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,                  T=2πlg⇒              T∝1gHere, in given condition,                   T1=Kg                                         …(i)and             T2=Kg1-dR                           …(ii)On dividing Eq. (i) by Eq. (ii), we get                   T1T2=K/gK/g1-dR   or                 T1T2=1-dR   or   T12T22=1-dR⇒               d=1-T12T22R