Question

Two rods of different metals have coefficients of linear expansion $\alpha_1$ and $\alpha_2$ respectively. Their respective lengths are $L_1$ and $L_2$. If at all temperatures $(L_2 - L_1)$ remains the same, the correct relation is

• A. $L_1 \alpha_1^2 = L_2 \alpha_2^2$
• B. $L_1^2 \alpha_1^2 = L_2^2 \alpha_2^2$
• C. $L_1 \alpha_2 = L_2 \alpha_1$
• D. $L_1 \alpha_1 = L_2 \alpha_2$

Hint:

To ensure that a physical gap or difference remains constant over time or temperature, both components must grow or shrink at identical absolute rates. Since growth rate depends on the product of initial length and its expansion coefficient ($L\alpha$), these products must be equalized directly: $L_1\alpha_1 = L_2\alpha_2$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question presents two rods of lengths $L_1$ and $L_2$ with different coefficients of linear expansion. We are given that the difference in their lengths $(L_2 - L_1)$ remains invariant regardless of temperature changes, and we need to determine the correct relationship between their lengths and expansion coefficients.

Step 2: Key Formula or Approach:
The change in length ($\Delta L$) of a material due to a temperature change $\Delta T$ is given by the linear expansion formula:
$$\Delta L = L \alpha \Delta T$$ For the difference between two lengths to remain completely constant at all temperatures, the change in length of the first rod must exactly equal the change in length of the second rod for any given variation in temperature:
$$\Delta L_1 = \Delta L_2$$

Step 3: Detailed Explanation:
Let's write out the individual linear expansion equations for both rods under a temperature change $\Delta T$:
$$\Delta L_1 = L_1 \alpha_1 \Delta T$$ $$\Delta L_2 = L_2 \alpha_2 \Delta T$$ Since the structural difference $(L_2 - L_1)$ is constant across all temperature scales, the differential expansion rate must satisfy:
$$\Delta (L_2 - L_1) = 0 \implies \Delta L_2 - \Delta L_1 = 0 \implies \Delta L_1 = \Delta L_2$$ Substitute our linear expansion expressions into this equality:
$$L_1 \alpha_1 \Delta T = L_2 \alpha_2 \Delta T$$ Since the change in temperature $\Delta T$ is non-zero and identical for both rods, we can divide both sides by $\Delta T$ to cancel it completely:
$$L_1 \alpha_1 = L_2 \alpha_2$$ This gives us the final direct algebraic relationship matching option (D).

Step 4: Final Answer:
The correct relation is $L_1 \alpha_1 = L_2 \alpha_2$, which corresponds to option (D).