Question

The surface mass density of a disc of radius $a$ varies with radial distance $r$ as $\sigma =A+Br$ where A & B are positive constants. Then, moment of inertia of the disc about an axis passing through its centre and perpendicular to the plane (in ${\text{kg m}}^2$) is:

• A. $2\pi a^4(\frac{A}{4}+\frac{Ba}{5})$
• B. $2\pi a^4(\frac{Aa}{4}+\frac{B}{5})$
• C. $\pi a^4(\frac{A}{4}+\frac{Ba}{5})$
• D. $2\pi a^4(\frac{A}{5}+\frac{Ba}{4})$

Answer 1

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


Given, $\sigma =A+Br$
Then, total mass of disc $={\int }_0^{a}dm={\int }_0^a(A+Br)2\pi rdr$
Moment of inertia of the disc about an axis passing through its centre and perpendicular to the plane is
$I={\int }_0^{a}dm{r}^2$
$={\int }_0^a(A+Br)2\pi r^{3}dr$
$=2\pi (A\frac{a^4}{4}+B\frac{a^5}{5})$
$=2\pi a^4(\frac{A}{4}+\frac{Ba}{5})$
Thus, option (a) is the correct answer.