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 error involved)
`l= 31.4 cm and Delta l= 1mm = 0.1cm`
`T= (112.0)/(100) = 1.12s and Delta T = 0.01s`
Formula for g is `g= 4pi^(2) ((l)/(T^(2)))`
`(Delta g)/(g) = ((Delta l)/(l)) + 2 ((Delta T)/(T))`
`=((0.1)/(31.4)) + 2 ((0.01)/(1.12)) = 0.003 + 0.02 = 0.023`
Relative error in determing g is 0.023 and percentage error `= 0.023 xx 100= 2.3%`