Revolutions from path length: A circular disc with area 0.49π m^2 rolls without slipping through 1.76 km. How many full revolutions does it make?

Introduction / Context:
To convert a rolling distance into a number of revolutions, we need the wheel’s circumference. When area is given, we must back out the radius first and then use the circumference-to-distance relation.

Given Data / Assumptions:

• Area of disc A = 0.49π m^2
• Travel distance L = 1.76 km = 1760 m
• No slipping, so L = N * circumference

Concept / Approach:
From A = π r^2, obtain r. Then circumference C = 2 * π * r. Finally, revolutions N = L / C.

Step-by-Step Solution:
π r^2 = 0.49π → r^2 = 0.49 → r = 0.7 mC = 2 * π * 0.7 = 1.4π ≈ 4.398 mN = 1760 / (1.4π) ≈ 1760 / 4.398 ≈ 400

Verification / Alternative check:
Using π = 22/7, C = 1.4 * 22/7 = 4.4 m; N = 1760 / 4.4 = 400 exactly—confirms the computation.

Why Other Options Are Wrong:
300 and 350 are too few; 600 and 4000 are too many compared to the circumference-derived distance per turn.

Common Pitfalls:
Mistaking diameter for radius, not converting km to m, or forgetting that area gives r via r = √(A/π).

Final Answer:
400