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Error Analysis

Practice Questions:

Question 21: easy

Taking into account of the significant figures, what is the value of \(9.99 \text{m} – 0.0099 \text{m}\)?

[NEET 2020]

1. \(9.98 m\)

2. \(9.980 \text{m}\)

3. \(9.9 \text{m}\)

4. \(9.9801 \text{m}\)

View Answer

The numbers are \(9.99 \text{m}\) (2 decimal places) and \(0.0099 \text{m}\) (4 decimal places). For subtraction, the result should have the same number of decimal places as the number with the fewest decimal places (2 in this case). \(9.9900 - 0.0099 = 9.9801\). Rounding \(9.9801\) to 2 decimal places gives \(9.98 \text{m}\).

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Question 22: easy

Following observations were taken with a vernier calliper while measuring the length of a cylinder: \(3.29 \text{cm}\), \(3.28 \text{cm}\), \(3.29 \text{cm}\), \(3.31 \text{cm}\), \(3.28 \text{cm}\). The most accurate length of the cylinder will be

1. \(3.28 \text{cm}\)

2. \(3.29 \text{cm}\)

3. \(3.32 \text{cm}\)

4. \(3.21 \text{cm}\)

View Answer

The most accurate value is the mean value of the observations. Mean length = \[\frac{3.29 + 3.28 + 3.29 + 3.31 + 3.28}{5} = \frac{16.45}{5} = 3.29 \text{cm}\].

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