The length of the second pendulum on the surface of earth is 1 m. The length of seconds pendulum on the surface of moon, where g is 1/6th value of g on the surface of earth, is

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1 / 6 m

6 m

1 / 36 m

36 m

answer is A.

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Detailed Solution

T=2πlg⇒lg = constant⇒l∝g; ⇒lm1=16gg⇒lm=16m

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