A circular disc of radius b has a hole of radius a at its centre (see figure). If the mass per unit area of the disc varies as ((σ₀)/(r)), then the radius of gyration of the disc about its axis passing through the centre is: (a) √((a²+b²+ab)/(2)) (b) √(a+b/3) (c) √((a²+b²+ab)/(3)) (d) √(a+b/2)

Answer: (c). Ring element: dm = (σ₀)/(r) 2π r dr = 2πσ₀ dr. M = 2πσ₀(b-a) and I = ∫ab 2πσ₀ r² dr = (2πσ₀(b³-a³))/(3) K² = I/M = (b³-a³)/(3(b-a)) = (a²+ab+b²)/(3) So the answer is (c) √((a²+b²+ab)/(3)).

A circular disc of radius b has a hole of radius a at its centre (see… | JEE Main 2019 (12 Apr, Shift 1)

Q168JEE Main 2019 · 12 APR · SHIFT 1Difficult

Moment of Inertia & Radius of Gyration

A circular disc of radius b has a hole of radius a at its centre (see figure). If the mass per unit area of the disc varies as $\left(\frac{\sigma_0}{r}\right)$ , then the radius of gyration of the disc about its axis passing through the centre is:

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Answer: (c)

Ring element: $dm = \frac{\sigma_0}{r}\,2\pi r\,dr = 2\pi\sigma_0\,dr$ .
$M = 2\pi\sigma_0(b-a)$ and $I = \int_a^b 2\pi\sigma_0 r^2 dr = \frac{2\pi\sigma_0(b^3-a^3)}{3}$
$K^2 = \frac{I}{M} = \frac{b^3-a^3}{3(b-a)} = \frac{a^2+ab+b^2}{3}$
So the answer is (c) $\sqrt{\frac{a^2+b^2+ab}{3}}$ .

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