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. (a) $\alpha I\Delta T$
• A. (b) $2\alpha I\Delta T$
• A. (c) $\frac{\alpha I\Delta T}{2}$
• A. (d) $\frac{2I\Delta T}{\alpha}$

Answer 1

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

$I = \frac{1}{12}ML^{2}$ ….. (i)

When the temperatue of the rod is increased by$\Delta T,$ there is increases in length of the rod $\Delta T$is given by

$\Delta L = L\alpha\Delta T$

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

$I' = \frac{1}{12}M(L + \Delta L)^{2} = \frac{1}{12}M(L^{2} + (\Delta L)^{2} + 2L\Delta L)$

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

$\therefore I' = \frac{1}{12}M\lbrack L^{2} + 2L\Delta L\rbrack$

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

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

$\therefore$Increases in moment of inertia,

$= I' - I = 2\alpha I\Delta T$