A pendulum that beats seconds on the surface of the earth were taken to a depth of (1/4)th  The radius of the earth is R. What will be its time period of oscillation?

from gravitation, acceleration due to gravity at a depth d from surface of earth is, g1=g1-dR, here given d=R4, substitute g1=g1−R/4R=3g/4  = g2  (suppose) time period of simple pendulum is    T=2πlg  ⇒Tα1g, where length l of pendulum is constant ⇒Tg=constant T1g1=T2g2  substitute g1=g and g2=3g4 pendulum beats seconds means, T1=2 seconds 2g=T23g4 2=T232 ∴T2=43=2.309 seconds