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. An unpolarised beam of intensity I is incident on a pair of nicols making an angle
  2. A wire having resistance 12Ω is bent in the form of an equilateral triangle.
  3. A wire of length 1 m is placed in a uniform magnetic field of 1.5 tesla at an angle
  4. Which of the following is true for elastic potential energy density?
  5. A 120 m long train is moving in a direction with speed 20 m/s.A train B moving

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Your email address will not be published.


*