Category: Machine Kinematics Objective Questions

250+ TOP MCQs on Differential Equation of Simple Harmonic Motion and Answers

Tricky Machine Kinematics Questions and Answers on “Differential Equation of Simple Harmonic Motion”.

1. By how much angle in degrees does the velocity leads the displacement in a body undergoing SIMPLE HARMONIC MOTION?
a) 90
b) 45
c) 180
d) 0
Answer: a
Clarification: For a body undergoing Simple Harmonic Motion, the velocity leads the displacement by an angle of 90 degrees as shown by the differential equation of the motion.

2. For a body undergoing SIMPLE HARMONIC MOTION, the acceleration is always in the direction of the displacement.
a) True
b) False
Answer: b
Clarification: For a body undergoing SIMPLE HARMONIC MOTION, the acceleration is always in the opposite direction of the displacement. This is indicated by a negative sign in the equation.

3. The maximum displacement of the body under Simple Harmonic Motion from its mean position is known as_______
a) Amplitude
b) Frequency
c) Time period
d) Range
Answer: a
Clarification: The maximum displacement of a body undergoing Simple Harmonic Motion is known as Amplitude, it is generally denoted by ‘A’.

4. The piston of an engine moves with SIMPLE HARMONIC MOTION. The crank rotates at a speed of 120 r.p.m. with a stroke of 2 metres. Find the velocity of the piston in m/s, when it is at a distance of 0.75 metre from the centre.
a) 8.31
b) 7.33
c) 8.41
d) 9.02
Answer: a
Clarification: When a body is undergoing SIMPLE HARMONIC MOTION, its velocity is given by the equation
(v = omega sqrt{(A^2-x^2)})
substituting the values we get
v = 8.31 m/s.

5. The piston of an engine moves with SIMPLE HARMONIC MOTION. The crank rotates at a speed of 120 r.p.m. with a stroke of 2 metres. Find the velocity of the piston in m/s2, when it is at a distance of 0.75 metre from the centre.
a) 118.46
b) 117.33
c) 128.41
d) 119.02
Answer: a
Clarification: When a body is undergoing SIMPLE HARMONIC MOTION, its acceleration is given by the equation
a=ω2x
substituting the values we get
a = 118.46 m/s2

6. A point moves with SIMPLE HARMONIC MOTION. When this point is 0.75 m away from the mid path, it has a velocity of 11 m/s and when 2 m from the centre of its path its velocity is 3 m/s. Find its angular velocity in rad/s.
a) 5.7
b) 7.5
c) 6.7
d) 7.6
Answer: a
Clarification: When a body is undergoing SIMPLE HARMONIC MOTION, its velocity is given by the equation
v = (omega sqrt{r^2 – x^2})
Given values:
when point (x) = 0.75 m, then its velocity (v) = 11 m/s
when point (x) = 2 m, then its velocity (v) = 3 m/s
When point is 0.75 m away from the mid path (v) is,
v = (omega sqrt{r^2 – x^2})
11 = (omega sqrt{r^2 – (0.75)^2}) —>Eq(1)
Similarly, When point is 2 m away from the centre (v) is,
v = (omega sqrt{r^2 – x^2})
3 = (omega sqrt{r^2 – 2^2}) —>Eq(2)
Solving Eq(1) & Eq(2) we get,
(frac{11}{3} frac{omega sqrt{r^2 – (0.75)^2}}{omega sqrt{r^2 – 2^2}} = frac{sqrt{r^2 – (0.75)^2}}{sqrt{r^2 – 2^2}} )
Squaring on both sides, we get
(frac{121}{9} = frac{r^2 – 0.5625}{r^2 – 4})
121r2 – 484 = 9r2 – 5.0625
121r2 – 9r2 = 484 – 5.0625
112r2 = 478.9375
r2 = (frac{478.9375}{112})
r2 = 4.27622
r = 2.07m
Substituting the value of r in Eq(1) we get,
11 = ( omega sqrt{(2.07)^2 – (0.75)^2})
11 = ( omega sqrt{4.2849 – 0.5625})
11 = ( omega sqrt{3.7224})
11 = 1.9293 ω
ω = (frac{11}{1.9293}) = 5.7 rad/s.

