Home > General Studies (Hindi) > Quizzes > Machine Dynamics Practice Test: Balancing of Reciprocating Masses
Machine Dynamics Practice Test: Balancing of Reciprocating Masses
Fast practice, instant feedback. Timer auto-submits when time’s up.
Avg score: 82% Most missed: “In an outside cylinder locomotive, the position of the driving wheels is _______…”
Balancing of Reciprocating Masses topics include: Primary unbalanced forces for reciprocating masses, locomatives balancing, tractive force, swaying couple, multi cylinder inline and v-engines. Balancing of reciprocating masses is a mechanical engineering process that involves distributing the weight of all components in a machine evenly.  The purpose of balancing is to: Reduce vibrations Ensure smooth operation of rotating machinery Minimize dynamic forces and moments Prevent excessive wear, noise, and structural failure  An imbalance in these masses can lead to machine failure,... Show more
Machine Dynamics Practice Test: Balancing of Reciprocating Masses
Time left 00:00
25 Questions

1. Hammer blow is along the line of stroke of the unbalanced force.
2. Tractive force attains a maxima or minima when the crank angle is 90 degrees.
3. The effect of an unbalanced primary force along the line of stroke is to produce ________
4. In a locomotive, the ratio of the connecting rod length to the crank radius is kept very large in order to
5. A V-twin engine has the cylinder axes at 90 degrees and the connecting rods operate a common crank. The reciprocating mass per cylinder is 23 kg and the crank radius is 7.5 cm. The length of the connecting rod is 0.3 m. If the engine is rotating at the speed is 500 r.p.m. What is the value of maximum resultant secondary force in Newtons?
6. What is the total no. of cylinders in a locomotive having crank at right angles?
7. The swaying couple is maximum or minimum when the angle of inclination of the crank to the line of stroke ( θ ) is equal to
8. Which of the following has a tendency to sway engine in alternately in clockwise and anticlockwise directions?
9. What is the effect of variation of pressure on the rails?
10. At which of the following angles in degrees does the swaying couple attains a maximum value?
11. In an inside cylinder locomotive, the position of the driving wheels is _________
12. Swaying couple attains a maxima or minima when the crank angle is 45 degrees.
13. What is the effect of partial balancing of the reciprocating parts?
14. If c be the fraction of the reciprocating parts of mass m to be balanced per cyclinder of a steam locomotive with crank radius r, angular speed ω, distance between centre lines of two cylinders a, then the magnitude of the maximum swaying couple is given by
15. The tractive force is maximum or minimum when the angle of inclination of the crank to the line of stroke ( θ ) is equal to
16. At which of the following angles in degrees does the tractive force attains a minimum value?
17. A V-twin engine has the cylinder axes at 90 degrees and the connecting rods operate a common crank. The reciprocating mass per cylinder is 11.5 kg and the crank radius is 7.5 cm. The length of the connecting rod is 0.3 m. If the engine is rotating at the speed is 500 r.p.m. What is the value of maximum resultant secondary force in Newtons?
18. The effect of an unbalanced primary force perpendicular to the line of stroke is to produce _____
19. Find the minimum value of tractive force in newtons from the following data:
m = 100Kg
ω = 300 rpm
radius = 0.6m
c = 2/3
20. Hammering action on the rails is caused by _______
21. From the following data of a 60 degree V-twin engine, determine the maximum value for secondary forces in newtons:
Reciprocating mass per cylinder = 1.5 Kg
Stroke length = 10 cm
Length of connecting rod = 25 cm
Engine speed = 2500 rpm
22. The swaying couple is due to the
23. Identify the type of locomotive from the given figure.
The figure is of an inside cylinder locomotive placed in between two driving wheels
24. Find the maximum value of swaying couple in newtons from the following data:
m = 200Kg
ω = 300 rpm
radius = 0.6m
c = 2/3
a = 2\(\sqrt{2}\)m
25. Due to partial balancing there is unbalanced force only along the line of stroke.