Question

The time period of a pendulum is given by `T = 2 pi sqrt((L)/(g))`. The length of pendulum is `20 cm` and is measured up to `1 mm` accuracy. The time period is about `0.6 s`. The time of `100` oscillations is measured with a watch of `1//10 s` resolution. What is the accuracy in the determination of `g`?

Answer 1

Correct Answer - `~ 0.8%`
`T^(2) = 4 pi^(2) (l)/(g) rArr g = 4 pi^(2) (l)/(T^(2))`
`(Delta g)/(g) = ( Delta l)/(l) + 2 (Delta T)/( T) = ( 0.1)/(20) + 2 ((1)/(600)) = (1)/(120)`
`(Delta g)/(g) xx 100 = 0.833 ~ 0.8%`