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JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
\(I _{ CM }\) is moment of inertia of a circular disc about an axis \(( CM )\) passing through its center and perpendicular to the plane of disc. \(I _{ AB }\) is it's moment of inertia about an axis \(A B\) perpendicular to plane and parallel to axis \(CM\) at a distance \(\frac{2}{3} R\) from center. Where \(R\) is the radius of the disc. The ratio of \(I _{ AB }\) and \(I _{ CM }\) is \(x: 9\). The value of \(x\) is \(........\)

- A \(15\)
- B \(16\)
- C \(18\)
- D \(17\)
Answer & Solution
Correct Answer
(D) \(17\)
Step-by-step Solution
Detailed explanation
\(I _{ cm }=\frac{ mR ^2}{2}\) \(I _{ AB }=\frac{ mR ^2}{2}+ m \left(\frac{2 R }{3}\right)^2=\frac{17}{18} mR ^2\) \(\frac{ I _{ AB }}{ I _{ cm }}=\frac{17}{9}\) \(\Rightarrow x =17\)
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