The correct option (B) t = (π/2)√[(Mℓ)/(2qE)
Explanation:
![](https://www.sarthaks.com/?qa=blob&qa_blobid=4900539320589117434)
It is an electric dipole.
τ = Force × perpendicular distance
= F × AC
= F × AB sinθ
= qE × ℓ × sinθ
as θ = 5° sin θ is very small
∴ τ = qEℓθ
τ = Rotating torque = – qEℓθ (1)
∵ τ = Iα
& I = M × (AO)2 + M(BO)2
= M(ℓ/2)2 + M(ℓ/2)2 = (Mℓ/2)2
∴ τ = (Mℓ2/2) α
i.e. α = (2τ/Mℓ2)
from (1)
α = [(– 2qEℓθ)/(Mℓ2)] = [(– 2qE)/(Mℓ)]θ
as α = – ω2θ
∴ ω2 = [(2qE)/(Mℓ)] (2)
Time period = T = (2π/ω)
T = [(2π)/{(2qE)/(Ml)}(1/2)] From (2)
Rotating in clockwise sense, the minimum time taken by rod to align itself parallel to field = time to complete (1/4)th of oscillation
∴ tmin = (T/4) = (π/2) [(Mℓ)/(2qE)](1/2)