JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Work, Power & Energy

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A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle (

θ

) of thread deflection in the extreme position will be :

(A) $tan^{- 1} (2)$
(B) $2 tan^{- 1} (\frac{1}{2})$ ✓ Correct
(C) $tan^{- 1} (\frac{1}{2})$
(D) $2 tan^{- 1} (\frac{1}{5})$

Answer: (B)

Solution

By energy conservation between the lowest point and the extreme point, where the bob rises a height

ℓ (1 - cos θ)

\frac{1}{2} v^{2} = g ℓ (1 - cos θ) ⟹ \frac{v ^{2}}{ℓ} = 2 g (1 - cos θ) .

At the lowest point the acceleration is purely centripetal,

a_{l o w} = v^{2} / ℓ

; at the extreme point the speed is momentarily zero, so the acceleration is purely tangential,

a_{e x t} = g sin θ

. Equating the two magnitudes,

g sin θ = 2 g (1 - cos θ) ⟹ sin θ = 2 (1 - cos θ) .

Writing

sin θ = 2 sin \frac{θ}{2} cos \frac{θ}{2}

and

1 - cos θ = 2 sin^{2} \frac{θ}{2}

2 sin \frac{θ}{2} cos \frac{θ}{2} = 4 sin^{2} \frac{θ}{2} ⟹ tan \frac{θ}{2} = \frac{1}{2} ⟹ θ = 2 tan^{- 1} (\frac{1}{2}) .

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Single correct option

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JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Work, Power & Energy

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