1. A student measures the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.56 s, 2.62 s, 2.70 s, 2.58 s and 2.54 s. The absolute error in the third measurement is:
- 0.10 s
- 0.05 s
- 0.02 s
- 0.08 s
Show answer and explanation
Answer: (A) 0.10 s
The mean value of the period is (2.56 + 2.62 + 2.70 + 2.58 + 2.54) / 5 = 13.00 / 5 = 2.60 s. The absolute error in the third measurement is |2.70 - 2.60| = 0.10 s.