Question

Mass per unit area of a circular disc of radius ${}^{\prime }{a}^{\prime }$ 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. $2\pi a^4(\frac{A}{4}+\frac{aB}{5})$
• B. $2\pi a^4(\frac{aA}{4}+\frac{B}{5})$
• C. $\pi a^4(\frac{A}{4}+\frac{aB}{5})$
• D. $2\pi a^4(\frac{A}{4}+\frac{B}{5})$

Answer 1

The correct option is A $2\pi a^4(\frac{A}{4}+\frac{aB}{5})$

Given,
mass per unit area of circular disc, $\sigma =A+Br$

Consider a small elemental ring of thickness $dr$ at a distance $r$ from the centre,

Area of the element $=2\pi rdr$
Mass of the element, $dm=\sigma 2\pi rdr$

The moment of inertia of the ring about an axis, perpendicular to the plane and passing through its centre, is given by,

$I=\int dm{r}^2=\int \sigma 2\pi rdr.r^2$

$\Rightarrow I=2\pi {\int }_0^a(A+Br)r^{3}dr$

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

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

Hence, option $\left(A\right)$ is correct.