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

The Correct Option is B

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} $