250+ TOP MCQs on Velocity and Acceleration of a Particle Moving with Simple Harmonic Motion and Answers

Machine Kinematics Questions on “Velocity and Acceleration of a Particle Moving with Simple Harmonic Motion”.

1. A body is said to vibrate with simple harmonic motion if its acceleration is proportional to the distance from the mean position.
a) True
b) False

Answer: a
Clarification: A body is said to move with simple harmonic motion, if it satisfies the following two conditions:
a) Its acceleration is always directed towards the center, known as point of reference or mean position.
b) Its acceleration is proportional to the distance from that point.

2. The maximum displacement of a body, from its mean position is called amplitude.
a) True
b) False

Answer: a
Clarification: The time taken for one complete revolution of the particle is called periodic time.
The maximum displacement of a body from its mean position is called amplitude.

3. Frequency of vibrations is usually expressed in
a) number of cycles per hour
b) number of cycles per minute
c) number of cycles per second
d) none of the mentioned

Answer: c
Clarification: The number of cycles per second is called frequency. It is the reciprocal of periodic time.

4. The amplitude of vibrations is always ______________ the radius of the circle.
a) equal to
b) less than
c) greater than
d) none of the mentioned

Answer: a
Clarification: None.

5. The time taken by a particle for one complete oscillation is known as periodic time.
a) True
b) False

Answer: a
Clarification: The time taken for one complete revolution of the particle is called periodic time.
The maximum displacement of a body from its mean position is called amplitude.

6. The periodic time is given by
a) ω/2п
b) 2п/ω
c) ω x 2п
d) п/ω

Answer: b
Clarification: Periodic time, tp = 2п/ω seconds
where ω = Angular velocity of the particle in rad/s.

7. When a body moves with simple harmonic motion, the product of its periodic time and frequency is equal to
a) zero
b) one
c) п/2
d) п

Answer: b
Clarification: The number of cycles per second is called frequency. It is the reciprocal of periodic time. Hence, when it is multiplied it is equal to one.

8. The acceleration of the particle moving with simple harmonic motion is ____________ at the mean position.
a) zero
b) minimum
c) maximum
d) none of the mentioned

Answer: a
Clarification: The acceleration of a body is zero at the mean position and maximum when x = r.

9. The maximum velocity of a particle moving with simple harmonic motion is
a) ω
b) ωr
c) ω2r
d) ω/r

Answer: b
Clarification:The velocity of a moving body with simple harmonic motion at any instant is given by
v = ω√r2 – x2
The velocity is maximum at the mean position i.e. when x = 0.

Hence, v = ωr.

10. When a particle moves round the circumference of a circle of radius r with ω rad/s, then its maximum acceleration is ω2r.
a) True
b) False

Answer: a
Clarification: The acceleration of a body moving with simple harmonic motion at any instant is given by
a = ω2r.

11. If a simple pendulum oscillates with an amplitude 50 mm and time period 2s, then its maximum velocity is
a) 0.1 m/s
b) 0.15 m/s
c) 0.8 m/s
d) 0.16 m/s

Answer: b
Clarification: Maximum velocity vmax = ωA where ‘ω’ is the angular frequency and ‘A’ is the amplitude. Therefore vmax = (2π/T)A = (2π/2)×50×10-3 = 0.157 m/s.

12. A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
a) 1/ 2π√3
b) 2π√3
c) 2π/√3
d) √3/2π

Answer: b
Clarification: The magnitudes of the velocity and acceleration of the particle when its displacement is ‘y’ are ω√(A2 –y2) and ω2y respectively. Equating them, ω√(A2 –y2) = ω2y, from which ω = [√(A2 –y2)]/y = √(4 –1) = √3. Period T = 2π/ω = 2π/√3.

13. Suppose you place a sphere of mass ‘m’ and radius ‘r’ inside a smooth, heavy hemispherical bowl of radius of 37r placed on a horizontal table. If the sphere is given a small displacement, what is its period of oscillation?
a) 2π√(m/37rg)
b) 2π√(m/rg)
c) 12π√(r/g)
d) 2π√(r/g)

Answer: c
Clarification: The arrangement depicted in this question is similar to that of a simple pendulum. Instead of the usual string, you have a concave surface to confine the bob (sphere) to its path along the arc of a circle. The usual expression for the period, T = 2π√(L/g) holds here also, where the length L = 36r since the length of the pendulum is measured from the centre of gravity of the bob. The point of ‘suspension’ is evidently at the centre of the hemispherical bowl. The correct option is 12π√(r/g).

