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A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60⁰. The torque needed to maintain the needle in this position will be:
Options
(a) 2 √3 J
(b) 3 J
(c) √3 J
(d) 3/2 J
Correct Answer:
3 J
Explanation:
According to work energy theorem
W = U final – U initial = MB (cos 0 – cos 60⁰)
W = MB/2 = √3J …(i)
? = M⃗ x B⃗ = MB sin 60⁰ = (MB √3 / 2) …(ii)
From eq (i) and (ii)
? = 2 √3 x √3 / 2 = 3 J
Related Questions: - If the terminal speed of a sphere of gold (density=19.5 kg/m³) is 0.2 m/s
- A uranium nucleus ₉₂ U ²³⁸ emits an α-particle and a β-particle in succession.
- If light emitted from sodium bulb is passed through sodium vapour,
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- Calculate the charge on equivalent capacitance of the conbination shown in figure
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If the terminal speed of a sphere of gold (density=19.5 kg/m³) is 0.2 m/s
- A uranium nucleus ₉₂ U ²³⁸ emits an α-particle and a β-particle in succession.
- If light emitted from sodium bulb is passed through sodium vapour,
- A proton is projected with a speed of 3×10⁶ m/s horizontally from east to west.
- Calculate the charge on equivalent capacitance of the conbination shown in figure
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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