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

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Solution

$L_{s t} = L_{a l}$ at $20^{\circ}$

$α_{a l} = 2.3 \times 10^{- 5} /^{\circ} C$

$α_{s t} = 1.1 \times 10^{- 5} /^{\circ} C$

So $L_{0 s t} (1 - α_{s t} \times 20)$

$= L_{0 a l} (1 - α_{a l} \times 20)$

(a) $\frac{L_{0 s t}}{L_{0 a l}} = \frac{(1 - α_{a l} \times 20)}{(1 - α_{s t} \times 20)}$

$= \frac{1 - 2.3 \times 10^{- 5} \times 20}{1 - 1.1 \times 10^{- 5} \times 20}$

$= \frac{0.99954}{0.99978} = 0.999759$

(b) $\frac{L_{40 A l}}{L_{40 s t}} = \frac{L_{0 A l} (1 + α_{A l} \times 40)}{L_{0 s t} (1 + α_{s t} \times 40)}$

$\Rightarrow \frac{L_{40 A l}}{L_{40 s t}} = \frac{L_{0} A l}{L_{0 s t}} \times \frac{(1 + 2.3 \times 10^{- 5} \times 40)}{(1 + 1.1 \times 10^{- 5} \times 40)}$

$= \frac{0.99977 \times 1.00092}{1.00044} = 1.0002496$

(c) $\frac{L_{100 A l}}{L_{100 s t}} = \frac{L_{0 A l} (1 + α_{A l} \times 100)}{L_{0 s t} (1 + α_{s t} \times 100)}$

$= \frac{0.99977 \times 1.0023}{1.0023} = 1.00096$

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Two metre scales, one of steel and the other of aluminium, agree at 20∘C. Calculate the ratio aluminium-centimetre/steel-centimetre at (a) 0∘C, (b) 40∘C and (c) 100∘C. α for steel = 1.1×10−5 ∘C−1 and for aluminium = 2.3×10−5 ∘C−1.