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 $0.1s$. For which of the following data, the measurement of $g$ will be most accurate?

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

Answer 1

Answer: (D)

$\Delta l=1mm$, $\Delta T=0.1s$, $n=50$, Amplitude of oscillation$=1mm$

we have to calculate the error in gravity using the formulae

$\frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2\Delta t}{nt}$

where g = acceleration due to gravity

l = length

n = number of oscillations

So, after hit and trial method we can say that with option D we will get the least error