Question

From a circular ring of mass ‘M’ and radius ‘R’ an arc corresponding to a 90\(^{\circ}\) sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is ‘K’ times ‘MR² ’. Then the value of ‘K’ is

• A. \(\frac{1}{8}\)
• B. \(\frac{3}{4}\)
• C. \(\frac{7}{8}\)
• D. \(\frac{1}{4}\)

Answer 1

The Correct Option is B

The problem requires determining the value of 'K' for which the moment of inertia of the remaining part of a circular ring is given by \( K \times MR^2 \). Let's solve this step by step.

1. Start by calculating the total moment of inertia of the complete ring about an axis passing through the center and perpendicular to the plane.

The moment of inertia \( I \) of a complete circular ring of mass \( M \) and radius \( R \) is given by:

\(I_{\text{ring}} = MR^2\)

1. Since an arc corresponding to a 90\(^\circ\) (or a quarter of the circle) is removed, the mass of the remaining part is three-quarters of the total mass.

Mass of the complete ring \(= M\), so mass of the removed arc is \(\frac{M}{4}\) and the mass of the remaining part is \(\frac{3M}{4}\).

1. The moment of inertia is directly proportional to the mass; hence the moment of inertia of the remaining arc is:

\(I_{\text{remaining}} = \frac{3M}{4}R^2\)

1. By defining \( I_{\text{remaining}} \) as \( K \times MR^2 \), we equate the expressions:

\(K \times MR^2 = \frac{3M}{4}R^2\)

Canceling \( MR^2 \) from both sides gives:

\(K = \frac{3}{4}\)

1. Thus, the value of 'K' is \(\frac{3}{4}\).

Therefore, the correct answer is:

\(\frac{3}{4}\)