# The acceleration due to gravity on the surface of the moon is 1.7ms^(-2). What is the time perioid of a simple pendulum on the surface of the moon,

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The acceleration due to gravity on the surface of the moon is 1.7ms^(-2). What is the time perioid of a simple pendulum on the surface of the moon, if its time period on the surface of earth is 3.5s ? Take g=9.8ms^(-2) on the surface of the earth.
A. 8.4 s
B. 8.2 s
C. 7.4 s
D. 6.4 s

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Given, g_(m)=1.7 ms^(-2),g_(e)=9.8 ms^(-2)
T_(m) = ? and T = 3.5 s
As T_(e)=2pi sqrt((l)/(g_(e)))
and T_(m)=2pi sqrt((l)/(g_(m)))
therefore" "(T_(m))/(T_(e))=sqrt(g_(e)/(g_(m))) or T_(m)=T_(e) sqrt((g_(e))/(g_(m)))
rArr" "T_(m) = 3.5 sqrt((9.8)/(1.7))=8.4 s