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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 _______.

The correct answer is
$\Delta L_1[1 + \alpha \Delta T]$

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 coefficient of linear expansion of metal strip: \(\alpha\)
  • Temperature initially increases from \(T_1\) to \(T_2\), leading to an increase in length by \(\Delta L_1\).
  • Temperature further increases from \(T_2\) to \(T_3\), leading to an increase in length by \(\Delta L_2\).
  • It's given that \(T_3 + T_1 = 2T_2\) and \(T_2 - T_1 = \Delta T\).

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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Important Questions from Physical and Thermal Properties of Bulk Matter

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