Question

Two rods of length $l_{1}$ and $l_{2}$ are made of materials whose coefficients of linear expansion are $\alpha_{1}$and $\alpha_{2}$. If the difference between two lengths is independent of temperature then,

• A. $\frac{l_{1}}{l_{2}} = \frac{\alpha_{1}}{\alpha_{2}}$
• B. $\frac{l_{1}}{l_{2}} = \frac{\alpha_{2}}{\alpha_{1}}$
• C. $l_{2}^{2}\alpha_{1} = l_{1}^{2}\alpha_{2}$
• D. $\frac{\alpha_{1}^{2}}{l_{1}} = \frac{\alpha_{2}^{2}}{l_{2}}$

Answer 1

Answer: (A)

$\frac{l_{1}}{l_{2}} = \frac{\alpha_{1}}{\alpha_{2}}$

Answer 2

f change in length is $\Delta l$

Then ∆I₁ = L₀₁ α₁∆T

∆I₂ = L₀₂ α₂∆T

difference in length is $l_{2} - l_{1} = \left( L_{02} + \Delta l_{2} \right) - \left( L_{01} + \Delta l_{1} \right)$

=(L₀₂ − L₀₁) + (∆I₂ − ∆I₁)

L₀₂ − L₀₁ is independent of temperature for I₂ − I₁ to be independent of temperature.

∆I₂ − ∆I₁ must be equal to zero

i.e. L₁ α₁∆T = L₂ α₂∆T

i.e. L₁ α₁ = L₂ α₂

i.e. $\frac{L_{1}}{L_{2}} = \frac{\alpha_{2}}{\alpha_{1}}$.