The time period of oscillation of simple pendulum is given by `t = 2pisqrt(I//g)` What is the accurancy in the determination of 'g' if 10cm length is knownj to 1mm accuracy and 0.5 s time period is measured form time of 100 oscillations with a wastch of 1 sec. resolution.

Answer

`+-5%`

Explanation

Here `(DeltaI)/(I) = (0.1)/(10), Deltat =1sec, and `
time of 100 oscill.
`t =100xx0.5 =50s`
`:. (Deltat)/(t) = (1)/(50)`
From `t =2pisqrt((i)/(g)), t^2 = 4pi^2(1)/(g)`
`g=4pi^2(I)/(t^2)`
`:. (Deltag)/(g) = +-((DeltaI)/(l) = 2(Deltat)/(t))`
`%error, (Deltag)/(g)xx100 = +-((0.1)/(10)+(2xx1)/(50))xx100`