7. A point moves with SIMPLE HARMONIC MOTION. When this point is 0.75 m away from the mid path, it has a velocity of 11 m/s and when 2 m from the centre of its path its velocity is 3 m/s. Find its time period in s.
a) 1.1
b) 1.2
c) 1.3
d) 1.4
Answer: a
Clarification: When a body is undergoing SIMPLE HARMONIC MOTION, its velocity is given by the equation
v = (omega sqrt{r^2 – x^2})
Given values:
when point (x) = 0.75 m, then its velocity (v) = 11 m/s
when point (x) = 2 m, then its velocity (v) = 3 m/s
When point is 0.75 m away from the mid path (v) is,
v = (omega sqrt{r^2 – x^2})
11 = (omega sqrt{r^2 – (0.75)^2}) —>Eq(1)
Similarly, When point is 2 m away from the centre (v) is,
v = (omega sqrt{r^2 – x^2})
3 = (omega sqrt{r^2 – 2^2}) —>Eq(2)
Solving Eq(1) & Eq(2) we get,
(frac{11}{3} frac{omega sqrt{r^2 – (0.75)^2}}{omega sqrt{r^2 – 2^2}} = frac{sqrt{r^2 – (0.75)^2}}{sqrt{r^2 – 2^2}} )
Squaring on both sides, we get
(frac{121}{9} = frac{r^2 – 0.5625}{r^2 – 4})
121r2 – 484 = 9r2 – 5.0625
121r2 – 9r2 = 484 – 5.0625
112r2 = 478.9375
r2 = (frac{478.9375}{112})
r2 = 4.27622
r = 2.07m
Substituting the value of r in Eq(1) we get,
11 = ( omega sqrt{(2.07)^2 – (0.75)^2})
11 = ( omega sqrt{4.2849 – 0.5625})
11 = ( omega sqrt{3.7224})
11 = 1.9293 ω
ω = (frac{11}{1.9293}) = 5.7 rad/s.
We know periodic time is,
Tp = (frac{2 pi}{omega} = frac{2 pi}{5.7} = frac{2 times 3.14}{5.7} = frac{6.28}{5.7}) = 1.1 s.

8. A point moves with SIMPLE HARMONIC MOTION. When this point is 0.75 m away from the mid path, it has a velocity of 11 m/s and when 2 m from the centre of its path its velocity is 3 m/s. Find its maximum acceleration in m/s2.
a) 61.1
b) 67.2
c) 51.3
d) 41.4
Answer: b
Clarification: When a body is undergoing SIMPLE HARMONIC MOTION, its velocity is given by the equation
v = (omega sqrt{r^2 – x^2})
Given values:
when point (x) = 0.75 m, then its velocity (v) = 11 m/s
when point (x) = 2 m, then its velocity (v) = 3 m/s
When point is 0.75 m away from the mid path (v) is,
v = (omega sqrt{r^2 – x^2})
11 = (omega sqrt{r^2 – (0.75)^2}) —>Eq(1)
Similarly, When point is 2 m away from the centre (v) is,
v = (omega sqrt{r^2 – x^2})
3 = (omega sqrt{r^2 – 2^2}) —>Eq(2)
Solving Eq(1) & Eq(2) we get,
(frac{11}{3} frac{omega sqrt{r^2 – (0.75)^2}}{omega sqrt{r^2 – 2^2}} = frac{sqrt{r^2 – (0.75)^2}}{sqrt{r^2 – 2^2}} )
Squaring on both sides, we get
(frac{121}{9} = frac{r^2 – 0.5625}{r^2 – 4})
121r2 – 484 = 9r2 – 5.0625
121r2 – 9r2 = 484 – 5.0625
112r2 = 478.9375
r2 = (frac{478.9375}{112})
r2 = 4.27622
r = 2.07m
Substituting the value of r in Eq(1) we get,
11 = ( omega sqrt{(2.07)^2 – (0.75)^2})
11 = ( omega sqrt{4.2849 – 0.5625})
11 = ( omega sqrt{3.7224})
11 = 1.9293 ω
ω = (frac{11}{1.9293}) = 5.7 rad/s.
Maximum acceleration is
Amax = ω2r = (5.7)2 × 2.07 = 32.49 × 2.07
= 67.25 m/s2.

9. If V is the maximum velocity of a body undergoing SIMPLE HARMONIC MOTION, then what is the average velocity of motion from one extreme to other extreme is?
a) 2V/π
b) 4V/π
c) V/2π
d) 2V/3π
Answer: a
Clarification: V = 2πA/T
V av = 2A/T÷2 = 4A/T
A/T = V/2π
Vav = 2V/π

10. If V is the maximum velocity of a body undergoing SIMPLE HARMONIC MOTION, then what is the average velocity of motion?
a) 2V/π
b) 4V/π
c) V/2π
d) 2V/3π
Answer: a
Clarification: V = Aω
= 4A/T
= 2aω/π
= 2V/π

