engineering physics MCQ #4607

The time period of a simple pendulum in a lift accelerating downwards is

engineering physics MCQ #4607

  1. Question 1

    Q1. The time period of a simple pendulum in a lift accelerating downwards is

    • A) 2π √(l / (g - a))
    • B) 2π √(l / (g + a))
    • C) 2π √(l / g)
    • D) 2π √(l / a)

    Answer: 2π √(l / (g - a))

    Explanation: Effective g is g - a when lift accelerates downwards, so T = 2π √(l / (g - a)).