JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Current Electricity

For a conductor, resistance $R=\rho \frac{l}{A}$, where $l$ is the length of current path and $A$ is the cross-sectional area. Between the two shaded opposite faces, current flows through thickness $t$, so $l=t$. The cross-sectional area is $A=L^2$. Therefore, $R=\rho \frac{t}{L^2}$. This shows $R$ is inversely proportional to $L^2$, not independent of $L$. However, examining the geometry: if the shaded faces are the top and bottom (area $L\times L$), then $l=t$ and $A=L^2$, giving $R=\rho t/L^2$. But the question asks about resistance between opposite faces shown as shaded. Looking at the standard interpretation: for a square sheet with current between opposite edges (not top/bottom), if we consider the sheet as having many parallel paths, the resistance becomes independent of $L$. For a square, $R=\rho /t$ (sheet resistance concept), which is independent of $L$. The answer is (C) independent of L.