Question

A thin rectangular magnet suspended freely has a period of oscillation equal to $T$. Now, it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. 1f its period of oscillation is $T'$, the ratio $T'/T$ is

• A. $\frac{1}{2 \sqrt2}$
• B. $\frac{1}{2}$
• C. 2
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

$\frac{1}{2}$

Answer 2

When magnet is divided into two equal parts, the magnetic dipole moment $M'$ = pole strength $\times \frac{l}{2} = \frac{M}{2}$ (pole strength remains same) Also, the mass of magnet becomes half, ie, $m' = \frac{m}{2}$ Moment of inertia of magnet $I = \frac{ml^2}{12}$ New moment of inertia $I' = \frac{1}{12} (\frac{m}{2}) (\frac{l}{2})^2 = \frac{ml^2}{12 \times 8}$ $\therefore \, I' = \frac{I}{8}$ Now $T = 2 \pi \sqrt{(\frac{I}{MB})}$ $T' = 2 \pi \sqrt{(\frac{I'}{M'B})} = 2 \pi \sqrt{(\frac{I/8}{MB/2})}$ $\therefore T' = \frac{T}{2} \Rightarrow \frac{T'}{T} = \frac{1}{2}$