Question 74 of 280

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 $\left(\dfrac{{\sigma }_{0}}{r}\right)$ then, the radius of gyration of the disc about its axis passing through the center is

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

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Answer: (B) $\sqrt{\dfrac{{a}^{2}+{b}^{2}+ab}{3}}$

Solution: (B)
$dm=\sigma 2\pi rdr$
$=\dfrac{{\sigma }_{0}}{r}2 \pi rdr$
$m=2\pi {\sigma }_{0}{\int }_{a}^{b}dr=2\pi {\sigma }_{0}(b-a)$

$I=2\pi {\sigma }_{0}{\int }_{a}^{b}{r}^{2}dr=\dfrac{2}{3}\pi {\sigma }_{0}\left({b}^{3}-{a}^{3}\right)$
Radius of gyration $=\sqrt{\dfrac{I}{m}}=\sqrt{\dfrac{\dfrac{2}{3}\pi {\sigma }_{0}\left({b}^{3}-{a}^{3}\right)}{2\pi {\sigma }_{0}\left(b-a\right)}}$
$=\sqrt{\dfrac{{b}^{3}-{a}^{3}}{3(b-a)}}=\sqrt{\dfrac{(b-a)({b}^{2}+{a}^{2}+ab)}{3 (b-a)}}$
$=\sqrt{\dfrac{{b}^{2}+{a}^{2}+ab}{3}}$.