JEE Main 2023 · 6 April · morning · Single correct · +4 / −1

A small block of mass 100 g is tied to a spring of spring constant 7.5 N / m and length 20 cm. The other end of spring is fixed at a particular point A. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity 5 rad / s about point A, then tension in the spring is -

  1. (a) 0.50 N
  2. (b) 1.5 N
  3. (c) 0.75 N
  4. (d) 0.25 N
Answer and solution

Answer (c) 0.75 N

Solution

Physics sir · 4 steps

The spring stretches as the block goes round, so the radius is the natural length plus the stretch.

  1. Spring provides force. The surface is smooth and horizontal, so the spring tension alone supplies the centripetal force. If the spring stretches by x, the tension is kx and the radius becomes L + x, not L.

    kx = mω² (L + x)

  2. Put in numbers. mω² = 0.1 × 25 = 2.5 N m⁻¹, so 7.5x = 2.5(0.2 + x) = 0.5 + 2.5x. This gives 5x = 0.5, so x = 0.1 m.
  3. Check. The radius is 0.2 + 0.1 = 0.3 m, and mω² r = 2.5 × 0.3 = 0.75 N. This matches kx, as it must.
  4. So. Tension = kx = 7.5 × 0.1 = 0.75 N, option (c).

The idea to keep. When a spring holds a body in circular motion, it stretches. Write kx = mω²(L + x) and solve for x. Never take the natural length as the radius.

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.