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Question:

In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is T=2π√7(R−r)/5g. The values of R and r are measured to be (60±1)mm and (10±1)mm, respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01 s. Which of the following statement(s) is(are) true?

The error in the measurement of T is 2

The error in the determined value of g is 11

The error in the measurement of T is 3.57

The error in the measurement of r is 10

Solution:

The observed values of time period, Ti = 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s
The mean value of time period, T = ΣTi/5 = 2.78/5 = 0.56 s
Magnitude of absolute error in each observation, |ΔT1| = |0.56 − 0.52| = |0.04|s
Similarly, |ΔT2| = |0.0|s
|ΔT3| = |0.01|s
|ΔT4| = |0.02|s
|ΔT5| = |0.03|s
Mean absolute error in time period, ΔTm = (0.04 + 0.00 + 0.01 + 0.02 + 0.03)/5 = 0.02 s
∴Error in T, ΔTm/T × 100 = 0.02/0.56 × 100 = 3.57