Torque acting on a bar magnet kept in a magnetic field \\\\(\\vec{B}\\\\) is \\\\(\ au=\\vec{m}\ imes \\vec{B}=mBsin\ heta\\\\)Hence work done in moving from \\\\(\ heta=0^{\\circ}\\\\) to \\\\(\ heta=60^{\\circ}\\\\) is \\\\(W=mBcos 60^{\\circ}=\\dfrac{1}{2}mB\\\\)Hence torque acting when magnet is placed at \\\\(60^{\\circ}=\\dfrac{\\sqrt{3}}{2}mB=\\sqrt{3}W\\\\)