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: First expansion
\[ \Delta L_1=L_1\alpha \Delta T \] New length: \[ L_2=L_1+\Delta L_1 \] \[ =L_1(1+\alpha \Delta T) \] Step 2: Second expansion
Since \[ T_3+T_1=2T_2 \] we get \[ T_3-T_2=T_2-T_1=\Delta T \] So second temperature rise is also $\Delta T$. \[ \Delta L_2=L_2\alpha \Delta T \] Substitute $L_2$: \[ \Delta L_2=L_1(1+\alpha \Delta T)\alpha \Delta T \] Since \[ L_1\alpha \Delta T=\Delta L_1 \] \[ \Delta L_2=\Delta L_1(1+\alpha \Delta T) \] Final Answer: \[ \boxed{\Delta L_1(1+\alpha \Delta T)} \]