Two metre scales, one of steel and the other of aluminium, agree … - Vidyadip

Question

Two metre scales, one of steel and the other of aluminium, agree at $20^{\circ} \mathrm{C}$. Calculate the ratio aluminium-centimetre/ steel-centimetre at:
a. $0^{\circ} \mathrm{C}$,
b. $40^{\circ} \mathrm{C}$
c. $100 \mathrm{C} . \alpha$ for steel $=1.1 \times 10^{-5}{ }^{\circ} \mathrm{C}{ }^{-1}$ and for aluminium $=2.3 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$.

✓

Answer

$\text{L}_{\text{s}\text{t}}=\text{L}_{\text{Al}}\ \text{at}\ 20^\circ\text{C}$$\alpha_\text{Al}=2.3\times10^{-5}/^\circ\text{C}$
so, $\text{Lo}_{\text{st}}(1-\alpha_{\text{st}}\times20)=\text{Lo}_\text{Al}(1-\alpha_{\text{Al}}\times20)$
$\alpha_\text{st}=1.1\times10^{-5}/^\circ\text{C}$

$\frac{\text{Lo}_\text{st}}{\text{Lo}_\text{Al}}=\frac{(1-\alpha_\text{Al}\times20)}{(1-\alpha_\text{st}\times20)}$

$=\frac{1-2.3\times10^{-5}\times20}{1-1.1\times10^{-5}\times20}$
$=\frac{0.99954}{0.99978}=0.999$

$\frac{\text{Lo}_{40\text{st}}}{\text{Lo}_{40\text{Al}}}=\frac{(1-\alpha_\text{Al}\times40)}{(1-\alpha_\text{st}\times40)}$

$=\frac{1-2.3\times10^{-5}\times20}{1-1.1\times10^{-5}\times20}$
$=\frac{0.99954}{0.99978}=0.999$
$=\frac{\text{Lo}_\text{Al}}{\text{Lo}_\text{st}}\times\frac{1+2.3\times10^{-5}\times10}{273}$
$=\frac{0.99977\times1.00092}{1.00044}$
$=1.0002496\approx1.00025$
$\frac{\text{Lo}_\text{100Al}}{\text{Lo}_\text{100st}}=\frac{(1+\alpha_\text{Al}\times100)}{(1+\alpha_\text{st}\times100)}$
$=\frac{0.99977\times1.00092}{1.00011}=1.00096$

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