A simple pendulum has a time period of  2.0sec at the earth's surface. It is taken to a height  Ro/2 above the earth's surface, where  Ro is the radius of the earth. What is the time period in sec?

To solve this problem, we need to use the formula for the period of a simple pendulum and the formula for the acceleration due to gravity at a height h above the Earth's surface.

1. The formula for the period of a simple pendulum is T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity.

2. The formula for the acceleration due to gravity at a height h above the Earth's surface is g' = g*(Re/(Re+h))^2, where Re is the radius of the Earth.

3. We are told that the pendulum has a period of 2.0 sec at the Earth's surface, so we can use this to find the length of the pendulum. Rearranging the formula for the period of a simple pendulum gives L = (T^2g)/(4π^2) = (2.0^29.8)/(4π^2) = 0.99 m.

4. We are told that the pendulum is taken to a height Ro/2 above the Earth's surface, where Ro is the radius of the Earth. So h = Ro/2.

5. We can substitute h = Ro/2 into the formula for g' to find the acceleration due to gravity at this height. This gives g' = g*(Re/(Re+Ro/2))^2 = 9.8*(6371/(6371+6371/2))^2 = 6.53 m/s^2.

6. Finally, we can substitute L = 0.99 m and g' = 6.53 m/s^2 into the formula for the period of a simple pendulum to find the new period. This gives T' = 2π√(L/g') = 2π√(0.99/6.53) = 2.44 sec.

So, the period of the pendulum at a height Ro/2 above the Earth's surface is 2.44 sec.