A magnet makes `40` oscillations per minute at a place having magnetic field intensity of `0.1xx10^(-5)T`. At another place, it takes 2.5 sec to complete one vibrating. The value of earth's horizontal field at that place is

Options

A. `0.25 xx 10^(-6)T`
B. `0.36 xx 10^(-6)T`
C. `0.66 xx 10^(-8)T`
D. `1.2 xx 10^(-6)T`

Answer

B

Explanation

`T= 2pi sqrt((I)/(MB_(H))) rArr T prop (1)/(sqrt(B_(H)))`
`rArr (T_(1))/(T_(2))= sqrt((B_(H_(2)))/(B_(H_(1)))) rArr ((60)/(40))/(2.5) = sqrt((B_(H_(2)))/(0.1 xx 10^(-5)))`
`rArr ((3)/(2) xx (2)/(5))^(2)=(B_(H_(2)))/(0.1 xx 10^(-5))`
`rArr B_(H_(2))= (9)/(25) xx 0.1 xx 10^(-5) = 0.36 xx 10^(-6)T`