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JEE Main 2021 (26 Aug, Shift 2) · Conical Pendulum & Funnel

Conical Pendulum & Funnel question from JEE Main 2021 (26 Aug, Shift 2)

Q13JEE Main 2021 · 26 AUG · SHIFT 2Medium

Conical Pendulum & Funnel

A particle of mass m is suspended from a ceiling through a string of length L. The particle moves in a horizontal circle of radius r such that $r = \frac{L}{\sqrt{2}}$ . The speed of particle will be:

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Answer: (a)

$\sin\theta = \frac{r}{L} = \frac{1}{\sqrt{2}} \Rightarrow \theta = 45^\circ$
Conical pendulum: $\tan\theta = \frac{v^2}{rg} \Rightarrow v^2 = rg\tan 45^\circ = rg$ .
So the answer is (a) $\sqrt{rg}$ .

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