Question:medium

What is the moment of inertia for two specific (solid) spheres about their common diameter?

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Moment of inertia of a solid sphere about its diameter is \( \frac{2}{5}MR^2 \). When multiple spheres are involved, their contributions add according to the axis and configuration.
Updated On: Apr 28, 2026
  • \( \frac{2}{5}MR^2 \)
  • \( \frac{7}{5}MR^2 \)
  • \( \frac{14}{5}MR^2 \)
  • \( \frac{4}{5}MR^2 \)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Question:
The question asks for the total Moment of Inertia (M.I.) of a system consisting of two solid spheres. The "common diameter" phrasing in competitive exams often refers to an axis passing through the point of contact, parallel to their diameters, or a specific composite configuration (like two spheres touching each other with an axis tangent to both).
Step 2: Key Formula or Approach:
1. M.I. of a solid sphere about its center (diameter): \( I_{cm} = \frac{2}{5}MR^2 \)

2. Parallel Axis Theorem: \( I = I_{cm} + Mh^2 \), where \(h\) is the distance between the axes.
Step 3: Detailed Explanation:
For a system where two spheres are touching and we calculate the M.I. about a common tangent axis passing through the point of contact:
For one sphere, the distance from the center to the tangent is \(R\).
Using the parallel axis theorem for one sphere:
\[ I_1 = \frac{2}{5}MR^2 + M(R)^2 = \frac{7}{5}MR^2 \]
For two such spheres arranged symmetrically about this axis:
\[ I_{total} = I_1 + I_2 = \frac{7}{5}MR^2 + \frac{7}{5}MR^2 = \frac{14}{5}MR^2 \]
Step 4: Final Answer:
The total moment of inertia for the described system is \( \frac{14}{5}MR^2 \).
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