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. `Delta L = 0.5, Delta T - 0.1, n = 20`
B. `Delta L = 0.5, Delta T = 0.1,n = 50`
C. `DeltaL = 0.5, Delta T = 0.01, n = 20`
D. `Delta L= 0.1, Delta T = 0.05, n = 50`

Answer

D

Explanation

`g=(4pi^(2)l)/(T^(2))rArr(Deltag)/(g)=(Deltal)/(l)+(2 DeltaT)/(T)`Here `T = 2sec`
For (A) `(Deltag)/(g)=(0.1)/(1)+(0.2(0.1))/(2) = 0.6`
For (B) `(Deltag)/(g)=0.5+(0.2)/(2)=0.6`
For (C) `(Deltag)/(g)=0.5+(0.02)/(2)=0.51`
For (D) `(Deltag")/(g)=0.1+(0.1)/(2)=0.15`
`rArr(Deltag)/(g)` minimum for (D). Also number of observations are maximum in (D)