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. What will be its time period oscillation?

acceleration due to gravity at a point at a depth from the surface of earth is g1=g1-dR , d=depth from surface of earth, R=radius of earth given d=R/4 g1=g1−R/4R=3g/4 time period of simple pendulum is T=2πlg here l=length of pendulum, g= acceleration due to gravity ⇒  Tα1g ⇒T1g1=T2g2 2g=T23g4 2=T232 ∴T2=43=2.309 seconds