The period of oscillation of a simple pendulum is `T = 2 pi sqrt((L)/(g)) .L` is about `10 cm` and is known to `1mm` accuracy . The period of oscillation is about `0.5 s`. The time of 100 oscillation is measured with a wrist watch of `1 s` resolution . What is the accuracy in the determination of `g` ?

Explanation

`T=2pisqrt((l)/(g))rArrT^(2)=(4pi^(2)l)/(g)rArrg=(4pi^(2)l)/(T^(2))=(Deltag)/(g)=(Deltal)/(l)+(2DeltaT)/(T)`
Here % error in `l=(1mm)/(100cm)xx100=(0.1)/(100)xx100=0.1%, % "error in" T=(0.1)2xx100)xx100=0.05%`
`therefore "error in" g=% "error in" l+2(% "error in" T)=0.1+2xx0.05rArr0.2%`