A five-tine cultivator having tine spacing 8cm, working depth of 5cm and speed is 3km/hr. Turning loss is 10%. Soil resistance is 0.6 kg/cm². Width of furrow is 5 cm. What will be the maximum draft?
a) 106 kg
b) 120 kg
c) 186 kg
d) 116 kg

Answer: b
Explanation: Total width of cultivator = 8*5 = 40 cm = 0.40 m
Speed of travel = 3 km/hr = 3000 m/hr
Area covered/hr = \(\frac{0.40*3000}{10000}*\frac{90}{100}\) = 0.108 ha
Time required = 1/0.108 = 9.25 hr
Cross section of 5 furrows = 5*8*5 = 200 cm²
Maximum draft= 200 * 0.6 = 120 kg.