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

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

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

JEE Mains · Physics · STD 11 - System of particles and rotational motion

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

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Verified Solution

Correct Answer

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

Detailed explanation

\(\mathrm{dI}=\mathrm{dm}^{2}\)

\(\mathrm{dI}=\sigma 2 \pi \mathrm{r}\) dr \(\mathrm{r}^{2}\)

\(\mathrm{dI}=2 \pi(\mathrm{A}+\mathrm{Br}) \mathrm{r}^{3} \mathrm{dr}\)

\(\int \mathrm{dI}=2 \pi \int_{0}^{\mathrm{a}}\left(\mathrm{Ar}^{3}+\mathrm{Br}^{4}\right) \mathrm{dr}\)

\(I=2 \pi a^{4}\left(\frac{A}{4}+\frac{B a}{5}\right)\)

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