A bob of heavy mass m is suspended by a light string of length l. The bob is given a horizontal velocity v₀ as shown in figure. If the string gets slack at some point P making an angle θ from the horizontal, the ratio of the speed v of the bob at point P to its initial speed v₀ is:

Question

Question

This question was previously asked in

NEET 2025 Official Paper (Held On: 04 May, 2025)

Answer 1

Correct option is: (4)

At Point P, mg sin θ = m v² / l ... (1)

By conservation of mechanical energy at point P ∈ Q

(1/2) mv₀² = (1/2) mv² + mg (I + I sin θ)

⇒ v₀² / 2 = v² / 2 + gl(1 + sin θ)

Put gl = v² / l sin θ using (1)

v₀² / 2 = v² / 2 + (v² / sin θ)(1 + sin θ)

v02 / 2 = (3v2/2) + (v2 / sin θ)

⇒v / v₀ = (sin θ /(2 + 3 sin θ))^1/2