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)
`1=31.4 cm and Delta1 = 1mm = 0.1 cm`
`T= (112.0)/(100) = 1.12s and Delta T == 0.01s`
Formula for g is `g= 4pi^(2) ((l)/(T^2) )`
`(Deltag)/( 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 determining g is 0.023 and percentage error `= 0.023 xx 100 = 2.3%`