14. The instantaneous displacement of a simple harmonic oscillator is given by y = A cos(ωt + π/4). Its speed will be maximum at the time
a) 2π/ω
b) ω/2π
c) ω/π
d) π/4ω

Answer: d
Clarification: The velocity is the time derivative of displacement: v = dy/dt = -Aω(sin ωt + π/4). Its maximum magnitude equal to Aω is obtained when ωt = π/4, from which t = π/4ω.

15. A particle of mass 5 g is executing simple harmonic motion with an amplitude 0.3 m and time period π/5 s. The maximum value of the force acting on the particle is
a) 5 N
b) 4 N
c) 0.5 N
d) 0.15 N

Answer: d
Clarification: T = 2π√(m/k) where ‘k’ is the force constant, the solution becomes quite easy. From this, k = 4π2m/T2 = 4π2 ×5×10-3/(π/5)2 = 0.5. Since ‘k’ is the force for unit displacement, the maximum force is k times the maximum displacement (amplitude). Therefore maximum force = kA = 0.5×0.3 = 0.15N.

250+ TOP MCQs on Mechanism – 2 and Answers

Machine Kinematics Quiz on “Mechanism – 2”.

1. In a coupling rod of a locomotive, each of the four pairs is a _____________ pair.
a) sliding
b) turning
c) rolling
d) screw
Answer: b
Clarification: In a locomotive, one rod of the coupling rod can only turn or revolve about a fixed axis of another rod. This takes place only in case of turning pair.

2. Whitworth quick return motion mechanism consists of three turning pairs and one sliding pair.
a) True
b) False
Answer: a
Clarification: Whitworth quick return motion mechanism is a form of single slider crank chain and it consists of one sliding pair and three turning pairs.

3. In a single slider crank chain
a) each of the four pairs is a turning pair
b) one is a turning pair and three are sliding pairs
c) two are turning pairs and two are sliding pairs
d) three are turning pairs and one is a sliding pair
Answer: d
Clarification: Single slider crank chain consists of one sliding pair and three turning pairs.
Four bar chain consists of four links and each of them forms a turning pair.

4. The mechanism consisting of three turning pairs and one sliding pair, is called a
a) single slider crank chain
b) whitworth quick return motion mechanism
c) crank and slotted lever quick return motion mechanism
d) all of the mentioned
Answer: d
Clarification: All the mechanisms mentioned above are forms of single slider crank chain only and hence, they all contain three turning pairs and one sliding pair. The inversion of a single slider crank chain are found in the following mechanism:
a) Pendulum pump or Bull engine
b) Rotary internal combustion engine
c) Oscillating cylinder engine
d) Crank and slotted lever quick return motion mechanism
e) Whitworth quick return motion mechanism

5. Which of the following is an inversion of a single slider crank chain?
a) Pendulum pump
b) Oscillating cylinder engine
c) Rotary internal combustion engine
d) All of the mentioned
Answer: d
Clarification: All the mechanisms mentioned above are forms of single slider crank chain only and hence, they all contain three turning pairs and one sliding pair. The inversion of a single slider crank chain are found in the following mechanism:
a) Pendulum pump or Bull engine
b) Rotary internal combustion engine
c) Oscillating cylinder engine
d) Crank and slotted lever quick return motion mechanism
e) Whitworth quick return motion mechanism.

6. A point on a connecting link of a double slider crank mechanism traces a
a) straight line path
b) hyperbolic path
c) parabolic path
d) elliptical path
Answer: d
Clarification: None.

7. The whitworth quick return motion mechanism is formed in a slider crank chain when the
a) coupler link is fixed
b) longest link is a fixed link
c) slider is a fixed link
d) smallest link is a fixed link
Answer: a
Clarification: None.

8. Scotch yoke mechanism is used to generate
a) sine functions
b) square roots
c) logarithms
d) inversions
Answer: a
Clarification: None.

