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 \(..........\).
Pick an option to check your answer.
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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