Question

In an aluminum $\left(Al\right)$ bar of square cross section, a square hole is drilled and is filled with iron $\left(Fe\right)$ 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

The correct option is B

$\frac{1875}{64}\mu \Omega$
$R_{Al}$ and $R_{Fe}$ are parallel to each other.
where resistance of system is given by, $R=\frac{\rho l}{A}$

Resistance for $Al$:
$Area=7^2-2^2=45m{m}^2$
$\rho =2.7\times {10}^{-8}\Omega m$ (given)
$l=50mm$
So, $R=\frac{2.7\times {10}^{-8}\times 50\times {10}^{-3}}{45\times {10}^{-4}}=30\mu \Omega$

Resistance for $Fe$:
$Area=2^2=4m{m}^2$
$\rho =1.0\times {10}^{-7}\Omega m$ (given)
$l=50mm$
So, $R=\frac{1.0\times {10}^{-7}\times 50\times {10}^{-3}}{4\times {10}^{-4}}=1250\mu \Omega$

we already found,
$R_{Al}=30\mu \Omega$
$R_{Fe}=1250\mu \Omega$
due to both rods are joined parallelly then,
$R_{total}=\frac{R_{Al}{R}_{Fe}}{R_{Al}+R_{Fe}}\Rightarrow \frac{30\times 1250}{30+1250}\Rightarrow \frac{1875}{64}\mu \Omega$