GATE Mechanical Engineering (ME) 2019 Shift 2 Solved Paper

$\frac{\mathit{c}}{\sqrt{\mathit{k}\mathit{m}}}=?;$ When System is critically damped
Suppose that we apply a external force downward at horizontal road at spring constant 3K and leave it freely for to and from motion.
Therefore, moment about point "O"; $\sum {\mathit{M}}_0=0$ Now, moment of inertia of rod about the center of the rod is ${\mathit{I}}_{\mathit{c}}=\mathit{x}\mathit{m}{\mathit{l}}^2$
And moment of inertia about the hinged point is ${\mathit{I}}_0={\mathit{I}}_{\mathit{C}}+\mathit{m}{\mathit{r}}^2=\frac{\mathit{m}{\mathit{L}}^2}{12}+\mathit{m}{\left(\frac{\mathit{L}}{4}\right)}^2=\frac{7}{48}\mathit{m}{\mathit{L}}^2$
$\frac{7}{48}\mathit{m}{\mathit{L}}^2\mathit{\theta }\stackrel{..}{\mathrm{\theta }}+\frac{1}{16}\mathit{C}{\mathit{L}}^2\stackrel{.}{\mathit{\theta }}+\frac{12}{16}\mathit{K}{\mathit{L}}^2\mathit{\theta }=0$
As we know that for critical dumping, the root of the quadratic equation should be real and equal.
$\therefore {\mathit{B}}^2-4\mathit{A}\mathit{C}=0$
${\left(\frac{1}{16}\mathit{C}{\mathit{L}}^2\right)}^2-4\left(\frac{7}{48}\mathit{m}{\mathit{L}}^2\right)\left(\frac{12}{16}\mathit{K}{\mathit{L}}^2\right)=0$
$\mathit{C}=\sqrt{4\left(\frac{7}{1}\mathit{m}\right)\left(\frac{4}{1}\mathit{K}\right)}=4\sqrt{7\mathit{k}\mathit{m}}$
$\therefore \frac{\mathit{C}}{\sqrt{\mathit{k}\mathit{m}}}=4\sqrt{7}$