Physics Circle home Student Login

Home›Chapters›Rotational Motion PYQs›Moment of Inertia & Radius of Gyration›Question 3

MHT CET 2026 (11 Apr, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2026 (11 Apr, Shift 2)

Q3MHT CET 2026 · 11 APR · SHIFT 2Easy

Moment of Inertia & Radius of Gyration

Let M and L be the mass and length of thin uniform rod respectively. In first case, axis of rotation is passing through centre and perpendicular to length of rod. In second case axis of rotation is passing through one end and perpendicular to length of rod. The ratio of radius of gyration in first case to second case is

Show answer and solution

Answer: (b)

  1. About the centre: I1=ML212⇒K1=L12I_1 = \frac{ML^2}{12} \Rightarrow K_1 = \frac{L}{\sqrt{12}}.
  2. About one end: I2=ML23⇒K2=L3I_2 = \frac{ML^2}{3} \Rightarrow K_2 = \frac{L}{\sqrt{3}}.
  3. K1K2=312=12\frac{K_1}{K_2} = \sqrt{\frac{3}{12}} = \frac{1}{2}
  4. So the answer is (b) 12\frac{1}{2}.
Report a mistake

More questions on Moment of Inertia & Radius of Gyration