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 $2.7\times 10^{-8}\Omega m$ and $1.0\times 10^{-7}\Omega m$, 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)

For Aluminium
$R_{Al}=\frac{\rho A_1\ell }{A_{Al}}=\frac{2.7\times 10^{-8}\times 50\times 10^{-3}}{\left(49-4\right)\times 10^{-6}}$
$R_{Al}=30\times 10^{-6}\Omega$
For iron Fe
$R_{Fe}=\frac{{\rho }_{Fe}\ell }{A_{Fe}}$
$R_{Fe}=\frac{10^{-7}\times 50\times 10^{-3}}{4\times 10^{-6}}$
$R_{Fe}=1250\times 10^{-6}\Omega$
$R_{eq.}=\frac{R_{Al}{R}_{Fe}}{R_{Al}+R_{Fe}}$
$R_{eq.}=\frac{30\times 10^{-6}\times 1250\times 10^{-6}}{\left(30+1250\right)\times 10^{-6}}$
$R_{eq.}=\frac{30\times 1250}{1280}\times 10^{-6}$
$R_{eq.}=\frac{1875}{85}\mu \Omega$