Question

A student performs an experiment for determination of $g(=\frac{4{\pi }^{2}l}{T^2}),l\approx 1m$, 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 the commits a human error of $0.1$ sec. For which of the following data, the measurement of g will be most accurate?

• A. $\Delta l$ - $5mm$, $\Delta T$-$0.2$sec, $n-10$, Amplitude of oscillation-$5$mm
• B. $\Delta l$-$5mm$, $\Delta T$-$0.2$ sec, $n-20$, Amplitude of oscillation- $5$mm
• C. $\Delta l$- $5mm$, $\Delta T$ -$0.1$sec, $n-20$, Amplitude of oscillation-$1$mm
• D. $\Delta l$-$1mm$, $\Delta T$-$0.1$sec, $n-50$, Amplitude of oscillation-$1$mm

Answer 1

The correct option is D $\Delta l$-$1mm$, $\Delta T$-$0.1$sec, $n-50$, Amplitude of oscillation-$1$mm

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

Taking log and differentiating,

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

Now, $g$ is more accurate in the case in which error in $g$ (i.e $\Delta g$) is minimum.

In option (D), number of repetitions to perform the experiment is maximum and $\Delta g$ is minimum. Hence $g$ is more accurate in option (D).