Question

A student performs an experiment for determination of $g(=\frac{4{\pi }^2\ell }{T^2}),\ell \approx 1m$, and he commits an error of $\Delta \ell$. For $T$ he takes the time of $n$ oscillations with the stop watch of least count $\Delta T$ and he commits a human error of 0.lsec. For which of the following data, the measurement of $g$ will be most accurate?

$\Delta \ell ,\Delta T_n$ and Amplitude of oscillation are:

• A. $5mm,0.2\sec$ and $105mm$
• B. $5mm,0.2\sec$ and $205mm$
• C. $5mm,0.1$$\sec$ and $202mm$
• D. $1mm,0.1\sec$ and $501mm$

Answer 1

The correct option is D $1mm,0.1\sec$ and $501mm$

$\frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{\Delta T}{T}+\frac{0.1}{T}\Rightarrow \frac{\Delta g}{g}=\Delta l+\frac{\Delta T}{nT}+constant\Delta l+\frac{\Delta T}{nT}\left(1\right)$

(1) is minimum for combination of $\Delta l,\Delta T,n$ given in option D. Thus $\Delta g$ is minimum in D.