MHT CET202613 April 2026Evening ShiftPhysicsRotational MotionActual
Moment of inertia of a disc of mass M and radius 'R' about any of its diameter is MR^2/4 . The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, (x/2) MR^2 . The value of x is
Options
- A1
- B3
- C5
- D7
Correct answer
B. 3
Step-by-step solution
Given the moment of inertia of the disc about its diameter is I_ d = MR^2 4 . By the perpendicular axis theorem, the moment of inertia of the disc about an axis passing through its center and perpendicular to its plane is I_ c = I_ d + I_ d = MR^2 4 + MR^2 4 = MR^2 2 . By the parallel axis theorem, the moment of inertia of the disc about an axis perpendicular to its plane and passing through a point on its edge is I = I_ c + MR^2 . Substituting the value of I_ c , we get I = MR^2 2 + MR^2 = 3 2 MR^2 . Comparing thi