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Question
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.
Solution
Given:
At 20°C, length of the metre scale made up of steel, Lst= length of the metre scale made up of aluminium, Lal
Coefficient of linear expansion for aluminium, αal = 2.3 × 10–5 °C-1
Coefficient of linear expansion for steel, αst = 1.1 × 10–5 °C-1
Let the length of the aluminium scale at 0°C, 40°C and 100°C be L0al, L40al and L100al.
And let the length of the steel scale at 0°C, 40°C and 100°C be L0st, L40st and L100st.
(a) So, L0st(1 – αst × 20) = L0al(1 – αal × 20)
\[\frac{L_{0st}}{L_{0al}} = \frac{\left( 1 - \alpha_{al} \times 20 \right)}{\left( 1 - \alpha_{st} \times 20 \right)}\]
\[ \Rightarrow \frac{L_{0st}}{L_{0al}} = \frac{1 - 2 . 3 \times {{10}^-}^5 \times 20}{1 - 1 . 1 \times {10}^{- 5} \times 20}\]
\[ \Rightarrow \frac{L_{0st}}{L_{0al}} = \frac{0 . 99954}{0 . 99978}\]
\[ \Rightarrow \frac{L_{0st}}{L_{0al}} = 0 . 999759\]
(b) \[\frac{L_{40 al}}{L_{40 st}} = \frac{L_{0 al} \left( 1 + \alpha {}_{al} \times 40 \right)}{L_{0 st} \left( 1 + \alpha_{st} \times 40 \right)}\]
\[ \Rightarrow \frac{L_{40 al}}{L_{40 st}} = \frac{L_{0 al}}{L_{0 st}} \times \frac{\left( 1 + 2 . 3 \times {10}^{- 5} \times 40 \right)}{\left( 1 + 1 . 1 \times {10}^{- 5} \times 40 \right)}\]
\[ \Rightarrow \frac{L_{40 al}}{L_{40 st}} = \frac{0 . 99977 \times 1 . 00092}{1 . 00044}\]
\[ \Rightarrow \frac{L_{40 al}}{L_{40 st}} = 1 . 0002496\]
(c) \[\frac{L_{100 al}}{L_{100 st}} = \frac{L_{0 al} \left( 1 + \alpha {}_{al} \times 100 \right)}{L_{0 st} \left( 1 + \alpha_{st} \times 100 \right)}\]
\[ \Rightarrow \frac{L_{100 al}}{L_{100 st}} = \frac{0 . 99977 \times 1 . 0023}{1 . 0023}\]
\[ \Rightarrow \frac{L_{100 al}}{L_{100 st}} = 1 . 00096\]
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