11. Which of the following is the correct differential equation of the SIMPLE HARMONIC MOTION?
a) d2x/dt2 + ω2x = 0
b) d2x/dt2 – ω2x = 0
c) d2x/dt + ω2x = 0
d) d2x/dt – ω2x = 0
Answer: a
Clarification: For body undergoing Simple Harmonic Motion, it’s motion can be represented as projected uniform circular motion with radius equal to the amplitude of motion.
Therefore x =Acosωt
dx/dt = -Aωsinωt
d2x/dt2 = -aω2cosωt
therefore
d2x/dt2 + ω2x = 0

12. For a body undergoing Simple Harmonic Motion, the acceleration is maximum at the extreme.
a) True
b) False
Answer: a
Clarification: For a body undergoing SIMPLE HARMONIC MOTION, the acceleration is always in the opposite direction of the displacement. This is indicated by a negative sign in the equation and at an extreme position, the acceleration attains a maximum value.

13. Which of the following is the solution of the differential equation of the SIMPLE HARMONIC MOTION?
a) x = Acosωt + B sinωt
b) x = (A+B)cosωt
c) x = (A+B)sinωt
d) x = Atanωt + B sinωt
Answer: a
Clarification: We know that dx/dt = -Aωsinωt
d2x/dt2 = -aω2cosωt
therefore
d2x/dt2 + ω2x = 0
is the standard differential equation of Simple Harmonic Motion
It’s solution is/are:
x = Acosωt + B sinωt

To practice Tricky questions and answers on all areas of Machine Kinematics,

250+ TOP MCQs on Number of Degrees of Freedom for Plane Mechanisms and Answers

Machine Kinematics Multiple Choice Questions on “Number of Degrees of Freedom for Plane Mechanisms”.

1. The total number of instantaneous centres for a mechanism consisting of n links are
a) n/2
b) n
c) n-1
d) n(n-1)/2
Answer: d
Clarification: The number of pairs of links or the number of instantaneous centres is the number of combinations of n links taken two at a time. Mathematically, number of instantaneous centres, n(n-1)/2.

2. According to Kennedy’s theorem, if three bodies move relatively to each other, their instantaneous centres will lie on
a) straight line
b) parabolic curve
c) triangle
d) rectangle
Answer: a
Clarification: The Aronhold Kennedy’s theorem states that if three bodies move relatively to each other,
they have three instantaneous centres and lie on a straight line.

3. Which of the following property of the instantaneous center is correct?
a) A rigid link rotates instantaneously relative to another link at the instantaneous centre for the configuration of the mechanism considered.
b) The two rigid links have no linear velocity relative to each other at the instantaneous centre.
c) The velocity of the instantaneous centre relative to any third link is same whether the instantaneous centre is regarded as a point on the first link or on the second rigid link.
d) all of the mentioned
Answer: d
Clarification: None.

4. The magnitude of velocities of the points on a rigid link is
a) directly proportional to the distance from the points to the instantaneous centre and is parallel to the line joining the point to the instantaneous centre.
b) directly proportional to the distance from the points to the instantaneous centre and is perpendicular to the line joining the point to the instantaneous centre.
c) inversely proportional to the distance from the points to the instantaneous centre and is parallel to the line joining the point to the instantaneous centre.
d) inversely proportional to the distance from the points to the instantaneous centre and is perpendicular to the line joining the point to the instantaneous centre.
Answer: d
Clarification: None.

5. In a mechanism, the fixed instantaneous centres are those centres which
a) remain in the same place for all configurations of the mechanism
b) vary with the configuration of the mechanism
c) moves as the mechanism moves, but joints are of permanent nature
d) none of the mentioned
Answer: a
Clarification: Fixed instantaneous centres remain in the same place for all configurations of the mechanism.

6. The instantaneous centres, which moves as the mechanism moves but joints are of permanent nature, are called permanent instantaneous centres.
a) True
b) False
Answer: a
Clarification: The permanent instantaneous centres move when the mechanism moves, but the joints are of permanent nature.
Fixed instantaneous centres remain in the same place for all configurations of the mechanism.

7. The instantaneous centres which vary with the configuration of mechanism, are called
a) permanent instantaneous centres
b) fixed instantaneous centres
c) neither fixed nor permanent instantaneous centres
d) none of the mentioned
Answer: c
Clarification: Neither fixed nor permanent instantaneous centres vary with the configuration of the mechanism.

8. When two links are connected by a pin joint, their instantaneous centre lies
a) on their point of contact
b) at the centre of curvature
c) at the centre of circle
d) at the pin joint
Answer: d
Clarification: None.

