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 us to find the moment of inertia of the remaining part of a circular ring after cutting out an arc corresponding to a 90° sector.

Step 1: Understand the distribution of mass.

The full circle is initially a ring with a uniform mass distribution of 'M' and radius 'R'. The moment of inertia of a complete ring about an axis passing through its center and perpendicular to its plane is given by:

\[ I_{\text{total}} = MR^2 \]

A 90° sector corresponds to \(\frac{1}{4}\) of the circle. Thus, the mass of the removed sector is:

\[ M_{\text{removed}} = \frac{M}{4} \]

The moment of inertia of the removed part (the 90° sector) about the same axis, considering it has the same radius 'R', is:

\[ I_{\text{removed}} = \frac{1}{4} \times MR^2 = \frac{MR^2}{4} \]

Step 2: Calculate the moment of inertia of the remaining part of the ring.

Since the remaining mass is \( M - \frac{M}{4} = \frac{3M}{4} \), and it's distributed in the remaining 270° of the ring, we need to find the adjusted moment of inertia:

\[ I_{\text{remaining}} = I_{\text{total}} - I_{\text{removed}} = MR^2 - \frac{MR^2}{4} = \frac{3MR^2}{4} \]

Step 3: Conclusion.

The moment of inertia of the remaining part is \(\frac{3}{4} \times MR^2\). Comparing with the given expression \(K \times MR^2\), we find that the value of \(K\) is:

\[ K = \frac{3}{4} \]

Thus, the correct answer is \(\frac{3}{4}\).