JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 18)
$$\mathrm{I_{CM}}$$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc. $$\mathrm{I_{AB}}$$ is it's moment of inertia about an axis AB 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 $$\mathrm{I_{AB}}$$ and $$\mathrm{I_{CM}}$$ is $$x:9$$. The value of $$x$$ is _____________.
_25th_January_Morning_Shift_en_18_1.png)
Answer
17
Explanation
_25th_January_Morning_Shift_en_18_2.png)
$$ =\frac{1}{2} M R^{2}+\frac{4}{9} M R^{2} $$
$$ \begin{aligned} & =\frac{(9+8) M R^{2}}{18}=\left(\frac{17}{18}\right) M R^{2} \end{aligned} $$
$\frac{I_{A B}}{I_{\mathrm{cm}}}=\frac{17 / 18}{1 / 2}=\left(\frac{17}{9}\right)$
Value of $x=17$
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