McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Mechanical Engineering .
Chapters
| 51. |
For a vibration system having different shaft parameters, calculate which of the following cannot be the diameter of the equivalent shaft if the diameters of shafts in m are: 0.05, 0.06, 0.07. |
| A. | 0.05 |
| B. | 0.06 |
| C. | 0.07 |
| D. | 0.08 |
| Answer» D. 0.08 | |
| 52. |
A single cylinder oil engine works with a three rotor system, the shaft length is 2.5m and 70mm in diameter, the middle rotor is at a distance 1.5m from one end. Calculate the free torsional vibration frequency for a single node system in Hz if the mass moment of inertia of rotors in Kg-m2 are: 0.15, 0.3 and 0.09. C=84 kN/mm2 |
| A. | 171 |
| B. | 181 |
| C. | 191 |
| D. | 201 |
| Answer» A. 171 | |
| 53. |
At a nodal point in the shaft, the frequency of vibration is _________ |
| A. | Zero |
| B. | Double than at the ends |
| C. | Minimum |
| D. | Maximum |
| Answer» A. Zero | |
| 54. |
For a gearing system, the equivalent system is made consisting of two rotors. How many nodes will this new equivalent system will have? |
| A. | 0 |
| B. | 1 |
| C. | 2 |
| D. | 3 |
| Answer» B. 1 | |
| 55. |
In a gearing system, pump speed is one third of the motor. Shaft from motor is 6 cm in diameter and 30cm long, the impellar shaft is 10cm diameter and 60cm long. Mass moment of inertia is 1500 Kgm2, C = 80Gn/m2. Neglecting the inertia of shaft and gears calculate the frequency of free torsional vibrations in Hz. |
| A. | 4.7 |
| B. | 5.7 |
| C. | 4.5 |
| D. | 5.5 |
| Answer» A. 4.7 | |
| 56. |
When inertia of gearing is taken into consideration, then which of the following should be taken into account. |
| A. | Addition of rotor |
| B. | Addition of gear |
| C. | Addition of shaft |
| D. | Addition of pump |
| Answer» A. Addition of rotor | |
| 57. |
Consider P and Q as the shaft having two rotors at the end of it, what is the point N known as in the given figure? |
| A. | Node |
| B. | Elastic point |
| C. | Inelastic point |
| D. | Breaking point |
| Answer» A. Node | |
| 58. |
For occurrence of free torsional vibration which of the condition is necessary? |
| A. | Rotors moving in same direction |
| B. | Rotors having same frequency |
| C. | Rotors having different frequency |
| D. | Rotors rotate in the same sense |
| Answer» B. Rotors having same frequency | |
| 59. |
n the given figure if N is the node then NQ acts as which of the following system? |
| A. | Single rotor system |
| B. | Two rotor system |
| C. | Three rotor system |
| D. | Four rotor system |
| Answer» A. Single rotor system | |
| 60. |
Keeping the mass moment of inertia of both the shafts in a two rotor system same, if the length of one shaft is doubled what should be the effect on the length of other shaft? |
| A. | Doubled |
| B. | Halved |
| C. | Constant |
| D. | increased to 4 times |
| Answer» A. Doubled | |
| 61. |
In the figure given below, the points N1 and N2 are known as_______ |
| A. | Nodes |
| B. | Elastic points |
| C. | Inelastic points |
| D. | Breaking points |
| Answer» A. Nodes | |
| 62. |
In which of the following condition torsional vibration will not take place, considering 3 rotors A, B and C. A is rotating in clockwise direction. |
| A. | B in clockwise C in anticlockwise |
| B. | C in clockwise B in anticlockwise |
| C. | B and C in clockwise |
| D. | B and C in anticlockwise |
| Answer» C. B and C in clockwise | |
| 63. |
For occurrence of free torsional vibration in a three rotor system which of the condition is necessary? |
| A. | Rotors moving in same direction |
| B. | Rotors having same frequency |
| C. | Rotors having different frequency |