9. Which of the following is an inversion of a double slider crank chain?
a) Oldham’s coupling
b) Elliptical trammel
c) Scotch yoke mechanism
d) All of the mentioned
Answer: d
Clarification: A double slider crank chain consists of two sliding pairs and two turning pairs. The inversions of a double slider crank chain are as follows:
a) elliptical trammel
b) scotch yoke mechanism
c) oldham’s coupling.

10. Whitworth quick return motion mechanism is an inversion of a double slider crank chain.
a) True
b) False
Answer: b
Clarification: Whitworth quick return motion mechanism is a form of single slider crank chain and it consists of one sliding pair and three turning pairs.

11. A rotary internal combustion engine has __________ cylinders.
a) four
b) five
c) six
d) seven
Answer: d
Clarification: None.

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250+ TOP MCQs on Relative Velocity of Two Bodies Moving in Straight Lines and Answers

Machine Kinematics online quiz on “Relative Velocity of Two Bodies Moving in Straight Lines”.

1. A machine raised a load of 360 N through a distance of 200 mm. The effort, a force of 60 N moved 1.8 m during the process. Calculate mechanical advantage.
a) 6
b) 7
c) 8
d) 9
Answer: a
Clarification: Given, load raised, W = 360 N
Effort applied, P = 60 N
Distance moved by the effort, y = 1.8 m
Distance moved by the load, x = 200 mm = 0.2 m
Mechanical advantage, M.A. = W/P = 360/60 = 6.

2. A machine raised a load of 360 N through a distance of 200 mm. The effort, a force of 60 N moved 1.8 m during the process. Calculate velocity ratio.
a) 6
b) 7
c) 8
d) 9
Answer: d
Clarification: Given, load raised, W = 360 N
Effort applied, P = 60 N
Distance moved by the effort, y = 1.8 m
Distance moved by the load, x = 200 mm = 0.2 m
Velocity ratio, V.R. = y/x = 1.8/0.2 = 9.

3. A machine raised a load of 360 N through a distance of 200 mm. The effort, a force of 60 N moved 1.8 m during the process. Calculate efficiency at this load.
a) 44.44%
b) 55.55%
c) 66.66%
d) 77.77%
Answer: c
Clarification: Given, load raised, W = 360 N
Effort applied, P = 60 N
Distance moved by the effort, y = 1.8 m
Distance moved by the load, x = 200 mm = 0.2 m
Mechanical advantage, M.A. = W/P = 360/60 = 6
Velocity ratio, V.R. = y/x = 1.8/0.2 = 9
Efficiency at the load, ȵ = M.A./V.R. x 100 = 6/9 x 100 = 66.66%.

4. A machine raised a load of 360 N through a distance of 200 mm. The effort, a force of 60 N moved 1.8 m during the process. Calculate effect of friction.
a) 10 N
b) 20 N
c) 30 N
d) 40 N
Answer: b
Clarification: Given, load raised, W = 360 N
Effort applied, P = 60 N
Distance moved by the effort, y = 1.8 m
Distance moved by the load, x = 200 mm = 0.2 m
Mechanical advantage, M.A. = W/P = 360/60 = 6
Velocity ratio, V.R. = y/x = 1.8/0.2 = 9
Effort lost in friction, FP = P – W/V.R. = 60 – 360/9 = 20 N.

5. In a lifting machine, the effort required to lift loads of 200N and 300N were 50N and 60N respectively. If the velocity ratio of the machine is 20 determine law of the machine.
a) P = 1/10W +30
b) P = 1/20W +30
c) P = 1/30W +30
d) P = 1/40W +30
Answer: a
Clarification: Let the law of machine be P = mW + C
where P = effort applied, W = load lifted and m and C being constants.
when W = 200 N P = 50 N
when W = 300 N P = 60 N
Putting these values in the law of machine.
50 = 200m + C …………(i)
60 = 300m + C …………(ii)

Subtracting (i) and (ii), we get
1 = 10 m
or, m = 1/10
Putting this value in equation (i), we get
50 = 200 x 1/10 + C
C = 30

Hence, the machine follows the laws
P = 1/10W +30.

