Question

When the temperature of the body increases from $T$ to $T + \Delta T$, its moment of inertia increases from $I$ to $I + \Delta I$. If $\alpha$ is the coefficient of linear expansion of the material of the body, then $\dfrac{{\Delta I}}{I}$ is (neglect higher orders of $\alpha$):
A. $\alpha \Delta T$
B. $2\alpha \Delta T$
C. $\dfrac{{\Delta T}}{\alpha }$
D. $\dfrac{{2\alpha }}{{\Delta T}}$

Answer 1

Since we have given the linear expansion coefficient, the body must be linear like metal rod. Recall the expression for the moment of inertia of the rod. Take the derivative of the moment of inertia with respect to length of the rod L. Take the ratio of change in moment of inertia and initial moment of inertia of the rod. Use the expression for linear expansion of material to get the change in length of the rod with temperature.

Formula used:
$I = \dfrac{{M{L^2}}}{{12}}$
Here, M is the mass of the rod and L is the length of the rod.
$\Delta L = L\alpha \Delta T$
Here, L is the original length, $\alpha$ is the linear expansion coefficient, $\Delta T$ is the change in temperature.

Complete step by step solution:
Since $\alpha$ is the linear expansion coefficient, the body must be linear like a metal rod. The moment of inertia of rod is given as,
$I = \dfrac{{M{L^2}}}{{12}}$ …… (1)
Here, M is the mass of the rod and L is the length of the rod.
We have given that this rod undergoes linear expansion when the temperature increases from $T$ to $T + \Delta T$. Since the length of the rod changes, the moment of inertia also changes. Therefore, we can express the change in moment of inertia of the rod by taking the derivative of equation (1) as follows,
$\Delta I = \dfrac{1}{{12}}\left( {2ML\Delta L} \right)$ ……. (2)
Dividing equation (2) by equation (1), we get,
$\dfrac{{\Delta I}}{I} = \dfrac{{\dfrac{1}{{12}}\left( {2ML\Delta L} \right)}}{{\dfrac{{M{L^2}}}{{12}}}}$
$\Rightarrow \dfrac{{\Delta I}}{I} = \dfrac{{2\Delta L}}{L}$
As the temperature increases, the length of the rod also increases. We have the expression for the change in the length of the rod with change in temperature as follows,
$\Delta L = L\alpha \Delta T$
Here, $\alpha$ is the linear expansion coefficient.
Substituting the above equation in equation (3), we get,
$\dfrac{{\Delta I}}{I} = \dfrac{{2L\alpha \Delta T}}{L}$
$\therefore \dfrac{{\Delta I}}{I} = 2\alpha \Delta T$

So, the correct answer is option (B).

Note: The final length of the rod with change in temperature is given as, ${L_f} = {L_i} + \alpha {L_i}\Delta T$, where, ${L_i}$ is the initial length of the rod. Therefore, the change in length can be expressed as, $\Delta L = L\alpha \Delta T$. There is a very small increase in the length of the rod with increase in temperature. Therefore, we have taken the derivation of the moment of inertia instead of taking the difference.