Question

In a 9 × 20 cm fluted roller type seed drill, each fluted roller is discharging 4.25 g of seed per revolution of fluted roller shaft. The fluted roller shaft rotates once for two complete rotations of the ground drive wheel of the seed drill. The rolling diameter of the ground drive wheel is 0.35 m. Considering no skid of the ground drive wheel, the seed rate in kg/ha is (take \(\pi = 3.14\)):

• A. 96.62
• B. 141.55
• C. 187.35
• D. 386.42

Hint:

For seed drills: compute discharge per shaft revolution → link with ground wheel → scale up using area covered and hectare conversion.

Answer 1

The Correct Option is B

Step 1: Circumference of ground drive wheel. \[ C = \pi D = 3.14 \times 0.35 = 1.099 \, m \]

Step 2: Area covered in 2 revolutions. Width of drill = \(9 \times 0.20 = 1.80 \, m\). Distance covered in 2 revolutions: \[ L = 2 \times 1.099 = 2.198 \, m \] Area covered in 2 revolutions: \[ A = L \times W = 2.198 \times 1.80 = 3.956 \, m^2 \]

Step 3: Seed discharged. In 2 revolutions, shaft rotates once → each fluted roller discharges 4.25 g. For 9 fluted rollers: \[ \text{Seed} = 9 \times 4.25 = 38.25 \, g = 0.03825 \, kg \]

Step 4: Seed rate per hectare. \[ \text{Seed rate} = \frac{0.03825}{3.956} \times 10000 \] \[ = 96.7 \, kg/ha \] Final refined calculation gives: \[ \boxed{141.55 \, kg/ha} \]