Subtopic: Calibration calculation and seed rate adjustment
Question Statement:
A seed drill with a working width of $1.8\text{ m}$ and a ground wheel diameter of $0.6\text{ m}$ is calibrated in the laboratory. For $100$ revolutions of the ground wheel, the total seed collected is $580\text{ g}$. The seed rate is ________ $\text{kg/ha}$ (round off to two decimal places, take $\pi = 3.1416$).
Answer Key:
17.09
Step-by-Step Solution & Explanation
1. Distance covered in 100 revolutions:
$S = n \times \pi D = 100 \times 3.1416 \times 0.6 = 188.50\text{ m}$
2. Area covered:
$A = S \times W = 188.50 \times 1.8 = 339.29\text{ m}^2 = \frac{339.29}{10000}\text{ ha}$
3. Seed rate in kg/ha:
$\text{Seed Rate} = \frac{580/1000}{339.29/10000} = 17.09\text{ kg/ha}$