A three-tine cultivator having tine spacing 6 cm, working depth of 3 cm and speed is 2 km/hr. Turning loss is 10%. Soil resistance is 0.6 kg/cm². Width of furrow is 6 cm. What will be the required power?
a) 0.54 KW
b) 0.23 KW
c) 0.17 KW
d) 1 KW

Answer: b
Explanation: Total width of cultivator = 6*3 = 18 cm = 0.18 m
Speed of travel = 2 km/hr = 2000 m/hr
Area covered/hr = \(\frac{0.18*2000}{10000}*\frac{90}{100}\) = 0.0324 ha
Time required = 1/0.0324 = 30.86 hr
Cross section of 3 furrows = 3*8*3 = 72 cm²
Maximum draft = 72 * 0.6 = 43.2 kg
Required power = \(\frac{Draft*Speed}{1000} = \frac{43.2*9.8*2*1000}{60*60*1000}\) = 0.23 KW.