A uniform rod of mass m is bent into the form of a semicircle of radius R. The moment of inertia of the rod about an axis passing through A and perpendicular to the plane of the paper is

Options

A. `2/3 mR^(2)`
B. `mR^(2)`
C. `5/pi mR^(2)`
D. `2m R^(2)`

Answer

D

Explanation

If an identical semicircle is joined to form a complete ring of mass 2m, then `I_(A)` of complete ring by parallel axis theorem is `(2m)R^(2) + 2mR^(2)=4mR^(2)` . And if I is required moment of inertia then`2I= 4mR^(2) rArr I = 2mR^(2)`