A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in the equilibrium state. The energy required to rotate it by \(60^{\circ}\) is \(W\). Now the torque required to keep the magnet in this new position is:
1. \(\frac{W}{\sqrt{3}}\) 
2. \(\sqrt{3} W\)
3. \(\frac{\sqrt{3} W}{2}\) 
4. \(\frac{2 W}{\sqrt{3}}\)

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

A closely wound solenoid of \(2000\) turns and area of cross-section \(1.5\times10^{-4}\) m² carries a current of \(2.0\) A. It is suspended through its center and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field \(5\times 10^{-2}\) tesla making an angle of \(30^{\circ}\) with the axis of the solenoid. The torque on the solenoid will be:
1. \(3\times 10^{-3}\) Nm 
2. \(1.5\times 10^{-3}\) Nm 
3. \(1.5\times 10^{-2}\) Nm 
4. \(3\times 10^{-2}\) Nm

Magnets \(A\) and \(B\) are geometrically similar but the magnetic moment of \(A\) is twice that of \(B\). If \(T_1\) and \(T_2\) be the time periods of the oscillation when their like poles and unlike poles are kept together respectively, then \(\frac{T_1}{T_2}\) will be:
1. \(\frac{1}{3}\)
2. \(\frac{1}{2}\)
3. \(\frac{1}{\sqrt{3}}\)
4. \(\sqrt{3}\)

A current-carrying loop is placed in a uniform magnetic field in four different orientations, I, II, III & IV. The decreasing order of potential energy is:

A thin rectangular magnet suspended freely has a period of oscillation equal to \(T\). Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is \(T'\), then ratio \(\frac{T'}{T}\) is:
1. \(\frac{1}{4}\)
2. \(\frac{1}{2\sqrt{2}}\)
3. \(\frac{1}{2}\)
4. \(2\)