| ⇦ | 
| ⇨ |
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: - If the alternating current I=I₁ cosωt+I₂ sinωt, then the rms current is given by
- Consider a two particle system with particle having masses m₁ and m₂. If the first
- A stone is tied to a string of length l and whirled in a vertical circle with the other end
- A photoelectric surface is illuminated successively by monochromatic light of wavelength
- The speed of an electron having a wavelength of 10⁻¹⁰ m is
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If the alternating current I=I₁ cosωt+I₂ sinωt, then the rms current is given by
- Consider a two particle system with particle having masses m₁ and m₂. If the first
- A stone is tied to a string of length l and whirled in a vertical circle with the other end
- A photoelectric surface is illuminated successively by monochromatic light of wavelength
- The speed of an electron having a wavelength of 10⁻¹⁰ m is
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.
Share this page with your friends
Leave a Reply