Question 913: Thermodynamics & KTG

Question

Two rods of different metals have coefficients of linear expansion '$$\alpha_1$$' and '$$\alpha_2$$' respectvely. Their respective lengths are '$$\mathrm{L}_1$$' and '$$\mathrm{L}_2$$'. At all temperatures ($$\mathrm{L}_2-\mathrm{L}_1$$) is same. The correct relation is

Accepted Answer

To solve this problem, we need to understand the relationship between the coefficients of linear expansion, initial lengths, and the condition given that ($$\mathrm{L}_2 - \mathrm{L}_1$$) is the same at all temperatures.
Linear expansion can be described by the formula for the change in length, $$\Delta \mathrm{L}$$, which is proportional to the initial length $$\mathrm{L}$$ and the change in temperature $$\Delta \mathrm{T}$$. For a given material, this can be written as:
$$\Delta \mathrm{L}_1 = \mathrm{L}_1 \alpha_1 \Delta \mathrm{T}$$
$$\Delta \mathrm{L}_2 = \mathrm{L}_2 \alpha_2 \Delta \mathrm{T}$$
Given that ($$\mathrm{L}_2 - \mathrm{L}_1$$) is the same at all temperatures, the difference in the change in lengths due to temperature changes should also be equal at all temperatures. This implies:
$$\Delta \mathrm{L}_2 - \Delta \mathrm{L}_1 = 0$$
Substitute the expressions for $$\Delta \mathrm{L}_1$$ and $$\Delta \mathrm{L}_2$$:
$$\mathrm{L}_2 \alpha_2 \Delta \mathrm{T} - \mathrm{L}_1 \alpha_1 \Delta \mathrm{T} = 0$$
Since $$\Delta \mathrm{T}$$ is common and non-zero, we can divide both sides by $$\Delta \mathrm{T}$$:
$$\mathrm{L}_2 \alpha_2 - \mathrm{L}_1 \alpha_1 = 0$$
Therefore, the correct relation is:
$$\mathrm{L}_1 \alpha_1 = \mathrm{L}_2 \alpha_2$$
So the correct option is:
Option D

Answer

$$\mathrm{L}_1 \alpha_1^2=\mathrm{L}_2 \alpha_2^2$$

Answer

$$\mathrm{L}_1^2 \alpha_1^2=\mathrm{L}_2^2 \alpha_2^2$$

Answer

$$\mathrm{L}_1 \alpha_2=\mathrm{L}_2 \alpha_1$$

Answer

$$\mathrm{L}_1 \alpha_1=\mathrm{L}_2 \alpha_2$$

Thermodynamics & KTG

Two rods of different metals have coefficients of linear expansion '$\alpha_1' and '\alpha_2' respectvely. Their respective lengths are '\mathrm{L}_1' and '\mathrm{L}_2'. At all temperatures (\mathrm{L}_2-\mathrm{L}_1$) is same. The correct relation is

Choose the correct answer: