A magnetic needle suspended parallel to a magnetic field requires $\sqrt{3}$ J of work to turn it through $60°$. The torque needed to maintain the needle in this position will be:

1.  3 J 

2. $\sqrt{3} J$

3.  $\frac{3}{2}J$

4. $2\sqrt{3} J$

A short bar magnet of magnet moment 0.4 ${\mathrm{JT}}^{-1}$ is placed in a uniform magnetic field of 0.16 $\mathrm{T}$. The magnet is in stable equilibrium when the potential energy is :
1. 0.064 J
2. – 0.064 J
3. Zero
4. – 0.082 J