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Step-by-Step Solution
Step 1: Understand the Physical Situation
A magnetic needle, which can be treated as a magnetic dipole of magnetic moment $M$, is placed in a uniform magnetic field $B$. Initially, the needle is aligned parallel to the field (making an angle $0^\circ$ with the field). We rotate it to an angle $60^\circ$.
Step 2: Express the Work Done in Turning the Needle
The work done $W$ in rotating a magnetic dipole in a uniform field from an initial angle $\theta_1$ to a final angle $\theta_2$ is given by:
$$
W = MB \bigl(\cos\theta_1 - \cos\theta_2\bigr).
$$
In this problem, $\theta_1 = 0^\circ$ and $\theta_2 = 60^\circ$, so
$$
W = MB \bigl(\cos 0^\circ - \cos 60^\circ\bigr) = MB \bigl(1 - \tfrac12\bigr) = \tfrac{MB}{2}.
$$
Step 3: Relate $MB$ to the Work Done
From the above expression, we see that
$$
W = \frac{MB}{2} \quad \Longrightarrow \quad MB = 2W.
$$
Step 4: Find the Torque at $60^\circ$
The torque $\tau$ acting on a magnetic dipole of magnetic moment $M$ in a uniform field $B$, when the dipole makes an angle $\theta$ with the field, is:
$$
\tau = MB \,\sin\theta.
$$
For $\theta = 60^\circ$, we get:
$$
\tau = MB \,\sin(60^\circ) = MB \times \frac{\sqrt{3}}{2}.
$$
Using $MB = 2W$ from Step 3,
$$
\tau = \left(2W\right)\,\frac{\sqrt{3}}{2} = \sqrt{3}\,W.
$$
Step 5: Conclude the Correct Option
The torque needed to maintain the needle at $60^\circ$ is therefore
$$
\tau = \sqrt{3}\,W.
$$
Hence, the correct answer is $\sqrt{3}\,W$.
