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JEE Main202324 Jan 2023Evening ShiftPhysicsLaws of MotionActual

A body of mass 200 g is tied to a spring of spring constant 12 . 5 N m - 1 , while the other end of spring is fixed at point O . If the body moves about O in a circular path on a smooth horizontal surface with constant angular speed 5 rad s - 1 , then the ratio of extension in the spring to its natural length will be :

Options

  1. A1 : 2
  2. B1 : 1
  3. C2 : 3
  4. D2 : 5

Correct answer

C. 2 : 3

Step-by-step solution

Let the extension in spring be x and the natural length of spring be l . Here the centrifugal force will stretch the spring. So, centrifugal force = spring force k x = m l + x ω 2 ⇒ 12 . 5 x = 1 5 l + x 25 ⇒ 1 . 5 x = l ⇒ x l = 1 1 . 5 = 2 3 ⇒ x : l = 2 : 3

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