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)

$T=2 \pi \sqrt{\frac{I}{M B_{H}}}=4\, s$
When magnet is cut into two equal halves, then New magnetic moment, $M'=\frac{M}{2}$
New moment of inertia, $I'=\frac{(w / 2)(l / 2)^{2}}{12}$
$=\frac{1}{8} \cdot \frac{w l^{2}}{12}$
Where $w$ is the initial mass of the magnet
But $I=\frac{w l^{2}}{12} ;$
$\therefore I'=\frac{I}{8}$
$\therefore $ New time period, $T'=2 \pi \sqrt{\frac{I'}{M' B_{H}}}$
$=2 \pi \sqrt{\frac{I / 8}{(M / 2) B_{H}}}$
$=\frac{1}{2} 2 \pi \sqrt{\frac{I}{M B_{H}}}$
$=\frac{1}{2} \times T=\frac{1}{2} \times 4=2\, s$