A pendulum is oscillating with frequency ' n ' on the surface of earth. If it is taken to a depth ^ R ^ 3 below the surface of earth, new frequency of oscillation is ( R= radius of earth)

Step-by-step solution

The frequency of oscillation of a simple pendulum is given by n = 1 2 g L , where g is gravitational acceleration and L is the pendulum length. At depth d = R 3 below Earth's surface, the reduced gravitational acceleration is g' = g (1 - d R ) = g (1 - 1 3 ) = 2 3 g . The new frequency becomes n' = 1 2 g' L = 1 2 2g 3L = 2 3 1 2 g L . Since 1 2 g L = n , it follows that n' = 2 3 n . The result corresponds to option A.