Question

1 cm of rainfall over a catchment area of 1 \(\text{km}^2\) represents a volume of water equal to:

• A. \(10^4 \text{ m}^3\)
• B. \(10^4 \text{ m}^2\)
• C. \(10^4 \text{ m}^4\)
• D. \(10^3 \text{ m}^3\)

Hint:

Keep these direct conversion shortcuts handy for hydrological exams:
- \(1\text{ mm of rain over 1 ha} = 10\text{ m}^3\).
- \(1\text{ cm of rain over 1 ha} = 100\text{ m}^3\).
- \(1\text{ cm of rain over 1 km}^2 = 10,000\text{ m}^3 = 10^4\text{ m}^3\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The volume of water deposited by a rainfall event over a specific catchment is computed as the product of the flat projected catchment surface area and the uniform depth of rainfall.

Step 2: Key Formula or Approach:
\[ \text{Volume } (V) = \text{Area } (A) \times \text{Depth } (d) \] Convert all dimensional units into standard SI base units (\(\text{m}^2\) and m) to obtain volume in cubic meters (\(\text{m}^3\)).

Step 3: Detailed Explanation:
1. Convert the catchment area from square kilometers to square meters:
\[ 1\text{ km} = 1000\text{ m} \] \[ A = 1\text{ km}^2 = (1000\text{ m})^2 = 1,000,000\text{ m}^2 = 10^6\text{ m}^2 \] 2. Convert the rainfall depth from centimeters to meters:
\[ d = 1\text{ cm} = \frac{1}{100}\text{ m} = 0.01\text{ m} = 10^{-2}\text{ m} \] 3. Calculate the total volume:
\[ V = A \times d \] \[ V = 10^6 \text{ m}^2 \times 10^{-2} \text{ m} \] \[ V = 10^{6 - 2} \text{ m}^3 = 10^4 \text{ m}^3 \]

Step 4: Final Answer:
The correct option is 1, which corresponds to \(10^4 \text{ m}^3\).