A magnetic needle is in the most equilibrium state in a uniform magnetic field requires "W" units of work to turn through 60°. The work done to turn the magnetic needle from the position of most stable equilibrium to the most unstable equilibrium position is

4W

The correct answer is Option (4) → 4W

The potential energy is: $U(θ)=-mB\cos θ$

The work done in rotating the needle from $θ_1$ to $θ_2$,

$W=U(θ_1)-U(θ_2)$

$=U(0°)-U(60°)$

$=-mB(1)+mB(\frac{1}{2})=\frac{mB}{2}=W$

$W_{unstable}=U(0°)-U(180°)$

$=(-mB\cos 0°)-(-mB(-1))$

$=2mB$

$∴W_{unstable}=2mB=2×2W=4W$   $[mB=2W]$