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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: - A charged particle with a velocity 2×10³ ms⁻¹ passes undeflected through electric field
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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
- A charged particle with a velocity 2×10³ ms⁻¹ passes undeflected through electric field
- The potential difference applied to an X-ray tube is 5 kV and the current through it
- The drift speed of electrons in copper wire of diameter and length l is v.
- Angle of minimum deviation for a prism of reractive index 1.5 equal to the angle
- Which of the following statements is correct?
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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