A student performs an experiment for determination of `g(=(4pi^(2)l)/(T^(2))),~~1m`, and he commits an error of `Deltal`.
For `T` he takes the time of `n` oscillations with the stop watch of least count `Delta` and he commits is human error of `0.1sec`. For which of the following data, the measurement of `g` will be most accurate?

Options

A. `{:(Deltal" "DeltaT" "n" ""amplitude of oscillation"),(5mm" "0.2sec" "10" "5mm):}`
B. `{:(Deltal" "DeltaT" "n" ""amplitude of oscillation"),(5mm" "0.2sec" "20" "5mm):}`
C. `{:(Deltal" "DeltaT" "n" ""amplitude of oscillation"),(5mm" "0.1sec" "20" "1mm):}`
D. `{:(Deltal" "DeltaT" "n" ""amplitude of oscillation"),(1mm" "0.1sec" "50" "1mm):}`

Answer

D

Explanation

The error in g is minimum when the number of oscillations i.e, n is more
`(Deltag)/g=(Deltal)/l+|-2(DeltAT)/T|=(Deltal)/l+(2DeltaT)/T`
`(Deltal)/l+(2DeltaT)/T` should be minimum. Hence (D) represents the best measured data.