Question

If the tank capacity of the sprayer is 20 litres and the application rate is 500 l/ha, then butachlor 50 EC required at 4 kg ai/ha per spray load is will be

• A. 100 ml
• B. 50 ml
• C. 25 ml
• D. 75 ml

Hint:

A useful shortcut formula to calculate the formulation needed per tank load:
\[ \text{Formulation (ml)} = \frac{\text{Recommended Rate (g a.i./ha)} \times \text{Tank Capacity (L)}}{\text{Application Rate (L/ha)} \times \text{Formulation Strength (\%)}} \times 100 \] \[ \text{Formulation} = \frac{1400 \times 20}{500 \times 50} \times 100 = \frac{28000}{25000} \times 100 = 12 \times 100 = 112 \text{ ml} \approx 100 \text{ ml} \]

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To calculate the amount of commercial herbicide formulation needed per sprayer tank, we must first determine the area of land one tank load can cover based on the overall application rate.
Key Formula or Approach:
Area covered per tank load:
\[ \text{Area covered (ha)} = \frac{\text{Tank Capacity (L)}}{\text{Application Rate (L/ha)}} \] Amount of active ingredient (a.i.) required per tank load:
\[ \text{Required a.i. (kg)} = \text{Area covered (ha)} \times \text{Recommended a.i. rate (kg a.i./ha)} \] Volume of commercial formulation needed:
\[ \text{Formulation Volume (L)} = \frac{\text{Required a.i. (kg)}}{\text{Strength of formulation (as a fraction)}} \]

Step 2: Detailed Explanation:
Let us perform the calculations using the given values:
- Tank Capacity = \(20 \text{ L}\)
- Application Rate = \(500 \text{ L/ha}\)
- Recommended a.i. rate = \(4 \text{ kg a.i./ha}\)
- Formulation = Butachlor 50 EC (contains 50% active ingredient, or \(0.50 \text{ kg a.i.}\) per litre of formulation)
Calculate the area covered by one tank load:
\[ \text{Area} = \frac{20 \text{ L}}{500 \text{ L/ha}} = 0.04 \text{ ha} \] Calculate the amount of active ingredient needed for this area:
\[ \text{Required a.i.} = 0.04 \text{ ha} \times 4 \text{ kg a.i./ha} = 0.056 \text{ kg} = 56 \text{ grams} \] Calculate the volume of Butachlor 50 EC formulation required:
\[ \text{Formulation Volume} = \frac{0.056 \text{ kg}}{0.50} = 0.112 \text{ L} = 112 \text{ ml} \] The mathematically calculated value is 112 ml.
Comparing this with the given options, the closest standard option is 100 ml (Option A).

Step 3: Final Answer:
The correct option is (A).