6. In a lifting machine, the effort required to lift loads of 200N and 300N were 50N and 60N respectively. If the velocity ratio of the machine is 20 determine efficiency to load of 200 N.
a) 10 %
b) 15 %
c) 20 %
d) 25 %
Answer: c
Clarification: When W = 200 N, P = 50 N
M.A. = W/P = 200/50 = 4
V.R. = 20
Efficiency at this load ȵ = M.A./V.R. x 100 = 4/20 x 100 = 20 %.

7. In a lifting machine, the effort required to lift loads of 200N and 300N were 50N and 60N respectively. If the velocity ratio of the machine is 20 determine efficiency to load of 300 N.
a) 10 %
b) 15 %
c) 20 %
d) 25 %
Answer: d
Clarification: When W = 300 N, P = 60 N
M.A. = W/P = 300/60 = 5
V.R. = 20
Efficiency at this load ȵ = M.A./V.R. x 100 = 5/20 x 100 = 25 %.

8. In a lifting machine, the effort required to lift loads of 200N and 300N were 50N and 60N respectively. If the velocity ratio of the machine is 20 determine effort loss in friction at 200 N.
a) 30 N
b) 35 N
c) 40 N
d) 45 N
Answer: c
Clarification: When W = 200 N, P = 50 N
Effort lost in friction, FP = P – W/ V.R. = 50 – 200/20 = 40 N.

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250+ TOP MCQs on Limiting Friction and Answers

Machine Kinematics Multiple Choice Questions on “Limiting Friction”.

1. What is the direction in which the limiting friction acts?
a) Opposite to the direction in which the body tends to move
b) Opposite to the direction in which the body moves
c) In the direction in which the body tends to move
d) In the direction in which the body moves
Answer: a
Clarification: Limiting friction acts in the opposite direction in which the body tends to move. It has a maximum value beyond which it cannot increase.

2. The force of friction is independent of the roughness of the surface.
a) True
b) False
Answer: b
Clarification: The force of friction is dependent on the roughness of the surface. There is a coefficient of friction on which the frictional force depends.

3. If the area of contact between the two surfaces is increased by two times, what will be the effect of it on the force of friction?
a) Increases by two times
b) Decreases by two times
c) Remains same
d) Depends on the surface
Answer: c
Clarification: If the area of contact between the two surfaces is increased by two times then there will be no effect on the force of friction as it is independent of the area of contact surface.

4. The ratio of frictional force to normal reaction is known as _______
a) Coefficient of friction
b) Coefficient of restitution
c) Coefficient of inertia
d) Coefficient of force
Answer: a
Clarification: The ratio of frictional force to the normal reaction at the surface of contact bears a constant ratio, this ratio is known as the coefficient of friction.

5. Which of the following is true regarding the limiting friction?
a) Its value is equal to the force applied in the opposite direction which tends to move the body
b) Its value is equal to the force applied in the opposite direction which moves the body
c) Its value is equal to μ.N
d) It is equal to the normal reaction
Answer: a
Clarification: For the limiting friction, the maximum value it can attain is μ.N, until then it has the same value as the force applied in the opposite direction which tends to move the body.

6. The angle at which the body just begins to slide down an incline is known as ________
a) Angle of inclination
b) Angle of motion
c) Angle of repose
d) Angle of stability
Answer: c
Clarification: When a body is kept on an incline, upto a certain angle it remains at rest but at a specific angle the body begins to slide down if the angle is increased further, this angle is known as the angle of repose.

7. For a body on an inclined plane, the value of coefficient of friction is equal to _____ (θ is the angle of friction)
a) Tanθ
b) Cotθ
c) Sinθ
d) Cosθ
Answer: a
Clarification: For a body on an inclined plane, the coefficient of friction can also be expressed in the form of the inclination as μ = Tanθ.

8. Static frictional force is equal to the coefficient of friction if the normal reaction is unity.
a) True
b) False
Answer: b
Clarification: The above statement talks about static frictional force, this force is always equal to the force which is applied on the body and tends to move away.

9. What is the maximum value of static friction?
a) Limiting friction
b) 0
c) Rolling friction
d) Kinetic friction
Answer: a
Clarification: When the applied force is less than the limiting friction, the body remains at rest, and the friction into play is called static friction which may have any value between zero and limiting friction.

