| ⇦ | 
| ⇨ |
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: - A ball is projected upwards from the foot of a tower. The ball crosses the top
- The ratio of radii of planets A and B is K₁ and ratio of accelerations
- When ₉₀Th²²⁸ transforms to ₈₃Bi²¹², the number of emitted α and β particles
- The magnetic field at the centre of a circular coil carrying current I ampere is B.
- If in a p-n junction, a square input signal of 10 V is applied as shown, then the output
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A ball is projected upwards from the foot of a tower. The ball crosses the top
- The ratio of radii of planets A and B is K₁ and ratio of accelerations
- When ₉₀Th²²⁸ transforms to ₈₃Bi²¹², the number of emitted α and β particles
- The magnetic field at the centre of a circular coil carrying current I ampere is B.
- If in a p-n junction, a square input signal of 10 V is applied as shown, then the output
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