\(I _{ CM }\) 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 \(........\)

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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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