Quantrex Academy · Free NEET PYQ solutions
NEET Physics Oscillations 2026 NEET 2026 (Cancelled)

NEET Physics Question (2026) — Solution

Question

The sum of kinetic energy and potential energy of a simple pendulum bob is 0.02 joule. The speed of the simple pendulum bob at equilibrium position is approximately : (Consider mass of the bob = 20 g)

Options

  1. A. 0.2 m/s
  2. B. 1.41 m/s
  3. C. 14.1 m/s
  4. D. 2.0 m/s

Answer

B. 1.41 m/s

Step-by-step solution

Total energy of the simple pendulum is given as E = 0.02 J. Mass of the bob, m = 20 g = 0.02 kg. At the equilibrium position, the potential energy is zero and the total energy is purely kinetic. E = 1 2 m v^2 0.02 = 1 2 0.02 v^2 v^2 = 2 v = 2 1.41 m/s. Answer: 1.41 m/s

Practice more on Quantrex App →

Related: Physics — Oscillations · All PYQ Banks