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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.
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Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- IF A⃗ and B⃗ are two vectors and θ is the angle between them
- Perfectly black body radiates the energy 18 J/s at 300K. Another ordinary body of e=0.8
- A ray of light is incident on a transparent glass slab of refractive index 1.62
- A stone of mass 1kg is tied to a string 4m long and is rotated at constant speed
- Two spherical bodies of mass M and 5 M and radii R and 2 R released in free space
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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