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Time for 20 oscillations of a pendulum is measured as t1 = 39.6 s; t2 = 39.9 s; t3 = 39.5 s. What is the precision in the measurements? What is the accuracy of the measurement? - Physics

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Question

Time for 20 oscillations of a pendulum is measured as t1 = 39.6 s; t2 = 39.9 s; t3 = 39.5 s. What is the precision in the measurements? What is the accuracy of the measurement?

Short Note

Solution

Given, t1 = 39.6 s, t2 = 39.9 s and t3 = 39.5 s

The least count of measuring instruments = 0.1 s  ......(As measurements have only one decimal place)

Precision in the measurement = Least count of the measuring instrument = 0.1 s

The mean value of time for 20 oscillations is given by

t = `(t_1 + t_2 + t_3)/3`

= `(39.6 + 39.9 + 39.5)/5`

= 39.7 s

Absolute errors in the measurements

Δt1 = t – t1 = 39.7 – 39.6 = 0.1 s

Δt2 = t – t2 = 39.7 – 39.9 = – 0.2 s

Δt3 = t – t3 = 39.7 – 39.5 = 0.2 s

Mean absolute error = `(|Δt_1| + |Δt_2| + |Δt_3|)/3`

= `(0.1 + 0.2 + 0.2)/3`

= `0.5/3` = 0.17 ≈ 0.2  .....(Rounding off up to one decimal place)

∴ Accuracy of measurement = ± 0.2 s

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Chapter 2: Units and Measurements - Exercises [Page 10]

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NCERT Exemplar Physics [English] Class 11
Chapter 2 Units and Measurements
Exercises | Q 2.35 | Page 10

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