Subtopic: Functional requirements, principles of working, construction and operation of manual, animal, tractor and renewable energy operated equipment for tillage, sowing, planting, fertilizer application, inter-cultivation, spraying, mowing, chaff cutting, harvesting and threshing
Question Statement:
A 9-row tractor seed drill with row spacing of $20\text{ cm}$ has a ground drive wheel of effective diameter $D = 70\text{ cm}$. During a stationary calibration test, the drive wheel is rotated through 50 revolutions, and the total seed collected from all 9 furrow openers is $1.98\text{ kg}$. The calibrated seed rate of the drill in $\text{kg/ha}$ is ________ (round off to 1 decimal place). Take $\pi = 3.1416$.
Answer Key:
100.0
Step-by-Step Solution & Explanation
1. Working width of the seed drill:
$W = 9 \times 0.20\text{ m} = 1.80\text{ m}$
2. Drive wheel circumference:
$C = \pi D = 3.1416 \times 0.70\text{ m} = 2.1991\text{ m}$
3. Distance traveled in 50 revolutions:
$S = 50 \times 2.1991 = 109.956\text{ m}$
4. Area covered:
$A = W \times S = 1.80\text{ m} \times 109.956\text{ m} = 197.92\text{ m}^2 = 0.019792\text{ ha}$
5. Seed rate in $\text{kg/ha}$:
$\text{Seed Rate} = \frac{1.98\text{ kg}}{0.019792\text{ ha}} \approx 100.04\text{ kg/ha} \approx 100.0\text{ kg/ha}$