Question

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.

Answer 1

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