Question

Time period for a magnet is $T$. If it is divided in four equal parts along its axis and perpendicular to its axis as shown, then time period for each part will be

• A. 4 T
• B. T / 4
• C. T / 2
• D. T

Answer 1

Answer: (C)

When magnet of length $l$ is cut into four equal parts, then
$m'=\frac{m}{2}$ and $l'=\frac{l}{2} ;$
$\therefore M'=\frac{m}{2} \times \frac{l}{2}=\frac{m l}{4}=\frac{M}{4}$
New moment of inertia
$I'=\frac{w' \ell^{' 2}}{12}=\frac{\frac{w}{4} \cdot\left(\frac{\ell}{2}\right)^{2}}{12}$
$=\frac{1}{16} \cdot \frac{w l^{2}}{12}$
Here $w$ is the mass of magnet.
$\therefore I'=\frac{1}{16} I$
Time period of each part,
$T'=2 \pi \sqrt{\frac{I'}{M' B_{H}}}$
$=2 \pi \sqrt{\frac{I / 16}{(M / 4) B_{H}}}$
$=2 \pi \sqrt{\frac{I}{4 M B_{H}}}=\frac{T}{2}$