Consider a bob of mass m and having charge q attached with a light string of length l and pivoted at 
point O. It is released at rest at ${60}^0$  with vertical. There are two regions– Region-I (left of line PQ) has a uniform and constant magnetic field B directed inside plane of paper. Region-II (right of line PQ) has a constant and uniform electric field E directed vertically up as shown. Consider no effect of gravity in both the regions.

Using work-energy theorem till string become taut
Work done by qE = Change in K.E.
 $\Rightarrow qEl=\frac{1}{2}m{v}^2-0$
 $\Rightarrow v=\sqrt{\frac{2qEl}{2m}}$
When string become taut an impulse is received and speed of the bob becomes
 $v\cos {30}^0=\sqrt{\frac{3qEl}{2m}}$
Further using work energy theorem speed of the bob becomes
$v^{\text{'}}=\sqrt{\frac{5qEl}{2m}}$
$\therefore$ time period $=\frac{2\pi l}{v^1}=2\pi \sqrt{\frac{2ml}{5qE}}$
After 1st half revolution in magnetic field work done by qE =Change in K.E., from lowermost to uppermost point.
$qE.2l=\frac{m{v}_1^2}{2}-\frac{m}{2}\frac{5qEl}{2m}$
$\therefore$Angular speed of 1st  revolution  $=\sqrt{\frac{5qE}{2ml}}$
$\therefore$ Angular speed of 2nd  revolution  $=\sqrt{\frac{13qE}{2ml}}$
$\therefore \text{ABC}$