9. The two links are said to have a pure rolling contact, when their instantaneous centre __________ on their point of contact.
a) lies
b) does not lie
Answer: a
Clarification: None

10. When a slider moves on a fixed link having ____________ their instantaneous center lies at infinity.
a) straight surface
b) curved surface
c) oval surface
d) none of the mentioned
Answer: a
Clarification: When a slider moves on a fixed link having curved surface, their instantaneous centre lies at the centre of curvature.
When a slider moves on a fixed link having straight surface their instantaneous center lies at infinity.

11. When a slider moves on a fixed link having curved surface, their instantaneous centre lies
a) on their point of contact
b) at the centre of curvature
c) at the centre of circle
d) at the pin joint
Answer: b
Clarification: When a slider moves on a fixed link having curved surface, their instantaneous centre lies at the centre of curvature.
When a slider moves on a fixed link having straight surface their instantaneous center lies at infinity.

12. A slider moving on a fixed link having constant radius of curvature will have its instantaneous centre at the center of the circle.
a) True
b) False
Answer: a
Clarification: None.

13. The instantaneous center of a rigid thin disc rolling on a plane rigid surface is located at
a) the centre of the disc
b) the point of contact
c) an infinite distance on the plane surface
d) the point on the circumference situated vertically opposite to the contact point
Answer: b
Clarification: None.

250+ TOP MCQs on Motion of a Link and Answers

Machine Kinematics Multiple Choice Questions on “Motion of a Link”.

1. What is the direction of velocity of a point in a link relative to another point on the same link rotating in a specific direction.
a) Perpendicular to line joining both the links
b) Parallel to line joining both the links
c) Perpendicular to the surface of the link
d) Parallel to the surface of the link
Answer: a
Clarification: Velocity of any point on a link with respect to another point on the same link is always in the direction perpendicular to the line joining these points on the space diagram.

2. The direction of velocity is parallel if the rotation is anticlockwise and perpendicular to the line joining links if the rotation is clockwise.
a) True
b) False
Answer: b
Clarification: Velocity of any point on a link with respect to another point on the same link is always in the direction perpendicular to the line joining these points on the space diagram both during anticlockwise and clockwise rotations.

3. What is the correct representation of velocity of point A with respect to B in a link?
a) Vab
b) Vba
c) Va-b
d) Vb-a
Answer: a
Clarification: When there are two points on a link, the velocity of a point A wrt to other point on the same link is represented by Vab.

4. What is the velocity of point C with respect to A in the given figure?
machine-kinematics-questions-answers-velocity-mechanisms-q1
a) Perpendicular to line joining BC
b) Perpendicular to line joining AC
c) Parallel to line joining BC
d) Parallel to line joining AC
Answer: b
Clarification: In the above figure we can assume an imaginary link between A and C, as per the rule, the velocity of a point with respect to any other point on the same link is perpendicular to the line joining the links.

5. Which type of pair formed by two elements which are so connected that one is constrained to turn or revolve about a fixed axis of another element?
a) Turning pair
b) Rolling pair
c) Sliding pair
d) Higher pair
Answer: a
Clarification: The type of pair formed by two elements which are so connected that one is constrained to turn or revolve about a fixed axis of another element is known as turning pair.

6. Which of the following is a lower pair?
a) Pulleys in belt drive
b) Cam and follower
c) Belt drive
d) Ball and socket joint
Answer: d
Clarification: If two moving elements do not have surface contact in motion, then the pair is known as lower pair, in the given question ball and socket joint is a lower pair.

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 effort loss in friction at 200 N.
a) 30 N
b) 35 N
c) 40 N
d) 45 N
Answer: d
Clarification: When W = 300 N, P = 60 N
Effort lost in friction, FP = P – W/V.R. = 60 – 300/20 = 45 N.

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 maximum efficiency which can be expected from this machine.
a) 30 %
b) 40 %
c) 50 %
d) 60 %
Answer: c
Clarification: Maximum possible efficiency of any machine = 1/m x V.R. = 1/1/10 x 20 = 0.5 = 50 %.

250+ TOP MCQs on Laws of Solid Friction and Limiting Friction and Answers

Machine Kinematics Multiple Choice Questions on “Laws of Solid Friction and Limiting Friction”.

1. The force of friction always acts in a direction, ___________ to that in which the body tends to move.
a) same
b) opposite
c) both of the mentioned
d) none of the mentioned
Answer: b
Clarification: The force of friction always acts in a direction, opposite to that in which the body tends to move.

2. The magnitude of the force of friction is ____________ to the force, which tends the body to move.
a) equal
b) different
c) both of the mentioned
d) none of the mentioned
Answer: a
Clarification: The magnitude of the force of friction is exactly equal to the force, which tends the body to move.

