Question

In an aluminum $(Al)$ 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 $2.7 ? 10^{-8 }\, \Omega m$ and $1.0 ? 10^{-7}\, \Omega m$, respectively. The electrical resistance between the two faces $P$ and $Q$ of the composite bar is

• A. $\frac{2475}{64}\mu\Omega$
• B. $\frac{1875}{64}\mu\Omega$
• C. $\frac{1875}{49}\mu\Omega$
• D. $\frac{2475}{132}\mu\Omega$

Answer 1

Answer: (B)

$\frac{1875}{64}\mu\Omega$

Answer 2

For Aluminium
$R_{Al} = \frac{\rho A_{1}\ell}{A_{Al}} = \frac{2.7\times10^{-8}\times50\times10^{-3}}{\left(49-4\right)\times10^{-6}}$
$R_{Al }= 30 ? 10^{-6} O$
For iron Fe
$R_{Fe} = \frac{\rho_{Fe}\ell}{A_{Fe}}$
$R_{Fe} = \frac{10^{-7}\times50\times10^{-3}}{4\times10^{-6}}$
$R_{Fe} =1250 ?10^{-6} O$
$R_{e} =\frac{R_{Al}R_{Fe}}{R_{Al}+R_{Fe}}$
$R_{e} =\frac{30\times10^{-6}\times1250\times10^{-6}}{\left(30+1250\right)\times10^{-6}}$
$R_{e} = \frac{30\times1250}{1280}\times10^{-6}$
$R_{e} = \frac{1875}{85}\mu\Omega$