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

View Answer

Explanation: Unit Draft = \(\frac{Total \, draft}{Area} = \frac{1700}{4*40*17.5}\) = 0.607 kg/cm

^{2}*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/cm^{2}. 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

View Answer

Explanation: Furrow cross section = 4*35*10=1400 cm

^{2}

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

View Answer

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

View Answer

Explanation: Area covered/hr = \(\frac{35}{100}*\frac{4}{1}*\frac{6*1000}{1}\) = 8400 m

^{2}

Area to be covered in 8 hours = 8400*8 m

^{2}

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

View Answer

Explanation: Total furrow cross section = 4*40*15 = 2400 cm

^{2}

Unit draft = 1600/2400 = 0.66 kg/cm

^{2}

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

View Answer

Explanation: Total cross-sectional area = 40*15*3 = 1800 cm

^{2}

Unit draft = 1200/1800 = 0.66 kg/cm

^{2}= 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

View Answer

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

View Answer

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.

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