \frac{4}{3}\pi c^{3\ }\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots\ldots.(ii)$
Now as mass is conserved.
$\mu_{a} + \mu_{b} = \mu_{c}$
$ie,\ \frac{p_{a}V_{a}}{R\ T_{a}} + \frac{p_{b}V_{b}}{R\ T_{b}} + \frac{p_{c}V_{c}}{R\ T_{c}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (as\ p\ V = \mu R\ T)$
As temperature is constant,
$ie,\ T_{a} = T_{b} = T_{c}$
$So,\ p_{a}V_{a} + p_{b}V_{b} = p_{c}V_{c}$
Which in the light of Eqs. (i) and (ii) becomes,
$\left( p + \frac{4T}{a} \right)\ \left( \frac{4}{3}\pi a^{3} \right) + \left( p + \frac{4T}{b} \right)\ \left( \frac{4}{3}\pi b^{3} \right)$
$\ \ = \left( p + \frac{4T}{c} \right)\ \left( \frac{4}{3}\pi c^{3} \right)$
$ie,\ \ \ \ 4T\left( a^{2} + b^{2} - c^{2} \right) = p\left( c^{3} - a^{3} - b^{3} \right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots\ldots.(iii)$
$Now,\ V = \frac{4}{3}\pi\left( a^{3} + b^{3} - c^{3} \right)$
$and\ \ A = 4\pi\left( a^{2} + b^{2} - c^{2} \right)$
$\therefore\ \ \frac{TA}{\pi} = - \frac{3}{4\pi} = Vp$
$or\ \ 4TA + 3pV = 0$