The moment of inertia of a circular ring abou | vedclass

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(D) According to the perpendicular axis theorem,for a planar body,the moment of inertia about an axis perpendicular to its plane $(I_z)$ is equal to t

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(D) According to the perpendicular axis theorem,for a planar body,the moment of inertia about an axis perpendicular to its plane $(I_z)$ is equal to the sum of the moments of inertia about two mutually perpendicular axes in its plane ($I_x$ and $I_y$): $I_z = I_x + I_y$ For a circular ring,by symmetry,the moment of inertia about any diameter is the same,so $I_x = I_y = I_D$. Given $I_z = 200 \, gm \cdot cm^2$. Substituting these into the theorem: $200 = I_D + I_D = 2I_D$ $I_D = \frac{200}{2} = 100 \, gm \cdot cm^2$. Thus,the correct option is $D$.

The moment of inertia of a circular ring about an axis perpendicular to its plane and passing through its center is $200 \, gm \cdot cm^2$. The moment of inertia about its diameter is ....... $gm \cdot cm^2$.

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