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MHT CET 2024 (10 May, Shift 2) · Conical Pendulum & Funnel

Conical Pendulum & Funnel question from MHT CET 2024 (10 May, Shift 2)

Q7MHT CET 2024 · 10 MAY · SHIFT 2Easy

Conical Pendulum & Funnel

A particle rotates in a horizontal circle of radius 'R' in a conical funnel with constant speed 'V'. The inner surface of the funnel is smooth. The height of the plane of the circle from the vertex of the funnel is (g-acceleration due to gravity)

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Answer: (d)

  1. Normal reaction: vertical part balances mgmg, horizontal part gives mV2R\frac{mV^2}{R}, so tan⁡θ=gRV2\tan\theta = \frac{gR}{V^2} (θ\theta = semi-vertical angle).
  2. Geometry of the cone: R=htan⁡θ=hgRV2R = h\tan\theta = h\frac{gR}{V^2}.
  3. h=V2gh = \frac{V^2}{g}
  4. So the answer is (d).
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