Explanation

(c) Time period of magnet in vibration magnetometer
$T = 2\pi\sqrt{\frac{I}{MH}}$
First case $T_{1} = 2\pi\sqrt{\frac{I_{1} + \ I_{2}}{M'H}}$
Where M = resultant magnetic moment of two magnets
Here, two identical magnets are placed perpendicular to each other.
$\therefore I_{1} = I_{2} = I(Let)$
And $M^{'} = \sqrt{M^{2} + M^{2}} = M\sqrt{2}$
$\therefore\ \ \ \ \ \ T_{1} = 2\pi\sqrt{\frac{2I}{\sqrt{2}MH}}$
$\ \ \ \ \ \ 2^{5/4} = 2\pi\sqrt{\frac{2I}{\sqrt{2}\ MH}}\ \ \ \ \ \ \ \ \ldots(i)$
When one magnet is removed, then time period
$T_{2} = 2\pi\sqrt{\frac{I}{MH}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Dividing Eq. (i) by Eq.(ii)
$\frac{2^{5/4}}{T_{2}} = \sqrt{\frac{2}{\sqrt{2}}}$
$\frac{2^{5/4}}{T_{2}} = 2^{1/4}$
$T_{2} = \frac{2^{5/4}}{2^{1/4}} = 2s$