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 ${60}^{\circ }$ 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.

$\Rightarrow qEl=\frac{1}{2}m{v}^2-0$

$\Rightarrow v=\sqrt{\frac{2qEl}{m}}$

When string become taut an impulse is received and speed of the bob becomes

$v\text{'}=\sqrt{\frac{5qEl}{2m}}$

$\therefore$                time period $=\frac{\pi l}{v\text{'}}=\pi \sqrt{\frac{2ml}{5qE}}$

After 1^st half revolution in magnetic field.

Work done by $qE=$ Change in K.E., from lowermost to uppermost point.

$qE\cdot 2l=\frac{m{v}_1^2}{2}-\frac{m}{2}\frac{5qEl}{2m}$

$v_1=\sqrt{\frac{13qEl}{2m}}$

$\therefore$       Time period of 2^nd revolution $=\frac{\pi l}{v_1}=\pi \sqrt{\frac{2ml}{13qE}}$

$\therefore$       Angular speed of 1^st revolution $=\sqrt{\frac{5qE}{2ml}}$

$\therefore$       Angular speed of 2^nd revolution $=\sqrt{\frac{13qE}{2ml}}$