Correct option is (B)

$T=2\pi \sqrt{\frac{1}{M\times B}}$ where $I=\frac{1}{12}m{\ell }^2$

When the magnet is cut into three pieces the pole strength will remain the same and

$M.\mathrm{I}.\left(I^{\prime }\right)=\frac{1}{12}\left(\frac{m}{3}\right){\left(\frac{\ell }{3}\right)}^2\times 3=\frac{I}{9}$

We have, Magnetic moment $(M)$

$=$ Pole strength $\left(m\right)\times \ell$

$\therefore$ New magnetic moment,

$M^{\prime }=m\times \left(\frac{\ell }{3}\right)\times 3=m\ell =M$

$\therefore$ $T^{\prime }=\frac{T}{\sqrt{9}}=\frac{2}{3}s.$

Correct option is (B)

KEY CONCEPT : The time period of a rectangular magnet oscillating in earth's magnetic field is given by

$T=2\pi \sqrt{\frac{I}{\mu B_H}}$

where $I=$ Moment of inertia of the rectangular magnet

$\mu =$ Magnetic moment

$B_H=$ Horizontal component of the earth's magnetic field

Case 1 : $T=2\pi \sqrt{\frac{I}{\mu B_H}}$ where $I=\frac{1}{12}M{\ell }^2$

Case 2 : Magnet is cut into two identical pieces such that each piece has half the original length. Then

$T^{\prime }=2\pi \sqrt{\frac{I^{\prime }}{{\mu }^{\prime }{B}_H}}$

where $I^{\prime }=\frac{1}{12}\left(\frac{M}{2}\right){\left(\frac{\ell }{2}\right)}^2=\frac{I}{8}$ and ${\mu }^{\prime }=\frac{\mu }{2}$

$\therefore$ $\frac{T^{\prime }}{T}=\sqrt{\frac{I^{\prime }}{{\mu }^{\prime }}\times \frac{\mu }{I}}$

$=\sqrt{\frac{I/8}{\mu /2}\times \frac{\mu }{I}}$

$=\sqrt{\frac{1}{4}}=\frac{1}{2}$

Correct option is (A)

$W=MB\left(\cos {\theta }_1-\cos {\theta }_2\right)$

$=MB\left(\cos {\theta }^{\circ }-\cos {60}^{\circ }\right)$

$=MB\left(1-\frac{1}{2}\right)=\frac{MB}{2}$

$\therefore$ $\tau =MB\sin \theta =MB\sin {60}^{\circ }$

$=\sqrt{3}\frac{MB}{2}=\sqrt{3}W$

Correct option is (D)

As shown in the figure, the magnetic lines of force are directed from south to north inside a bar magnet.