Question

When the temperature of a rod increases from $T$ to $T+\Delta T$, the moment of inertia of the rod about an axis increases from $I$ to $I+\Delta I$. If the coefficient of linear expansion of rod is $\alpha$, then find the ratio $\frac{\Delta I}{I}$

• A. $\frac{\Delta T}{T}$
• B. $\frac{2\Delta T}{T}$
• C. $\alpha \Delta T$
• D. $2\alpha \Delta T$

Answer 1

The correct option is D $2\alpha \Delta T$

Let us consider the axis passing through centre of mass of the rod and perpendicular to the rod.
We know that, moment of inertia of rod about an axis passing through the centre of mass and perpendicular to the rod is given by,
$I=KM{L}^2\dots \dots \left(1\right)$
where $K=\frac{1}{12}$
Applying logarithm on both sides, we get
$\ln I=\ln \left(KM\right)+2\ln L$
Differentiating on both sides,
$\frac{dI}{I}=2\frac{dL}{L}.......\left(2\right)$
[mass $M$ is constant]

From the definition of linear expansion,
$\frac{dL}{L}=\alpha \left(dT\right)......\left(3\right)$
Substituting $\left(3\right)$ in $\left(2\right)$, we get
$\frac{dI}{I}=2\alpha \left(dT\right)$
or $\frac{\Delta I}{I}=2\alpha \Delta T$
Thus, option (d) is the correct answer.