1.A magnetic needle lying parallel to the magnetic field requires WWW units of work to turn it through an angle 45∘45^\circ45∘. The torque required to maintain the needle in this position will bea.2W\sqrt{2}W2Wb.13W\frac{1}{\sqrt{3}}W31Wc.(2−1)W(\sqrt{2} - 1)W(2−1)Wd.W2−1\frac{W}{\sqrt{2}-1}2−1WLogin to continueOnly logged in users canattempt or see the solution.