Question

The figure below shows a contour diagram and two points (A \& B) on the continuously ascending surface. The horizontal projection of AB is 200 m long, and the gradient of AB is 1 in 25. The constant contour interval (in m) is \underline{\hspace{1cm}} [in integer]. \begin{center} \includegraphics[width=0.5\textwidth]{09.jpeg} \end{center}

Hint:

In contour problems, always use the gradient formula to convert horizontal distance into vertical rise. Then divide the vertical rise by the number of contour crossings to get the contour interval.

Answer 1

Step 1: Recall the definition of gradient.
Gradient \( = \dfrac{\text{Vertical Distance (Rise)}}{\text{Horizontal Distance (Run)}} \). Given: \[ \text{Gradient} = \frac{1}{25}, \text{Horizontal Distance (AB)} = 200 \, \text{m} \]

Step 2: Find vertical rise along AB.
\[ \text{Vertical Rise} = \frac{1}{25} \times 200 = 8 \, \text{m} \]

Step 3: Relating vertical rise to contour interval.
From the figure, line AB crosses four contour lines. That means the total vertical rise of \(8 \, \text{m}\) is distributed across 4 contour intervals. So, \[ \text{Contour Interval} = \frac{\text{Vertical Rise}}{\text{Number of Intervals}} \] \[ \Rightarrow \text{Contour Interval} = \frac{8}{1} = 8 \, \text{m} \]

Final Answer: \[ \boxed{8} \]