Question

$I$ is moment of inertia of a circular ring about an axis passing through its diameter. This ring is cut then unfolded into a straight rod. The moment of inertia of rod about an axis perpendicular to its length and passing through its end is :

• A. $\frac{2{\pi }^{2}I}{3}$
• B. $\frac{4{\pi }^{2}I}{3}$
• C. $\frac{8{\pi }^{2}I}{3}$
• D. $\frac{9{\pi }^{2}I}{3}$

Answer 1

The correct option is C $\frac{8{\pi }^{2}I}{3}$

Let mass of ring be $m$ and radius be $r$.

So MI of ring about an axis passing through its diameter is $I=m{r}^2/2$

Now if it is unfolded into rod, the length of rod so obtained is $l=2\pi r$

its MI about an axis perpendicular to its length and passing through its end is $I_1=m{l}^2/3=m(2\pi r{)}^2/3=4{\pi }^{2}m{r}^2/3=8{\pi }^2/3\times (m{r}^2/2)=8{\pi }^{2}I/3$