A student measures the time period of 100 oscillations of a simpl… - Vidyadip

MCQ

A student measures the time period of $100$ oscillations of a simple pendulum four times. The data set is $90\;s$ ,$91\;s $, $95\;s$ and $92\;s$. If the minimum division in the measuring clock is $1\;s$, then the reported mean time should be

✓
$92\pm 2\;s$
B
$92\pm 3\;s$
C
$92\pm 1.8\;s$
D
$92\pm 5\;s$

✓

Answer

Correct option: A.

$92\pm 2\;s$

a
$\Delta T = \frac{{\left[ {\Delta {T_1}| + |\Delta {T_2}| + |\Delta {T_3}| + |\Delta {T_4}} \right]}}{4}$
$ = \frac{{2 + 1 + 3 + 0}}{4} = 1.5$
As the resolutions of measuring clock is $1.5$ therefore the mean time should be $92$ pm $1.5$
but Least count is $1\; sec$

then answer should be $92\, \pm \,2\;\;sec$

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