Question

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/cm^2. Width of furrow is 5 cm. What will be the maximum draft?

(a) 106 kg

(b) 120 kg

(c) 186 kg

(d) 116 kg

The question was asked during an online exam.

My enquiry is from Cultivators topic in section Harrows, Cultivators and Intercultural Implements of Farm Machinery

Answer 1

The correct choice is (b) 120 kg

For explanation I would say: 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 cm^2

Maximum draft= 200 * 0.6 = 120 kg.