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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.
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- A particle moves in a straight line with a constant acceleration
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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
- When the speed of electron beam used in Young’s double slit experiment is increased,
- A particle moves in a straight line with a constant acceleration
- A block of mass 5 kg is moving horizontally at a speed of 1.5 m/s. A perpendicular force
- The rest mass of a body is m1. It moves with a velocity of 0.6c, then its relativistic
- A short bar magnet of magnetic moment 0.4 JT⁻¹ is placed in a uniform magnetic field
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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