1. What is the formula for computing depreciation in straight line method?
a) D = \(\frac{P-S}{L*H} \)
b) D = \(\frac{P+S}{L*H} \)
c) D = \(\frac{P-S}{L+H} \)
d) D = (P-S)*(L*H)
Discussion
Explanation: Depreciation is the decrease in cost of machines with passage of time, whether it is used or not. Simple line method is commonly used for determining the depreciation of agricultural machines.
2. What are the ways of doing round and round ploughing?
a) Starting at the outer end
b) Starting at the centre
c) Back furrowing
d) Starting at the outer end & starting at the centre
Discussion
Explanation: In starting at the centre method a small plot of land is marked in the middle of the field and it is ploughed first. After that, the plough works round this small plot and the entire plot is completed. In starting at the outer end tractor starts ploughing at one end of the field and then moves on all sides of the plot and comes gradually from the sides to the centre of the field.
3. What is distance of centre of resistance from share wing?
a) 3/4th size of plough
b) 3/4th size of share
c) 3/4th size of tractor
d) 3/4th size of frog
Discussion
Explanation: Centre of resistance is the point at which the resultant of all the horizontals and vertical force acts. The centre lies at a distance equal to 3/4th size of plough from the share wing.
4. 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
Discussion
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.
5. 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
Discussion
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.
6. 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
Discussion
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.
7. 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
Discussion
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.
8. 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
Discussion
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.
9. 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
Discussion
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.
10. 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
Discussion
Explanation: Depreciation = \(\frac{C-S}{L*H}=\frac{[300000-(\frac{300000}{10})]}{10*1000} = \frac{270000}{10000}\) = Rs 27.