A cubical block of side  30cm is moving with velocity 2m/s  on a smooth horizontal surface. The surface has a bump at a point O as shown in figure. The angular velocity (in rad/sec) of the block immediately after it hits the bump, is:

1. 
Given Side of block  $a=30\text{\hspace{0.17em}cm=0.3\hspace{0.17em}m}$
Velocity  $v=2\text{\hspace{0.17em}m}/\text{s}$
Before hitting point O ,We know angular momentum $L=mvr=mv\frac{a}{2}$
Also, after hitting point O, $L=I\omega$
2. Using conservation of angular momentum,  $mv\frac{a}{2}=I\omega$
3. Moment of inertia (I) of cubical block about one edge is given by
$I=m\left(\frac{a^2}{6}+{\left(\frac{a}{\sqrt{2}}\right)}^2\right)$ 
$\therefore \omega =\frac{m.2\frac{a}{2}}{m\left[\frac{a^2}{6}+{\left(\frac{a}{\sqrt{2}}\right)}^2\right]}$ 
$\Rightarrow \omega =\frac{12}{8a}=\frac{3}{2\times 0.3}=\frac{10}{2}=5\text{\hspace{0.17em}rad}/\text{sec}$ 
Therefore, the correct answer is (B).