Q. A student performs an experiment for determination of $g\left( = \frac{4\pi^{2}l}{T^{2}} \right),\ l \approx 1m,$ and he commits an error of $\mathrm{\Delta}l.$ For $T$ he takes the time of $n$ oscillations with the stop watch of least count $\mathrm{\Delta}T$ 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) $\mathrm{\Delta}L = 0.5,\ \mathrm{\Delta}T = 0.1,\ n = 20$
B) $\mathrm{\Delta}L = 0.5,\ \mathrm{\Delta}T = 0.1,\ n = 50$
C) $\mathrm{\Delta}L = 0.5,\ \mathrm{\Delta}T = 0.01,\ n = 20$
D) $\mathrm{\Delta}L = 0.5,\ \mathrm{\Delta}T = 0.05,\ n = 50$
Explanation
$\frac{\mathrm{\Delta}g}{g} = \frac{\mathrm{\Delta}l}{l} + 2\frac{\mathrm{\Delta}T}{T}$ In option (d) error in $\mathrm{\Delta}g$ is minimum and number of observations made are maximum. Hence, in this case error in g will be minimum.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.