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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 oscillations | Time period $(s)$ |
| $1.$ | $64.0$ | $8$ | $128.0$ | $16.0$ |
| $2.$ | $64.0$ | $4$ | $64.0$ | $16.0$ |
| $3.$ | $20.0$ | $4$ | $36.0$ | $9.0$ |
(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. ....... .
$4$
$3$
$1$
same in all
Solution
$T=2 \pi \sqrt{\frac{\ell}{g}} \Rightarrow g=\frac{4 \pi^{2} \ell}{T^{2}}$
$\frac{\Delta g}{g}=\frac{\Delta \ell}{\ell}+\frac{2 \Delta T}{T}$
$\Delta T=\frac{\text { least count of time }\left(\Delta T_{0}\right)}{\text { number of oscillations(n) }}$
$\frac{\Delta g}{g}=\frac{\Delta \ell}{\ell}+\frac{2 \Delta T_{0}}{n T}$
As $\Delta \ell$ and $\Delta T_{0}$ same for all observations so $\frac{\Delta g}{g}$ is minimum for highest value of,$n$ and $T$
Minimum percentage error in $g$ is for student number$-1$
