The magnetic needle of an oscillation magnetometer makes 10 oscillations per minut under the action of earth's magnetic field along. When a bar magnet is placed at some distance along the axis of the needle it makes 14 oscillations per minute. If the bar magnet is turned so that its poles interchange their position, then the new frequency of oscillation of the needle is:

Options

A. `10` vibrations per minute
B. `14` vibrations per minute
C. `4` vibrations per minute
D. `2` vibration per minute

Answer

D

Explanation

`T=2pisqrt(I/(MH))`
For Ist case: `60/10=2pisqrt(I/(MH)) …(i)`
For Iind case: `30/7=2pisqrt(I/(M(H+R))) ...(ii)`
(Divide (2) by (1))
`(30/7)/6=sqrt(H/(H+R))
R=(24/25)H ...(iii)`
For IIIrd case: `60/n=2pisqrt(I/(M(H-R)))`
`=2pixx5sqrt(I/(M(H-24/25H))) ( :.R=24/25H)`
`=2pixx5xxsqrt(I/(MH)) ...(iv)`
[From equationss (i) and (iv)]
We get `60/n=30impliesn=2` per minute.