Discussion Forum

 In an aluminium (Al) bar of square cross-section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are  $2.7\times 10^{-8}$ Ω m and  $10\times 10^{-7}$ Ω m, respectively. The electrical  resistance between  the two faces P and Q of the composite bar is

Answer:

Explanation:

  $\frac{1}{R}=\frac{1}{R_{Al}}+\frac{1}{R_{Fe}}=\left(\frac{A_{Al}}{\rho_{Al}}+\frac{A_{Fe}}{\rho_{Fe}}\right)\frac{1}{l}$

   $= \left[\frac{(7^{2}-2^{2})}{2.7}+\frac{2^{2}}{10}\right]\frac{10^{-6}}{10^{-8}}\times\frac{1}{50\times 10^{-3}}$

  Solving we get,

            $R=\frac{1875}{64}\times 10^{-6}$Ω

              $=\frac{1875}{64}\mu$ Ω