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Correct option is D)

Time period of a simple pendulum $=2πgl $

$⇒g$ on earth's surface $=R_{2}GM $

$g$ on a height $R$ above earth's surface $=(2R)_{2}GM $

$⇒T_{1}$ (time period on earth's surface ) $=2πGMl(R_{2}) $

$T_{2}$ (time period at a height $_{′}h_{′}$ above earth's surface ) $=2πGMl(2R)_{2} $

$⇒T_{1}T_{2} =4 =2.$

Hence, the answer is $2.$

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