A copper rod, 5m long, when heated through 10ºC, expands by 1mm. If a second copper rod, 2.5m long, is heated through 5ºC, by how much will it expand?

1. Analyze the Question

We are given the initial length ($L_1$), the temperature change ($\Delta\theta_1$), and the expansion ($\Delta L_1$) for a copper rod. We need to find the expansion ($\Delta L_2$) for a second copper rod with a different length ($L_2$) and a different temperature change ($\Delta\theta_2$).

2. Explain the Concept

The linear expansion of a material is governed by the formula:

Where:

• $\Delta L$ = Change in length (expansion)
• $\alpha$ = Coefficient of linear expansion (constant for the same material)
• $L$ = Original length
• $\Delta\theta$ = Change in temperature

Since both rods are made of copper, the coefficient of linear expansion ($\alpha$) is the same for both. We can set up a ratio:

3. Show the Solution

Given for Rod 1:

• $L_1 = 5\text{ m}$
• $\Delta\theta_1 = 10^\circ\text{C}$
• $\Delta L_1 = 1\text{ mm}$

Given for Rod 2:

• $L_2 = 2.5\text{ m}$
• $\Delta\theta_2 = 5^\circ\text{C}$
• $\Delta L_2 = ?$

Step-by-step calculation: Using the ratio formula:

Substitute the values:

Simplify the fractions:

4. Identify the Answer

The expansion of the second rod is 0.25mm. This matches option B.

5. Exam Tip

When dealing with expansion problems, always check if the material is the same. If it is, $\alpha$ cancels out, and you can simply use the proportionality relationship $\Delta L \propto L \cdot \Delta\theta$. This saves you from having to calculate the actual value of $\alpha$, which is often a very small, messy decimal!