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.

Answer 1

Given:
At 20°C, length of the metre scale made up of steel, L_st= length of the metre scale made up of aluminium, L_al​
Coefficient of linear expansion for aluminium, α_al = 2.3 × 10^–5 °C^-1
Coefficient of linear expansion for steel, α_st = 1.1 × 10^–5 °C^-1
Let the length of the aluminium scale at 0°C, 40°C and 100°C be L_0al​, L​_40al and L_10​0al.
And let the length of the steel scale at 0°C, 40°C and 100°C be L_0st​, L​_40st and L_10​0st.
(a) So, L_0st(1 – α_st × 20) = L_0al(1 – α_al × 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)  \[\frac{L_{100 al}}{L_{100 st}} = \frac{L_{0 al} \left( 1 + \alpha {}_{al} \times 100 \right)}{L_{0 st} \left( 1 + \alpha_{st} \times 100 \right)}\]

\[ \Rightarrow \frac{L_{100 al}}{L_{100 st}} = \frac{0 . 99977 \times 1 . 0023}{1 . 0023}\]

\[ \Rightarrow \frac{L_{100 al}}{L_{100 st}} = 1 . 00096\]