Physics Circle home Student Login

Home›Chapters›Rotational Motion PYQs›Moment of Inertia & Radius of Gyration›Question 204

NEET 1990 (AIPMT) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from NEET 1990 (AIPMT)

Q204NEET 1990 · AIPMTEasy

Moment of Inertia & Radius of Gyration

Moment of inertia of a uniform circular disc about a diameter is II. Its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim will be

Show answer and solution

Answer: (c)

  1. About a diameter: I=14MR2I = \frac{1}{4}MR^2, so MR2=4IMR^2 = 4I.
  2. About the rim, normal to the plane: I′=12MR2+MR2=32MR2I' = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2.
  3. I′=32(4I)=6II' = \frac{3}{2}(4I) = 6I
  4. So the answer is (c) 6I6I.
Report a mistake

More questions on Moment of Inertia & Radius of Gyration