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A steel scale measures the length of a copper wire as 80.0 cm, when both are at $20^{\circ} \mathrm{C}$, the calibration temperature for the scale. What would the scale read for the length of the rod when both are at $40^{\circ} \mathrm{C}$ ?
Given: $\alpha$ for steel $=11 \times 10^{-6}$ per ${ }^{\circ} \mathrm{C}$ and $\alpha$ for $\mathrm{Cu}=17 \times 10^{-6}$ per $^{\circ} \mathrm{C}$
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Given: $\alpha$ for steel $=11 \times 10^{-6}$ per ${ }^{\circ} \mathrm{C}$ and $\alpha$ for $\mathrm{Cu}=17 \times 10^{-6}$ per $^{\circ} \mathrm{C}$
Solution:
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Verified Answer
The correct answer is:
80.0096 cm
Using the relation $l_t=l_0(1+\alpha t)$
$\begin{aligned} & =1 \times\left[1+11 \times 10^{-6} \times\left(40^{\circ}-20^{\circ}\right)\right] \\ & =1.00022 \mathrm{~cm}\end{aligned}$
Now length of copper rod at $40^{\circ} \mathrm{C}$
$\begin{aligned} R_t{ }^{\prime} & =l_0{ }^{\prime}\left(1+\alpha^{\prime} t\right) \\ & =80\left[1+17 \times 10^{-6}\left(40^{\circ}-20^{\circ}\right)\right] \\ & =80.0272 \mathrm{~cm}\end{aligned}$
Now number of cms observed on the scale
$=\frac{80.0272}{1.00022}=80.0096$
$\begin{aligned} & =1 \times\left[1+11 \times 10^{-6} \times\left(40^{\circ}-20^{\circ}\right)\right] \\ & =1.00022 \mathrm{~cm}\end{aligned}$
Now length of copper rod at $40^{\circ} \mathrm{C}$
$\begin{aligned} R_t{ }^{\prime} & =l_0{ }^{\prime}\left(1+\alpha^{\prime} t\right) \\ & =80\left[1+17 \times 10^{-6}\left(40^{\circ}-20^{\circ}\right)\right] \\ & =80.0272 \mathrm{~cm}\end{aligned}$
Now number of cms observed on the scale
$=\frac{80.0272}{1.00022}=80.0096$
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