A bar magnet of magnetic moment `M_(1)` is suspended by a wire in a magnetic field. The upper end of the wire is rotated through `180^(@)`, then the magnet is rotated through `45^(@)`. Under similar conditions another magnet of magnetic moment `M_(2)` is rotated through `30^(0)`. Then find the ratio of `M_(1)&M_(2)`.

Explanation

`C(alpha-theta)=MBsintheta`
For first magnet, `C(180-45)=M_(1)Bsin45^(@)" "...(1)`
For second magnet, `C(180-30)=M_(2)Bsin30^(@)" "...(2)`
Diving equation (1) by equation (2)
`135/150=M_(1)/M_(2)xxsqrt2rArrM_(1)/M_(2)=9/(10sqrt2)`