| ⇦ | 
| ⇨ |
The acceleration due to gravity at a place is π² m/sec². Then the time period of a simple pendulum of length one metre is
Options
(a) 2/π sec
(b) 2π sec
(c) 2 sec
(d) π sec
Correct Answer:
2 sec
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - A body of mass m is accelerated uniformly from rest to a speed v in a time T
- Two sources P and Q produce notes of frequency 660 Hz each.
- A stone of mass 1 kg tied to a light inexensible string of length L=10/3 is whirling
- The dimensional formula for Young’s modulus is
- A parallel plate capacitor of a capacitance 1 pF has seperation between the plates
Topics: Oscillations
(58)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A body of mass m is accelerated uniformly from rest to a speed v in a time T
- Two sources P and Q produce notes of frequency 660 Hz each.
- A stone of mass 1 kg tied to a light inexensible string of length L=10/3 is whirling
- The dimensional formula for Young’s modulus is
- A parallel plate capacitor of a capacitance 1 pF has seperation between the plates
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