⇦ | ⇨ |
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: - The 6563 Å line emitted by hydrogen atom in a star is found to be red shifted by 5Å
- The wheel of a car is rotating at the rate of 1200 rev/min. On pressing
- The de-Broglie wavelength of an electron having 80 eV of energy is nearly
- The ionisation energy of hydrogen is 13.6 eV. The energy of the photon released
- The electirc field in a certain region is acting radially outward and is given by E=Ar.
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The 6563 Å line emitted by hydrogen atom in a star is found to be red shifted by 5Å
- The wheel of a car is rotating at the rate of 1200 rev/min. On pressing
- The de-Broglie wavelength of an electron having 80 eV of energy is nearly
- The ionisation energy of hydrogen is 13.6 eV. The energy of the photon released
- The electirc field in a certain region is acting radially outward and is given by E=Ar.
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