Explore topic-wise MCQs in Farm Machinery.

This section includes 8 Mcqs, each offering curated multiple-choice questions to sharpen your Farm Machinery knowledge and support exam preparation. Choose a topic below to get started.

1.

Engines of different cylinder dimensions, power and speed can be compared on the basis of __________

A. maximum pressure
B. fuel consumption
C. mean effective pressure
D. unit power
Answer» D. unit power
2.

Which of the following is not a type of transmission dynamometer?

A. Epicyclic train dynamometer
B. Belt transmission dynamometer
C. Rope break dynamometer
D. Torsion dynamometer
Answer» D. Torsion dynamometer
3.

A four cylinder two- stroke cycle engine with a swept volume of 750 cm3 per cylinder gives a specific fuel consumption of 0.25 kg KW-1 h-1. When developing a brake mean effective mean pressure 650 kPa at 1500 rpm. What is the brake thermal efficiency, if the calorific value is 40 MJ kg-1?

A. 9%
B. 18%
C. 32%
D. 36%
Answer» E.
4.

What is the formula of controlling force (Fc) of a governor?

A. Fc = n2 ω r
B. Fc = m ω r
C. of a governor?a) Fc = n2 ω rb) Fc = m ω rc) Fc = m ω2 r
D. Fc = m ω r2
Answer» D. Fc = m ω r2
5.

What is the name of the phenomenon caused in a governor in which the speed of the engines fluctuates continuously above and below the mean speed?

A. Firing interval
B. Air-Supply measurement
C. Sensitiveness of Governor
D. Governor hunting
Answer» E.
6.

What is the formula for specific fuel consumption (i.s.f.c.)?

A. i.s.f.c.=\(\frac{60 I.P.}{Wf} \)
B. i.s.f.c.=\(\frac{60 Wf}{I.P.} \)
C. i.s.f.c.=\(\frac{120 Wf}{I.P.} \)
D. i.s.f.c.=\(\frac{120 I.P.}{Wf} \)
Answer» C. i.s.f.c.=\(\frac{120 Wf}{I.P.} \)
7.

What is value of ‘n’, if there is no missing cycle, in brake power when it is applied as double acting, two stroke engines?

A. 2N
B. N
C. N/2
D. 4N
Answer» B. N
8.

What is the formula for indicated mean effective pressure?

A. Pi = \(\frac{(Area \, of \, indicator \, diagram)}{Length \, of \, indicator \, diagram \, * \, spring \, scale \, index} \)
B. Pi = \(\frac{(Length\, of \, indicator \, diagram \, * \, spring\, scale \,index)}{Area\, of \,indicator \, diagram} \)
C. Pi = \(\frac{Length \, of \, indicator \, diagram}{Area \,of \,indicator\, diagram \, * \, spring \, scale \, index} \)
D. Pi = \(\frac{Spring \, scale \, index}{Area \, of \, indicator \, diagram \, * \, Length \, of \, indicator \, diagram} \)
Answer» B. Pi = \(\frac{(Length\, of \, indicator \, diagram \, * \, spring\, scale \,index)}{Area\, of \,indicator \, diagram} \)