A) Decreases
B) Increases
C) Remains the same
D) Decreases and then increases
Correct Answer: A
Solution :
As we go from equator to pole the value of g increases. Therefore time period of simple pendulum \[\left( T\propto \frac{1}{\sqrt{g}} \right)\] decreases. \[\left( \because \,T\propto \frac{1}{\sqrt{g}} \right)\]
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