Question

$I$ is moment of inertia of a thin circular ring about an axis perpendicular to the plane of ring and passing through its centre. The same ring is folded into 2 turns coil. The moment of inertia of circular coil about an axis perpendicular to the plane of coil and passing through its centre is:

• A. $2I$
• B. $4I$
• C. $\frac{I}{2}$
• D. $\frac{I}{4}$

Answer 1

The correct option is D $\frac{I}{4}$

$I=M{r}^2$,

When same ring is folded into 2 turns coils , then the new radius $r_1=r/2$ and m=M/2

So, $I=2\times m{r}_1^2=\frac{M{r}^2}{4}=\frac{I}{4}$