$\text{ Moseley's equation is represented }\mathrm{as}\sqrt{\mathrm{v}}=\mathrm{a}(\mathrm{Z}-\mathrm{b})\text{ where, }\mathrm{a}\text{ and }\mathrm{b}\text{ are constants. }$

If OA =1,then atomic number of the element showing frequency of 400 Hz is 

$\sqrt{\mathrm{v}}=\mathrm{a}(\mathrm{Z}-\mathrm{b})=\mathrm{aZ}-\mathrm{ab}$

Thus, graph is a straight line

$\begin{array}{l}\text{ At }\sqrt{\mathrm{v}}=0,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{aZ}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\mathrm{ab}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{Z}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\mathrm{b}=\mathrm{OA}=1\\ \text{ Thus, }\mathrm{b}=1\text{ and at }\mathrm{Z}=0,\sqrt{\mathrm{v}}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=-\mathrm{ab}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ Slope }=\mathrm{a}=\tan {45}^{\circ }=1\\ \text{ Thus, }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\sqrt{\mathrm{v}}=(\mathrm{Z}-1)\\ \text{ If }\mathrm{v}=400\mathrm{Hz},\text{ then }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ \therefore \sqrt{400}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\mathrm{Z}-1\\ 20\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\mathrm{Z}-1\\ \therefore \mathrm{Z}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=21\end{array}$