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If the work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the magnetic meridian is n times the corresponding work done to turn it through an angle of $60^{\circ}$, then the value of $n$ is
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2
Wehave, $\mathrm{W}=-\mathrm{MB}\left(\cos \theta_{2}-\cos \theta_{1}\right)$
So, $\mathrm{W}_{1}=-\mathrm{MB}\left(\cos 90^{\circ}-\cos 0^{\circ}\right)=\mathrm{MB}$
and $\mathrm{W}_{2}=-\mathrm{MB}\left(\cos 60^{\circ}-\cos 0^{\circ}\right)=\frac{1}{2} \mathrm{MB}$
As $W_{1}=n W_{2}$
$\therefore \mathrm{n}=\frac{\mathrm{W}_{1}}{\mathrm{~W}_{2}}=\frac{\mathrm{MB}}{\frac{1}{2} \mathrm{MB}}=2$
So, $\mathrm{W}_{1}=-\mathrm{MB}\left(\cos 90^{\circ}-\cos 0^{\circ}\right)=\mathrm{MB}$
and $\mathrm{W}_{2}=-\mathrm{MB}\left(\cos 60^{\circ}-\cos 0^{\circ}\right)=\frac{1}{2} \mathrm{MB}$
As $W_{1}=n W_{2}$
$\therefore \mathrm{n}=\frac{\mathrm{W}_{1}}{\mathrm{~W}_{2}}=\frac{\mathrm{MB}}{\frac{1}{2} \mathrm{MB}}=2$
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