Physics Circle home Student Login

Home›Chapters›Rotational Motion PYQs›Moment of Inertia & Radius of Gyration›Question 66

MHT CET 2024 (15 May, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2024 (15 May, Shift 2)

Q66MHT CET 2024 · 15 MAY · SHIFT 2Easy

Moment of Inertia & Radius of Gyration

The radius of gyration of a circular disc of radius R and mass m rotating about diameter as axis is

Show answer and solution

Answer: (c)

  1. Disc about a diameter: I=mR24I = \frac{mR^2}{4} (half of mR22\frac{mR^2}{2} by the perpendicular axis theorem).
  2. K=Im=R24=R2K = \sqrt{\frac{I}{m}} = \sqrt{\frac{R^2}{4}} = \frac{R}{2}
  3. So the answer is (c) R/2R/2.
Report a mistake

More questions on Moment of Inertia & Radius of Gyration