The coefficients of thermal expansion of stee | vedclass

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(A) Let $\alpha_s = 12 	imes 10^{-6} /^{\circ}C$ be the coefficient of thermal expansion of steel and $\alpha_x = 2 	imes 10^{-6} /^{\circ}C$ be tha

Vedclass · Accepted Answer

$> 100 \, mm$

Vedclass · Answer

(A) Let $\alpha_s = 12 	imes 10^{-6} /^{\circ}C$ be the coefficient of thermal expansion of steel and $\alpha_x = 2 	imes 10^{-6} /^{\circ}C$ be that of metal $X$.At temperature $T = 40^{\circ}C$,the measured length $L_m = 100 \, mm$.The steel scale expands as temperature increases. The length of the scale at $40^{\circ}C$ is $L_s = L_0(1 + \alpha_s \Delta T)$.The cube of metal $X$ also expands: $L_x = L_{0x}(1 + \alpha_x \Delta T)$.The measured value is the ratio of the object length to the scale length: $L_m = L_x / (1 + \alpha_s \Delta T) = L_{0x}(1 + \alpha_x \Delta T) / (1 + \alpha_s \Delta T)$.Using the binomial approximation $(1+x)^{-1} \approx 1-x$,we get $L_m \approx L_{0x}(1 + \alpha_x \Delta T)(1 - \alpha_s \Delta T) \approx L_{0x}(1 + (\alpha_x - \alpha_s) \Delta T)$.Given $L_m = 100 \, mm$,$\Delta T = 40^{\circ}C$,$\alpha_x - \alpha_s = (2 - 12) 	imes 10^{-6} = -10 	imes 10^{-6} /^{\circ}C$.$100 = L_{0x}(1 - 10 	imes 10^{-6} 	imes 40) = L_{0x}(1 - 400 	imes 10^{-6}) = L_{0x}(1 - 0.0004)$.$L_{0x} = 100 / 0.9996 > 100 \, mm$.

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