| ⇦ | 
| ⇨ |
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: - Dimensions of resistance in an electrical circuit, in terms of dimension of mass M,
- A man goes 10 m towards north, then 20m towards east, the displacement is
- Young’s modulus of the material of a wire is 18×10¹¹ dyne cm⁻².its value in SI is
- An object placed at 20 cm in front of a concave mirror produces three times
- If (range)² is 48 times (maximum height)², then angle of projection is
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Dimensions of resistance in an electrical circuit, in terms of dimension of mass M,
- A man goes 10 m towards north, then 20m towards east, the displacement is
- Young’s modulus of the material of a wire is 18×10¹¹ dyne cm⁻².its value in SI is
- An object placed at 20 cm in front of a concave mirror produces three times
- If (range)² is 48 times (maximum height)², then angle of projection 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