To find the depth of a cylindrical tank given its capacity and the diameter of its base, we can use the formula for the volume of a cylinder: \[ V = \pi r^2 h \] Where: - \( V \) is the volume of the cylinder, - \( r \) is the radius of the base, - \( h \) is the height (or depth) of the cylinder, - \( \pi \) is a constant approximately equal to 3.14, but in this case, we will use \( \frac{22}{7} \) as specified. ### Step-by-Step Solution 1. **Identify the given values:** - Volume \( V = 3080 \, m^3 \) - Diameter of the base \( d = 14 \, m \) 2. **Calculate the radius:** The radius \( r \) is half of the diameter: \[ r = \frac{d}{2} = \frac{14}{2} = 7 \, m \] 3. **Substitute the values into the volume formula:** We can rearrange the volume formula to solve for height \( h \): \[ h = \frac{V}{\pi r^2} \] Now, substituting the known values: \[ h = \frac{3080}{\frac{22}{7} \cdot (7^2)} \] 4. **Calculate \( r^2 \):** \[ r^2 = 7^2 = 49 \] 5. **Substitute \( r^2 \) into the formula:** \[ h = \frac{3080}{\frac{22}{7} \cdot 49} \] 6. **Calculate \( \frac{22}{7} \cdot 49 \):** \[ \frac{22}{7} \cdot 49 = \frac{22 \cdot 49}{7} = 22 \cdot 7 = 154 \] 7. **Now substitute back to find \( h \):** \[ h = \frac{3080}{154} \] 8. **Perform the division:** \[ h = 20 \, m \] ### Conclusion The depth of the tank is **20 m**. Therefore, the correct option is **C**. ### Explanation of Other Options - **Option A (23 m):** This option is incorrect because it does not satisfy the volume equation when substituting back into the volume formula. - **Option B (25 m):** Similar to option A, this value would yield a volume greater than 3080 m³ when substituted back into the volume formula. - **Option D (22 m):** This option also results in a volume that exceeds the given capacity of 3080 m³, making it incorrect. ### Revision Summary - The volume of a cylinder is calculated using \( V = \pi r^2 h \). - The radius is half of the diameter. - To find the height, rearrange the volume formula to \( h = \frac{V}{\pi r^2} \). - Always check your calculations to ensure the volume matches the given capacity.