Physics Circle home Student Login

Home›Chapters›Rotational Motion PYQs›Moment of Inertia & Radius of Gyration›Question 89

MHT CET 2023 (12 May, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2023 (12 May, Shift 1)

Q89MHT CET 2023 · 12 MAY · SHIFT 1Easy

Moment of Inertia & Radius of Gyration

I1I_1 is the moment of inertia of a circular disc about an axis passing through its centre and perpendicular to the plane of disc. I2I_2 is its moment of inertia about an axis ABAB perpendicular to plane and parallel to axis CM\text{CM} at a distance 2R3\frac{2R}{3} from centre. The ratio of I1I_1 and I2I_2 is x:17x : 17. The value of 'xx' is (R = radius of the disc)

Figure

Show answer and solution

Answer: (a)

  1. I1=MR22I_1 = \frac{MR^2}{2}.
  2. Parallel axis at 2R3\frac{2R}{3}: I2=MR22+4MR29=1718MR2I_2 = \frac{MR^2}{2} + \frac{4MR^2}{9} = \frac{17}{18}MR^2
  3. I1I2=917⇒x=9\frac{I_1}{I_2} = \frac{9}{17} \Rightarrow x = 9
  4. So the answer is (a) 9.
Report a mistake

More questions on Moment of Inertia & Radius of Gyration