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: Understand the condition.
Two rods of lengths $L_1$ and $L_2$ expand by different amounts when heated, yet the gap $(L_2 - L_1)$ must stay fixed at every temperature. We need the relation this forces between their lengths and expansion coefficients.
Step 2: Write the expansion of each rod.
For a temperature rise $\Delta T$, the increase in length is $\Delta L = L\alpha\,\Delta T$, so $$\Delta L_1 = L_1\alpha_1\,\Delta T,\qquad \Delta L_2 = L_2\alpha_2\,\Delta T.$$
Step 3: Translate the constancy condition.
If $(L_2 - L_1)$ never changes, then its change must be zero: $$\Delta L_2 - \Delta L_1 = 0.$$
Step 4: Set the expansions equal.
This means both rods must lengthen by the same amount: $$\Delta L_1 = \Delta L_2.$$
Step 5: Substitute the expressions.
$$L_1\alpha_1\,\Delta T = L_2\alpha_2\,\Delta T.$$
Step 6: Cancel the common factor.
Since $\Delta T$ is the same and non-zero for both, dividing it out gives $$L_1\alpha_1 = L_2\alpha_2.$$
\[ \boxed{L_1\alpha_1 = L_2\alpha_2} \]