A Horizontal electric field E = mg/q exists as shown in the figure and a mass m is attached at the end of a light rod. if mass m is released from the position as shown in the figure. find the angular velocity at the bottom-most position

Hello Student

According to the work energy theorem we have
W_e​ + W_{g ​}= ​(mv²)/2
We have work done by electrostatic force as
W_e​= qElsinθ
and work done by the gravitational force as
W_g​ = mg(l−lcosθ)
Thus we get
qElsinθ + mg(l−lcosθ) =​ (mv²)/2
Thus we get
mgsinθ + mgl − mglcosθ = ​(mv²)/2
asθ = 45^o, we get
mgl =​ (mv²)/2
also as v = ωl
we get
ω =√ (2g/l)

Hence option B is correct.

I hope this answer will help you.