JEE Main 2023 · 24 January · evening · Single correct · +4 / −1

A body of mass 200g is tied to a spring of spring constant 12.5 N/m, 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. Then the ratio of extension in the spring to its natural length will be :

  1. (a) 1 : 2
  2. (b) 2 : 3
  3. (c) 2 : 5
  4. (d) 1 : 1
Answer and solution

Answer (b) 2 : 3

Solution

Physics sir · 4 steps

The stretched spring pulls the body around, so the radius is the natural length plus the extension.

  1. Spring supplies force. On a smooth horizontal surface the only horizontal force is the spring force kx, pointing towards O. It must provide the centripetal force mω² r.
  2. Radius of circle. The body sits at the end of the stretched spring, so r = l + x, where l is the natural length and x is the extension.

    kx = mω² (l + x)

  3. Put in numbers. mω² = 0.2 × 5² = 5 N m⁻¹. So 12.5x = 5(l + x), which gives 7.5x = 5l and x/l = 5/7.5 = 2/3.
  4. So. The ratio of extension to natural length is 2 : 3, option (b).

The idea to keep. For a mass whirled on a spring, always take the radius as natural length plus extension. Set kx = mω²(l + x) and solve for x/l.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Circular Motion

Asked about 70 times in JEE Main since 2020. This question is from 2023.