Physics Circle home Student Login

Home›Chapters›Rotational Motion PYQs›Moment of Inertia & Radius of Gyration›Question 164

JEE Main 2019 (10 Apr, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2019 (10 Apr, Shift 1)

Q164JEE Main 2019 · 10 APR · SHIFT 1Medium

Moment of Inertia & Radius of Gyration

A thin disc of mass M and radius R has mass per unit area σ\sigma(r) = kr2 where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is:

Show answer and solution

Answer: (c)

  1. M=∫0Rkr2⋅2πr dr=πkR42M = \int_0^R kr^2 \cdot 2\pi r\,dr = \frac{\pi k R^4}{2}
  2. I=∫0Rkr2⋅r2⋅2πr dr=πkR63I = \int_0^R kr^2 \cdot r^2 \cdot 2\pi r\,dr = \frac{\pi k R^6}{3}
  3. IM=2R23⇒I=2MR23\frac{I}{M} = \frac{2R^2}{3} \Rightarrow I = \frac{2MR^2}{3}
  4. So the answer is (c).
Report a mistake

More questions on Moment of Inertia & Radius of Gyration