Animated Solution for Physics - Current Electricity: 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 $2.7\times 10^{-8}\Omega \text{m}$ and $1.0\times 10^{-7}\Omega \text{m}$, respectively. The electrical resistance between the two faces P and Q of the composite bar is

The journey to solving this problem begins with a fundamental realization about how current flows through composite materials. Imagine you are standing at face P, and you are an electron trying to make your way to face Q. You have two distinct paths you can take: you can either travel through the dense iron core, or you can travel through the surrounding aluminum shell.

Because both of these paths start at the exact same surface (face P) and end at the exact same surface (face Q), they share the exact same potential difference. This is the hallmark of a parallel circuit.

Before we can calculate the equivalent resistance, we need to understand the physical dimensions of our two parallel resistors. The problem gives us the outer dimensions of the bar and the inner dimensions of the hole.

The total cross-sectional area of the square face is simply the outer width squared:

The iron core occupies the square hole in the center, so its area is:

Now, the aluminum shell is everything that isn't the iron core. So, we subtract the core's area from the total area:

It is absolutely critical to convert these areas into standard SI units (square meters) to match the units of resistivity ($\Omega \text{m}$). Since $1\text{ mm}=10^{-3}\text{ m}$, it follows that $1{\text{ mm}}^2=10^{-6}{\text{ m}}^2$.

The resistance of any uniform conductor is given by the classic formula:

Let's apply this to our aluminum shell first. We substitute its specific resistivity, the length of the bar ($50\text{ mm}$), and the area we just calculated:

Simplifying the powers of 10 and the fractions:

Next, we do the exact same thing for the iron core:

Notice how much higher the resistance of the iron core is! Even though iron is a decent conductor, its smaller cross-sectional area and higher resistivity make it a much tougher path for the electrons compared to the thick aluminum shell.

Now that we have our two individual resistances, we can find the equivalent resistance of the entire composite bar using the parallel combination formula:

Substituting our values:

To add these fractions, we find a common denominator:

Finally, we take the reciprocal to find $R_{\text{eq}}$:

By canceling out the trailing zeros and dividing the numerator and denominator by 2, we arrive at our final, elegant answer:

This perfectly matches option (b). The beauty of this problem lies in recognizing the parallel nature of the physical setup and carefully managing the geometric subtractions and unit conversions.