A magnet is suspended in the magnetic meridian with an untwisted wire. The upper end of wire is rotated through `180^(@)` to deflect the magnet by `30^(@)` from magnetic meridian. When this magnet is replaced by another magnet, the upper end of wire is rotated through `270^(@)` to deflect the magnet `30^(@)` from magnetic meridian. The ratio of magnetic moment of magnets is

Options

A. `1:5`
B. `1:8`
C. `5:8`
D. `8:5`

Answer

C

Explanation

Let `M_(1)` and `M_(2)` be the magnetic moments of magnets and `H` the horizontal component of earth's field.
We have `tau=MH sintheta`. If phi is the twist of wire, then `t=C varphi`, `C` being restoring couple per unit twist of wire
`implies C varphi=MH sintheta`
Here `varphi_(1)=(180^(@)-30^(@))=150^(@)=150xxpi/180` rad
`varphi_(2)=(270^(@)-30^(@))=240^(@)=240xxpi/180` rad
So, `C varphi_(1) =M_(1)H sin theta` (For deflection `theta=30^(@)` of I magnet)
`C varphi_(2)=M_(2)H sin theta` (For deflection `theta=30^(@)` of II magnet)
Dividing `varphi_(1)/varphi_(2)=M_(1)/M_(2)`
`impliesM_(1)/M_(2)=varphi_(1)/varphi_(2)=(150xx(pi/180))/(240xx(pi/180))=15/24=5/8`
`impliesM_(1):M_(2)=5:8`