Question

Moment of inertia of ring about its diameter is $I.$ Then, moment of inertia about an axis passing through centre perpendicular to its plane is

• A. $2I$
• B. $\frac{I}{2}$
• C. $\frac{3}{2}I$
• D. $I$

Answer 1

Answer: (A)

$2I$

Answer 2

By the theorem of perpendicular axes, the moment of inertia about the central axis $I_{C}$, will be equal to the sum of its moments of inertia about two mutually perpendicular diameters lying in its plane. Thus, $I_{d}=I=\frac{1}{2} M R^{2}$ $\therefore I_{C}=I+I=\frac{1}{2} M R^{2}+\frac{1}{2} M R^{2}$ $=I+I=2 I$