This set of Farm Machinery Multiple Choice Questions & Answers (MCQs) focuses on “Harvesting Equipments – Power Transmitting Unit Mower”.
1. Which part operates the tractor drawn semi-mounted or mounted type mowers?
a) PTO shaft
b) Cutter
c) Pitman
d) Grass board
View Answer
Explanation: Tractor drawn semi mounted or mounted type mowers are operated by PTO shaft. In this case, the cutting mechanism is driven independently of the forward speed of the mower. A shaft is connected with the PTO shaft which drives a pulley with the help of a universal joint.
2. Which clutch is used in the driving unit of a mower?
a) Fluid coupling
b) Dog clutch
c) Friction clutch
d) Single plate
View Answer
Explanation: In bullock drawn mowers, the power transmitting unit consists of axle, gears, crank wheel, crankshaft and pitman. The transport wheel gives power to the axle from where the power is transmitted through the gear box. There is a ball and socket arrangement to connect the knife with the pitman. The operator controls the driving unit with the help of a dog clutch.
3. What is the difference between each knife clips?
a) 5-15 cm
b) 40-50 cm
c) 35-45 cm
d) 20-30 cm
View Answer
Explanation: Cutter bar is an assembly comprising of fingers, knife guides, on wearing plates and shoes. It is used for cutting grasses and forge. It is made of high grade steel. It works like a knife. There are ledger plates provided with the knife guard, on which the knife moves. Knife clips hold the sections down against the ledger plates. Knife clips are placed with wearing plates spaced 20 to 30 cm apart.
4. When the knife section stops in the centre of its guard on every stroke, it is known as?
a) Re-embankment
b) Reinforcement
c) Registration
d) Rescue
View Answer
Explanation: A mower knife is said to be in proper registration when the knife section stops in the centre of its guard on every stroke i.e. the centre of the knife section is at the centre of the guard, when it is in opening condition.
5. Calculate the total time required to harvest 2.5 hectares of grass by means of a 2 metre mower being operated at 4 km/hr. (Field efficiency=80%)
a) 2.5 hrs
b) 1.2 hrs
c) 3.9 hrs
d) 7 hrs
View Answer
Explanation: Theoretical area covered/hr = 2*4*1000 m2
Actual area covered/hr = \(\frac{2*4*1000*80}{100}\)m2 = \(\frac{(2*4*1000*80)}{100*1000}\) ha
Time required for 2.5 ha = 2.5/0.64 = 3.9 hrs
6. What power is required to pull 1.2 mete mower working at a speed of 4.8 km/hr, if there is a load of 50 kg per metre length of mower and mechanical efficiency is 80%?
a) 1 KW
b) 0.98 KW
c) 0.23 KW
d) 2 KW
View Answer
Explanation: Total load = 1.2*50 = 60 kg
= 60*9.8 N
Power = \(\frac{N.\frac{m}{s}}{1000} \)
Total power = \(\frac{60*9.8*4.8*1000}{60*60*1000}\) = 0.784 KW
Power required to pull the mower = 0.784/0.80 = 0.98 KW
7. How many hectares per day of 10 hours can be cut by a combine with 4 metre cutter bar, when it is running at 4 k/hr?
a) 16ha
b) 20 ha
c) 28 ha
d) 8 ha
View Answer
Explanation: Width of cutter bar = 4m
Speed of combine = 4km/hr = 4*1000 m/hr
Area covered per hour = 4*4*1000 m2 = 1.6 ha
Area covered in 10 hrs = 1.6*10 = 16 ha
8. A mower has drive wheel of 60 cm diameter. The crank of the mower makes 600 rev/min when it is driven by a tractor, moving at a speed of 2.3 km/hr. If the speed ratio between the crank wheel and land wheel is changed to 27:1, calculate the increase in speed of mower to maintain same speed of crank.
a) 0.19 km/hr
b) 0.67 km/hr
c) 0.11 km/hr
d) 0.21 km/hr
View Answer
Explanation: Revolution of the drive wheel = 600/27 = 22.2 rev/min
Linear speed of mower = π * \(\frac{60}{100}*\frac{60}{27}\)*60 m = 2.51 km/hr
Increase in speed of mower = 2.51-2.30 = 0.21 km/hr
9. Which part of mower is used to regulate the height of cut above the ground?
a) Ledger plate
b) Wearing plate
c) Shoe
d) Grass board
View Answer
Explanation: A shoe on each end of the cutter bar is always provided to regulate the height of cut above the ground. The inner shoe is larger in section and is placed at the inner end of the cutter bar. The outer shoe is placed at the outer end is smaller in section.
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