Q67 / 90 · JEE Main 2022 (24 Jun Shift 1) · Physics

A boy ties a stone of mass 100 g to the end of a 2 m long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of 80 N. If the maximum speed with which the stone can revolve is Kπ rev min-1. The value of K is :
(Assume the string is massless and un-stretchable)

  • A
    400
  • B
    300
  • Correct answer
    C
    600
  • D
    800

Solution

The maximum angular velocity of the stone is ω=rpm×2π60=K30 rad s-1

The tension in the string will provide the required centripetal force. Therefore,

T=mω2l⇒80=0.1×K302×2⇒K2=360000⇒K=600

Hence, the correct answer is option: 'C'.

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