Q.A negligibly small current is passed through a wire of length $15\ \text{m}$ and uniform cross-section $6.0 \times 10^{-7}\ \text{m}^2$, and its resistance is measured to be $5.0\ \Omega$. What is the resistivity of the material at the temperature of the experiment?

The key idea is that resistance depends on the material’s resistivity, length, and cross-sectional area through $R = \rho \frac{L}{A}$.

Step 1: Write the formula for resistance in terms of resistivity:

Step 2: Rearrange to solve for resistivity:

Step 3: Substitute the given values — $R = 5.0\ \Omega$, $A = 6.0 \times 10^{-7}\ \text{m}^2$, $L = 15\ \text{m}$: …

The key idea here is that resistance depends on both the material's intrinsic property (resistivity) and the geometry of the wire. When we pass a negligibly small current, we avoid heating the wire, so the measurement is at the ambient temperature — exactly what the question asks for.

Why this approach works:

For a uniform conductor, resistance $R$ is directly proportional to length $L$ and inversely proportional to cross-sectional area $A$. The constant of proportionality is the resistivity $\rho$, a material property that changes with temperature. Since the current is tiny, Joule heating is negligible, so the measured resistance corresponds to the temperature of the experiment. We simply plug into the formula.

Let’s work through it step by step.

1. Write the fundamental relation For a wire of uniform cross-section,

This is the standard formula — memorize it. Resistivity $\rho$ has units $\Omega \cdot \text{m}$.

2. Rearrange for resistivity Multiply both sides by $A$ and divide by $L$:

3. Substitute the given values

   • $R = 5.0\ \Omega$
   • $A = 6.0 \times 10^{-7}\ \text{m}^2$
   • $L = 15\ \text{m}$

   So

4. Simplify step by step First multiply numerator: $5.0 \times 6.0 \times 10^{-7} = 30 \times 10^{-7} = 3.0 \times 10^{-6}$ …

Method: Direct Application of the Resistivity Formula

This is a straightforward problem using the resistivity definition from Ohm's law in terms of material geometry.

Concept (Why this works)

Resistance $R$ of a wire depends on three things:

• Material (resistivity $\rho$)
• Length $L$ (longer → more resistance)
• Cross-sectional area $A$ (thicker → less resistance)

The relationship is:

We know $R$, $L$, and $A$ — so we can solve for $\rho$.

Steps

Step 1: Write the formula

Step 2: Substitute the given values

• $R = 5.0\ \Omega$
• $L = 15\ \text{m}$
• $A = 6.0 \times 10^{-7}\ \text{m}^2$

Step 3: Simplify

$\rho = 5.0 \times 0.4 \times 10^{-7}$ …

Here are the common mistakes students make on this exact problem, along with how to avoid each one.

Mistake 1: Forgetting to convert units

The error:

Students often plug in the numbers as given without checking if all units are in SI. Here, length is in meters ($15\ \text{m}$) and area is in $\text{m}^2$ ($6.0 \times 10^{-7}\ \text{m}^2$), so no conversion is needed. But in other problems, length might be in cm or area in $\text{mm}^2$, and forgetting to convert leads to a wrong resistivity.

How to avoid:

Always write the SI base units next to each quantity before substituting. If any value is not in SI (e.g., cm, mm²), convert it first. For this problem, confirm:

• $L = 15\ \text{m}$ ✓
• $A = 6.0 \times 10^{-7}\ \text{m}^2$ ✓
• $R = 5.0\ \Omega$ ✓

Mistake 2: Using the wrong formula

The error:

Some students confuse resistivity ($\rho$) with resistance ($R$) and write $R = \rho \frac{L}{A}$ but then solve incorrectly — e.g., they might write $\rho = R \cdot A \cdot L$ instead of $\rho = R \cdot \frac{A}{L}$.

How to avoid:

Memorise the definition:

Resistivity is the resistance of a unit cube of the material.

The formula is:

Rearranging:

Write the rearrangement step explicitly before plugging in numbers.

Mistake 3: Arithmetic errors with powers of ten

The error:

When substituting $A = 6.0 \times 10^{-7}$ and $L = 15$, students often mishandle the exponent — e.g., they might compute $10^{-7} / 15$ incorrectly or forget to multiply by $R$.

How to avoid:

Do the arithmetic in two steps:

1. Compute the numeric part: $5.0 \times \frac{6.0}{15} = 5.0 \times 0.4 = 2.0$
2. Attach the power of ten: $10^{-7}$

So:

Check:

$\frac{A}{L} = \frac{6.0 \times 10^{-7}}{15} = 0.4 \times 10^{-7} = 4.0 \times 10^{-8}$

Then $\rho = 5.0 \times 4.0 \times 10^{-8} = 20 \times 10^{-8} = 2.0 \times 10^{-7}$ ✓

Always re-check the exponent by estimating orders of magnitude.

Mistake 4: Ignoring the phrase “at the temperature of the experiment”

The error:

Students treat the result as the standard resistivity of the material (like from a table at 0°C or 20°C). But the problem clearly states the measurement is at the experiment’s temperature, which may not be room temperature. …