A student uses a simple pendulum of exactly `1m` length to determine `g`, the acceleration due ti gravity. He uses a stop watch with the least count of `1sec` for this and record `40 seconds` for `20` oscillations for this observation, which of the following statement `(s) is (are)` true?

Options

A. Error `DetaT` in measureing `T` , the time period is `0.05 s`.
B. Error `DeltaT` in measuring `T`, the time perios is `1s`
C. Percentage error in the determination of `g` is `5%`
D. Percentage error in the determination of `g` is `3.5%`

Answer

C

Explanation

Time period `T=40/20=2s`
Further, `t=nT=20T`
or `Deltat=20DeltaT`
`(Deltat)/t=(DeltaT)/T`or
`DeltaT=T/t . Deltat=(2/40)(1)=0.05s`
Further `T=2pisqrt(l/g)`
or `Tpropg^(-1//2)`
`:.(Delta)/Txx100=-1/2xx(Deltag)/gxx100`
or % error in determination of `g` is
`(Deltag)/gxx100=-200xx(DeltaT)/T`
`=-(200xx0.05)/2=-5%`