Question 72 of 280

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

A$2\pi {a}^{4}\left(\dfrac{A}{4}+\dfrac{aB}{5}\right)$B$2\pi {a}^{4}\left(\dfrac{aA}{4}+\dfrac{B}{5}\right)$C$\pi {a}^{4}\left(\dfrac{A}{4}+\dfrac{aB}{5}\right)$D$2\pi {\text{a}}^{4}\left(\dfrac{\text{A}}{4}+\dfrac{\text{B}}{5}\right)$

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Answer: (A) $2\pi {a}^{4}\left(\dfrac{A}{4}+\dfrac{aB}{5}\right)$

$\sigma =A+Br$
$\int dm=\int \left(A+Br\right)2\pi rdr$
$I=\int dm{r}^{2}$
$\int _{0}^{a}\left(A+Br\right)2\pi {r}^{3}dr$
$=2\pi \left(A\dfrac{{a}^{4}}{4}+B\dfrac{{a}^{5}}{5}\right)$
$=2\pi {a}^{4}\left(\dfrac{A}{4}+\dfrac{aB}{5}\right)$