JEE Main 2021 · 16 March · morning · Single correct · +4 / −1

A block of 200 g mass moves with a uniform speed in a horizontal circular groove, with vertical side walls of radius 20 cm. If the block takes 40 s to complete one round, the normal force by the side walls of the groove is :

  1. (a) 9.859 × 10⁻² N
  2. (b) 0.0314 N
  3. (c) 9.859 × 10⁻⁴ N
  4. (d) 6.28 × 10⁻³ N
Answer and solution

Answer (c) 9.859 × 10⁻⁴ N

Solution

Physics sir · 4 steps

The side wall is the only thing pushing the block inward, so its normal force is just mω² r.

  1. Who supplies the pull. The groove is horizontal. The floor balances gravity, and the vertical side wall pushes the block toward the centre. So the wall's normal force is the whole centripetal force.

    N = mω² r = m(2π/T)² r

  2. Put in SI units. Use m = 0.2 kg, r = 0.2 m and T = 40 s. Then ω = 2π/40 = π/20 rad s⁻¹.
  3. Calculate. N = 0.2 × π²/400 × 0.2 = π² × 10⁻⁴ N ≈ 9.87 × 10⁻⁴ N.
  4. So. The normal force by the side wall is about 9.86 × 10⁻⁴ N, which is option (c).

The idea to keep. In circular motion, first find which force points toward the centre. Set that force equal to mω² r, and change every quantity to SI units before you multiply.

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 2021.