Dynamics Of Machines Set 6

Mechanical Engineering

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This set of Dynamics of Machines Multiple Choice Questions & Answers (MCQs) focuses on Dynamics Of Machines Set 6

Q1 | In latitude 25.0 S, SA (spin axis) of a FG (free gyro) is in position S40E and horizontal. Find the tilt after 6 hours.
  • 61.16 up
  • 61.16 down c) 51.15 up
  • d) none of the mentioned
Answer: 61.16 up
Q2 | The steering of a ship means
  • movement of a complete ship up and down in vertical plane about transverse axis
  • turning of a complete ship in a curve towards right r left, while it moves forward
  • rolling of a complete ship sideways
  • none of the mentioned
Answer: turning of a complete ship in a curve towards right r left, while it moves forward
Q3 | When the pitching of a ship is upward, the effect of gyroscopic couple acting on it will be
  • to move the ship towards starboard
  • to move the ship towards port side
  • to raise the bow and lower the stern
  • to raise the stern and lower the bow
Answer: to move the ship towards starboard
Q4 | 24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m
  • 1.35 m
  • 1.42 m
  • 1.48 m
  • 1.50 m
Answer: 1.42 m
Q5 | 24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m
  • 52.7
  • 49.5
  • 59.5
  • 56.5
Answer: 59.5
Q6 | 24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m
  • 16.782
  • 17.824
  • 15.142
  • 17.161
Answer: 17.824
Q7 | 24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m
  • 19000 n
  • 19064 n
  • 19032 n
  • 20064 n
Answer: 19064 n
Q8 | 66.
  • 4/3
  • 3/2 c) 3/4
  • d) 1
Answer: 3/2 c) 3/4
Q9 | The starting torque of the steam engine is 1500 N-m and may be assumed constant, using this data find the angular acceleration of the flywheel in rad/s2.
  • 0.4
  • 0.6
  • 0.3
  • 1.2
Answer: 0.6
Q10 | 1m then find the other balancing mass situated at a distance of 0.2m.
  • 80
  • 40
  • 20
  • 10
Answer: 10
Q11 | 1m. One of the balancing mass is 30 Kg with RoR of 10cm.
  • 70
  • 35
  • 20
  • 10
Answer: 35
Q12 | 2m and r=0.3m, l=1m, find l2.
  • 0.26m
  • 0.52m c) 1.04m d)0.13m
Answer: 0.26m
Q13 | 2m and r=0.3m, l=1m, find l2.
  • 0.26m
  • 0.52m
  • 1.04m
  • 0.13m
Answer: 0.13m
Q14 | 3m, l=1m, find l2 = 0.5m, find r1 in m.
  • 1.5
  • 0.75
  • 3
  • 6
Answer: 0.75
Q15 | 2m, 0.15m, 0.25m and 0.3m. Angles between consecutive masses = 45, 75 and 135 degrees.
  • 116
  • 58 c) 232 d) 140
Answer: 116
Q16 | 2m, 0.15m, 0.25m and 0.3m. Angles between consecutive masses = 45, 75 and 135 degrees.
  • 201.48
  • 200.32
  • 210.34
  • 202.88
Answer: 201.48
Q17 | The system is automatically statically balanced.
  • 1, 2, 3 and 4
  • 1, 2, and 3 only
  • 2, 3 and 4 only
  • 1, 3 and 4 only
Answer: 1, 3 and 4 only
Q18 | 6m c = 2/3
  • 6978
  • 7574
  • 6568
  • 7374
Answer: 6978
Q19 | 6m c = 2/3
  • 13956
  • 17574
  • 16568
  • 17374
Answer: 13956
Q20 | 6m c = 2/3
  • -6978
  • -7574
  • -6568
  • -7374
Answer: -6978
Q21 | 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?
  • 736
  • 836
  • 936
  • 636
Answer: 836
Q22 | 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?
  • 7172
  • 1672
  • 1122
  • 1272
Answer: 1672
Q23 | 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?
  • 2.238
  • 2.508
  • 2.754
  • 2.908
Answer: 2.508
Q24 | 6 kN/m and the damping constant of the damper is 400 Ns/m. If the mass is 50 kg, then the damping factor (d ) and damped natural frequency (fn), respectively, are
  • 0.471 and 1.19 hz
  • 0.471 and 7.48 hz
  • 0.666 and 1.35 hz
  • 0.666 and 8.50 hz
Answer: 0.471 and 1.19 hz
Q25 | 1 mm.
  • 0.1
  • 0.2
  • 0.5
  • 0.6
Answer: 0.6
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