Question:
A steel rod of length 5 m and radius 2 cm is kept at room temperature and heated to 10°C above room temperature. The change in the cross-sectional area of the rod is (coefficient of linear expansion = \(10 \times 10^{-6} \, ^\circ\text{C}^{-1}\)).
A steel rod of length 5 m and radius 2 cm is kept at room temperature and heated to 10°C above room temperature. The change in the cross-sectional area of the rod is (coefficient of linear expansion = \(10 \times 10^{-6} \, ^\circ\text{C}^{-1}\)).
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For solids: linear expansion = αΔT, area expansion = 2αΔT, volume expansion = 3αΔT.
Updated On: Jun 20, 2026
- 0.01%
- 0.02%
- 0.03%
- 0.09%
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The Correct Option is B
Solution and Explanation
Step 1: Concept of area expansion.
For isotropic solids, fractional change in area is: \[ \frac{\Delta A}{A} = 2\alpha \Delta T \]
Step 2: Substitute values.
\[ \alpha = 10 \times 10^{-6}, \quad \Delta T = 10 \] \[ \frac{\Delta A}{A} = 2 \times 10 \times 10 \times 10^{-6} \] \[ = 200 \times 10^{-6} \] \[ = 2 \times 10^{-4} \]
Step 3: Convert into percentage.
\[ 2 \times 10^{-4} \times 100 = 2 \times 10^{-2} = 0.02\% \]
Step 4: Final interpretation.
Area increases slightly due to thermal expansion in both perpendicular directions.
\[ \boxed{0.02\%} \]
For isotropic solids, fractional change in area is: \[ \frac{\Delta A}{A} = 2\alpha \Delta T \]
Step 2: Substitute values.
\[ \alpha = 10 \times 10^{-6}, \quad \Delta T = 10 \] \[ \frac{\Delta A}{A} = 2 \times 10 \times 10 \times 10^{-6} \] \[ = 200 \times 10^{-6} \] \[ = 2 \times 10^{-4} \]
Step 3: Convert into percentage.
\[ 2 \times 10^{-4} \times 100 = 2 \times 10^{-2} = 0.02\% \]
Step 4: Final interpretation.
Area increases slightly due to thermal expansion in both perpendicular directions.
\[ \boxed{0.02\%} \]
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