Question:hard

An error of 1 % in measuring head (H) will produce

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The percentage error in discharge is simply the exponent of the head \(H\) multiplied by the percentage error in the head measurement.
Since the exponent for a triangular weir is 2.5 and for a rectangular weir is 1.5, the triangular weir will always exhibit a higher percentage error.
  • Same error in discharge over a rectangular weir and triangular weir
  • More error in discharge over triangular weir compared to rectangular weir
  • More error in discharge over rectangular weir compared to triangular weir
  • No error in discharge over a rectangular weir and triangular weir
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Write the discharge formula for each weir type.
A rectangular weir has discharge proportional to head raised to the power 3/2, \( Q \propto H^{3/2} \), while a triangular, or V-notch, weir has discharge proportional to head raised to the power 5/2, \( Q \propto H^{5/2} \).
Step 2: Turn each formula into a percentage error relation.
Taking logarithms and differentiating a relation of the form \( Q \propto H^n \) gives \( \dfrac{dQ}{Q} = n\dfrac{dH}{H} \), so the percentage error in discharge is simply n times the percentage error in head.
Step 3: Substitute the exponents for each weir.
For the rectangular weir, n = 3/2, so a 1 percent error in H produces \( 1.5\times 1\% = 1.5\% \) error in Q. For the triangular weir, n = 5/2, so the same 1 percent error in H produces \( 2.5\times 1\% = 2.5\% \) error in Q.
Step 4: Compare the two results.
Since 2.5 percent is larger than 1.5 percent, the same small error in measuring head always produces a bigger discharge error on a triangular weir than on a rectangular weir.
\[ \boxed{\text{More error in discharge over triangular weir compared to rectangular weir}} \]
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