4. Four persons measure the length of a rod as 20.00 cm,19.75 cm, 17.01 cm and 18.25 cm. The relative error in the measurement of average length of the rod is
Let absolute error, is ΔL,
Average length = 20+19.75+17.01+18.25/4
= 18.72525 cm
ΔL_1 = |20.00 - 18.7525| = 1.2475 cm
ΔL_2 = |19.75 - 18.7525| = 0.9975 cm
ΔL_3 = |17.01 - 18.7525| = 1.7425 cm
ΔL_4 = |18.25- 18.7525| = 0.5025 cm
Mean absolute error = 1.2475 + 0.9975 + 1.7425 + 0.5025 / 4
= 1.1225 cm
Relative error = 1.1225 cm/18.7525 cm = 0.0598 ≈ 0.06
Let absolute error, is ΔL,
Average length = 20+19.75+17.01+18.25/4
= 18.72525 cm
ΔL_1 = |20.00 - 18.7525| = 1.2475 cm
ΔL_2 = |19.75 - 18.7525| = 0.9975 cm
ΔL_3 = |17.01 - 18.7525| = 1.7425 cm
ΔL_4 = |18.25- 18.7525| = 0.5025 cm
Mean absolute error = 1.2475 + 0.9975 + 1.7425 + 0.5025 / 4
= 1.1225 cm
Relative error = 1.1225 cm/18.7525 cm = 0.0598 ≈ 0.06
Let absolute error, is ΔL,
Average length = 20+19.75+17.01+18.25/4
= 18.72525 cm
ΔL_1 = |20.00 - 18.7525| = 1.2475 cm
ΔL_2 = |19.75 - 18.7525| = 0.9975 cm
ΔL_3 = |17.01 - 18.7525| = 1.7425 cm
ΔL_4 = |18.25- 18.7525| = 0.5025 cm
Mean absolute error = 1.2475 + 0.9975 + 1.7425 + 0.5025 / 4
= 1.1225 cm
Relative error = 1.1225 cm/18.7525 cm = 0.0598 ≈ 0.06