Step 1: Fertilizer requirement per hectare.
Application = 200 kg/ha = 200,000 g/ha
Step 2: Fertilizer delivered per revolution of metering wheel.
One revolution delivers 38 g per row.
Row spacing = 40 cm → 1 ha has \(10000 / 0.40 = 25000\) row-meters.
Step 3: Total revolutions needed.
Total revolutions = 200000 / 38 = 5263.16
Step 4: Ground wheel circumference (D=60 cm).
\[
C = \pi D = 3.14 \times 0.60 = 1.884\ \text{m}
\]
Total distance = 25000 m.
Ground wheel revolutions = 25000 / 1.884 = 13271.7
Step 5: Speed ratio.
\[
\text{Speed ratio} = \frac{\text{Ground wheel rev}}{\text{Metering wheel rev}} = \frac{13271.7}{5263.16} = 2.52
\]
Final Answer: (B) 2.52 : 1
An engine’s torque-speed characteristics is given below:
\[ T_{maxP} = 125 \, \text{N.m}, \, N_{maxP} = 2400 \, \text{rpm}, \, N_{HI} = 2600 \, \text{rpm}, \, T_{max} = 160 \, \text{N.m}, \, N_{maxT} = 1450 \, \text{rpm} \] Where:
The Governor’s regulation is _________% (Rounded off to 2 decimal places).