Question

The period of oscillation of a freely suspended long bar magnet is 4 seconds. If it is cut into two equal parts lengthwise, then the time period of each part will be

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

Answer 1

Answer: (A)

4 sec

Given that: $T=2\pi \sqrt{\frac{I}{MH}}=4sec$
When magnet is cut into two equal halves lengthwise, then
New magnetic moment $M=\frac{M}{2}$
New moment of inertia $I^{\prime }=\frac{\left(\frac{m}{2}\right)(l{)}^2}{12}=\frac{1}{2}\frac{m{l}^2}{12}$
Where m is the initial mass of the magnet, $l$ is length of magnet
But, $I=\frac{m{l}^2}{12}$ hence, $I^{\prime }=\frac{I}{2}$
Therefore, new time period
$T^{\prime }=2\pi \sqrt{\frac{I^{\prime }}{M^{\prime }H}}=(\sqrt{\frac{1}{2}}\times \sqrt{\frac{2}{1}})T=\sqrt{1}\times 4=4sec$