Question

The time period of a freely suspended bar magnet in a field is $2\, \sec$. It is cut into two equal parts along its axis, (hen the time period is

• A. 4 s
• B. 0.5 s
• C. 2 s
• D. 0.25 s

Answer 1

Answer: (C)

2 s

Answer 2

$T=2 \pi \sqrt{\frac{I}{M B}}$ $T=2 s, I'=\frac{I}{2}, M'=\frac{M}{2}$ $\therefore T'=T$ $\Rightarrow T'=2 s$