TS EAMCET · Physics · Rotational Motion
The moment of inertia of a thin circular disc about an axis passing through its centre and perpendicular to its plane is \(I\). Then, the moment of inertia of the disc about an axis parallel to its diameter and touching the edge of the rim is
- A \(I\)
- B \(2 I\)
- C \(\frac{3}{2} I\)
- D \(\frac{5}{2} I\)
Answer & Solution
Correct Answer
(D) \(\frac{5}{2} I\)
Step-by-step Solution
Detailed explanation
Moment of inertia of a circular disc about an axis passing through centre of gravity and perpendicular to its plane \[ I=\frac{1}{2} M R^2 \] From Eq. (i) \(\quad M R^2=2 I\) Then, moment of inertia of disc about tangent…
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