The correct option (A) √{2gℓ(1 – cos θ)}
Explanation:
cos θ = (h/ℓ) i.e. h = ℓ cos θ
∴ ℓ – h = ℓ – ℓ cos θ = ℓ (1 – cos θ) (1)
KE at 0 = PE at A
(1/2) mV2 = mgh'
V2 = 2gh'
here h' = ℓ (1 – cos θ) ----- from (1)
∴ V2 = 2gℓ (1 – cos θ) V = √{2gℓ(1 – cos θ)}