Q64 / 90 · JEE Main 2022 (27 Jun Shift 2) · Physics

One end of a massless spring of spring constant k and natural length l0 is fixed while the other end is connected to a small object of mass m lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity ω about an axis passing trough fixed end, then the elongation of the spring will be

  • A
    k-mω2l0mω2
  • B
    mω2l0k+mω2
  • Correct answer
    C
    mω2l0k-mω2
  • D
    k+mω2l0mω2

Solution

The particle is moving in a horizontal circle, so it is accelerated towards the centre with magnitude v2r. The horizontal force on the particle is due to the spring and is given by kx, where x is the elongation and k is the spring constant.

kx=mv2r=mω2r=mω2l0+x

⇒k-mω2x=mω2l0

⇒x=mω2l0k-mω2

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

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