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(B) The formula for linear thermal expansion is given by $\Delta L = L \alpha \Delta T$, where $\Delta L$ is the change in length, $L$ is the original

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(B) The formula for linear thermal expansion is given by $\Delta L = L \alpha \Delta T$, where $\Delta L$ is the change in length, $L$ is the original length, $\alpha$ is the coefficient of linear expansion, and $\Delta T$ is the change in temperature.Given:$\Delta L = 5 	imes 10^{-5} \ m$$L = 1 \ m$$\alpha = 10 	imes 10^{-6} \ K^{-1}$Rearranging the formula to solve for $\Delta T$:$\Delta T = \frac{\Delta L}{L \alpha}$Substituting the values:$\Delta T = \frac{5 	imes 10^{-5}}{1 	imes 10 	imes 10^{-6}}$$\Delta T = \frac{5 	imes 10^{-5}}{10^{-5}}$$\Delta T = 5^{\circ} C$Therefore, the maximum temperature variation allowable is $5^{\circ} C$.

$A$ steel meter scale is to be ruled so that millimeter intervals are accurate within about $5 \times 10^{-5} \ m$ at a certain temperature. The maximum temperature variation allowable during the ruling is (Coefficient of linear expansion of steel $= 10 \times 10^{-6} \ K^{-1}$) (in $^{\circ} C$)

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