Question

How many number of cells of a 30 m spatial resolution DEM would be required to cover a 1:50,000 topographic map of Survey of India, assuming that 1 minute = 1.85 km?

• A. 855,625
• B. 855,525
• C. 855,425
• D. 855,325

Hint:

To calculate the number of DEM cells, divide the area to be covered by the area of each cell. Always convert units to be consistent (e.g., meters in this case).

Answer 1

The Correct Option is A

To solve this, we need to calculate the total number of cells required for the DEM. A 30 m spatial resolution means each cell represents an area of 30 m × 30 m. Step 1: Calculate the area of one cell.
The area of one cell is: \[ \text{Area of one cell} = 30 \, \text{m} \times 30 \, \text{m} = 900 \, \text{m}^2. \] Step 2: Determine the area of the topographic map.
The topographic map scale is 1:50,000, which means 1 unit on the map represents 50,000 units on the ground. We are given that 1 minute (of latitude) corresponds to 1.85 km, which is the length of one side of the map. Since 1 minute of latitude corresponds to 1.85 km, the length of the map in kilometers is 1.85 km. In meters, this is: \[ \text{Length of the map} = 1.85 \, \text{km} = 1850 \, \text{m}. \] Now, the area of the map is: \[ \text{Area of map} = 1850 \, \text{m} \times 1850 \, \text{m} = 3,422,500 \, \text{m}^2. \] Step 3: Calculate the total number of cells.
To cover the entire area of the map, the number of cells required is: \[ \text{Number of cells} = \frac{\text{Area of map}}{\text{Area of one cell}} = \frac{3,422,500 \, \text{m}^2}{900 \, \text{m}^2} = 855,625. \] Thus, the correct answer is (A) 855,625.