A student performs an experiment for determination of ` g [ = ( 4 pi^(2) L)/(T^(2)) ] , L ~~ 1m` , and he commits an error of `Delta L`. For `T` he takes the time of `n` oscillations with the stop watch of least count ` Delta T` . For which of the following data , the measurement of `g` will be most accurate ?

Options

A. `Deltal=0.5`, `Deltat=0.1`,`n=20`
B. `Deltal=0.5`, `Deltat=0.1`,`n=50`
C. `Deltal=0.5`, `Deltat=0.01`,`n=20`
D. `Deltal=0.1`, `Deltat=0.05`,`n=50`

Answer

D

Explanation

Here `T=("total time")/("total oscillation")=(t)/(n)` so `dT=(dt)/(n)`
`(Delta g)/(g)=(DeltaL)/(L)+2(DeltaT)/(T)`
In option (`D`) error in `L(DeltaL)` is minimum and number of repetition of measurement are maximum so `dT` will be less. So in this case, the error in `g` is minimum.