Question

The time period of a freely suspended magnet is $4$s. If it is broken in length into two equal parts and one part is suspended in the same way. Then its time period will be.

• A. $2$ s
• B. $4$ s
• C. $0.5$ s
• D. $0.25$ s

Answer 1

The correct option is A $2$ s

Let a magnet of Magnetic moment M is kept in a uniform magnetic field B.

Torque on the magnet $=\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}$

$⟹\tau =MB\sin \theta$

Angular accleration $=\alpha =\frac{\tau }{I}$ where $I$=moment of Inertia

For small angle $\theta$, $\alpha =-\frac{MB\theta }{I}$ because the torque is restoring

Since $\alpha \propto \theta ⟹\alpha =-{\omega }^2\theta$

$⟹\omega =\sqrt{\frac{MB}{I}}$

We know that $T\propto \frac{1}{\omega }$

We know that $I\propto L^3$and $L^{\prime }=\frac{L}{2}$

$⟹I^{\prime }=\frac{I}{8}$

$M\propto L$

$⟹M^{\prime }=\frac{M}{2}$

${\omega }^{\prime }=\sqrt{\frac{M^{\prime }B}{I^{\prime }}}=2\sqrt{\frac{MB}{I}}=2\omega$

$T^{\prime }=\frac{T}{2}$

$⟹T=2s$

Answer-(A)