| D. | Rotors rotate in the same sense |
| Answer» B. Rotors having same frequency | |
| 64. |
A vertical circular disc is supported by a horizontal stepped shaft as shown below. Determine equivalent length of shaft when equivalent diameter is 20 mm. |
| A. | 1.559 m |
| B. | 0.559 m |
| C. | 0.633 m |
| D. | None of the above |
| Answer» B. 0.559 m | |
| 65. |
In a three rotor system, for the middle rotor, if the stiffness of both the length either side of the rotor is increased to two times what will |
| A. | Remains constant |
| B. | Decreases by two times |
| C. | Increases by two times |
| D. | Increases by 4 times |
| Answer» C. Increases by two times | |
| 66. |
A single cylinder oil engine works with a three rotor system, the shaft length is 2.5m and 70mm in diameter, the middle rotor is at a distance 1.5m from one end. Calculate the free torsional vibration frequency for a two node system in Hz if the mass moment of inertia of rotors in Kg-m2 are: 0.15, 0.3 and 0.09. C=84 kN/mm2 |
| A. | 257 |
| B. | 281 |
| C. | 197 |
| D. | 277 |
| Answer» D. 277 | |
| 67. |
Increasing which of the following factor would result in increase of free torsional vibration? |
| A. | Radius of gyration |
| B. | Mass moment of inertia |
| C. | Polar moment of inertia |
| D. | Length |
| Answer» C. Polar moment of inertia | |
| 68. |
The frequency of the free torsional vibration depends on….... |
| A. | number of rotors |
| B. | the number of nodes |
| C. | Both a and b |
| D. | Only b |
| Answer» C. Both a and b | |
| 69. |
For a two rotor system, the mass moment of inertia of one shaft(A) is twice the other(B), then what is the relation between the length of the shafts. |
| A. | 2L(A) = L(B) |
| B. | L(A) = 2L(B) |
| C. | L(A) = L(B) |
| D. | 2L(A) = 3L(B) |
| Answer» A. 2L(A) = L(B) | |
| 70. |
Which of the following is true for centrifugal force causing unbalance? |
| A. | Direction changes with rotation |
| B. | Magnitude changes with rotation |
| C. | Direction and magnitude both change with rotation |
| D. | Direction and magnitude both remain unchanged with rotation |
| Answer» A. Direction changes with rotation | |
| 71. |
If the unbalanced system is not set right then. |
| A. | Static forces develop |
| B. | Dynamic forces develop |
| C. | Tangential forces develop |
| D. | Radial forces develop |
| Answer» A. Static forces develop | |
| 72. |
What is the effect of a rotating mass of a shaft on a system? |
| A. | Bend the shaft |
| B. | Twist the shaft |
| C. | Extend the shaft |
| D. | Compress the shaft |
| Answer» A. Bend the shaft | |
| 73. |
In a revolving rotor, the centrifugal force remains balanced as long as the centre of the mass of the rotor lies ___________ |
| A. | Below the axis of shaft |
| B. | On the axis of the shaft |
| C. | Above the axis of shaft |
| D. | Away from the axis of shaft |
| Answer» B. On the axis of the shaft | |
| 74. |
Often an unbalance of forces is produced in rotary or reciprocating machinery due to the ______ |
| A. | Centripetal forces |
| B. | Centrifugal forces |
| C. | Friction forces |
| D. | Inertia forces |
| Answer» D. Inertia forces | |
| 75. |
The mass used to balance the mass defect is known as ______ |
| A. | Balancing mass |
| B. | Defect mass |
| C. | Replacement mass |
| D. | Fixing mass |
| Answer» A. Balancing mass | |
| 76. |
In balancing of single-cylinder engine, the rotating unbalance is ____________ |
| A. | completely made zero and so also the reciprocating unbalance |
| B. | completely made zero and the reciprocating unbalance is partially reduced |
| C. | partially reduced and the reciprocating unbalance is completely made zero |
