A steel scale measures the length of a copper | vedclass

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(A) Let the initial length be $L_0 = 80.0\,cm$ at $T_1 = 20^\circ C$. The temperature change is $\Delta T = 40^\circ C - 20^\circ C = 20^\circ C$.The

Vedclass · Accepted Answer

$80.0096$

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(A) Let the initial length be $L_0 = 80.0\,cm$ at $T_1 = 20^\circ C$. The temperature change is $\Delta T = 40^\circ C - 20^\circ C = 20^\circ C$.The actual length of the copper wire at $40^\circ C$ is $L_{Cu} = L_0(1 + \alpha_{Cu} \Delta T)$.The length marked on the steel scale at $40^\circ C$ is $L_{steel} = L_0(1 + \alpha_{steel} \Delta T)$.The scale reading is the ratio of the actual length of the wire to the length of a unit division on the scale at $40^\circ C$. Since the scale itself expands,the reading $L'$ is given by:$L' = \frac{L_{Cu}}{1 + \alpha_{steel} \Delta T} = \frac{L_0(1 + \alpha_{Cu} \Delta T)}{1 + \alpha_{steel} \Delta T}$.Using the binomial approximation $(1+x)^{-1} \approx 1-x$ for small $\alpha \Delta T$:$L' \approx L_0(1 + \alpha_{Cu} \Delta T)(1 - \alpha_{steel} \Delta T) \approx L_0(1 + (\alpha_{Cu} - \alpha_{steel}) \Delta T)$.Substituting the values:$L' = 80.0 	imes (1 + (17 	imes 10^{-6} - 11 	imes 10^{-6}) 	imes 20)$$L' = 80.0 	imes (1 + 6 	imes 10^{-6} 	imes 20) = 80.0 	imes (1 + 120 	imes 10^{-6})$$L' = 80.0 + 80.0 	imes 0.00012 = 80.0 + 0.0096 = 80.0096\,cm$.

$A$ steel scale measures the length of a copper wire as $80.0\,cm$ when both are at $20^\circ C$ (the calibration temperature for the scale). What would be the scale reading for the length of the wire when both are at $40^\circ C$ (in $,cm$)?

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