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MHT CET 2023 (11 May, Shift 2) · Conical Pendulum & Funnel

Conical Pendulum & Funnel question from MHT CET 2023 (11 May, Shift 2)

Q8MHT CET 2023 · 11 MAY · SHIFT 2Easy

Conical Pendulum & Funnel

In a conical pendulum the bob of mass 'mm' moves in a horizontal circle of radius 'rr' with uniform speed 'VV'. The string of length 'LL' describes a cone of semi vertical angle 'θ\theta'. The centripetal force acting on the bob is ( g=g = acceleration due to gravity)

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Answer: (a)

  1. Tcos⁡θ=mgT\cos\theta = mg and Tsin⁡θ=FcT\sin\theta = F_c, so Fc=mgtan⁡θF_c = mg\tan\theta.
  2. Geometry: sin⁡θ=rL\sin\theta = \frac{r}{L}, height of cone =L2−r2= \sqrt{L^2 - r^2}, so tan⁡θ=rL2−r2\tan\theta = \frac{r}{\sqrt{L^2-r^2}}.
  3. Fc=mgrL2−r2F_c = \frac{mgr}{\sqrt{L^2-r^2}}
  4. So the answer is (a).
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