Discussion
A 5 × 20 cm seed drill has a ground drive wheel of rolling diameter 0.5 m. While testing under laboratory condition, 320 g of seeds were collected in 20 revolutions of the ground drive wheel. The same seed drill when operated in a 2 ha field, amount of seeds dropped was found to be 185 kg. The variation in the seed dropped between laboratory and field conditions due to skid of ground drive wheel is ________
(Take π = 3.14)
Answer: B
The variation due to skid is 9.23%.
The drill has 5 furrow openers at 20 cm spacing, a width of 1.0 m. In the laboratory, 20 turns of the 0.5 m wheel cover
\[ \begin{aligned} L &= 20 \times 3.14 \times 0.5 = 31.4\ \text{m} \\ \text{area} &= 31.4 \times 1.0 = 31.4\ \text{m}^{2} \end{aligned} \]
so the calibrated rate, and the seed it should drop on 2 ha, are
\[ \begin{aligned} \text{rate} &= \frac{0.320 \times 10000}{31.4} = 101.9\ \text{kg/ha} \\ 2\ \text{ha} &: 203.8\ \text{kg} \end{aligned} \]
The field actually received 185 kg, because the drive wheel skidded and turned the metering mechanism less than the distance travelled. The variation is
\[ \frac{203.8 - 185}{203.8} \times 100 = 9.23\% \]
10.17% is the same difference divided by the field quantity (185 kg) instead of the expected quantity. The other values come from slips in the width or circumference.
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