A bar magnet is held perpendicular to a uniform field. If the couple acting on the magnet is to be halved by rotating it, the angle by which it is to be rotated is

Given that ${\mathrm{\theta }}_1={90}^{\circ }$
$\therefore$Initial torque ${\mathrm{\tau }}_1=\mathrm{\tau }$
Final torque ${\mathrm{\tau }}_2=\mathrm{\tau }/2$
$$\begin{gathered} \text{ Now }\mathrm{\tau }={\mathrm{\tau }}_1=\mathrm{MBsin}{\mathrm{\theta }}_1=\mathrm{MBsin}{90}^{\circ }=\mathrm{MB} \\ \begin{array}{r}\text{ and }{\mathrm{\tau }}_2=\mathrm{MBsin}{\mathrm{\theta }}_2\\ \text{ or }\mathrm{\tau }/2=\mathrm{MBsin}{\mathrm{\theta }}_2\end{array} \\ \therefore \frac{\mathrm{MB}}{2}=\mathrm{MBsin}{\mathrm{\theta }}_2\text{ or }\sin {\mathrm{\theta }}_2=\frac{1}{2} \\ \therefore {\mathrm{\theta }}_2={30}^{\circ } \end{gathered}$$
So, the angle by which the bar magnet is to be rotated $\left({\mathrm{\theta }}_2\right)={30}^{\mathrm{o}}$. $d\theta ={\theta }_1-{\theta }_2={90}^0-{30}^0={60}^0$