Q#17 (Past Exam Paper – November 2017 Paper 22 Q1)
One end of a wire is connected to a fixed point. A load is attached to the other end so that the wire hangs vertically.
The diameter d of the wire and the load F are measured as
d = 0.40 ± 0.02 mm,
F = 25.0 ± 0.5 N.
(a) For the measurement of the diameter of the wire, state
(i) the name of a suitable measuring instrument, [1]
(ii) how random errors may be reduced when using the instrument in (i). [2]
(b) The stress σ in the wire is calculated by using the expression
σ = 4F / πd2 .
(i) Show that the value of σ is 1.99 × 108 N m-2. [1]
(ii) Determine the percentage uncertainty in σ. [2]
(iii) Use the information in (b)(i) and your answer in (b)(ii) to determine the value of σ, with its absolute uncertainty, to an appropriate number of significant figures. [2]
[Total: 8]
Solution:
(a)
(i) micrometer (screw gauge)/digital calipers
(ii) Random errors can be reduced by taking several readings along the wire and then averaging them.
(b)
(i)
σ = 4 × 25 / [π × (0.40×10-3)2] = 1.99 × 108 N m-2
or
{σ = 4F / πd2 = 4F / π(2r)2 = F / πr2 }
σ = 25 / [π × (0.20 × 10-3)2] = 1.99 × 108 N m-2
(ii)
EITHER
{%F = (ΔF/F) × 100% and %d = (Δd/d) × 100%
%d = (0.02/0.40) × 100% = 5%
%F = (0.5/25) × 100% = 2%}
%F = 2% and %d = 5%
{%σ = %F + 2(%d)}
%σ = 2% + (2 × 5%)
%σ = 12%
OR
{Δσ / σ = ΔF/F + 2(Δd/d)}
ΔF / F = 0.5 / 25 and Δd / d = 0.02 / 0
{%σ = (Δσ / σ) × 100%}
%σ = [0.02 + (2×0.05)] ×100 %
%σ = 12%
(iii)
{ Δσ / σ = 12 % = 12 / 100
Δσ = 12 % × σ = (12 / 100) × σ }
absolute uncertainty Δσ = (12 / 100) × 1.99×108 = 2.4×107
{uncertainty should be given to only 1 s.f.
absolute uncertainty Δσ = 2×107 = 0.2×108}
σ = 2.0 × 108 ± 0.2 × 108 N m-2
or (2.0 ± 0.2) × 108 N m-2
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