Answer:
Explanation:
Given;
original length of steel, L₁ = 1.00 m
original length of invar, L₁ = 1.00 m
coefficients of volume expansion for steel, [tex]\gamma_{st.}[/tex] = 3.6 × 10⁻⁵ /°C
coefficients of volume expansion for invar, [tex]\gamma_{in.}[/tex] = 2.7 × 10⁻⁶ /°C
temperature rise in both meter stick, θ = 20.5°C
Difference in length, can be calculated as:
L₂ = L₁ (1 + αθ)
L₂ = L₁ + L₁αθ
L₂ - L₁ = L₁αθ
ΔL = L₁αθ
Where;
ΔL is difference in length
α is linear expansivity = [tex]\frac{\gamma}{3}[/tex]
Difference in length, for steel at 20.5°C:
ΔL = L₁αθ
Given;
L₁ = 1.00 m
θ = 20.5°C
[tex]\alpha = \frac{\gamma}{3} = \frac{3.6*10^{-5}}{3} = 1.2*10^{-5} /^oC[/tex]
ΔL = 1 x 1.2 x 10⁻⁵ x 20.5 = 2.46 x 10⁻⁴ m
Difference in length, for invar at 20.5°C:
ΔL = L₁αθ
Given;
L₁ = 1.00 m
θ = 20.5°C
[tex]\alpha = \frac{\gamma}{3} = \frac{2.7*10^{-6}}{3} = 0.9*10^{-6}/^oC[/tex]
ΔL = 1 x 0.9 x 10⁻⁶ x 20.5 = 1.845 x 10⁻⁵ m