10. A body of mass 1 kg is kept on a rough surface having coefficient of friction = 0.25 being pulled by a force of 2N, what will be the value of friction force in N?
a) 2
b) 2.5
c) 0
d) 4
Answer: a
Clarification: Since the applied force of 2N has a value less than the limiting friction 0.25×9.81
The acting friction will be static having a value of 2N.

11. A body of mass 1 kg is kept on a rough surface having coefficient of friction = 0.25 being pulled by a force of 2N, which frictional force will be acting on the body?
a) Limiting friction
b) Static friction
c) Kinetic friction
d) Rolling friction
Answer: b
Clarification: Since the applied force of 2N has a value less than the limiting friction 0.25×9.81
The acting friction will be static friction and the body will stay at rest.

12. A body of mass 1 kg is kept on a rough surface having coefficient of friction = 0.25 being pulled by a force of 2N, after how long will the body come to rest again?
a) 2s
b) 2.5s
c) 0s
d) 4s
Answer: c
Clarification: Since the applied force of 2N has a value less than the limiting friction 0.25×9.81
The acting friction will be static friction and the body will stay at rest and will not undergo any motion until the applied force exceeds the value of limiting friction.

13. A body, resting on a rough horizontal plane required a pull of 181 N inclined at 30° to the plane just to move it. It was recorded that a push of 220 N inclined at 30° to the plane just moved the body. Estimate the weight of the body in N.
a) 2000N
b) 2500N
c) 1000N
d) 4000N
Answer: c
Clarification: For vertical equilibrium we have reaction R = W – 181sin30
horizontally,
f = 181cos30
Now considering a push of 220N
horizontally
f =220cos30
vertically
R = W + 220sin30
solving the equations we get
W = 1000N.

14. A body, resting on a rough horizontal plane required a pull of 181 N inclined at 30° to the plane just to move it. It was recorded that a push of 220 N inclined at 30°° to the plane just moved the body. Estimate the coefficient of friction.
a) 0.17
b) 0.2
c) 0.5
d) 0.75
Answer: a
Clarification: For vertical equilibrium we have reaction R = W – 181sin30
horizontally,
f = 181cos30
Now considering a push of 220N
horizontally
f = 220cos30
vertically
R = W + 220sin30
solving the equations we get
μ = 0.171.

15. A body of mass 1 kg is kept on a rough surface is being pulled by a force of 2N, what will be the value of coefficient of friction if it just begins to move. Take g = 10m/s2.
a) 0.2
b) 0.25
c) 0
d) 0.4
Answer: a
Clarification: The acting friction will be limiting having a value of 2N.
Since the body just begins to move, we have
f = μN = 2N
therefore
μx10 = 2N
μ = 0.2.

250+ TOP MCQs on Advantages and Disadvantages of Chain Drive Over Belt or Rope Drive and Answers

Machine Kinematics Multiple Choice Questions on “Advantages and Disadvantages of Chain Drive Over Belt or Rope Drive”.

1. The advantages of the V-belt drive over flat belt drive are
a) The V-belt drive gives compactness due to the small distance between the centres of pulleys.
b) The drive is positive, because the slip between the belt and the pulley groove is negligible.
c) Since the V-belts are made endless and there is no joint trouble, therefore the drive is smooth.
d) all of the mentioned
Answer: d
Clarification: Following are the advantages of the V-belt drive over flat belt drive:
a) The V-belt drive gives compactness due to the small distance between the centres of pulleys.
b) The drive is positive, because the slip between the belt and the pulley groove is negligible.
c) Since the V-belts are made endless and there is no joint trouble, therefore the drive is smooth.
d) It provides longer life, 3 to 5 years.
e) It can be easily installed and removed.
f) The operation of the belt and pulley is quiet.
g) The belts have the ability to cushion the shock when machines are started.
h) The high velocity ratio (maximum 10) may be obtained.
i) The wedging action of the belt in the groove gives high value of limiting ratio of tensions.

