Skip to content

Random Error And Mean

Easyphysics

A student records the period of an oscillating spring four times as 2.01 s, 2.05 s, 1.99 s and 2.03 s during a laboratory session. What is the best estimate of the true period from these readings?

Select the correct option:

🔒 Solution Hidden from View

Submit your answer to unlock the detailed step-by-step solution.

About This Question

Subject
physics
Chapter
physics and measurement
Topic
random error and mean
Difficulty
Easy
Year
2025
Tags
random errorarithmetic meanbest estimaterepeated readingsoscillation period

Solution

Correct Answer:

2.02 s

NCERT Class 11, Chapter 2 (Units and Measurements) explains that random errors fluctuate in sign and magnitude between trials, and the best estimate of the true value is obtained by taking the arithmetic mean of repeated readings. Averaging the four periods gives (2.01 + 2.05 + 1.99 + 2.03) / 4 = 8.08 / 4 = 2.02 s. The option 2.05 s is wrong because it selects only the largest single reading rather than averaging. The option 1.99 s is wrong as it picks only the smallest reading, ignoring the others. The option 2.10 s is wrong since it exceeds every individual measurement and cannot be an average of them. A final plausibility check confirms that the mean value 2.02 s lies comfortably within the spread of the readings, which is exactly what we expect when averaging measurements affected only by random error.

This easy difficulty physics question is from the chapter physics and measurement, covering the topic of random error and mean. It appeared in the 2025 exam.

Looking for more practice? Explore all physics questions or browse physics and measurement questions on RankGuru.