Magnetic Properties of Matter

JEE Main level, single correct, medium

33 of 57

Question 33 of 57

A magnetic needle lying parallel to a magnetic field requires W\mathrm{W} units of work to turn it through 60∘60^{\circ}. The torque needed to maintain the needle in this position will be

Show answer and solution

Answer: (A) 3 W\sqrt{3} \mathrm{~W}

τ=(H)tan⁡60∘=W⋅3\tau=(\mathrm{H}) \tan 60^{\circ} \quad=\mathrm{W} \cdot \sqrt{3}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Magnetic Properties of Matter (Magnetism and Properties of Magnet), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.