The period of oscillation of a simple pendulum is `T = 2pisqrt(L//g)`. Measured value of L is `20.0 cm` known to `1mm` accuracy and time for 100 oscillations of the pendulum is found to be 90 s 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 errors 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 in 3%.