# Farm Machinery Questions and Answers – Numericals on Mouldboard Plough

This set of Farm Machinery Multiple Choice Questions & Answers (MCQs) focuses 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

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

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

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

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

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.
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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

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

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

Explanation: Interest/hr = $$\frac{[12000+12000*\frac{10}{100}]}{[2*300]} * \frac{10}{100} = \frac{13200}{2} * \frac{10}{100*300}$$ = Rs 2.20/hr.

Sanfoundry Global Education & Learning Series – Farm Machinery.