Question

I is the moment of inertia of a thin circular ring about an axis perpendicular to the plane of the ring and passing through its centre. The same ring is folded into 2 turns coils. The moment of inertia of circular coil about an axis perpendicular to the plane of coil and passing through its centre is:
$
  {\text{A}}{\text{. 2}}I \\
  {\text{B}}{\text{. 4}}I \\
  {\text{C}}{\text{. }}\dfrac{I}{2} \\
  {\text{D}}{\text{. }}\dfrac{I}{4} \\
$

Answer 1

Hint: The moment of inertia is dependent on the mass distribution of the body about an axis. Greater the mass and greater the distance of that mass from the axis, larger is the moment of inertia about that axis.

Complete step by step answer:

Moment of inertia of a body about an axis is defined as the sum of products of masses and distance of masses from the axis.
$I = \sum\limits_i {{m_i}r_i^2} = M{r^2}$

Where I is the moment of inertia, is the ith mass of the body and is distance of ith mass from the axis of rotation while M is the total mass and r is the distance of that mass from axis of rotation.
We are given that a ring has a moment of inertia $I$ about an axis passing through its centre.
$I = M{r^2}{\text{ }}...{\text{(i)}}$

where M is the mass of the ring and r is the radius of the ring
This ring is converted into a coil of 2 turns and we need to find a new moment of inertia of this coil. Mass remains the same here but gets distributed equally between two coils and radius gets reduced by half since the rings are thin.
$\therefore I' = 2\left( {\dfrac{M}{2}} \right){\left( {\dfrac{r}{2}} \right)^2} = \dfrac{{M{r^2}}}{4}{\text{ }}...{\text{(ii)}}$

Factor 2 comes because there are two turns in the coil. Using equation (i) here, we get
$I' = \dfrac{I}{4}$

Hence, the correct answer is option D.

Note: Moment of inertia plays a role only when the body is rotating about an axis. That is why, the axis is always related to the moment of inertia. Moment of inertia is analogous to inertia of the body when it is at rest or in translational motion.