2. The disadvantages of the V-belt drive over flat belt drive are
a) The V-belt drive cannot be used with large centre distances.
b) The V-belts are not so durable as flat belts.
c) The construction of pulleys for V-belts is more complicated than pulleys for flat belts.
d) all of the mentioned
Answer: d
Clarification: Following are the disadvantages of the V-belt drive over flat belt drive :
a) The V-belt drive cannot be used with large centre distances.
b) The V-belts are not so durable as flat belts.
c) The construction of pulleys for V-belts is more complicated than pulleys for flat belts.
d) Since the V-belts are subjected to certain amount of creep, therefore these are not suitable for constant speed application such as synchronous machines, and timing devices.
e) The belt life is greatly influenced with temperature changes, improper belt tension and mismatching of belt lengths.
f) The centrifugal tension prevents the use of V-belts at speeds below 5 m/s and above 50m/s.

3. The advantages of the V-belt drive over flat belt drive are
a) It provides longer life, 3 to 5 years.
b) It can be easily installed and removed.
c) The operation of the belt and pulley is quiet.
d) all of the mentioned
Answer: d
Clarification: Following are the advantages of the V-belt drive over flat belt drive:
a) The V-belt drive gives compactness due to the small distance between the centres of pulleys.
b) The drive is positive, because the slip between the belt and the pulley groove is negligible.
c) Since the V-belts are made endless and there is no joint trouble, therefore the drive is smooth.
d) It provides longer life, 3 to 5 years.
e) It can be easily installed and removed.
f) The operation of the belt and pulley is quiet.
g) The belts have the ability to cushion the shock when machines are started.
h) The high velocity ratio (maximum 10) may be obtained.
i) The wedging action of the belt in the groove gives high value of limiting ratio of tensions.

4. The disadvantages of the V-belt drive over flat belt drive are
a) Since the V-belts are subjected to certain amount of creep, therefore these are not suitable for constant speed application such as synchronous machines, and timing devices.
b) The belt life is greatly influenced with temperature changes, improper belt tension and mismatching of belt lengths.
c) The centrifugal tension prevents the use of V-belts at speeds below 5 m/s and above 50m/s.
d) all of the mentioned
Answer: d
Clarification: Following are the disadvantages of the V-belt drive over flat belt drive :
a) The V-belt drive cannot be used with large centre distances.
b) The V-belts are not so durable as flat belts.
c) The construction of pulleys for V-belts is more complicated than pulleys for flat belts.
d) Since the V-belts are subjected to certain amount of creep, therefore these are not suitable for constant speed application such as synchronous machines, and timing devices.
e) The belt life is greatly influenced with temperature changes, improper belt tension and mismatching of belt lengths.
f) The centrifugal tension prevents the use of V-belts at speeds below 5 m/s and above 50m/s.

5. The distance between the hinge centre of a link and the corresponding hinge centre of the adjacent link is called
a) pitch of the chain
b) bush roller chain
c) block chain
d) none of the mentioned
Answer: a
Clarification: A bush roller chain, consists of outer plates or pin link plates, inner plates or roller link plates, pins, bushes and rollers.
The distance between the hinge centre of a link and the corresponding hinge centre of the adjacent link is called pitch of the chain.

6. Industrial rotors will not have uniform diameter throughout their lengths.
a) True
b) False
Answer: a
Clarification: Industrial rotors are not uniform, because of fitment of pulleys, gears, belt etc.

7. For cutting multi-start threads, the speed ratio is expressed in terms of the lead of the job thread and lead of the lead screw threads.
a) True
b) False
Answer: a
Clarification: During multi-start thread cutting operation, speed is reduced to one-third to one-fourth of that in turning operation.

8. Which one of the following is a positive drive?
a) Crossed flat belt drive
b) Rope drive
c) V-belt drive
d) Chain drive
Answer: d
Clarification: The chains are mostly used to transmit motion and power from one shaft to another, when the centre distance between their shafts is short such as in bicycles, motor cycles, agricultural machinery, conveyors, rolling mills, road rollers etc.

9. The chain drive transmits ____________ power as compared to belt drive.
a) more
b) less
c) equal
d) none of the mentioned
Answer: a
Clarification: Following are the advantages and disadvantages of chain drive over belt or rope drive:
1. As no slip takes place during chain drive, hence perfect velocity ratio is obtained.
2. Since the chains are made of metal, therefore they occupy less space in width than a belt or rope drive.
3. It may be used for both long as well as short distances.
4. It gives a high transmission efficiency (upto 98 percent).
5. It gives less load on the shafts.
6. It has the ability to transmit motion to several shafts by one chain only.
7. It transmits more power than belts.
8. It permits high speed ratio of 8 to 10 in one step.
9. It can be operated under adverse temperature and atmospheric conditions.

