Question

The diameter of a wire as measured by screw gauge was found to be 2.620, 2.625, 2.630, 2.628 and 2.626 cm. Calculate (a) mean value of diameter (b) absolute error in each measurement (c) mean absolute error (d) fractional error (e) percentage error (d) (f) Express the result in terms of percentage error

Answer 1

(a) Mean value of diameter

$d_{m}$ = (2.620 + 2.625 + 2.630 + 2.628 + 2.626) / 5

$d_{m}$ = 13.129 / 5

$d_{m}$ = 2.6258 cm ≈ 2.626 cm

(b) absolute error in each measurement

Δd₁ = $d_{m}$ - d₁ = 2.626 - 2.620 = +0.006 cm

Δd₂ = $d_{m}$ - d₂ = 2.626 - 2.625 = +0.001 cm

Δd₃ = $d_{m}$ - d₃ = 2.626 - 2.630 = -0.004 cm

Δd₄ = $d_{m}$ - d₄ = 2.626 - 2.628 = -0.002 cm

Δd₅ = $d_{m}$ - d₅ = 2.626 - 2.626 = +0.000 cm

(c) Mean Absolute error

Δ$d_{m}$ = (IΔd₁I + IΔd₂I + IΔd₃I + IΔd₄I + IΔd₅I) / 5

Δ$d_{m}$ = (0.006 + +0.001 + -0.004 + 0.002 + 0.000) / 5

Δ$d_{m}$ = 0.013/5 = 0.0026 ≈ 0.003

(d) Fractional error

δd = Δ$d_{m}$ / $d_{m}$

δd = ± 0.003 / 2.626

δd = ± 0.001

(e) Percentage error

% error = Δ$d_{m}$ / $d_{m}$ × 100

% error = ± 0.001 × 100 = ± 0.1 %

(f) The diameter in terms of percentage error

d = (2.626 ± 0.1 %) cm