⇦ | ⇨ |
The point of suspension λ of a simple pendulum with normal time period T₁ is moving upward according to equation y=kt² where k=1 m/s². If new time period is T² the ratio T₁²/T₂² will be
Options
(a) (2/3)
(b) (5/6)
(c) (6/5)
(d) (3/2)
Correct Answer:
(6/5)
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - The current gain of a transistor in common base configuration is 0.96.
- Two satellites of earth, S₁ and S₂ are moving in the same orbit. The mass of S₁
- Surface tension of water will be minimum at
- Find out the e.m.f. produced when the current changes from 0 to 1 A in 10 sec.
- A particle of mass ‘m’ is kept at rest at a height 3R from the surface of earth
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The current gain of a transistor in common base configuration is 0.96.
- Two satellites of earth, S₁ and S₂ are moving in the same orbit. The mass of S₁
- Surface tension of water will be minimum at
- Find out the e.m.f. produced when the current changes from 0 to 1 A in 10 sec.
- A particle of mass ‘m’ is kept at rest at a height 3R from the surface of earth
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