Given, \(\theta=60^{\circ}=\frac{\pi}{3} \mathrm{rad}\)
The new magnetic length of the dipole will be, \(I^{\prime}=r\)
Now, \(M=m \cdot I\) and \(M^{\prime}=m \cdot I^{\prime}\)
\(\therefore \frac{\mathrm{M}}{\mathrm{M}^{\prime}}=\frac{\mathrm{I}}{\mathrm{I}^{\prime}} \ldots \quad (1)\)
Now, angle in radians \(=\frac{\text { arc length }}{\text { radius }}\)
\(\therefore \frac{\pi}{3}=\frac{1}{r} \Rightarrow 1=\frac{\pi r}{3}\)
In the equilateral triangle, marked by dotted lines, \(I^{\prime}=r\)
\(\therefore \frac{1}{I^{\prime}}=\frac{\pi}{3} \ldots \quad (2)\)
From and, we get,
\(\begin{aligned}
& \frac{M}{M^{\prime}}=\frac{\pi}{3} \\
& \therefore M^{\prime}=\frac{3 M}{\pi}
\end{aligned}\)
Hence, \((C)\) is the correct answer.
Alternative method:
Using the short cut formula about bending of a bar magnet, \(M=\frac{2 M \sin \left(\frac{\theta}{2}\right)}{\theta}\)
\(=\frac{2 M \sin 30^{\circ}}{\frac{\pi}{3}}=\frac{3 M}{\pi}\)
Hence, (C) is the correct answer.