Physics - Error Question with Solution | TestHub

Given,                             t₁= 39.6 s

t₂= 39.9 s

t₃= 39.5 s

Least count of measuring instrument = 0.1 s

[As measurements have only one decimal place] Precision in the measurement = Least count of the measuring instrument = 0.1 s

Mean value of time for 20 oscillations

$t=\frac{t_1+t_2+t_3}{3}=\frac{39\text{.}6+39\text{.}9+39\text{.}5}{3}=39\text{.}7\text{s}$

Absolute errors in the measurements

$\begin{array}{c}\Delta t_1=t-t_1=39\text{.}7-39\text{.}6=0\text{.}1\text{s}\\ \Delta t_2=t-t_2=39\text{.}7-39\text{.}9=-0\text{.}2\text{s}\\ \Delta t_3=t-t_3=39\text{.}7-39\text{.}5=0\text{.}2\text{s}\end{array}$

$\text{Mean absolute error}=\frac{\left|\Delta t_1\right|+\left|\Delta t_2\right|+\left|\Delta t_3\right|}{3}$

$=\frac{0\text{.}1+0\text{.}2+0\text{.}2}{3}=\frac{0\text{.}5}{3}=0\text{.}17\approx 0\text{.}2\left(\text{rounding-off upto one decimal place}\right)\text{.}$

∴ Accuracy of measurement$=\pm 0\text{.}2\text{s}$