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.

Question 17

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]

Reference: Past Exam Paper – November 2017 Paper 22 Q1

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