Question

The time period of a freely suspended magnet is $4$ seconds. 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. 4 s
• B. 2 s
• C. 0.5 s
• D. 0.25 s

Answer 1

Answer: (B)

$T=2\pi \sqrt{\frac{I}{M{B}_H}}=4s$
When magnet is cut into two equal halves, then New magnetic moment, $M^{\prime }=\frac{M}{2}$
New moment of inertia, $I^{\prime }=\frac{(w/2)(l/2{)}^2}{12}$
$=\frac{1}{8}\cdot \frac{w{l}^2}{12}$
Where $w$ is the initial mass of the magnet
But $I=\frac{w{l}^2}{12};$
$\therefore I^{\prime }=\frac{I}{8}$
$\therefore$ New time period, $T^{\prime }=2\pi \sqrt{\frac{I^{\prime }}{M^{\prime }{B}_H}}$
$=2\pi \sqrt{\frac{I/8}{(M/2)B_H}}$
$=\frac{1}{2}2\pi \sqrt{\frac{I}{M{B}_H}}$
$=\frac{1}{2}\times T=\frac{1}{2}\times 4=2s$