The period of oscillation of a simple pendulum is T =`2pisqrt(L/g)`. Measuted value of L is 20.0 cm known to 1mm accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g ?

Explanation

`g = 4pi^(2)L//T^(2)`
Here, T = `(t)/(n)` and `DeltaT = (Deltat)/(n)`. Therefore, `(DeltaT)/(T) = (Deltat)/(t)`
The error in both L and t are the least count errors. Therefore,
`(Deltag//g) = (DeltaL//L) +2(DeltaT//T)`
`=(0.1)/(20.0)+2(1)/(90) = 0.027`
Thus, the percentage error in g is
`100(Deltag//g) = 100(DeltaL//L)+2xx100(DeltaT//T)`
=3%