| D. | partially reduced and so also the reciprocating unbalance |
| Answer» B. completely made zero and the reciprocating unbalance is partially reduced | |
| 77. |
Let the disturbing mass be 100 kg and the radius of rotation be 10 cm and the rotation speed be 50 rad/s, then calculate the centrifugal force in kN. |
| A. | 50 |
| B. | 25 |
| C. | 50000 |
| D. | 25000 |
| Answer» B. 25 | |
| 78. |
Which of the following statement is correct? |
| A. | In any engine, 100% of the reciprocating masses can be balanced dynamically |
| B. | In the case of balancing of multicylinder engine, the value of secondary force is higher than the value of the primary force |
| C. | In the case of balancing of multimass rotating systems, dynamic balancing can be directly started without static balancing done to the system |
| D. | none of the mentioned |
| Answer» C. In the case of balancing of multimass rotating systems, dynamic balancing can be directly started without static balancing done to the system | |
| 79. |
If all the masses are in one plane, then what is the maximum no. of masses which can be placed in the same plane? |
| A. | 3 |
| B. | 4 |
| C. | 6 |
| D. | No limitation |
| Answer» D. No limitation | |
| 80. |
Which of the following statements is correct about the balancing of a mechanical system? |
| A. | If it is under static balance, then there will be dynamic balance also |
| B. | If it is under dynamic balance, then there will be static balance also |
| C. | Both static as well as dynamic balance have to be achieved separately |
| D. | None of the mentioned |
| Answer» C. Both static as well as dynamic balance have to be achieved separately | |
| 81. |
In a locomotive, the ratio of the connecting rod length to the crank radius is kept very large in order to |
| A. | minimize the effect of primary forces |
| B. | minimize the effect of secondary forces |
| C. | have perfect balancing |
| D. | start the locomotive conveniently |
| Answer» B. minimize the effect of secondary forces | |
| 82. |
Secondary forces in reciprocating mass on engine frame are |
| A. | of same frequency as of primary forces |
| B. | twice the frequency as of primary forces |
| C. | four times the frequency as of primary forces |
| D. | none of the mentioned |
| Answer» B. twice the frequency as of primary forces | |
| 83. |
For a V-twin engine, which of the following means can be used to balance the primary forces? |
| A. | Revolving balance mass |
| B. | Rotating balance mass |
| C. | Reciprocating balance mass |
| D. | By the means of secondary forces |
| Answer» A. Revolving balance mass | |
| 84. |
The primary unbalanced force is maximum when the angle of inclination of the crank with the line of stroke is |
| A. | 0° |
| B. | 90° |
| C. | 180° |
| D. | 360° |
| Answer» C. 180° | |
| 85. |
From the following data of a 60 degree V-twin engine, determine the minimum value for primary forces in newtons:Reciprocating mass per cylinder = 1.5 Kg Stroke length = 10 cm Length of connecting rod = 25 cm Engine speed = 2500 rpm |
| A. | 7711 |
| B. | 4546 |
| C. | 2570 |
| D. | 8764 |
| Answer» C. 2570 | |
| 86. |
From the following data of a 60 degree V-twin engine, determine the maximum value for primary forces in newtons:Reciprocating mass per cylinder = 1.5 Kg Stroke length = 10 cm Length of connecting rod = 25 cm Engine speed = 2500 rpm |
| A. | 7711 |
| B. | 4546 |
| C. | 2570 |
| D. | 8764 |
| Answer» A. 7711 | |
| 87. |
In the given figure, m1=10 Kg, m2=30Kg and m=50 Kg, if r=0.3m, l=1m, find l2 = 0.5m, find r1 in m. |
| A. | 1.5 |
| B. | 0.75 |
| C. | 3 |
| D. | 6 |
| Answer» B. 0.75 | |
| 88. |
