[" 27.At what depth below the surface of the earth should a simple pendulum "],[" be taken so that its time period becomes double "?" (Radius of earth "],[6400km" .) "]

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 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 )

If the length of a simple pendulum is equal to the radius of the earth, its time period will be

If the length of a simple pendulum is equal to the radius of the earth, its time period will be

If the length of a simple pendulum is equal to the radius of the earth, its time period will be

The acceleration due to gravity at a depth d below the surface of the earth is 20% of its value at the surface. What is the value of d if the radius of the earth=6400km ?

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.

How much below the surface of the earth does the acceleration due to gravity become 1% of its value at the earth's surface ? Radius of the earth=6400 km.

A satellite is at a height of 25, 600km from the surface of the earth. If its orbital speed is 3.536km/s find its time period. (Radius of the earth = 6400km)

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