A student performs an experiment for determination of $g\left(=\frac{4 \pi^{2} l}{T^{2}}\right)$. Th

Question

A student performs an experiment for determination of $g\left(=\frac{4 \pi^{2} l}{T^{2}}\right)$. The error in length $l$ is $\Delta l$ and in time $T$ is $\Delta T$ and $n$ is number of times the reading is taken. The measurement of $g$ is most accurate for

Professor PC Thomas Classes · Answer

(d) : $g=\frac{4 \pi^{2} l}{T^{2}}$. Take $\log$ and differentiate\ $	herefore \frac{\Delta g}{g}=\frac{\Delta l}{l}+2 \frac{\Delta T}{T}$\ Since $\Delta l$ is minimum in option (d), $\Delta T$ is minimum in options (c) and (d). $n$ is maximum in option (d).\ $	herefore$ Error will be minimum in option (d).\ Hence $g$ is most accurate for option (d).\