Mass per unit area of a circular disc of radius a depends

Question:

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:

  1. $2 \pi a^{4}\left(\frac{A}{4}+\frac{a B}{5}\right)$

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

  3. $\pi a^{4}\left(\frac{A}{4}+\frac{a B}{5}\right)$

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


Correct Option: 1

Solution:

(1) Given,

mass per unit area of circular disc, $\sigma=A+B r$

Area of the ring $=2 \pi \mathrm{r} d r$

Mass of the ring, $d m=\sigma 2 \pi r d r$

Moment of inertia,

$I=\int d m r^{2}=\int \sigma 2 \pi r d r \cdot r^{2}$

$\Rightarrow \quad I=2 \pi \int_{0}^{a}(A+B r) r^{3} d r=2 \pi\left[\frac{A a^{4}}{4}+\frac{B a^{5}}{5}\right]$

$\Rightarrow I=2 \pi a^{4}\left[\frac{A}{4}+\frac{B a}{5}\right]$

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