Question:
A metallic bar of coefficient of linear expansion \(10^{-5}\,\text{K}^{-1}\) is heated from \(0^\circ C\) to \(100^\circ C\). The percentage increase in its length is
A metallic bar of coefficient of linear expansion \(10^{-5}\,\text{K}^{-1}\) is heated from \(0^\circ C\) to \(100^\circ C\). The percentage increase in its length is
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Always multiply fractional change by 100 to get percentage.
Updated On: May 8, 2026
- 0.1%
- 1%
- 10%
- 0.01%
- 0.001%
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The Correct Option is A
Solution and Explanation
Concept:
Linear expansion is given by:
\[
\Delta L = \alpha L \Delta T
\]
Percentage increase:
\[
\frac{\Delta L}{L} \times 100 = \alpha \Delta T \times 100
\]
Step 1: Identify given quantities. \[ \alpha = 10^{-5}\,\text{K}^{-1}, \quad \Delta T = 100^\circ C \]
Step 2: Substitute into formula. \[ \frac{\Delta L}{L} = 10^{-5} \times 100 = 10^{-3} \]
Step 3: Convert into percentage. \[ 10^{-3} \times 100 = 0.1% \]
Step 4: Conclusion. \[ \boxed{0.1%} \]
Step 1: Identify given quantities. \[ \alpha = 10^{-5}\,\text{K}^{-1}, \quad \Delta T = 100^\circ C \]
Step 2: Substitute into formula. \[ \frac{\Delta L}{L} = 10^{-5} \times 100 = 10^{-3} \]
Step 3: Convert into percentage. \[ 10^{-3} \times 100 = 0.1% \]
Step 4: Conclusion. \[ \boxed{0.1%} \]
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