The time period of a freely suspended bar magnet in a field is 2s. It is cut into two equal parts along its axis, then the time period is

Options

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

Answer

A

Explanation

According to the relation: `T = 2pi sqrt(I/(MB))`
`T =2sI' = 1/2, M=M/2`
`T_(1)/T_(2) = sqrt(1/(MB) xx (2 xx 2)/(1 xx M xx B))`
`T_(1)/T_(2) =2`
`(2 sec)/T_(2) =2`
`T_(2) = 1` sec