| ⇦ |
| ⇨ |
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: - Photocells convert
- A pellet of mass 1 g is moving with an angular velocity of 1 rad/s along a circle
- Three capacitors each of capacitance C and of breakdown voltage V are joined in series
- Which dimensions will be the same as that of time?
- If the magnetising field on a ferromagnetic material is increased, its permeability
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Photocells convert
- A pellet of mass 1 g is moving with an angular velocity of 1 rad/s along a circle
- Three capacitors each of capacitance C and of breakdown voltage V are joined in series
- Which dimensions will be the same as that of time?
- If the magnetising field on a ferromagnetic material is increased, its permeability
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