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

(B) $2 \pi \mathrm{a}^{4}\left(\frac{\mathrm{~A}}{4}+\frac{\mathrm{aB}}{5}\right)$

(A) $2 \pi \mathrm{a}^{4}\left(\frac{\mathrm{~A}}{4}+\frac{\mathrm{B}}{5}\right)$

(C) $2 \pi \mathrm{a}^{4}\left(\frac{\mathrm{aA}}{4}+\frac{\mathrm{B}}{5}\right)$

(D) $\pi \mathrm{a}^{4}\left(\frac{\mathrm{~A}}{4}+\frac{\mathrm{aB}}{5}\right)$

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Rotational Mechanics

Mass per unit area of a circular disc of radius a depends on the distance

r

from its centre as

σ (r) = A + B r

. The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its centre is:

Answer: (B)

Slice the disc into thin rings of radius

r

and width

d r

. Such a ring has mass

d m = σ (r) 2 π r d r = (A + B r) 2 π r d r

and contributes

d I = r^{2} d m

about the central axis. Integrating,

I = 2 π A \int_{0}^{a} r^{3} d r + 2 π B \int_{0}^{a} r^{4} d r = 2 π A \frac{a ^{4}}{4} + 2 π B \frac{a ^{5}}{5} = 2 π a^{4} (\frac{A}{4} + \frac{a B}{5})

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option