Question

The moment of inertia of a rod about its perpendicular bisector is I. When the temperature of the rod is increased by $△T$, 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. $\alpha I△T$
• B. $2\alpha I△T$
• C. $\frac{\alpha I△T}{2}$
• D. $\frac{2I△T}{\alpha }$

Answer 1

The correct option is B $2\alpha I△T$

Moment of inertia a 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 $△T,$ there is increase in length of the road $△L$is given by
$△L=L\alpha △T$................ (ii)
$\therefore$ New moment of inertia about the same axis is
$I^{\prime }=\frac{1}{12}m(l+△L{)}^2=\frac{1}{12}M[L^2+(△L{)}^2+2L△L]$
Since $△L$ is a small quantity , the term $△L{)}^2$ being very small can be neglected.
$\therefore I^{\prime }=\frac{1}{12}M[L^2+2L△L]$
$=\frac{1}{12}M{L}^2+\frac{1}{12}M{L}^{2}2\alpha △T$ ......Using (ii)
$=I+2\alpha I△T$ ......Using (i)
$\therefore$ Increase in moment of inertia,
$=I^{\prime }-I=2\alpha I△T$