In an oscillation magnetometer, a bar magnet completes 40...

The time period of oscillation of a bar magnet in a uniform magnetic field is given by $T = 2\pi \sqrt{\dfrac{I}{MB}}$, where $I$ is the moment of inertia and $M$ is the magnetic moment.

Let the initial moment of inertia be $I$ and the magnetic moment be $M$. The time taken for $40$ oscillations is $t_1 = 1$ minute.

When an identical demagnetized magnet is placed over the first one, the total moment of inertia becomes $I' = I + I = 2I$. The magnetic moment remains $M' = M$ since the second magnet is demagnetized and induced magnetism is neglected.

The new time period is $T' = 2\pi \sqrt{\dfrac{2I}{MB}} = \sqrt{2} T$.

Since the time required for a fixed number of oscillations is directly proportional to the time period, the new time for $40$ oscillations is $t_2 = \sqrt{2} t_1$.

Substituting $t_1 = 1$ minute, we get $t_2 = \sqrt{2}$ minutes.