Ncert Exercises & Solutions

35. Time for 20 oscillations of a pendulum is measured as $t_{1} = 39 .6 s$ ; $t_{2} = 39 .9 s$ and $t_{3} = 39 .5 s$ . What is the precision in the measurements? What is the accuracy of the measurement?

Hint: The precision of the measurement is given by the least count of the measurement and error gives the amount of accuracy.

Step 1: Find the precision of the measurement.

$Given, t_{1} = 39 .6 s, t_{2} = 39 .9 s and t_{3} = 39 .5 s$

Least count of measuring instrument = 0.1 sec

(As measurements have only one decimal place)
Precision in the measurement = Least count of the measuring instrument = 0.1 sec

Step 2: Find the mean absolute error of the measurement.
The mean value of time for 20 oscillations is given by,

$\begin{array}{l} t = \frac{t_{1} + t_{2} + t_{3}}{3} \\ = \frac{39 .6 + 39 .9 + 39 .5}{3} = 39 .7 s \end{array}$

Absolute errors in the measurements,

$\begin{array}{l} {Δt}_{1} = t - t_{1} = 39 .7 - 39 .6 = 0 .1 s \\ {Δt}_{2} = t - t_{2} = 39 .7 - 39 .9 = - 0 .2 s \\ {Δt}_{3} = t - t_{3} = 39 .7 - 39 .5 = 0 .2 s \end{array}$

$\begin{array}{l} Mean absolute error = \frac{| {Δt}_{1} | + | {Δt}_{2} | + | {Δt}_{3} |}{3} \\ = \frac{0 .1 + 0 .2 + 0 .2}{3} \\ = \frac{0 .5}{3} = 0 .17 \approx 0 .2 \end{array}$

Step 3: Find the accuracy of the measurement.

(rounding off up to one decimal place)

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

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