The acceleration due to gravity on the surface of the Moon is `1.7ms^(-2)`. What is the time period of a simple pendulum on the surface of the Moon, if its time period on the surface of Earth is 3.5 s? (g on the surface of the Earth is `9.8ms^(-2)`).

Explanation

Given, g on the Moon `=1.7ms^(-2)`
T on the surface of the Earth is 3.5 s.
`"g on the Earth"=9.8ms^(-2)`.
We know that `T^(2) prop (1)/(g)`
`therefore" "(T_("moon"))/(T_("Earth"))=sqrt((g_(B))/(g_(M)))=sqrt((9.8)/(1.7))=5.765`
Hence `T_("moon")=(5.765xx3.5)s=20.196s`
`"% "DeltaT=((16.676)/(3.5))xx100=476.4%`
Percentage increase in the T will be `476.4%`.