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The acceleration due to gravity on the surface of moon is 1.7 ms-2 .What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g = 9.8 ms-2 )

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Best answer

On earth time period,

\(T =2\pi\sqrt{\frac{l}{g}}\)

i.e.,  3.5 = \(2\pi\sqrt{\frac{l}{9.8}}\)

Moon g = 1.7ms-1

T' = \(\sqrt{\frac{l}{1.7}}\)

Dividing  \(\frac{T'}{3.5}= \frac{2\pi\sqrt{\frac{l}{1.7}}}{2\pi\sqrt{\frac{l}{9.8}}}\)

or  T' = \(\sqrt\frac{9.8}{1.7}\times3.5\)= 8.4 s.

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