Difficulty: Easy
Correct Answer: 2.50%
Explanation:
Introduction / Context:
Knowing how measurement errors propagate into discharge estimates helps set instrumentation accuracy. For a V-notch weir, Q depends strongly on head, so small head errors can produce larger percentage errors in Q.
Given Data / Assumptions:
Concept / Approach:
For Q ∝ H^n, the relative error in Q is n times the relative error in H: dQ/Q ≈ n * dH/H. Here, n = 5/2 = 2.5.
Step-by-Step Solution:
Verification / Alternative check:
More exact finite-difference check: ((H * 1.01)^(2.5) − H^(2.5)) / H^(2.5) ≈ 2.5% for small errors, confirming the differential result.
Why Other Options Are Wrong:
1.25% and 1.5% correspond to lower exponents; 2.0% assumes square dependence; 5.0% doubles the correct factor.
Common Pitfalls:
Using the rectangular-weir exponent (3/2) accidentally; mixing absolute and relative errors.
Final Answer:
2.50%
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