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

`T=2pi sqrt((l)/(g)) :. T^(2)=(4pi^(2)l)/(g)`
or `g=(4pi^(2)l)/(T^(2)) " " [ 4pi^(2)=` constant]
`:.(Deltag)/(g)=(Deltal)/(l)+2(Deltat)/(T)...(i)`
Now `Deltal=1 mm=0.1 cm, l=10 cm`
Total time `t=nT=0.5xx100=50 s and Deltat=1s`
Now `T=(t)/(n) and DeltaT=(Deltat)/(n)` hence `(DeltaT)/(T)=(Deltat)/(t)`
`:.(Deltag)/(g)=(Deltal)/(l)+2(Delta)/(t)` (Substituting `(DeltaT)/(T)` in eq. (i))
`:.(Deltag)/(g)=(0.1)/(10)+2xx(1)/(50)=0.05`
`:.` Percentage error in measurement of g
`=0.05 xx100=5%`