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Question

Physics Question on Moment Of Inertia

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

A

2I2I

B

I2\frac{I}{2}

C

32I\frac{3}{2}I

D

II

Answer

2I2I

Explanation

Solution

By the theorem of perpendicular axes, the moment of inertia about the central axis ICI_{C} , will be equal to the sum of its moments of inertia about two mutually perpendicular diameters lying in its plane.
Thus, Id=I=12MR2I_{d}=I=\frac{1}{2} M R^{2}
IC=I+I\therefore I_{C}=I+I
=12MR2+12MR2=\frac{1}{2} M R^{2}+\frac{1}{2} M R^{2}
=I+I=2I=I+I=2 I