Question

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

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

Answer 1

Answer: (B)

Time period of a suspended magnet is given by $T=2 \pi \sqrt{\frac{l}{M B}}$
Where, $I=$ moment of inertia of the magnet about the point of Suspension.
$M=$ magnetic moment
$B=$ magnetic field
When the magnet is broken in length
$T^{2'}=2 \pi \sqrt{\frac{l'}{M' B}}$
$\Rightarrow \frac{T}{T^{2'}}=\sqrt{\frac{1}{I'} \times} \sqrt{\frac{M'}{M}}$
$=\sqrt{\frac{m L^{2}}{m' L^{2}}} \times \sqrt{\frac{L'}{L}}$
Here, $m$ and $m$ ' are the masses of the magnet and $L$ and $L$ 's are their lengths.
$=\sqrt{\frac{m}{m'}} \times \sqrt{\frac{L}{L'}}=\sqrt{\frac{m}{m / 2}} \times \sqrt{\frac{L}{L / 2}}=\sqrt{2} \times \sqrt{2}=2$
$\Rightarrow T'=\frac{T}{2}=\frac{4}{2}=2\, s$