$(a)\;\large\frac{k\;q^2}{2a}\qquad(b)\;\large\frac{k\;q^2}{3a}\qquad(c)\;\large\frac{k\;q^2}{4a}\qquad(d)\;\large\frac{k\;q^2}{6a}$

Answer : (c) $\;\large\frac{k\;q^2}{4a}$

Explanation :

Capacitance of a spherical capacitor having inner radius of sphere a and radius of outer sphere b is :

$C=\large\frac{ab}{k\;(b-a)}$

As $\;b\;\to\;\infty \quad \; c=\large\frac{a}{k}$

And energy of capacitor = $\;\large\frac{q^2}{2C}$

Before closing the switch charge distribution is given by

Total energy of system = Energy of capacitor (1) + Energy of capacitor (2)

Capacitance of $\;1^{st}\;$ capacitor $\;C_{1}=\large\frac{a\times2a}{k\;(2a-a)}=\large\frac{2a}{k}$

Capacitance of $\;2^{nd}\;$ capacitor $\;C_{2}=\large\frac{3a}{k}$

Total energy = $\;\large\frac{q^2}{2(\large\frac{2a}{k})}+\large\frac{q^2}{2(\large\frac{3a}{k})}$

$T .E_{i}= \large\frac{k\;q^2}{4a}+\large\frac{k\;q^2}{6a}$

After closing the switch charge distribution is

Total energy = $\;0+\large\frac{q^2}{2(\large\frac{3a}{k})}$

$T . E_{f} = \large\frac{k\;q^2}{6a}$

$\bigtriangleup T .E = -\large\frac{k\;q^2}{4a} = T.E_{f} -T . E_{i}$

Thus heat produced = $+\large\frac{k\;q^2}{4a}\;.$

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