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MHT CET 2026 (16 Apr, Shift 1) · Conical Pendulum & Funnel

Conical Pendulum & Funnel question from MHT CET 2026 (16 Apr, Shift 1)

Q2MHT CET 2026 · 16 APR · SHIFT 1Easy

Conical Pendulum & Funnel

A particle describes a horizontal circle of radius 'r' in conical funnel with smooth inner surface with a speed of 0.50.5 m/s. The height of the plane of the circle from the vertex of the funnel is (acceleration due to gravity, g=10g = 10 m/s2{}^{2})

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Answer: (b)

  1. Smooth funnel, semi-vertical angle θ\theta: Nsin⁡θ=mgN\sin\theta = mg, Ncos⁡θ=mv2rN\cos\theta = \frac{mv^2}{r}, so tan⁡θ=grv2\tan\theta = \frac{gr}{v^2}.
  2. Geometry: r=htan⁡θr = h\tan\theta, which gives h=v2gh = \frac{v^2}{g}.
  3. h=(0.5)210=0.025h = \frac{(0.5)^2}{10} = 0.025 m
  4. So h=2.5h = 2.5 cm, the answer is (b).
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