Question

The time period of a freely suspended magnet is \(4\,s\). If it is broken in length into two equal parts and one part is suspended in the same way, then the time period will be:

• A. \(4\,s\)
• B. \(2\,s\)
• C. \(0.5\,s\)
• D. \(0.25\,s\)

Hint:

Cut along length: \(I\) decreases faster than \(M\) → time period decreases.

Answer 1

The Correct Option is B

Concept: \[ T = 2\pi \sqrt{\frac{I}{MB}} \]

Step 1:Changes after cutting along length

• Pole strength remains same
• Length halves \(\Rightarrow M = m \cdot l \Rightarrow M' = \frac{M}{2}\)
• Mass halves and length halves \(\Rightarrow I \propto m l^2 \Rightarrow I' = \frac{1}{2}\cdot\frac{1}{4}I = \frac{I}{8}\)

Step 2:Ratio \[ \frac{I'}{M'} = \frac{I/8}{M/2} = \frac{I}{4M} \]

Step 3:Time period \[ T' = 2\pi \sqrt{\frac{I'}{M'B}} = 2\pi \sqrt{\frac{I}{4MB}} = \frac{T}{2} \] \[ T' = \frac{4}{2} = 2\,s \]