A magnet makes 40 oscillations per minute at a place having magnetic field of `0.1 xx 10^(-5) T`. At another place, " it take `2.5 s` to complete one vibration. If the value of the earth's horizontal field at that place is `y xx 10^(-6) T`. then find `y`.

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

At a certain place, a magnet makes 30 oscillations per minute. At another place, the magnetic field is double. What is the its time period ?

At a certain place a magnet makes 30 oscillations per minute. At another place where the magnetic field is double, its time period will be

A magnet makes 12 oscillations per minutes at a place a where the horizontal component of earth's field is 3.2 xx 10^(-3)T . It is found to require 4 seconds per oscillation at another place B. Calculate the vertical component of earth's field at place B, if dip at B is 45^(@) .

A magnet has time period of oscillations T in earth's magnetic field at a place. If it is replaced by a magnet of magnetic moment six times the magnetic moment of the original one, the time period at the same place will be

A magnet makes 5 oscillations per minute in H=0.3times10^(-1) T .By what amount the field should be increased so that the number of oscillations is 10 in the same time?