Question
A metre scale is made up of steel and measures correct length at $16^{\circ} \mathrm{C}$. What will be the percentage error if this scale is used:
a. On a summer day when the temperature is $46^{\circ} \mathrm{C}$.
b. On a winter day when the temperature is $6^{\circ} \mathrm{C}$ ? Coefficient of linear expansion of steel $=11 \times 10^{-60} \mathrm{C}^{-1}$.

Answer

  1. Length at $16^\circ C = L$
L = ?
$T_1 = 16^\circ C$
$T_2 = 46^\circ C$
$\alpha=1.1\times10^{-5/^\circ}\text{C}$
$\Delta\text{L}=\text{L}\alpha\Delta\theta=\text{L}\times1.1\times10^{-5}\times30$
$\%\ \text{of error}$ $=\Big(\frac{\Delta\text{L}}{\text{L}}\times100\Big)\%$
$=\Big(\frac{\text{L}\alpha\Delta\theta}{2}\times100\Big)\%$
$=1.1\times10^{-5}\times30\times100\%=0.33\%$
  1. $\text{T}_2=6^\circ\text{C}$
$\%\ \text{of error}$ $=\Big(\frac{\Delta\text{L}}{\text{L}\times100}\Big)\%$
$=\Big(\frac{\text{L}\alpha\Delta\theta}{\text{L}}\times100\Big)\%$
$=-1.1\times10^{-5}\times10\times100=-0.011\%$

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