Physics Circle home Student Login

Home›Chapters›Circular Motion PYQs›Centripetal Acceleration & Force›Question 34

JEE Main 2021 (16 Mar, Shift 1) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from JEE Main 2021 (16 Mar, Shift 1)

Q34JEE Main 2021 · 16 MAR · SHIFT 1Easy

Centripetal Acceleration & Force

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:

Show answer and solution

Answer: (c)

  1. The side wall's normal force provides the centripetal force: N=mω2rN = m\omega^2 r.
  2. ω=2π40=0.157 rad/s\omega = \frac{2\pi}{40} = 0.157\text{ rad/s}, ω2=0.02467\omega^2 = 0.02467
  3. N=0.2×0.02467×0.2=9.87×10−4 NN = 0.2 \times 0.02467 \times 0.2 = 9.87 \times 10^{-4}\text{ N}
  4. So the answer is (c) 9.859×10−4 N9.859 \times 10^{-4}\text{ N}.
Report a mistake

More questions on Centripetal Acceleration & Force