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

The period of simple pendulum
`T= 2 pi sqrt((l)/(g))`
`:.T^(2)=(4pi^(2)l)/(g)`
`:.g=(4pi^(2)l)/(T^(2))`
`:.` Percentage error in g,
`(Deltag)/(g)xx100=(Deltal)/(l)xx100+(2DeltaT)/(T)xx100`
`=(Deltal)/(l)xx100+(2Deltat)/(t)xx100 [ :. (DeltaT)/(T)=(DeltaT)/(t)=(Deltat)/(t)]`
`=(0.1)/(20)xx100+(2xx1)/(90)xx100`
`=0.5+2.2`
`=2.7%`
`~~3%` (Approx)