Question

Maximum productivity of maize is to be obtained with a population of 40,000 plants per hectare at a row spacing of 80 cm. If the seed emergence is 80%, and the number of seeds dropped per hill is 2, determine the seed spacing.

• A. 40 cm
• B. 50 cm
• C. 60 cm
• D. 80 cm

Hint:

Fast shortcut formula: \(S = \frac{10,000 \times n \times E}{\text{Population} \times R} = \frac{10000 \times 2 \times 0.8}{40000 \times 0.8} = \frac{16000}{32000} = 0.5\text{ m} = 50\text{ cm}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Plant population per unit field area depends on row-to-row spacing, hill/seed spacing along the row, the number of seeds dropped per hill, and the percentage field emergence.
Key Formula or Approach:
\[ \text{Plant Population (plants/ha)} = \frac{10,000 \times n \times E}{R \times S} \]
where \(R\) is row spacing (m), \(S\) is seed/hill spacing (m), \(n\) is number of seeds per hill, and \(E\) is emergence rate (fraction).

Step 2: Detailed Explanation:
Given parameters:
- Desired plant population: \(P = 40,000\text{ plants/ha}\)
- Row spacing: \(R = 80\text{ cm} = 0.8\text{ m}\)
- Number of seeds dropped per hill: \(n = 2\)
- Seed emergence rate: \(E = 80\% = 0.80\)
Area of 1 hectare \(= 10,000\text{ m}^2\).
Effective plants produced per hill:
\[ \text{Plants per hill} = n \times E = 2 \times 0.80 = 1.6\text{ plants} \]
Number of hills required per hectare:
\[ N_{\text{hills}} = \frac{40,000}{1.6} = 25,000\text{ hills/ha} \]
Area occupied per hill:
\[ A_{\text{hill}} = \frac{10,000\text{ m}^2}{25,000} = 0.40\text{ m}^2 \]
Since \(A_{\text{hill}} = R \times S\):
\[ S = \frac{A_{\text{hill}}}{R} = \frac{0.40\text{ m}^2}{0.80\text{ m}} = 0.50\text{ m} = 50\text{ cm} \]

Step 3: Final Answer:
Thus, the required seed spacing is 50 cm, corresponding to option (B).