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A simple pendulum has a time period T1 when on the earth's surface and T2 when taken to a height R above the earth's surface, where R is the radius of the earth. The value of T2T1 is:

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Solution

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Time period of a simple pendulum =2π√lg

⇒g on earth's surface =GMR2

g on a height R above earth's surface =GM(2R)2

⇒T1 (time period on earth's surface ) =2π√l(R2)GM

T2 (time period at a height ′h′ above earth's surface ) =2π√l(2R)2GM

⇒T2T1=√4=2.

Hence, the answer is 2.

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