Resistivity

Worked out. A wire of resistance $20\ \Omega$ is stretched uniformly to 8 times its original length, and the new resistance is required. Since stretching keeps the wire's volume constant, $A_1l_1 = A_2l_2$ with $l_2 = 8l_1$ gives $A_1/A_2 = 8$. Now $R_1 = \rho l_1/A_1$ and $R_2 = \rho l_2/A_2 = \rho(8l_1)/A_2$, so dividing gives $R_2/R_1 = 8\times(A_1/A_2) = 8\times8 = 64$. Hence $R_2 = 64\times20 = 1280\ \Omega$. The lesson: because resistance depends on BOTH the increasing length and the shrinking area, stretching a wire increases its resistance far more steeply (as the square of the stretch …

Worked out. A rectangular metal block of height A, width B and length C can have a potential difference V applied either between its two AB faces (current $I_{AB}$ flows along the C-direction) or between its two BC faces (current $I_{BC}$ flows along the A-direction); the question asks for $I_{BC}$ in terms of $I_{AB}$. The resistance in the first case is $R_{AB} = \rho C/(AB)$, giving $I_{AB} = V\!AB/(\rho C)$. The resistance in the second case is $R_{BC} = \rho A/(BC)$, giving $I_{BC} = V\!BC/(\rho A)$. Dividing the second by the first, after multiplying and dividing by AC to compare them properly, gives $I_{BC} = (C/A)^2\, I_{AB}$. Since $C > A$ for a typical block, $I_{BC} > I_{AB}$: the same block conducts more current when measured across its shorter dimension, because that configuration has a shorter current pa …