Question

The moment of inertia of a rod about its perpendicular bisector is I. When the temperature of the rod is increased by $\Delta T$, the increase in the moment of inertia of the rod about the same axis is (Here, $\alpha $ is the coefficient of linear expansion of the rod)
(A) $\alpha I\Delta T$
(B) $2\alpha I\Delta T$
(C) $\dfrac{{\alpha I\Delta T}}{2}$
(D) $\dfrac{{2I\Delta T}}{\alpha }$

Answer 1

Hint: Begin by writing the expression for moment of inertia of a rod about its perpendicular axis passing through the centre. Use the coefficient of linear expansion to find the length of the rod after increase in temperature. Then find the moment of inertia of the rod at the final increased temperature using expanded length and subtract it from the moment of inertia at initial temperature.

Complete step by step solution
The moment of inertia of a rod about its perpendicular bisector is I.
$ \Rightarrow I = \dfrac{{M{L^2}}}{{12}}$
Where M is the mass of the rod and L is the length of the rod.
When the temperature of the rod is increased by$\Delta T$, there is some expansion in the length of the rod which is given by: $\Delta L = \alpha L\Delta T$.
$ \Rightarrow $ The length of the rod will now become, $L + \Delta L$ .
Now, the moment of inertia of the rod about the perpendicular bisector will be
\[
   \Rightarrow I' = \dfrac{{M{{(L + \Delta L)}^2}}}{{12}} \\
   \Rightarrow I\prime = \dfrac{M}{{12}}\left[ {{L^2} + {{(\Delta L)}^2} + 2L\Delta L} \right] \\
\]
Since $\Delta L$ is a very small quantity, its square will be even smaller and almost negligible.
$ \Rightarrow I\prime = \dfrac{M}{{12}}\left[ {{L^2} + 2L\Delta L} \right]$
Now substituting the value of \[\Delta L\]in the above equation,
\[ \Rightarrow I\prime = \dfrac{M}{{12}}\left[ {{L^2} + 2L\left( {\alpha L\Delta T} \right)} \right]\]
\[ \Rightarrow I\prime = \dfrac{{M{L^2}}}{{12}} + 2\alpha \Delta T\dfrac{{M{L^2}}}{{12}}\]
Therefore, the increase in moment of inertia of the rod is $I' - I$ ,
$
   \Rightarrow I' - I = \dfrac{{M{L^2}}}{{12}} + 2\alpha \Delta T\dfrac{{M{L^2}}}{{12}} - \dfrac{{M{L^2}}}{{12}} \\
   \Rightarrow I' - I = 2\alpha \Delta T\dfrac{{M{L^2}}}{{12}} \\
   \Rightarrow I' - I = 2\alpha I\Delta T \\
$

So, option (B) is correct.

Note: One must not get confused between the expressions of moment of inertia of the rod through its perpendicular bisector (i.e. through its centre of mass) and through one end of the rod.
Moment of inertia of the rod through an axis passing through the centre is $\dfrac{{M{L^2}}}{{12}}$. This can be derived from the centre of mass expression.
Moment of inertia of the rod through an axis passing through one end of the rod is $\dfrac{{M{L^2}}}{3}$. This can be derived using the parallel axis theorem.