Question

A steel rod of length \(5\,\text{m}\) and radius \(2\,\text{cm}\) is heated by \(10^\circ\text{C}\). Find the % change in volume \(\left(\alpha = 10 \times 10^{-6}\,^\circ\text{C}^{-1}\right)\):

• A. 0.01
• B. 0.03
• C. 0.9
• D. 1.2

Hint:

For solids, volumetric expansion coefficient = \(3\alpha\).

Answer 1

The Correct Option is B

Step 1: Derive the factor of 3 from first principles instead of quoting $\Delta V/V = 3\alpha\Delta T$ directly.
Think of a small cube of the rod's material with edge $L$. When heated by $\Delta T$, each of its three edges, length, width and height, expands by the same linear factor: \[ L' = L(1+\alpha\Delta T) \]
Step 2: Cube this to get the new volume.
\[ V' = (L')^3 = L^3(1+\alpha\Delta T)^3 \] Expanding $(1+\alpha\Delta T)^3$ with the binomial expansion and dropping the tiny squared and cubed terms, since $\alpha\Delta T$ is extremely small: \[ (1+\alpha\Delta T)^3 \approx 1 + 3\alpha\Delta T \]
Step 3: Read off the fractional change in volume.
\[ \frac{V'-V}{V} = \frac{\Delta V}{V} \approx 3\alpha\Delta T \] This shows directly why the factor of 3 appears, one contribution from each of the three dimensions expanding together.
Step 4: Substitute the given numbers.
\[ \alpha = 10\times10^{-6} \ ^{\circ}\text{C}^{-1}, \quad \Delta T = 10 \ ^{\circ}\text{C} \] \[ \frac{\Delta V}{V} = 3 \times 10\times10^{-6} \times 10 = 3\times10^{-4} \]
Step 5: Convert to a percentage.
\[ 3\times10^{-4} \times 100 = 0.03\% \]
Final Answer:
\[ \boxed{0.03\%} \]