The time period of a simple pendulum is given by `T = 2 pi sqrt(L//g)`, where `L` is length and `g` acceleration due to gravity. Measured value of length is `10 cm` known to `1 mm` accuracy and time for 50 oscillations of the pendulum is 80 s using a wrist watch of 1 s resloution. What is the accuracy in the determination of `g`?

Explanation

`T = 2 pi sqrt((L)/(g))`
`g = (4 pi^(2) L)/(T^(2))`
Here `T = (t)/(n), t:` total time, `n` : number of oscillaiton
`Delta T = (Delta t)/(n)`
`(Delta T)/(T) = (Delta t)/(t)`
Maximum relative errors in `g`
`(Delta g)/(g) = (Delta L)/(L) + 2 (Delta T)/(T)`
Maximum percentage error is `g`
`(Delta g)/(g) xx 100 = (Delta L)/(L) xx 100 + 2 (Delta T)/(T) xx 100`
`= (0.1)/(10) xx 100 + 2 (1)/(80) xx 100`
`= 1 + 2.3 = 3.5%`