A simple pendulum doing small oscillations at a place R height above earth surface has time period of T_1=4 ~s . T_2 would be it's time period if it is brought to a point which is at a height 2 R from earth surface. Choose the correct relation [ R = radius of Earth ]:

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Correct answer: C. 3 ~T _1=2 ~T _2

Time period of simple pendulum, T =2 l ~g \& ~g = GM ( R + h )^2 T ( R + h ) T _1 ~T _2 = ( R + h _1 ) ( R + h _2 ) = ( R + R ) ( R +2 R ) = 2 3 3 ~T _1=2 ~T _2

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