In an experiment to determine the value of acceleration due to gravity g using a simple pendulum , the measured value of lenth of the pendulum is 31.4 cm known to 1 mm accuracy and the time period for 100 oscillations of pendulum is 112.0 s known to 0.01 s accuracy . find the accuracy in determining the value of g.

Explanation

(Accuracy is to be taken as the eror involved)
`1=31.4cm ` and `Delta1 =1 mm = 0.1cm`
`T = (112.0)/(100) = 1.12s ` and `DeltaT= 0.01 s `
Formula for is g `= 4pi^(2) ((l)/(T^(2)))`
`(Deltag)/(g)=((Deltal)/(l))+2((DeltaT)/(T))`
`=((0.1)/(31.4))+2((0.01)/(1.12))`
=0.003 +0.02 = 0.023
Relative error in determining g is 0.023 and percentage error `=0.023 xx100 = 2.3%`