Question

A solid sphere of mass \(M\) and radius \(R\) is divided into two unequal parts. The smaller part having mass \(M/8\) is converted into a sphere of radius \(r\) and the larger part is converted into a circular disc of thickness \(t\) and radius \(2R\). If \(I_1\) is moment of inertia of a sphere having radius \(r\) about an axis through its centre and \(I_2\) is the moment of inertia of a disc about its diameter, the ratio of their moment of inertia \(I_2/I_1 =\) ______.

• A. 35
• B. 70
• C. 140
• D. 210

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We must calculate the moments of inertia for two distinct objects formed from the original sphere.
The smaller object is a solid sphere, and the larger object is a uniform circular disc.
We will use the standard formulas for moments of inertia of these standard geometries.
Step 2: Key Formula or Approach:
1. Moment of inertia of a solid sphere about its center: \(I = \frac{2}{5} m r^2\).
2. Moment of inertia of a circular disc about its diameter: \(I = \frac{1}{4} m r^2\).
To find the radius \(r\) of the new small sphere, we use the fact that density remains constant (mass is proportional to volume):
\(\frac{m_{\text{small}}}{M} = \frac{r^3}{R^3} \implies \frac{1}{8} = \left(\frac{r}{R}\right)^3 \implies r = \frac{R}{2}\).
Step 3: Detailed Explanation:
Let's analyze the smaller part (the new sphere):
Mass \(m_1 = \frac{M}{8}\).
Radius \(r = \frac{R}{2}\).
Its moment of inertia \(I_1\) is:
\(I_1 = \frac{2}{5} m_1 r^2 = \frac{2}{5} \left(\frac{M}{8}\right) \left(\frac{R}{2}\right)^2 = \frac{2}{5} \left(\frac{M}{8}\right) \left(\frac{R^2}{4}\right) = \frac{2MR^2}{160} = \frac{MR^2}{80}\).
Let's analyze the larger part (the disc):
Mass \(m_2 = M - \frac{M}{8} = \frac{7M}{8}\).
Radius of the disc \(R_{\text{disc}} = 2R\).
Its moment of inertia \(I_2\) about its diameter is:
\(I_2 = \frac{1}{4} m_2 R_{\text{disc}}^2 = \frac{1}{4} \left(\frac{7M}{8}\right) (2R)^2 = \frac{1}{4} \left(\frac{7M}{8}\right) (4R^2) = \frac{7MR^2}{8}\).
Step 4: Final Answer:
We are required to find the ratio \(I_2 / I_1\):
\(\frac{I_2}{I_1} = \frac{\frac{7MR^2}{8}}{\frac{MR^2}{80}} = \frac{7}{8} \times 80 = 7 \times 10 = 70\).