A student measures the length and width of a rectangular sheet of metal to be 24.5 ± 0.5 cm and 10.2 ± 0.5 cm respectively.

The best way to write the area and its uncertainty is 250 ± 17 cm2.

Question

A student measures the length and width of a rectangular sheet of metal to be 24.5 ± 0.5 cm  and 10.2 ± 0.5 cm  respectively.

The best way to write the area and its uncertainty is 250 ± 17 cm2.

BlackTom AI · Accepted Answer

A = L × W, so A = 24.5 cm × 10.2 cm = 249.9 cm^2 ≈ 250 cm^2.

1. Relevant knowledge points and concepts
- Area of a rectangle: A = L × W.
- Propagation of uncertainties for products: use partial derivatives. For absolute uncertainties ΔL, ΔW, the first-order change in A can be bounded by ΔA ≈ |∂A/∂L|ΔL + |∂A/∂W|ΔW = W·ΔL + L·ΔW (conservative, worst-case estimate). If uncertainties are independent random errors, one may instead combine contributions in quadrature: ΔA = sqrt((W·ΔL)^2 + (L·ΔW)^2).

2. Step-by-step reasoning and calculations
- Compute nominal area:
  A = 24.5 × 10.2 = 249.9 cm^2 → round to 250 cm^2 for reporting.
- Compute uncertainty using the conservative linear (worst-case) method:
  ΔA = W·ΔL + L·ΔW
  = 10.2×0.5 + 24.5×0.5
  = 5.1 + 12.25
  = 17.35 cm^2.
- Round the uncertainty reasonably (two significant digits here): ΔA ≈ 17 cm^2.

3. Intermediate steps shown above (multiplications and additions listed).

4. Why the correct answer is correct
Using the partial-derivative (linear) propagation gives ΔA = WΔL + LΔW = 17.35 cm^2, which rounded gives 17 cm^2. The nominal area 249.9 rounds to 250 cm^2 to have the same decimal place as the uncertainty. Thus 250 ± 17 cm^2 is the correct conservative statement of the result.

5. Why other plausible answers are incorrect
- The quadrature (root-sum-square) method yields ΔA = sqrt((10.2·0.5)^2 + (24.5·0.5)^2) = sqrt(26.01 + 150.0625) = 13.27 cm^2, so 250 ± 13 cm^2. That value is smaller because it assumes uncorrelated random errors; it is not the conservative (maximum-error) estimate used here. If the problem expects a worst-case absolute uncertainty (instrument tolerances ±0.5 cm), the 17 cm^2 result is appropriate.

6. Conclusion
The area and uncertainty written as 250 ± 17 cm^2 match the provided answer (250, 17) and are justified by computing A = 249.9 cm^2 and using ΔA = WΔL + LΔW = 17.35 cm^2 → 17 cm^2.

A student measures the length and width of a rectangular sheet of metal to be 24.5 ± 0.5 cm and 10.2 ± 0.5 cm respectively. The best way to write the area and its uncertainty is 250 ± 17 cm2.多重下拉选择题

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