A stone of mass 1 kg is tied to end of a massless string of length 1 m. If the breaking tension of the string is 400 N, then maximum linear velocity, the stone can have without breaking the string, while rotating in horizontal plane, is: (a) 20 ms⁻¹ (b) 40 ms⁻¹ (c) 400 ms⁻¹ (d) 10 ms⁻¹

Answer: (a). Tension supplies the centripetal force: T = mv²/L (weight neglected). 400 = (1 × v²)/(1) ⇒ v² = 400 v = 20 m s ⁻¹ So the answer is (a) 20 ms⁻¹.

A stone of mass 1 kg is tied to end of a massless string of length 1… | JEE Main 2023 (31 Jan, Shift 2)

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Conical Pendulum & Funnel

A stone of mass $1\text{ kg}$ is tied to end of a massless string of length $1\text{ m}$ . If the breaking tension of the string is $400\text{ N}$ , then maximum linear velocity, the stone can have without breaking the string, while rotating in horizontal plane, is:

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Answer: (a)

Tension supplies the centripetal force: $T = \frac{mv^2}{L}$ (weight neglected).
$400 = \frac{1 \times v^2}{1} \Rightarrow v^2 = 400$
$v = 20$ m s $^{-1}$
So the answer is (a) $20\text{ ms}^{-1}$ .

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