Question

The temperature of a metal strip having coefficient of linear expansion $\alpha$ is increased from $T_1$ to $T_2$ resulting in increase of its length by $\Delta L_1$. The temperature is further increased from $T_2$ to $T_3$ such that the increase in its length is $\Delta L_2$.
Given $T_3 + T_1 = 2T_2$ and $T_2 - T_1 = \Delta T$, the value of $\Delta L_2$ is ________.

• A. $\Delta L_1 [1 + 2\alpha^2(\Delta T)^2]$
• B. $\Delta L_1 [1 + \alpha^2(\Delta T)^2]$
• C. $\Delta L_1 [1 + 2\alpha\Delta T]$
• D. $\Delta L_1 [1 + \alpha\Delta T]$

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When an object is heated, its expansion is proportional to its *current* length prior to that specific heating phase. So the second expansion ($\Delta L_2$) will be slightly larger than the first ($\Delta L_1$) because it expands from the already elongated length $L_2$ rather than the original $L_1$.

Step 2: Key Formula or Approach:
Formula for linear expansion: $\Delta L = L_{initial} \cdot \alpha \cdot \Delta T$.

Step 3: Detailed Explanation:
We are given the relations for temperature:
$T_2 - T_1 = \Delta T$
$T_3 + T_1 = 2T_2 \implies T_3 - T_2 = T_2 - T_1 = \Delta T$.
So, the temperature increment is exactly $\Delta T$ in both stages.
Stage 1 (Expansion from $T_1$ to $T_2$):
Let initial length be $L_1$.
Increase in length: $\Delta L_1 = L_1 \alpha \Delta T$.
The new length is $L_2 = L_1 + \Delta L_1 = L_1(1 + \alpha \Delta T)$.
Stage 2 (Expansion from $T_2$ to $T_3$):
The strip expands from the new length $L_2$ under the exact same temperature difference $\Delta T$.
Increase in length: $\Delta L_2 = L_2 \alpha \Delta T$.
Substitute the expression for $L_2$:
$\Delta L_2 = [L_1(1 + \alpha \Delta T)] \cdot \alpha \Delta T$.
Rearranging terms:
$\Delta L_2 = (L_1 \alpha \Delta T) \cdot (1 + \alpha \Delta T)$.
Notice that the term in the first parenthesis is exactly $\Delta L_1$. Substituting this back:
$\Delta L_2 = \Delta L_1 (1 + \alpha \Delta T)$.

Step 4: Final Answer:
The value of $\Delta L_2$ is $\Delta L_1 [1 + \alpha\Delta T]$.