हम एक सरल लोलक कस दोलन - काल ज्ञात करते है । प्रयोग के क्रमिक मापनों में लिए गए पाठ्यांक हैं : 2.63 s, 2.56 s, 2.42 s, 2.71 s एवं 2.80 s । निरपेक्ष त्रुटि , सापेक्ष त्रुटि एवं प्रतिशत त्रुटि परिकलित कीजिए ।

Explanation

The mean value of length
` L= ((2.63+2.56+2.42+2.71+2.80)m)/(5)`
`=(13.12 m)/(5) = 2.624 m = 2.62 m`
As the lengths are measured to a resolution of 0.01 m, all lengths are given to the second place of decimal, it is proper to round off this mean length also to the second place of decimal. In the first measurement `"Error" = Deltax_(1) = 2.63 m - 2.62 m = +0.01 m`
Absolute error = 0.01 m
`"Relative error" = (0.01)/(2.62) = 0.0038`
`"Percentage error" = "relative error" xx 100 = 0.38%`
In the second measurement
`"Error" = Deltax_(2)= 2.56 m - 2.62 m = -0.06 m`
Absolute error = 0.06 m
`"Relative error" = 0.06//2.62 = 0.023`
`"Percentage error" = "relative error" xx 100 = 2.3%`
In the third measurement
`"Error" = Deltax_(3)= 2.42 m - 2.62 m = 0.2 m`
Absolute error = 0.2m
`"Relative error" = 0.2//2.62 = 0.076`
`"Percentage error" = "relative error" xx 100 = 7.6%`
In the fourth measurement
`"Error" =Deltax_(4), = 2.71 m - 2.62 m = -0.09 m`
Absolute error = 0.09 m
`"Relative error" = 0.09//2.62 = 0.034`
`"Percentage error" = "relative error" xx 100 = 3.4%`
In the fifth measurement
`"Error" = Deltax_(5) = 2.80 m - 2.62 m = 0.18 m`
Absolute error = 0.18 m
`"Relative error" = 0.18//2.62 = 0.068`
`"Percentage error" = "relative error" xx 100 = 6.8%`
`"Mean or final absolute error" = (sum_(i=1)^5 |Deltax_i|)/(5)`
`=((0.01+0.06+0.20+0.09+0.18)m)/(5)`
`= (0.54m)/(5) = 0.108m = 0.11m`
This means that the length is `(2.62 +-0.11m)` i.e, it lies between `(2.62 +- 0.11 m)` and (2.62 -0.11m)
i.e, between 2.73 m and 2.51 m