Question

A circular disc of radius $b$ has a hole of radius $a$ at its centre (see figure). If the mass per unit area of the disc varies as $\left(\frac{{\sigma }_0}{r}\right)$, then the radius of gyration of the disc about its axis passing through the centre is :

• A. $\sqrt{\frac{a^2+b^2+ab}{3}}$
• B. $\frac{a+b}{3}$
• C. $\frac{a+b}{2}$
• D. $\sqrt{\frac{a^2+b^2+ab}{2}}$

Answer 1

The correct option is A $\sqrt{\frac{a^2+b^2+ab}{3}}$

The moment of inertia about an axis passing through the centre is given by,

$I=\underset{a}{\overset{b}{\int }}\left(dm\right)r^2$

Considering a small element of mass $dm$ and thickness $dr$ at a distance $r$ from the centre,

$dm=\sigma \left(2\pi rdr\right)$

$\Rightarrow I=\underset{a}{\overset{b}{\int }}(\sigma \times 2\pi rdr)r^2$

$\Rightarrow I=\frac{2\pi {\sigma }_0}{3}{r}^3{|}_a^b[\because \sigma =\frac{{\sigma }_0}{r}]$

$\Rightarrow I=\frac{2\pi {\sigma }_0}{3}(b^3-a^3)$

Mass of the disc,

$m=\int dm=\underset{a}{\overset{b}{\int }}\frac{{\sigma }_0}{r}\times 2\pi rdr$

$\Rightarrow m=2\pi {\sigma }_0(b-a)$

Radius of gyration,

$k=\sqrt{\frac{I}{m}}$

$\Rightarrow k=\sqrt{\frac{\left(2\pi {\sigma }_0/3\right)(b^3-a^3)}{2\pi {\sigma }_0(b-a)}}$

$\Rightarrow k=\sqrt{\frac{a^2+b^2+ab}{3}}$

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