$ \Rightarrow {\rm{N}}i{\rm{AB}} = {\rm{C}}\frac{\pi }{2} \Rightarrow {\rm{C}} = \frac{{2{\rm{N}}i{\rm{AB}}}}{\pi }$         ........$(1)$

When $Q$ is passed,

$\int \tau  dt = \int {Ni} ABdt$

$\mathrm{I} \omega=\mathrm{N} \mathrm{A} \mathrm{B} \mathrm{Q}$       ........$(2)$

Also, maximum deflection happens when entire $\frac{1}{2} \mathrm{I} \omega^{2}$ converts to $\frac{1}{2} \mathrm{C} \theta_{\mathrm{max}}^{2}$

So ${\theta _{\max }} = \sqrt {\frac{{{\rm{I}}{\omega ^2}}}{{\rm{C}}}}  = \frac{1}{{\sqrt {{\rm{IC}}} }}{\rm{I}}\omega  = \frac{{{\rm{NABQ}}}}{{\sqrt {{\rm{IC}}} }}$        (from $(1)$)