Question

The moment of inertia of a solid sphere of mass 20 kg and diameter 20 cm about the tangent to the sphere is:

• A. \( 0.24 \, \text{kgm}^2 \)
• B. \( 0.14 \, \text{kgm}^2 \)
• C. \( 0.28 \, \text{kgm}^2 \)
• D. \( 0.08 \, \text{kgm}^2 \)

Hint:

N/A

Answer 1

The Correct Option is C

We are given the following: - Mass of the solid sphere \( M = 20 \, \text{kg} \), - Diameter of the sphere \( D = 20 \, \text{cm} = 0.2 \, \text{m} \), - The moment of inertia is to be calculated about the tangent to the sphere. The moment of inertia \( I_{\text{sphere}} \) of a solid sphere about its center of mass is given by the formula: \[ I_{\text{sphere}} = \frac{2}{5} M R^2, \] where \( M \) is the mass and \( R \) is the radius of the sphere. 
Step 1: Moment of inertia about the center For a sphere with mass \( M = 20 \, \text{kg} \) and radius \( R = \frac{D}{2} = \frac{0.2}{2} = 0.1 \, \text{m} \), the moment of inertia about the center is: \[ I_{\text{center}} = \frac{2}{5} \times 20 \times (0.1)^2 = \frac{2}{5} \times 20 \times 0.01 = 0.08 \, \text{kgm}^2. \] 
Step 2: Using the parallel axis theorem To find the moment of inertia about a tangent to the sphere, we use the parallel axis theorem: \[ I_{\text{tangent}} = I_{\text{center}} + M R^2. \] Substituting the values: \[ I_{\text{tangent}} = 0.08 + 20 \times (0.1)^2 = 0.08 + 20 \times 0.01 = 0.08 + 0.2 = 0.28 \, \text{kgm}^2. \] Thus, the moment of inertia about the tangent to the sphere is \( 0.28 \, \text{kgm}^2 \). 
Therefore, the correct answer is option (3).