If ${\text{T}}_1{\text{,T}}_2{\text{ and T}}_3$ are the time periods of a given simple pendulum on the surface of the earth, at a depth 'h' in a mine and at an altitude 'h' above the earth's surface respectively, then

${\text{g}}_{\text{d}}=\text{g}\left(1-\frac{\text{h}}{\text{R}}\right);{\text{g}}_{\text{A}}=\text{g}\left(1-\frac{\text{2h}}{\text{R}}\right)$

As we go upwards (or) downwards, the value of ‘g’ decreases. So time period increases when compared to surface of earth

$\therefore {\text{T}}_{\text{3}}\text{and}{\text{T}}_{\text{2}}$ must be greater than ${\text{T}}_{\text{1}}$