• # question_answer A solid metallic sphere of radius r is enclosed by a thin metallic shell of radius 2r, A charge q is given to the inner sphere- When the inner sphere is connected to the shell by a metal wire, the heat energy generated in it is given by A) $\frac{{{q}^{2}}}{\pi {{\varepsilon }_{0}}r}$ B) $\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}r}$ C) $\frac{{{q}^{2}}}{8\pi {{\varepsilon }_{0}}r}$ D)  $\frac{{{q}^{2}}}{16\pi {{\varepsilon }_{0}}r}$

 the capacitance of inner sphere is ${{C}_{1}}=4\pi {{\varepsilon }_{0}}r$ the capacitance of the outer shell is ${{C}_{2}}=4\pi {{\varepsilon }_{0}}(2r)=8\pi {{\varepsilon }_{0}}r$ Before connection, the total energy is ${{U}_{1}}=\frac{{{q}^{2}}}{2{{C}_{1}}}=\frac{{{q}^{2}}}{8\pi {{\varepsilon }_{0}}r}$
 After connection, the entire charge q of the inner sphere is transferred to the outer shell. Hence, energy after connection is ${{U}_{2}}=\frac{{{q}^{2}}}{2{{C}_{2}}}=\frac{{{q}^{2}}}{16\pi {{\varepsilon }_{0}}r}$$\therefore$Heat generated = ${{U}_{1}}-{{U}_{2}}$ $=\frac{{{q}^{2}}}{8\pi {{\varepsilon }_{0}}r}-\frac{{{q}^{2}}}{16\pi {{\varepsilon }_{0}}r}=\frac{{{q}^{2}}}{16\pi {{\varepsilon }_{0}}r}$