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JEE Main 2020 (7 Jan, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2020 (7 Jan, Shift 2)

Q155JEE Main 2020 · 7 JAN · SHIFT 2Difficult

Moment of Inertia & Radius of Gyration

Mass per unit area of a circular disc of radius $a$ depends on the distance r from its centre as $\sigma(r)$ = A + Br. The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its centre is:

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Answer: (a)

Take a ring of radius $r$ , width $dr$ : $dm = (A+Br)\,2\pi r\,dr$ , $dI = r^2 dm$ .
$I = 2\pi\int_0^a (Ar^3 + Br^4)dr = 2\pi\left(\frac{Aa^4}{4} + \frac{Ba^5}{5}\right)$
$I = 2\pi a^4\left(\frac{A}{4} + \frac{aB}{5}\right)$
So the answer is (a).

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