| ⇦ | 
| ⇨ |
A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equillibrium state. The energy required to rotate it by 60⁰ is W. Now the torque required to keep the magnet in this new position is
Options
(a) W/√3
(b) √3W
(c) √3W/2
(d) 2W/√3
Correct Answer:
√3W
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - A particle executes simple harmonic oscillations with an amplitude
- An α-particle of energy 5 MeV is scattered through 180⁰ by a fixed uranium nucleus.
- In a photoelectric effect experiment, the slope of the graph between the stopping
- The difference in the lengths of a mean solar day and a sidereal day is about
- If α (current gain) of transistor is 0.98, then what is the value of β
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A particle executes simple harmonic oscillations with an amplitude
- An α-particle of energy 5 MeV is scattered through 180⁰ by a fixed uranium nucleus.
- In a photoelectric effect experiment, the slope of the graph between the stopping
- The difference in the lengths of a mean solar day and a sidereal day is about
- If α (current gain) of transistor is 0.98, then what is the value of β
Topics: Magnetic Effects of Current and Magnetism (167)
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