The time period of a simple pendulum in a lift descending with constant acceleration g is

The Time Period Of A Simple Pendulum In A Lift Physics Question

The time period of a simple pendulum in a lift descending with constant acceleration g is

Options

(a) T=2π√l/g
(b) T=2π√l/2g
(c) zero
(d) Infinite

Correct Answer:

Infinite

Explanation:

When the lift falls freely under gravity, effective ‘g’ for pendulum in the lift = zero.
T = 2π √(l/g) = 2π√(l/o) = Infinite.

Related Questions:

  1. A ship A is moving westwards with a speed of 10 km/h .The time after
  2. Which of the following parameters does not characterize the thermodynamic state
  3. A source and an observer approach each other with same velocity 50 m/s. If the frequency
  4. If doubly ionised lithium atom is hydrogen like with atomic number 3, the wavelength
  5. Two particles P and Q start from origin and execute simple harmonic motion along X-axis

Topics: Oscillations (58)
Subject: Physics (2479)

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

NEETLab Mobile App

Share this page with your friends

Be the first to comment

Leave a Reply

Your email address will not be published.


*