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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: - An ideal spring with spring constant K is hung from the ceiling and a block of mass
- Two lenses of power 15 D and -3 D are placed in contact.
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Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An ideal spring with spring constant K is hung from the ceiling and a block of mass
- Two lenses of power 15 D and -3 D are placed in contact.
- A solid sphere of mass M, radius R and having moment of inertia about an axis passing
- The mean radius of the earth is R, its angular speed on its own axis is ?
- If a magnet is suspended at an angle 30⁰ to the magnetic meridian, the dip needle
Topics: Magnetic Effects of Current and Magnetism (167)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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