Question

A thin circular disk is in the $xy$ plane as shown in the figure. The ratio of its moment of inertia about $z$ and $z'$ axes will be :

• A. 1:03
• B. 1:04
• C. 1:05
• D. 1:02

Answer 1

The Correct Option is A

The problem asks for the ratio of the moment of inertia of a thin circular disk about two axes: the \(z\)-axis and the \(z'\)-axis. Let's solve this problem using the concept of moment of inertia for a circular disk.

For a thin circular disk of mass \(M\) and radius \(R\), the moment of inertia \(I_z\) about an axis perpendicular to the disk (like the \(z\)-axis through its center) is given by:

\(I_z = \frac{1}{2} M R^2\)

The moment of inertia about the \(z'\)-axis, which is perpendicular to the disk through a point on its edge, can be found using the parallel axis theorem. This theorem states that:

\(I_{z'} = I_{z} + M d^2\)

where \(d\) is the distance from the center of the disk to the new axis. Here, \(d = R\) (radius of the disk). Therefore,

\(I_{z'} = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2\)

Finally, the ratio of the moment of inertia about the \(z\) and \(z'\)-axes is:

\(\frac{I_z}{I_{z'}} = \frac{\frac{1}{2} M R^2}{\frac{3}{2} M R^2} = \frac{1}{3}\)

Therefore, the correct option is 1:03.