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A steel tape measures the length of a copper rod as $90\, cm$ when both are at $10\,^oC$. ........ $cm$ would the tape read for the length of rod when both are at $30\,^oC$ ? $[\alpha _{st} = 1.2\times10^{-5}/^oC$ and $\alpha _{cu} = 1.7\times10^{-5}/^oC]$
$89$
$90.21$
$89.80$
$90.01$
Solution
$\mathrm{L}_{\mathrm{c}}^{\prime}=\mathrm{L}_{\mathrm{c}}(1+\alpha \Delta \mathrm{T})$
$=90\left(1+1.7 \times 10^{-5} \times 20\right)$
$\mathrm{L}_{\text {steal }}=(1 \mathrm{cm})\left[1+\alpha_{\mathrm{st}} \Delta \mathrm{T}\right]$
[new per division length]
$=1\left[1+1.2 \times 10^{-5} \times 50\right]$
$\frac{\mathrm{L}_{\mathrm{C}}}{\mathrm{L}_{\mathrm{St}}}=\frac{90\left(1+1.7 \times 10^{-5} \times 20\right)}{1\left(1+1.2 \times 10^{-5} \times 20\right)}=90.01 \mathrm{cm}$