At what depth from the surface of earth the time period of a simple pendulum is `0.5%` more than that on the surface of the Earth? (Radius of earth is `6400 km`)

At what height above the surface of earth acceleration due to gravity will be (ii) 50% of its value on the surface of the earth. Given radius of earth =6400 km.

How much below the surface of the earth does the acceleration due to gravity become 70% of its value at the surface of the earth ? Radius of the earth is 6400 km

What is the time period of a simple pendulum if length of the pendulum is equal to the radius of Earth ?

Calculate the depth below the surface of the earth where acceleration due to gravity becomes half of its value at the surface of the earth . Radius of the earth = 6400 km.

The depth from the surface of the earth where acceleration due to gravity is 20% of its value on the surface of the earth is (R = 6400 km)

Find the depth at which the value of g becomes 25% of that at the surface of the earth. (Radius of the earth = 6400 km)