Question

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

• A. 5 mm, 0.2 sec, $n=$10
• B. 5 mm, 0.2 sec, $n=$20
• C. 5 mm, 0.1 sec, $n=$10
• D. 1 mm, 0.1 sec, $n=$50

Answer 1

The correct option is D 1 mm, 0.1 sec, $n=$50

Given: $g=\frac{4{\pi }^{2}l}{T^2}$
$\frac{\Delta g}{g}=\frac{\Delta l}{l}+2\frac{\Delta T}{T}$

Error in $g$ that is $\Delta g$ will be minimum for minimum $\Delta l$ and $\Delta T$ and more number of readings.