Problem Statement
A metal rod of length $2.0\,\text{m}$ has a linear expansion coefficient $1.2\times10^{-5}\,\text{K}^{-1}$. By how much does it lengthen when its temperature rises by $50\,\text{K}$? (HC Verma, Chapter 23 — Heat and Temperature (Thermal Expansion), Problem 41.)
Given Information
- $L_0 = 2.0\,\text{m}$ — original length of the rod
- $\alpha = 1.2\times10^{-5}\,\text{K}^{-1}$ — coefficient of linear expansion
- $\Delta T = 50\,\text{K}$ — rise in temperature
Physical Concepts & Formulas
Solids expand when heated because the mean interatomic spacing grows with the amplitude of thermal vibration. The linear expansion is proportional to the original length and the temperature change through the coefficient of expansion.
- $\Delta L=\alpha L_0\,\Delta T$ — linear thermal expansion
- $\Delta V=\gamma V_0\,\Delta T$ — volume expansion, $\gamma\approx3\alpha$
- $L=L_0(1+\alpha\,\Delta T)$ — length at the new temperature
Step-by-Step Solution
Step 1 — Write the linear-expansion law: The change in length is proportional to the original length and the temperature change.
$$\Delta L=\alpha L_0\,\Delta T$$
Step 2 — Substitute the data: Insert the expansion coefficient, original length and temperature rise.
$$\Delta L=(1.2\times10^{-5})(2.0)(50)$$
Step 3 — Evaluate the expansion: Multiply the three factors to obtain the increase in length.
$$\Delta L=1.2\times10^{-3}\,\text{m}$$
Worked Calculation
$$\Delta L=(1.2\times10^{-5})(2.0)(50)=1.2\times10^{-3}\,\text{m}$$
Answer
$$\boxed{\Delta L=1.2\,\text{mm}}$$
A two-metre rod lengthens by just over a millimetre for a 50 K rise.
Physical Interpretation
Millimetre-scale expansion over metres of metal is exactly why bridges and rail tracks need expansion joints; ignoring it would buckle the structure on a hot day.
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