Kinematics

JEE Main level, single correct, medium

290 of 308

Question 290 of 308

A stone of mass 1kg1kg is tied to end of a massless string of length 1m1m. If the breaking tension of the string is 400N400N, then maximum linear velocity, the stone can have without breaking the string, while rotating in horizontal plane, is:

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Answer: (A) 20ms−120m{s}^{-1}

In horizontal circle, tension towards the centre of the circular path at any moment will provide the required centripetal force.
Therefore,
Tmax=mv2r⇒400=1×v21⇒v=20ms−1{T}_{max}=\dfrac{m{v}^{2}}{r} \Rightarrow 400=\dfrac{1\times {v}^{2}}{1} \Rightarrow v=20m{s}^{-1}

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