Question

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.

Student No.	Length of pendulum (cm)	No. of oscillations (n)	Total time for $n$ oscillations	Time period (s)
1.	64.0	8	128.0	8
2.	64.0	4	64.0	16.0
3.	20.0	4	36.0	9.0

(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. _________.

Answer 1

$T=2 \pi \sqrt{\frac{l}{g}} \Rightarrow g=\frac{4 \pi^{2} l}{T^{2}}$
$\frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2 \Delta T}{T} $
$\Delta T=\frac{\text { least count of time }\left(\Delta T_{0}\right)}{\text { number of oscillations(n) }} $
$\frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2 \Delta T_{0}}{n T}$
As $\Delta l$ and $\Delta T_{0}$ are same for all observations
so $\frac{\Delta g}{g}$ is minimum for highest value of $l, n$ and $T$
$\Rightarrow$ Minimum percentage error in $g$ is for student number $-1$