If T₁ and T₂ are the time periods of oscillation of a simple pendulum on the surface of earth (of radius R) and at a depth d, the d is equal to

We know that, the time period of a simple pendulum,

                  $\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}}}$

$\Rightarrow \mathrm{T}\propto \frac{1}{\sqrt{\mathrm{g}}}$

Here, in given condition,

                   ${\mathrm{T}}_1=\frac{\mathrm{K}}{\sqrt{\mathrm{g}}}$                                         …(i)

and             ${\mathrm{T}}_2=\frac{\mathrm{K}}{\sqrt{\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)}}$                           …(ii)

On dividing Eq. (i) by Eq. (ii), we get

                   $\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}=\frac{\mathrm{K}/\sqrt{\mathrm{g}}}{\mathrm{K}/\sqrt{\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)}}$   

or                 $\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}=\sqrt{\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)}$   or   $\frac{{\mathrm{T}}_1^2}{{\mathrm{T}}_2^2}=1-\frac{\mathrm{d}}{\mathrm{R}}$

$\Rightarrow \mathrm{d}=\left(1-\frac{{\mathrm{T}}_1^2}{{\mathrm{T}}_2^2}\right)\mathrm{R}$