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.Multiple dropdown selections

Similar Questions

Which answer choice explains why furanoses and pyranoses can be reduced by sodium borohydride despite having no carbonyl functional groups.

Which of the following statements describes a glycoside?

Which of the following structures represents chitin?

In solution, glucose exists in

According to Xenophanes, what should be avoided at a symposium?

What is "kottabos"?

Why are there so many naked figures in vase iconography?

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler