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The length of a seconds pendulum on the surface of Earth is 1m. what will be the length of a second’s pendulum on the room?

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For Earth, \(T = 2\pi\sqrt{\frac{1}{g_{earth}}}\)

For Moon, T' = \(2\pi\sqrt{{\frac{1}{g_{moon}}}}\)

Since T = T’

So, \(\frac{l_{moon}}{g_{moon}}=\frac{l_{earth}}{g_{earth}}\) or lmoon = learth (\(\frac{g_{moon}}{g_{earth}}\))= 1 x \(\frac{1}{6}\)

as gmoon = \(\frac{1}{6}g_{earth}\) = \(\frac{1}{6}m\)

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