Question

A rod of length L is heated from temperature \( T_1 \) to \( T_2 \). Let \( T_2 - T_1 = \Delta T \) and the expansion of the rod be \( \Delta L_1 \). The rod is further heated from \( T_2 \) to \( T_3 \) such that \( T_1 + T_3 = 2T_2 \). Find the expansion of the rod \( \Delta L_2 \):

• A. \( \Delta L_2 = \Delta L_1 (1 + \alpha\Delta T) \)
• B. \( \Delta L_2 = \Delta L_1 (1 + 2\alpha\Delta T) \)
• C. \( \Delta L_2 = \Delta L_1 (1 + \alpha^2\Delta T^2) \)
• D. \( \Delta L_2 = \Delta L_1 (1 + 2\alpha^2\Delta T^2) \)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Linear expansion depends on the initial length of the rod at the start of the heating process and the change in temperature. When the rod is heated a second time, its "initial" length is the expanded length from the first stage.
Step 2: Key Formula or Approach:
1. \( \Delta L = L_{initial} \alpha \Delta T \)
2. \( L_{final} = L_{initial} (1 + \alpha \Delta T) \)
Step 3: Detailed Explanation:
For the first expansion: \( \Delta L_1 = L \alpha (T_2 - T_1) = L \alpha \Delta T \). The new length at \( T_2 \) is \( L' = L + \Delta L_1 = L(1 + \alpha \Delta T) \). For the second expansion (from \( T_2 \) to \( T_3 \)): Given \( T_1 + T_3 = 2T_2 \), we rearrange: \( T_3 - T_2 = T_2 - T_1 = \Delta T \). The change in temperature is the same. \( \Delta L_2 = L' \alpha (T_3 - T_2) = [L(1 + \alpha \Delta T)] \alpha \Delta T \). Since \( L \alpha \Delta T = \Delta L_1 \): \( \Delta L_2 = \Delta L_1 (1 + \alpha \Delta T) \).
Step 4: Final Answer:
The expansion \( \Delta L_2 \) is equal to \( \Delta L_1 (1 + \alpha\Delta T) \).