Three students S_ 1 , S_ 2 and S_ 3 perform an experiment for det… - Vidyadip

MCQ

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. ....... .

A
$4$
B
$3$
✓
$1$
D
same in all

✓

Answer

Correct option: C.

$1$

c
$T=2 \pi \sqrt{\frac{\ell}{g}} \Rightarrow g=\frac{4 \pi^{2} \ell}{T^{2}}$

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

$\Delta T=\frac{\text { least count of time }\left(\Delta T_{0}\right)}{\text { number of oscillations(n) }}$

$\frac{\Delta g}{g}=\frac{\Delta \ell}{\ell}+\frac{2 \Delta T_{0}}{n T}$

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$-1$

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