Question

a pendulum beats second's on the surface of earth m Determine its time period when it's taken to the surface of the moon where acceleration due to gravity in one sixth of that on the surface of the earth​

Answer 1

Answer:

Correct option is

A

2.3s

Gravitational acceleration at a depth d from the surface of earth is given by g ′

=g(1− Rd )

here d= 4R

⟹g ' = 43g

Time period of oscillation of pendulum is given as

T=2π gI

Hence TT

= g g = 34 ≈1.154

∴T =1.154×2=2.3 s