Figure shows a wedge on a smooth horizontal ground. two blocks of masses $m_1$ and $m_2$ are released from rest on the smooth and inclined surfaces of the wedge of the acceleration of the wedge is zero, the value of  $\left(\frac{m_2^2}{m_1^2}\right)=\frac{x}{3}$, then value of $x$ is

Let ${\mathrm{N}}_1, {\mathrm{N}}_2$ be the normal forces by to ${\mathrm{m}}_1, {\mathrm{m}}_2$ respectively on the wedge.

$\sum F_x=0$  on wedge and also  $\sum F_y=0$

$\Rightarrow {\mathrm{N}}_1{\mathrm{sin\theta }}_1={\mathrm{N}}_2{\mathrm{sin\theta }}_2$......(1)

Normal force due to either masses is

${\mathrm{N}}_1={\mathrm{m}}_1{\mathrm{gcos\theta }}_1 \mathrm{and} {\mathrm{N}}_2={\mathrm{m}}_2{\mathrm{gcos\theta }}_2$.......(2)

From (1) and (2)

${\mathrm{m}}_1{\mathrm{gcos\theta }}_1{\mathrm{sin\theta }}_1= {\mathrm{m}}_2{\mathrm{gcos\theta }}_2{\mathrm{sin\theta }}_2$

$\Rightarrow \frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}=\frac{\sin 2{\mathrm{\theta }}_2}{\sin 2{\mathrm{\theta }}_1}=\frac{\sqrt{3}}{2}$

$\Rightarrow \frac{{\mathrm{m}}_2^2}{{\mathrm{m}}_1^2}=\frac{4}{3}$