I_1 is the moment of inertia of a circular di | vedclass

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(D) Using the parallel axis theorem,$I_2 = I_{CM} + Mh^2$.For a circular disc,the moment of inertia about an axis passing through its centre and perpe

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(D) Using the parallel axis theorem,$I_2 = I_{CM} + Mh^2$.For a circular disc,the moment of inertia about an axis passing through its centre and perpendicular to its plane is $I_1 = I_{CM} = \frac{1}{2}MR^2$.The distance between the parallel axes is $h = \frac{2R}{3}$.Substituting these values into the parallel axis theorem:$I_2 = \frac{1}{2}MR^2 + M\left(\frac{2R}{3}ight)^2$$I_2 = \frac{1}{2}MR^2 + M\left(\frac{4R^2}{9}ight)$$I_2 = MR^2 \left(\frac{1}{2} + \frac{4}{9}ight) = MR^2 \left(\frac{9 + 8}{18}ight) = \frac{17}{18}MR^2$.Now,find the ratio $\frac{I_2}{I_1}$:$\frac{I_2}{I_1} = \frac{\frac{17}{18}MR^2}{\frac{1}{2}MR^2} = \frac{17}{18} 	imes 2 = \frac{17}{9}$.Comparing this with the given ratio $\frac{x}{9}$,we get $x = 17$.

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