Under isothermal conditions two soap bubbles of radii a and b coalesce to form a single bubble of radius c. If external pressure is P₀, the surface tension is equal to :

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$\frac{P_{0} (c^{3} - a^{3} - b^{3})}{4 (a^{2} + b^{2} - c^{2})}$

$\frac{P_{0} (a^{3} + b^{3} - c^{3})}{4 (a^{2} + b^{2} + c^{2})}$

$\frac{P_{0} (a^{3} + b^{3} - c^{3})}{4 (a^{2} - b^{2} - c^{2})}$

$\frac{4 P_{0} (a^{3} + b^{3} - c^{3})}{(a^{2} + b^{2} - c^{2})}$

answer is C.

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Detailed Solution

PV = P₁V₁ + P₂V₂

$(P_{0} + \frac{4 T}{R}) V = (P_{0} + \frac{4 T}{R_{1}}) V_{1} + (P_{0} + \frac{4 T}{R_{2}}) V_{2}$

$(P_{0} + \frac{4 T}{R}) \frac{4}{3} {πR}^{3} = (P_{0} + \frac{4 T}{R_{1}}) \frac{4}{3} {πR}_{1}^{3} + (P_{0} + \frac{4 T}{R_{2}}) \frac{4}{3} {πR}_{2}^{3}$

$P_{0} R^{3} + 4 {TR}^{2} = P_{0} R_{1}^{3} + 4 {TR}_{1}^{2} + P_{0} R_{2}^{3} + 4 {TR}_{2}^{2}$

$P_{0} (R^{3} - R_{1}^{3} - R_{2}^{3}) = 4 T (R_{1}^{2} + R_{2}^{2} - R^{2})$

$T = \frac{P_{0} (c^{3} - a^{3} - b^{3})}{4 (a^{2} + b^{2} - c^{2})}$

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