Question:
A metal tape is calibrated at \(25^\circ C\). On a cold day when the temperature is \(-15^\circ C\), the percentage error in the measurement of length is
Coefficient of linear expansion of metal \(=1\times10^{-5}\;^\circ C^{-1}\)
A metal tape is calibrated at \(25^\circ C\). On a cold day when the temperature is \(-15^\circ C\), the percentage error in the measurement of length is
Coefficient of linear expansion of metal \(=1\times10^{-5}\;^\circ C^{-1}\)
Coefficient of linear expansion of metal \(=1\times10^{-5}\;^\circ C^{-1}\)
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For thermal expansion problems, percentage change in length is given by
\[
\alpha \Delta T \times 100
\]
where \(\alpha\) is the coefficient of linear expansion.
Updated On: Jun 22, 2026
- \(0.04\%\)
- \(0.05\%\)
- \(0.1\%\)
- \(0.08\%\)
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The Correct Option is A
Solution and Explanation
Step 1: Find the change in temperature.
Initial calibration temperature: \[ T_1=25^\circ C \] Cold day temperature: \[ T_2=-15^\circ C \] Therefore, \[ \Delta T=T_2-T_1 \] \[ =-15-25 \] \[ =-40^\circ C \]
Step 2: Use linear expansion formula.
Fractional change in length is \[ \frac{\Delta L}{L}=\alpha \Delta T \] Given, \[ \alpha=1\times10^{-5}\;^\circ C^{-1} \] Hence, \[ \frac{\Delta L}{L} = 1\times10^{-5}\times(-40) \] \[ =-4\times10^{-4} \] Negative sign indicates contraction of the tape.
Step 3: Find percentage error.
Percentage error is \[ \left|\frac{\Delta L}{L}\right|\times100 \] \[ =4\times10^{-4}\times100 \] \[ =4\times10^{-2} \] \[ =0.04\% \]
Step 4: Final conclusion.
Hence, the percentage error in measurement of length is \[ \boxed{0.04\%} \]
Initial calibration temperature: \[ T_1=25^\circ C \] Cold day temperature: \[ T_2=-15^\circ C \] Therefore, \[ \Delta T=T_2-T_1 \] \[ =-15-25 \] \[ =-40^\circ C \]
Step 2: Use linear expansion formula.
Fractional change in length is \[ \frac{\Delta L}{L}=\alpha \Delta T \] Given, \[ \alpha=1\times10^{-5}\;^\circ C^{-1} \] Hence, \[ \frac{\Delta L}{L} = 1\times10^{-5}\times(-40) \] \[ =-4\times10^{-4} \] Negative sign indicates contraction of the tape.
Step 3: Find percentage error.
Percentage error is \[ \left|\frac{\Delta L}{L}\right|\times100 \] \[ =4\times10^{-4}\times100 \] \[ =4\times10^{-2} \] \[ =0.04\% \]
Step 4: Final conclusion.
Hence, the percentage error in measurement of length is \[ \boxed{0.04\%} \]
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