Time period of a simple pendulum on earth's surface is 'T'. It time period becomes 'xT' when taken to a height '2R' above earth's surface. The value of x will be ( R= radius of earth)

Step-by-step solution

The time period of a simple pendulum is given by T = 2 L g , where L is the pendulum length and g is Earth's surface gravity. At height h = 2R above Earth's surface, the gravitational acceleration becomes g' = g ( R R + h )^2 = g ( R 3R )^2 = g 9 . The time period at this height is T' = 2 L g' = 2 L g/9 = 3 2 L g = 3T . Since T' = xT by definition, we conclude that x = 3 . Answer: D