JEE Main 2008 — Current Electricity Question with Solution

Consider a block of conducting material of resistivity $${}^{\prime }{\rho }^{\prime }$$ shown in the figure. Current $${}^{\prime }{I}^{\prime }$$ enters at $${}^{\prime }{A}^{\prime }$$ and leaves from $${}^{\prime }{D}^{\prime }$$. We apply superposition principle to find voltage $${}^{\prime }\Delta V^{\prime }$$ developed between $${}^{\prime }{B}^{\prime }$$ and $${}^{\prime }{C}^{\prime }$$. The calculation is done in the following steps:
(i) Take current $${}^{\prime }{I}^{\prime }$$ entering from $${}^{\prime }{A}^{\prime }$$ and assume it to spread over a hemispherical surface in the block.
(ii) Calculate field $$E(r)$$ at distance $${}^{\prime }{r}^{\prime }$$ from A by using Ohm's law $$E=\rho j,$$ where $$j$$ is the current per unit area at $${}^{\prime }{r}^{\prime }$$.
(iii) From the $${}^{\prime }{r}^{\prime }$$ dependence of $$E(r)$$, obtain the potential $$V(r)$$ at $$r$$.
(iv) Repeat (i), (ii) and (iii) for current $${}^{\prime }{I}^{\prime }$$ leaving $${}^{\prime }{D}^{\prime }$$ and superpose results for $${}^{\prime }{A}^{\prime }$$ and $${}^{\prime }{D}^{\prime }.$$

$$\Delta V$$ measured between $$B$$ and $$C$$ is