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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2023
  • Medium
  • Numerical

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 \(..........\).

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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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