JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
Moment of inertia of a dise of mass \(M\) and radius ' \(R\) ' about any of its diameter is \(\frac{ MR ^2}{4}\). The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, \(\frac{ x }{2} MR ^2\). The value of \(x\) is \(..........\).
- A \(1.5\)
- B \(6\)
- C \(9\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
\(I = I _{ cm }+ Md ^2\) \(=\frac{ MR ^2}{2}+ MR ^2\) \(=\frac{3}{2} MR ^2\) \(x =3\)
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