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MHT CET 2025 (23 Apr, Shift 2) · Conical Pendulum & Funnel

Conical Pendulum & Funnel question from MHT CET 2025 (23 Apr, Shift 2)

Q3MHT CET 2025 · 23 APR · SHIFT 2Easy

Conical Pendulum & Funnel

An inextensible string of length ' $l$ ' fixed at one end, carries a mass ' $m$ ' at the other end. If the string makes $\frac{1}{\pi}$ revolutions per second around the vertical axis through the fixed end, the tension in the string is [The string makes an angle $\theta$ with the vertical]

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Answer: (c)

$\omega = 2\pi n = 2\pi \times \frac{1}{\pi} = 2$ rad/s
Horizontal: $T\sin\theta = m\omega^2 (l\sin\theta) \Rightarrow T = m\omega^2 l$ .
$T = 4\,ml$
So the answer is (c) 4 ml.

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