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 :

(B) $\sqrt{\frac{a^{2}+b^{2}+ab}{3}}$

(D) $\sqrt{\frac{a^{2}+b^{2}+ab}{2}}$

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Rotational Mechanics

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

(\frac{σ _{0}}{r})

, then the radius of gyration of the disc about its axis passing through the centre is :

Answer: (B)

Take a thin ring of radius

r

and thickness

d r

. Its mass is

d m = σ \cdot 2 π r d r = \frac{σ _{0}}{r} \cdot 2 π r d r = 2 π σ_{0} d r

, and it contributes

d I = r^{2} d m = 2 π σ_{0} r^{2} d r

to the moment of inertia. Integrating from

a

b

, the moment of inertia is

I = \frac{2 π σ _{0}}{3} (b^{3} - a^{3})

and the total mass is

m = \int_{a}^{b} 2 π σ_{0} d r = 2 π σ_{0} (b - a)

. The radius of gyration is then

k = \frac{I}{m} = \frac{b ^{3} - a ^{3}}{3 ( b - a )} = \frac{a ^{2} + b ^{2} + ab}{3}

, using

b^{3} - a^{3} = (b - a) (a^{2} + ab + b^{2})

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option