10. The relation between the pitch of the chain (p) and pitch circle diameter of the sprocket (D) is given by
a) p = D sin(900/T)
b) p = D sin(1200/T)
c) p = D sin(1800/T)
d) p = D sin(3600/T)
Answer: c
Clarification: None.

11. In order to have smooth operation, the minimum number of teeth on the smaller sprocket, for moderate speeds, should be
a) 15
b) 17
c) 21
d) 25
Answer: b
Clarification: In order to have smooth operation, the minimum number of teeth on the smaller sprocket or pinion may be taken as 17 for moderate speeds and 21 for high speeds.

12. The speed of the sprocket reduces as the chain pitch _____________ for a given number of teeth.
a) increases
b) decreases
c) remains same
d) none of the mentioned
Answer: a
Clarification: The r.p.m. of the sprocket reduces as the chain pitch increases for a given number of teeth.

13. The greater angle of articulation will increase the life of the chain.
a) True
b) False
Answer: b
Clarification: Greater angle of articulation will lead to breaking of chain & reduction in life of the chain.

250+ TOP MCQs on Bevel Gears and Answers

Machine Kinematics Multiple Choice Questions on “Bevel Gears”.

1. The bevel gears are used to connect
a) two parallel shafts
b) two intersecting shafts
c) two non intersecting shafts
d) none of the mentioned

Answer: b
Clarification: Two intersecting shafts are connected by bevel gears and pitch cylinders in spur gears.

2. Bevel gears with shafts angle of 900 are termed as
a) zerol gears
b) angular bevel gears
c) mitre gears
d) hypoid gears

Answer: c
Clarification: The bevel gears with shaft angle 900 are termed as Mitre gears. The bevel gears with any other shaft angle are termed as angular bevel gears.

3. The bevel gears used for connecting non intersecting shafts are
a) mitre gears
b) hypoid gears
c) spiral bevel gears
d) zerol gears

Answer: b
Clarification: The bevel gears used for connecting non intersecting shafts are hypoid gears. The bevel gears with shaft angle 900 are termed as Mitre gears.

4. Face width of the bevel gear is usually equal to
a) 10 modules
b) pitch cone radius/3
c) pitch cone radius/2
d) none of the mentioned

Answer: b
Clarification: For correctly designed gears the face width should not be more than 1/3 the slant length, i.e. pitch cone radius.

5. Formative number of teeth on bevel gears is equal to
a) 2 x actual number of teeth
b) actual number of teeth/ cosɸ
c) actual number of teeth/cosα
d) actual number of teeth/cosϴ

Answer: d
Clarification: None.

6. Lewis Equation for bevel gear is corrected for
a) variation in p.c.d.
b) variation in tooth thickness
c) taking care of axial thrust
d) variation in torque acting on the tooth

Answer: a
Clarification: As radius of bevel gear varies along the width, the torque does not produce the same tangential force.

7. Ratio factor Q in wear load equation of bevel gear is given by
a) 2gear ratio G/G + 1
b) 2ratio of formative number of teeth of gear and pinion G
c) G + 1/G +1
d) 2G/G

Answer: b
Clarification: None.

8. Bevel factor should not be less than
a) 0.75
b) 0.8
c) 0.67
d) 0.76

Answer: c
Clarification: None.

9. Pitch cone angle of pinion of straight tooth bevel gear pair with ratio 1.732 is
a) 250
b) 300
c) 600
d) none of the mentioned

Answer: c
Clarification: None.

10. The face width of the bevel gear is 0.3 times the radius of pitch cone. Here the bevel factor must be
a) 1.3
b) 3
c) 0.7
d) 1.7

Answer: c
Clarification: Face width = Pitch cone radius/3
therefore, bevel factor = 0.7.

11. Interchangibility is possible only in
a) bevel gear
b) helical gear
c) spur gear
d) mitre gear

Answer: c
Clarification: Spur gears are connected by pitch cylinders and therefore, they can interchange.