This set of Farm Machinery Multiple Choice Questions on “Numericals on Mouldboard Plough”.
1. Total draft of four bottom, 40cm Mb plough when ploughing 17.5 cm deep at 5.5 km/hr speed is 1700. Field efficiency is 75%. Calculate the actual power requirement.
a) 24.73KW
b) 25.45KW
c) 23.39KW
d) 22.41KW
Answer: b
Explanation: Unit Draft = (frac{Total , draft}{Area} = frac{1700}{4*40*17.5}) = 0.607 kg/cm2 *9.8 kPa = 59.48 kPa
Power requirement = (frac{1700*9.8*5.5*1000}{60*60})N.m/s = 25452.77 N.m/s = 25.45 KW.
2. Calculate the size of tractor to pull a four bottom 35 cm MB plough through a depth of 10 cm. the soil resistance is 0.5 kg/cm2. The speed of the tractor is 5.5 km/h, the transmission and tractive efficiency of the tractor being 85% and 30% respectively.
a) 41.1 KW
b) 43.9 KW
c) 42.7 KW
d) 44.8 KW
Answer: a
Explanation: Furrow cross section = 4*35*10=1400 cm2
Total draft = 1400 * 0.5 kg = 700 kg
700 * 9.8 = 6860 N
Power = 6860 * (frac{(5.5*1000)}{(60*60)}*frac{1}{0.85}*frac{1}{0.30})N.m/s = 41100.21 W
= 41.1 KW.
3. Line of pull of a MB plough is 15° with the horizontal and is in a vertical plane which at an angle of 12° with the direction of travel plane which is at an angle 12° with the direction off travel. Calculate the side draft.
a) 234.12 kg
b) 219.06 kg
c) 212.45 kg
d) 200.09 kg
Answer: c
Explanation: Draft = P * cos 15° * cos 12°
Side draft = P * cos 15° * sin 15°
Or, P = (frac{Draft}{cos15°*cos12°}=frac{1000}{cos15°*cos12°}) = 1058 kg
Side draft = 1058 * cos15° * sin 12°
= 212.45 kg.
4. What is the area covered per day of 8 hours by a tractor down four bottom 35 cm plough if the speed of the ploughing is 6km per hour, the time lost in turning is 6%?
a) 0.39 hectare
b) 0.67 hectare
c) 0.23 hectare
d) 0.40 hectare
Answer: d
Explanation: Area covered/hr = (frac{35}{100}*frac{4}{1}*frac{6*1000}{1}) = 8400 m2
Area to be covered in 8 hours = 8400*8 m2
Turning loss = (frac{6.72*6}{100}) = 0.40 hectare.
5. A four bottom 40 cm MB plough has a working depth of 15 cm and draft of 1600 kg. It is working at a speed of 15 cm and draft of 1600 kg. it is working at a speed of 4.5 km/hr with field efficiency of 70%. What will be the drawbar power?
a) 14.9 KW
b) 19.6 KW
c) 18.7 KW
d) 17.6 KW
Answer: b
Explanation: Total furrow cross section = 4*40*15 = 2400 cm2
Unit draft = 1600/2400 = 0.66 kg/cm2
Drawbar power = (frac{Draft*speed}{1000} = frac{1600*9.8*4.5*1000}{1000*60*60}) = 19.6KW.
6. A three bottom 40 cm MB plough has a working depth of 15cm, draft is 1200 kg, field efficiency is 80%, and working speed is 4km/hr. What will be actual field capacity?
a) 0.384 ha/hr
b) 0.400 ha/hr
c) 0.200 ha/hr
d) 0.319 ha/hr
Answer: a
Explanation: Total cross-sectional area = 40*15*3 = 1800 cm2
Unit draft = 1200/1800 = 0.66 kg/cm2 = 64.68 KPa
Actual field capacity = Theoretical field capacity * field efficiency
= (frac{3*40*4*1000}{100*10000}*frac{80}{100}) = 0.384 ha/hr.
7. What would be the depreciation cost of a tractor purchased by the farmer having 25 KW power at a total cost of Rs 300000/-?
a) Rs 25
b) Rs 24
c) Rs 27
d) Rs 29
Answer: c
Explanation: Depreciation = (frac{C-S}{L*H}=frac{[300000-(frac{300000}{10})]}{10*1000} = frac{270000}{10000}) = Rs 27.
8. What would be the interest/hr of a three-bottom plough purchased by a farmer having 30 cm bottom width at ₨ 12000/- only? (Ploughing speed is 5 km per hour. Interest rate @ 10% of capital cost/year and salvage value @ 10% capital cost per year.)
a) Rs 2.20/hr
b) Rs 2.40/hr
c) Rs 3.02/hr
d) Rs 1.50/hr