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  • Time for 20 oscillations of a pendulum is measured as t_{1}= 39.6 \; s; t_{2} = 39.9 \; s; t_{3} = 39.5\; s. What is the precision in the measurements? What is the accuracy of the measurement?

Time for 20 oscillations of a pendulum is measured as t_{1}= 39.6 \; s; t_{2} = 39.9 \; s; t_{3} = 39.5\; s. What is the precision in the measurements? What is the accuracy of the measurement?

Answers (1)

Given data :

t_{1}= 39.6 \; s

t_{2} = 39.9 \; s

and t_{3} = 39.5\; s

L.C. of the instrument is 0.1 sec, hence Precision (LC) = 0.1 sec.

Now,

For 20 oscillations, the mean value will be

=\frac{39.6+39.9+39.5}{3}

=\frac{119}{2}

=39.7 \; s

Now the absolute errors in measurement

\left | \Delta t_{1} \right |=\left | \bar{t}- t_{1}\right |=\left | 39.7-39.6 \right |=\left | 0.1 \right |=0.1\; s

\left | \Delta t_{2} \right |=\left | \bar{t}- t_{2}\right |=\left | 39.7-39.9 \right |=\left | 0.2 \right |=0.2\; s

\left | \Delta t_{3} \right |=\left | \bar{t}- t_{3}\right |=\left | 39.7-39.5 \right |=\left | 0.2 \right |=0.2\; s

Now, Mean absolute error =\frac{0.1+0.2+0.2}{3}

                                          =\frac{0.5}{3}\cong 0.2\; s

& Accuracy of measurement =\pm 0.2\; s

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