A soap bubble of radius $r$ and surface tension T is present under atmospheric pressure $P_{0}$ . Its radius becomes double when a charge $q = 8 π r \sqrt{ε_{0} r (a P_{0} r + b T)}$ is uniformly distributed on its surface. The value of b-a is

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answer is D.

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Detailed Solution

From mole conversion $P_{1} V = P_{2} 8 V a s r_{2} = 2 r_{1}$
$\Rightarrow P_{2} = \frac{P_{1}}{8}$
Now, $P_{1} - P_{0} = \frac{4 T}{r} a n d P_{2} - P_{0} = \frac{4 T}{2 r} - \frac{σ^{2}}{2 ε_{0}}$
$\Rightarrow \frac{P_{1}}{8} - P_{0} = \frac{4 T}{2 r} - \frac{σ^{2}}{2 ε_{0}}$
Or, $\frac{1}{8} (P_{0} + \frac{4 T}{r}) - P_{0} = \frac{4 T}{2 r} - \frac{σ^{2}}{2 ε_{0}}$
$\Rightarrow \frac{7 P_{0}}{8} + \frac{3 T}{2 r} = \frac{σ^{2}}{2 ε_{0}} = \frac{q^{2}}{{(4 π {(2 r)}^{2})}^{2} 2 ε_{0}} q^{2} = 16 π^{2} \times 16 r^{4} \times 2 ε_{0} . (\frac{7 P_{0}}{8} + \frac{3 T}{2 r}) \Rightarrow q = 8 π r \sqrt{ε_{0} r (7 P_{0} r + 12 T)}$