Three students $S_1$, $S_2$ and $S_3$ perform an experiment for determining the acceleration due to gravity ($g$) using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.
$$\begin{array}{|c|c|c|c|c|}\hline \text{Student No.} & \text{Length of pendulum (cm)} & \text{Number of oscillations } (n) & \text{Total time for } n \text{ oscillations (s)} & \text{Time period (s)} \\ \hline 1. & 64.0 & 8 & 128.0 & 16.0 \\ \hline 2. & 64.0 & 4 & 64.0 & 16.0 \\ \hline 3. & 20.0 & 4 & 36.0 & 9.0 \\ \hline \end{array}$$
(Least count of length = 0.1 m, least count for time = 0.1 s)
If $E_1$, $E_2$ and $E_3$ are the percentage errors in $g$ for students 1, 2 and 3, respectively, then the minimum percentage error is obtained by student no. ______.
Numerical value type. Enter your answer.
Answer: 1
$g = \dfrac{4\pi^2L}{T^2}$, so $\dfrac{\Delta g}{g} = \dfrac{\Delta L}{L} + 2\dfrac{\Delta t}{t}$, where $t$ is the total time measured. Taking the least count of length as 0.1 cm (the 0.1 m in the paper would make every error huge; the comparison is the same either way):
Student 1: $\dfrac{0.1}{64} + 2\cdot\dfrac{0.1}{128} = 0.31\%$
Student 2: $\dfrac{0.1}{64} + 2\cdot\dfrac{0.1}{64} = 0.47\%$
Student 3: $\dfrac{0.1}{20} + 2\cdot\dfrac{0.1}{36} = 1.06\%$
The minimum error is for student 1.
Solution by Sreeraj P, M.Sc Physics