Question

A thin rod having length L₀ at 0ºC and coefficient of linear expansion α has its two ends maintained at temperatures θ₁and θ₂ respectively. Find its new length.

Answer 1

The rod’s temperature varies from θ₁to θ₂ [from one and to another].

So, mean temperature of rod is θ = \(\frac{\theta_1+\theta_2}{2}\)(at C)

So, rate of flow heat A → C and C → B are equal.

θ₁ > θ > θ₂

Therefore, \(\frac{d\theta}{dt}=\frac{KA(\theta_1-\theta)}{\frac{L_0}{2}}= \frac{KA(\theta-\theta_2)}{\frac{L_0}{2}}\)

Here, K = Coefficient of thermal conductivity.

∴ θ₁ − θ = θ − θ₂

θ = \(\frac{\theta_1+\theta_2}{2}\)

Using relation. L = L₀(1 + αθ)

Thus,L = L₀ [1 + α ( \(\frac{\theta_1+\theta_2}{2}\) )]