Three identical rigid cylinders A, B and C are arranged on smooth inclined surfaces as shown in figure. If the least value of  $\theta$ that prevent the arrangement from collapse is  $\tan \theta =\frac{1}{x\sqrt{x}}$. Then the value of x is
                                            

For the first cylinder(top),
$mg=2N\cos \beta$
For the second cylinder,
Along Y axis,  $mg+N\cos \beta -N_0\cos \theta =0$
Along X axis,  $N_1+N\sin \beta -N_0\sin \theta =0$
When the system starts to destroy, $N_1 = 0$
$\beta ={30}^0$
From first equation,
$N=\frac{mg}{2\cos \beta }$
From second equation,
$N_0=\frac{3mg}{2\cos \beta }$
From third equation,
 $\frac{mg}{2}\tan \beta -\frac{3mg}{2}\tan \theta =0\Rightarrow \tan \theta =\frac{1}{3}\tan \beta =\frac{1}{3\sqrt{3}}$