\(\tau \,\Delta t = \Delta L\)

\(mg\frac{\ell }{2} \times 0.01 = \frac{{m{\ell ^2}}}{3}\omega \)

\(\omega  = \frac{{3g \times 0.01}}{{2\ell }} = \frac{{3 \times 10 \times 0.01}}{{2 \times 0.3}} = \frac{1}{2} = 0.5\,rad/s\)

Time taken by rod to hit the ground 

\(t\sqrt {\frac{{2h}}{g}}  = \sqrt {\frac{{2 \times 5}}{{10}}}  = 1\,\sec .\)

In this time angule rotate by rod

\(\theta  = \omega t = 0.5 \times 1 = 0.5\,radian\)