A small bar magnet `A` oscillates in a horizontal plane with a period `T` at a place where the angle of dip is `60^(@)`. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic merdian, its period will be

Options

A. `T/sqrt(2)`
B. `T`
C. `sqrt(2)T`
D. `2T`

Answer

A

Explanation

`T=2pisqrt(I/(MB))impliesT/T^(')=sqrt(B^(')/B)=sqrt(B/B_(H))`
`implies T/T^(')=sqrt(1/cos varphi)=sqrt(1/cos 60^(@))=sqrt(2)impliesimplies T^(')=T/sqrt(2)`