A steel tape measures the length of a copper | vedclass

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(D) Let $L_0 = 90 \ cm$ be the initial length at $T_1 = 10 \ ^oC$. The change in temperature is $\Delta T = 30 - 10 = 20 \ ^oC$.The actual length of t

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$90.01$

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(D) Let $L_0 = 90 \ cm$ be the initial length at $T_1 = 10 \ ^oC$. The change in temperature is $\Delta T = 30 - 10 = 20 \ ^oC$.The actual length of the copper rod at $30 \ ^oC$ is $L_{cu} = L_0(1 + \alpha_{cu} \Delta T) = 90(1 + 1.7 	imes 10^{-5} 	imes 20) = 90(1 + 3.4 	imes 10^{-4}) = 90.0306 \ cm$.The steel tape also expands. The length of each $1 \ cm$ division on the tape at $30 \ ^oC$ becomes $L'_{st} = 1(1 + \alpha_{st} \Delta T) = 1(1 + 1.2 	imes 10^{-5} 	imes 20) = 1(1 + 2.4 	imes 10^{-4}) = 1.00024 \ cm$.The reading on the tape is the ratio of the actual length of the rod to the length of one division on the tape: $	ext{Reading} = \frac{L_{cu}}{L'_{st}} = \frac{90(1 + 3.4 	imes 10^{-4})}{1 + 2.4 	imes 10^{-4}} \approx 90(1 + 3.4 	imes 10^{-4})(1 - 2.4 	imes 10^{-4}) \approx 90(1 + 1.0 	imes 10^{-4}) = 90.009 \ cm \approx 90.01 \ cm$.

$A$ steel tape measures the length of a copper rod as $90 \ cm$ when both are at $10 \ ^oC$. What length (in $cm$) would the tape read for the length of the rod when both are at $30 \ ^oC$? $[\alpha_{st} = 1.2 \times 10^{-5} / ^oC$ and $\alpha_{cu} = 1.7 \times 10^{-5} / ^oC]$

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