| ⇦ | 
| ⇨ |
The string of a pendulum is horizontal.The mass of bob attached to it is m.Now, the string is released.The tension in the string in the lowest position is
Options
(a) mg
(b) 2 mg
(c) 3 mg
(d) 4 mg
Correct Answer:
3 mg
Explanation:
When string is released, then according to conservation of energy,
(1/2) mv² = mgh
v² = 2gh = 2gR
Now, when it is at the bottom,
T – mg cos 0² = mv²/R
or T – mg = (m/R)(2gR)
T = mg + 2mg = 3mg.
Related Questions: - “On flowing current in a conducting wire the magnetic field is produced around it”.
- A small angled prism of refractive index 1.4 is combined with another small angled prism
- A remote-sensing satellite of earth revolves in a circular orbit at a height
- The half-life period of a radio-active element X is same as the mean life time
- An automobile engine develops 100 kW when rotating at a speed of 1800 rev/min
Topics: Laws of Motion
(103)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- “On flowing current in a conducting wire the magnetic field is produced around it”.
- A small angled prism of refractive index 1.4 is combined with another small angled prism
- A remote-sensing satellite of earth revolves in a circular orbit at a height
- The half-life period of a radio-active element X is same as the mean life time
- An automobile engine develops 100 kW when rotating at a speed of 1800 rev/min
Topics: Laws of Motion (103)
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