The refractive index of water as measured by a student are 1.29, 1.33, 1.34, 1.35, 1.32, 1.36, 1.30, 1.33. Calculate the relative error.
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The refractive index of water as measured by a student are 1.29, 1.33, 1.34, 1.35, 1.32, 1.36, 1.30, 1.33. Calculate the relative error.
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Answer:
Mean value of quantity measured,
(1.29 + 1.33 + 1.34 + 1.35 + 1.32+ 1.36 + 1.30 + 1.33)/8
x = 1.33(round off)
Δx1 = 1.33 – 1.29 = 0.04
Δx2 = 1.33 – 1.33= 0.00
Δx3 = 1.33 – 1.34= -0.01
Δx4 = 1.33 – 1.35= -0.02
Δx5 = 1.33 – 1.32= 0.01
Δx6 = 1.33 – 1.36= -0.03
Δx7 = 1.33 – 1.30= 0.3
Δx8 = 1.33 – 1.33= 0.00
So mean absolute value = [0.04+0.00+0.01+0.02+0.01+0.03+0.03+0.00]/8 = 0.0175 = 0.02 (rounded off)
Relative error = +- 0.02/1.33 =+- 0.015 = +- 0.02