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. The power obtained in a reactor using U²³⁵ disintegration is 1000 kW
  2. A metal rod of length l cuts across a uniform magnetic field B with a velocity v.
  3. A particle moves a distance x in time but according to equation x=(t+5)⁻¹
  4. If the alternating current I=I₁ cos⁡ωt+I₂ sin⁡ωt, then the rms current is given by
  5. A choke is preferred to a resistance for limiting current in AC circuit because

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Your email address will not be published.


*