Question

A student performs an experiment for determination of $⟮g=\frac{4{\pi }^{2}l}{T^2}⟯$, 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 $\Delta T$ and he commits a human error of 0.1 s. For which of the following data, the measurement of g will be most accurate?

• A. $\Delta L=0.5,\Delta T=0.1,n=20$
• B. $\Delta L=0.5,\Delta T=0.1,n=50$
• C. $\Delta L=0.5,\Delta T=0.01,n=20$
• D. $\Delta L=0.5,\Delta T=0.05,n=50$

Answer 1

The correct option is B $\Delta L=0.5,\Delta T=0.1,n=50$

It is given that $g$ is determined by $:g=\frac{4{\pi }^{2}l}{T^2}$

Taking log and differentiating$,$

$\Rightarrow \frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2\Delta T}{T}$

Now$,$ $g$ is most accurate in the case in which error in $g\left(l.e\Delta g\right)$ is minimum$.$ In case $B,$ number of repetitions to perform the experiment is maximum and $\Delta g$ is minimum$.$

Hence,

option $\left(B\right)$ is correct answer.