find the absolute error in the measurement. different trails are as follows, 1.2,1.4,1.3,1.6
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Explanation:
Mean value of quantity measured, V=
8
1.29+1.33+1.34+1.35+1.32+1.36+1.30+1.33
x=1.3275=1.33(round off to two places of decimal).Absolute errors in measurement are:Δx
1
=1.33−1.29=0.04; Δx
2
=1.33−1.33=0.00Δx
3
=1.33−1.34=−0.01; Δx
4
=1.33−1.35=−0.02Δx
5
=1.33−1.32=+0.01; Δx
6
=1.33−1.36=−0.03Δx
7
=1.33−1.30=+0.03; Δx
8
=1.33−1.33=0.00
Mean absolute error,
Δx
=
n
∑
i=l
i=n
∣(Δx)
i
∣
=
8
0.04+0.00+0.01+0.02+0.01+0.03+0.03+0.00
8
0.14
=0.0175=0.02
Relative error=±
x
Δx
=±
1.33
0.02
=±0.015=±0.02
Percentage error=±0.015×100=1.5%
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