From the given data, find the balancing mass’s inclination in degrees if r=0.2m required in the same plane.Masses = 200kg, 300kg, 240 kg, 260Kg, |
| A. | 201.48 |
| B. | 200.32 |
| C. | 210.34 |
| D. | 202.88 |
| Answer» A. 201.48 | |
| 89. |
Which of the following statements are associated with complete dynamic balancing of rotating systems? 1. Resultant couple due to all inertia forces is zero. 2. Support reactions due to forces are zero but not due to couples . 3. The system is automatically statically balanced. 4. Centre of masses of the system lies on the axis of rotation. |
| A. | 1, 2, 3 and 4 |
| B. | 1, 2, and 3 only |
| C. | 2, 3 and 4 only |
| D. | 1, 3 and 4 only |
| Answer» D. 1, 3 and 4 only | |
| 90. |
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? |
| A. | 736 |
| B. | 836 |
| C. | 936 |
| D. | 636 |
| Answer» B. 836 | |
| 91. |
Let the centrifugal force in kN be 25 and the radius of rotation be 20 cm and the rotation speed be 50 rad/s, then calculate the mass defect in Kg. |
| A. | 50 |
| B. | 25 |
| C. | 50000 |
| D. | 25000 |
| Answer» A. 50 | |
| 92. |
In a multicylinder inline engine, each imaginary secondary crank with a mass attached to the crankpin is inclined to the line of stroke at which angle? |
| A. | Twice the angle of crank |
| B. | Half the angle of crank |
| C. | Thrice the angle of crank |
| D. | Four times the angle of crank |
| Answer» A. Twice the angle of crank | |
| 93. |
In order to achieve stability, the sum of secondary forces and secondary couples must be ------- and their respective polygons must be …...... |
| A. | Zero, Closed |
| B. | one, Open |
| C. | Zero, open |
| D. | one, closed |
| Answer» A. Zero, Closed | |
| 94. |
For the secondary balancing of the engine, which of the condition is necessary? |
| A. | Secondary force polygon must be close |
| B. | Secondary force polygon must be open |
| C. | Primary force polygon must be close |
| D. | Primary force polygon must be open |
| Answer» A. Secondary force polygon must be close | |
| 95. |
The numerical values of the secondary forces and secondary couples couples may be obtained by considering the ___________ |
| A. | Revolving mass |
| B. | Reciprocating mass |
| C. | Translating mass |
| D. | Rotating mass |
| Answer» A. Revolving mass | |
| 96. |
Which of the following is the correct expression for secondary force? |
| A. | m𝜔^2r.cos2θ/n |
| B. | m 𝜔^2r.sin2θ/n |
| C. | m𝜔^2r.tan2θ/n |
| D. | m𝜔^2r.cosθ/n |
| Answer» A. m𝜔^2r.cos2θ/n | |
| 97. |
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? |
| A. | 7172 |
| B. | 1672 |
| C. | 1122 |
| D. | 1272 |
| Answer» B. 1672 | |
| 98. |
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 34.5 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 kN? |
| A. | 2.238 |
| B. | 2.508 |
| C. | 2.754 |
| D. | 2.908 |
| Answer» B. 2.508 | |
| 99. |
From the given data, find the balancing mass in Kg if r=0.2m required in the same plane.Masses = 200kg, 300kg, 240 kg, 260Kg, corresponding radii = 0.2m, 0.15m, 0.25m and 0.3m.Angles between consecutive masses = 45, 75 and 135 degrees. |
| A. | 116 |
| B. | 58 |
| C. | 232 |
| D. | 140 |
| Answer» A. 116 | |
| 100. |
Let the disturbing mass be 200 kg and the radius of rotation be 20 cm and the rotation speed be 50 rad/s, then calculate the centripetal force in kN. |
| A. | -50 |
| B. | -25 |
| C. | -100 |
| D. | -2500 |
| Answer» C. -100 | |
Done Studing? Take A Test.
Great job completing your study session! Now it's time to put your knowledge to the test. Challenge yourself, see how much you've learned, and identify areas for improvement. Don’t worry, this is all part of the journey to mastery. Ready for the next step? Take a quiz to solidify what you've just studied.