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
$m^{\prime }=\frac{m}{2}$ and $l^{\prime }=\frac{l}{2};$
$\therefore M^{\prime }=\frac{m}{2}\times \frac{l}{2}=\frac{ml}{4}=\frac{M}{4}$
New moment of inertia
$I^{\prime }=\frac{w^{\prime }{\ell }^{{}^{\prime }2}}{12}=\frac{\frac{w}{4}\cdot {\left(\frac{\ell }{2}\right)}^2}{12}$
$=\frac{1}{16}\cdot \frac{w{l}^2}{12}$
Here $w$ is the mass of magnet.
$\therefore I^{\prime }=\frac{1}{16}I$
Time period of each part,
$T^{\prime }=2\pi \sqrt{\frac{I^{\prime }}{M^{\prime }{B}_H}}$
$=2\pi \sqrt{\frac{I/16}{(M/4)B_H}}$
$=2\pi \sqrt{\frac{I}{4M{B}_H}}=\frac{T}{2}$