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. $\frac{\alpha I\Delta T}{2}$
• D. $\frac{2I\Delta T}{\alpha }$

Answer 1

Answer: (B)

$2\alpha I\Delta T$

Moment of inertia of a uniform rod of mass M, length L about its perpendicular bisector is

$I=\frac{1}{12}M{L}^2$ .....(i)

When the temperature of the rod is increased by $\Delta T$, there is an increase in the length of the rod $\Delta L$ is given by

$\Delta L=L\alpha \Delta T$ ....(ii)

$\therefore$ New moment of inertia about the same axis is

$I\prime =\frac{1}{12}M{(L+\Delta L)}^2=\frac{1}{12}M[L^2+{\left(\Delta L\right)}^2+2L\Delta L]$

Since $\Delta L$ is a small quantity, the term ${\left(\Delta L\right)}^2$ being very small can be neglected.

$\therefore I\prime =\frac{1}{12}M[L^2+2L\Delta L]$

$=\frac{1}{12}M{L}^2+\frac{1}{12}{ML}^{2}2\alpha \Delta T$ (Using (ii))

$=I+2\alpha I\Delta T$ (Using (i))

$\therefore$ Increase in moment of inertia,

$=I\prime -I=2\alpha I\Delta T$