The period of oscillation of a simple pendulum is given by `T=2pisqrt((l)/(g))` where l is about 100 cm and is known to have 1 mm accuracy. The period is about 2 s. The time of 100 oscillation is measrued by a stop watch of least count 0.1 s. The percentage error is g is

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%`