Two thin concentric conducting spherical shells of radii 0.1 m and 0.3 m carry charges +3 nC (inner) and -1 nC (outer) and are isolated. Use…

Before joining, $V_{in}=k\left(\dfrac{q}{a}+\dfrac{Q}{b}\right)=9\times10^{9}\left(\dfrac{3\times10^{-9}}{0.1}-\dfrac{10^{-9}}{0.3}\right)=240\ \mathrm{V}$, not $300\ \mathrm{V}$ (that ignores the negative outer charge), so (A) is wrong. Outer shell potential is $k(q+Q)/b=60\ \mathrm{V}$. Joining the shells makes them equipotential, and all charge of a conductor connected to an enclosing shell resides on the outer surface, so the inner shell is left with zero charge and the full $+3\ \mathrm{nC}$ moves outward; (B) is correct. The outer potential is still $k(2\ \mathrm{nC})/0.3=60\ \mathrm{V}$, unchanged, so (D) is wrong. Energy before is $\tfrac12(qV_{in}+QV_{out})=330\ \mathrm{nJ}$ and after is $\tfrac12(2\ \mathrm{nC})(60\ \mathrm{V})=60\ \mathrm{nJ}$, so the heat is $270\ \mathrm{nJ}$; $330\ \mathrm{nJ}$ is only the initial energy, so (C) is wrong. Correct option: B only.