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)

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,
$I_{d}=I=\frac{1}{2} M R^{2}$
$\therefore I_{C}=I+I=\frac{1}{2} M R^{2}+\frac{1}{2} M R^{2}$
$=I+I=2 I$