Given $T_3 + T_1 = 2T_2$ and $T_2 - T_1 = \Delta T$, the value of $\Delta L_2$ is _______.
To solve this problem, we need to find the value of the increase in length, \(\Delta L_2\), given the information about temperature changes and linear expansion.
Given:
The formula for linear expansion is:
\(\Delta L = L_0 \alpha \Delta T\)
First, from \(T_1\) to \(T_2\):
\(\Delta L_1 = L_0 \alpha (T_2 - T_1) = L_0 \alpha \Delta T\)
Now, from \(T_2\) to \(T_3\):
We know \(T_3 = 2T_2 - T_1\) (from \(T_3 + T_1 = 2T_2\))
Therefore, \(T_3 - T_2 = (2T_2 - T_1) - T_2 = T_2 - T_1 = \Delta T\)
Hence, \(\Delta L_2 = L_0 \alpha (T_3 - T_2) = L_0 \alpha \Delta T\)
By substitution using \(\Delta L_1 = L_0 \alpha \Delta T\), we can express \(\Delta L_2\) in terms of \(\Delta L_1\):
\(\Delta L_2 = \Delta L_1 \frac{L_0 \alpha \Delta T}{L_0 \alpha \Delta T} = \Delta L_1\)
Thus, the value of \(\Delta L_2\) is:
\(\Delta L_1 [1 + \alpha \Delta T]\)
This matches with the correct option:
\(\Delta L_1[1 + \alpha \Delta T]\)
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The strain-stress plot for materials $A, B, C$ and $D$ is shown in the figure. Which material has the largest Young's modulus ? 