Question

With 46% N content in urea and recommended dose of 150 kg N/ha for maize crop, how much urea is required to apply one-third nitrogen of the recommended dose?

• A. 150 kg/ha
• B. 109 kg/ha
• C. 124 kg/ha
• D. 96 kg/ha

Hint:

Fast shortcut: Quantity of Urea $= \frac{\text{N required}}{0.46} = 2.174 \times \text{N required}$. Here $2.174 \times 50 = 108.7 \approx 109\text{ kg/ha}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Fertilizer dose calculation determines the commercial fertilizer quantity needed based on the percentage of active nutrient content in the fertilizer carrier.
Key Formula or Approach:
\[ \text{Fertilizer Required (kg/ha)} = \frac{\text{Nutrient Required (kg/ha)}}{\%\text{ Nutrient Content}} \times 100 \]

Step 2: Detailed Explanation:
Given parameters:
- Recommended nitrogen dose: \(N_{\text{total}} = 150\text{ kg N/ha}\)
- Nitrogen fraction to apply: \(\frac{1}{3}\text{ of recommended dose}\)
- Nitrogen content in urea: \(46\% = 0.46\)

Step 1: Calculate the nitrogen quantity for one-third dose:
\[ N_{\text{applied}} = \frac{1}{3} \times 150\text{ kg N/ha} = 50\text{ kg N/ha} \]
Compute the required quantity of commercial urea:
\[ \text{Urea Required} = \frac{50\text{ kg N}}{0.46} = 108.695\text{ kg/ha} \approx 109\text{ kg/ha} \]

Step 3: Final Answer:
Therefore, the required urea amount is 109 kg/ha, matching option (B).