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JEE Main 2019 (8 Apr, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2019 (8 Apr, Shift 1)

Q165JEE Main 2019 · 8 APR · SHIFT 1Medium

Moment of Inertia & Radius of Gyration

A thin circular plate of mass M and radius R has its density varying as ρ\rho(r) = ρ\rho0r with ρ\rho0 as constant and r is the distance from its centre. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is I = aMR2. The value of the coefficient a is:

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Answer: (c)

  1. M=∫0Rρ0r⋅2πr dr=2πρ0R33M = \int_0^R \rho_0 r \cdot 2\pi r\,dr = \frac{2\pi\rho_0 R^3}{3}
  2. Ic=∫0Rρ0r⋅r2⋅2πr dr=2πρ0R55=35MR2I_c = \int_0^R \rho_0 r \cdot r^2 \cdot 2\pi r\,dr = \frac{2\pi\rho_0 R^5}{5} = \frac{3}{5}MR^2
  3. Edge axis (parallel axis): I=35MR2+MR2=85MR2I = \frac{3}{5}MR^2 + MR^2 = \frac{8}{5}MR^2
  4. So a=85a = \frac{8}{5}, answer (c).
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