CUET · PHYSICS · PYQ PAPER 2023
A magnetic needle lying parallel to a magnetic field requires \( W \) units of work to turn it through \( 60^\circ \). The torque needed to maintain the needle in its position, will be :
- A \( \sqrt{3}W \)
- B \( W \)
- C \( \sqrt{3}W/2 \)
- D \(2W \)
Answer & Solution
Correct Answer
(A) \( \sqrt{3}W \)
Step-by-step Solution
Detailed explanation
Work done: \( W = MB( \cos 0^\circ - \cos 60^\circ ) \) \( W = MB( 1 - 1/2 ) = MB/2 \) \( MB = 2W \) Torque: \( \tau = MB \sin 60^\circ \) \( \tau = (2W) (\sqrt{3}/2) = \sqrt{3}W \)
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