Question

The moment of inertia of a circular ring about an axis passing through its center and normal to its plane is $200gm-c{m}^2$. Then its moment of inertia about a diameter is

• A. 40 gm cm²
• B. 60gm cm²
• C. 80gm cm²
• D. 100gm cm²

Answer 1

The correct option is D

100gm cm²

Applying perpendicular axes theorem,

$I_z=I_x+I_y$ [where 1z = moment of inertia about natural axis]

$\Rightarrow As{I}_x=I_y=I$

$\Rightarrow I_z=2I$

$\Rightarrow I=\frac{I_z}{2}=\frac{200}{2}=100gm-c{m}^2$