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(C) Let $q_1$ and $q_2$ be the charges on the spheres of radii $r$ and $R$ respectively.Given $q_1 + q_2 = Q$.Since surface charge densities are equal

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$\frac{1}{4\pi \varepsilon_0} \frac{(R + r)Q}{(R^2 + r^2)}$

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(C) Let $q_1$ and $q_2$ be the charges on the spheres of radii $r$ and $R$ respectively.Given $q_1 + q_2 = Q$.Since surface charge densities are equal,$\sigma_1 = \sigma_2$.$\frac{q_1}{4\pi r^2} = \frac{q_2}{4\pi R^2} \implies \frac{q_1}{r^2} = \frac{q_2}{R^2}$.Using componendo and dividendo,$\frac{q_1}{r^2} = \frac{q_2}{R^2} = \frac{q_1 + q_2}{r^2 + R^2} = \frac{Q}{r^2 + R^2}$.Thus,$q_1 = \frac{Q r^2}{R^2 + r^2}$ and $q_2 = \frac{Q R^2}{R^2 + r^2}$.The electric potential at the common centre is $V = \frac{1}{4\pi \varepsilon_0} \left( \frac{q_1}{r} + \frac{q_2}{R} \right)$.Substituting the values of $q_1$ and $q_2$:$V = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q r^2}{r(R^2 + r^2)} + \frac{Q R^2}{R(R^2 + r^2)} \right) = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q r + Q R}{R^2 + r^2} \right)$.$V = \frac{1}{4\pi \varepsilon_0} \frac{(R + r)Q}{R^2 + r^2}$.

$A$ charge of total amount $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R$ $(R > r)$ such that the surface charge densities on the two spheres are equal. The electric potential at the common centre is

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