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)(i)For the measurement of the diameter of the wire, state the name of a suitable measuring instrument,[1]
- (a)(ii)how random errors may be reduced when using the instrument in (i).[2]
- (b)(i)The stress σ in the wire is calculated by using the expression σ= 4F πα². Show that the value of σ is 1.99 × 10⁸Nm⁻².[1]
- (b)(ii)Determine the percentage uncertainty in σ.[2]
- (b)(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]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.
- [B1] micrometer (screw gauge)/digital calipers
- [M1] take several readings (and average)
- [A1] along the wire or around the circumference
- [A1] σ= 4 × 25/ [π × (0.40 × 10⁻³)²] = 1.99 × 10⁸ N m⁻² or σ= 25/ [π × (0.20 × 10⁻³)²] = 1.99 × 10⁸ Nm⁻²
- [C1] %F = 2% and %d = 5% or ∆F/F = 0.5/25 and ∆d/d = 0.02/0.4
- [A1] %σ = 2% + (2 × 5%) or %σ= [0.02 + (2 × 0.05)] × 100 = 12%
- [C1] absolute uncertainty = (12/100) × 1.99 × 10⁸ = 2.4 x 10⁷
- [A1] σ = 2.0 × 10⁸ ± 0.2 × 10⁸ Nm⁻² or 2.0 ± 0.2 × 10⁸ Nm⁻²
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