The magnetic needle of a `V.M.M` completes `10` oscillations in `92`seconds. When a small magnet is placed in the magnetic meridian `10cm` due north of needle with north pole towards south completes `15` oscillations in `69`seconds. The magnetic moment of magnet

Options

A. `4.5Am^(2)`
B. `0.45Am^(2)`
C. `0.75Am^(2)`
D. `0.225Am^(2)`

Answer

C

Explanation

`T_(1)=9.2S,T_(2)=4.6SandT_(1)/T_(2)=sqrt(B_(2)/B_(1))`
`rArr9.2/4.6=sqrt((B-B_(H))/B_(H))rArrB=5B_(H)=1.5xx10^(-4)T`
`rArrM_(0)/(4pi)(2M)/d^(3)=1.5xx10^(-4)rArrM=0.75Am^(2)`