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MHT CET 2023 (11 May, Shift 1) · Conical Pendulum & Funnel

Conical Pendulum & Funnel question from MHT CET 2023 (11 May, Shift 1)

Q9MHT CET 2023 · 11 MAY · SHIFT 1Easy

Conical Pendulum & Funnel

A ball of mass 'mm' is attached to the free end of a string of length 'll'. The ball is moving in horizontal circular path about the vertical axis as shown in the diagram.

The angular velocity 'ω\omega' of the ball will be [ T=T = Tension in the string.]

Figure

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Answer: (d)

  1. Radius of the circle r=lsin⁡θr = l\sin\theta.
  2. Horizontal component of tension gives centripetal force: Tsin⁡θ=mω2lsin⁡θT\sin\theta = m\omega^2 l\sin\theta.
  3. T=mω2l⇒ω=TmlT = m\omega^2 l \Rightarrow \omega = \sqrt{\frac{T}{ml}}
  4. So the answer is (d).
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