Physics Circle home Student Login

Home›Chapters›Rotational Motion PYQs›Moment of Inertia & Radius of Gyration›Question 91

JEE Main 2023 (25 Jan, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2023 (25 Jan, Shift 1)

Q91JEE Main 2023 · 25 JAN · SHIFT 1Easy

Moment of Inertia & Radius of Gyration

ICMI_{\text{CM}} is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc. IABI_{\text{AB}} is its moment of inertia about an axis AB perpendicular to plane and parallel to axis CM at a distance 23\frac{2}{3} R from center. Where R is the radius of the disc. The ratio of IABI_{\text{AB}} and ICMI_{\text{CM}} is x:9x : 9. The value of xx is _____________.

Figure

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 17

  1. ICM=12MR2I_{CM} = \frac{1}{2}MR^2
  2. IAB=12MR2+M(2R3)2=9+818MR2=1718MR2I_{AB} = \frac{1}{2}MR^2 + M\left(\frac{2R}{3}\right)^2 = \frac{9+8}{18}MR^2 = \frac{17}{18}MR^2
  3. IABICM=179\frac{I_{AB}}{I_{CM}} = \frac{17}{9}, so x=17x = 17.
Report a mistake

More questions on Moment of Inertia & Radius of Gyration