3. The magnitude of the limiting friction (F ) bears a constant ratio to the normal reaction (RN) between the two surfaces.
a) True
b) False
Answer: a
Clarification: The magnitude of the limiting friction (F ) bears a constant ratio to the normal reaction (RRN) between the two surfaces. Mathematically
F/RRN = constant.

4. The force of friction is _____________ of the area of contact, between the two surfaces.
a) dependent
b) independent
c) both of the mentioned
d) none of the mentioned
Answer: b
Clarification: The force of friction is independent of the area of contact, between the two surfaces.

5. The force of friction does not depends upon the roughness of the surfaces.
a) True
b) False
Answer: b
Clarification: The force of friction depends upon the roughness of the surfaces.

6. The ratio of magnitude of the kinetic friction to the normal reaction between the two surfaces is_____________ than that in case of limiting friction.
a) greater
b) less
c) equal
d) none of the mentioned
Answer: b
Clarification: The magnitude of the kinetic friction bears a constant ratio to the normal reaction between the two surfaces. But this ratio is slightly less than that in case of limiting friction.

7. For moderate speeds, the force of friction
a) increases
b) decreases
c) remains constant
d) none of the mentioned
Answer: c
Clarification: For moderate speeds, the force of friction remains constant. But it decreases slightly with the increase of speed.

8. A body of mass 400 g slides on a rough horizontal surface. If the frictional force is 3.0 N, find the angle made by the contact force on the body with the vertical.
a) 350
b) 360
c) 370
d) 380
Answer: c
Clarification: Let the contact force on the block by the surface be F which makes an angle ϴ with the vertical.
The component of F perpendicular to the contact surface is the normal force N and the component of F parallel to the surface is the friction f As the surface is horizontal, N is vertically upward. For vertical equilibrium,
N = mg = (0.400 kg) (10 m/s2) = 4.0 N.
The frictional force is f = 3.0 N.
tan ϴ = f/N = 3/4
or, ϴ = tan-1 (3/4) = 370.

9. A body of mass 400 g slides on a rough horizontal surface. If the frictional force is 3.0 N, find the magnitude of the contact force.
a) 5 N
b) 10 N
c) 15 N
d) 20 N
Answer: a
Clarification: Let the contact force on the block by the surface be F which makes an angle ϴ with the vertical.
The component of F perpendicular to the contact surface is the normal force N and the component of F parallel to the surface is the friction f As the surface is horizontal, N is vertically upward. For vertical equilibrium,
N = mg = (0.400 kg) (10 m/s2) = 4.0 N.
The frictional force is f = 3.0 N.
F = √N2
= √42 + 32 = 5 N.

10. A heavy box of mass 20 kg is pulled on a horizontal surface by applying a horizontal force. If the coefficient of kinetic friction between the box and the horizontal surface is 0.25, find the force of friction exerted by the horizontal surface on the box.
a) 29 N
b) 39 N
c) 49 N
d) 59 N
Answer: c
Clarification: As the box slides on the horizontal surface, the surface exerts kinetic friction on the box. The magnitude of the kinetic friction is
f = μN
= μmg
= 0.25 x (20 kg) x (9.8 m/s 2) = 49 N.
This force acts in the direction opposite to the pull.

11. A boy (30 kg) sitting on his horse whips it. The horse speeds up at an average acceleration of 2.0 m/s 2.If the boy does not slide back, what is the force of friction exerted by the horse on the boy ?
a) 20 N
b) 30 N
c) 40 N
d) 60 N
Answer: d
Clarification: The forces acting on the boy are
(i) the weight Mg.
(ii) the normal contact force N and
(iii) the static friction fs

As the boy does not slide back, its acceleration a is equal to the acceleration of the horse. As friction is the only horizontal force, it must act along the acceleration and its magnitude is given by Newton’s second law
fs = Ma = (30 kg) (2.0 m/s2) = 60 N.

12. A boy (30 kg) sitting on his horse whips it. The horse speeds up at an average acceleration of 2.0 m/s 2.If the boy slides back during the acceleration, what can be said about the coefficient of static friction between the horse and the boy.
a) 0.10
b) 0.20
c) 0.30
d) 0.40
Answer: b
Clarification: If the boy slides back, the horse could not exert a
friction of 60 N on the boy. The maximum force of static
friction that the horse may exert on the boy is
fs = μs = μsMg

μs(30 kg) (10m/s2) = μs 300 N
where μs is the coefficient of static friction. Thus,
μs(300 N) <60 N
or, μs <60/300 = 0.20.

13. A wooden block is kept on a polished wooden plank and the inclination of the plank is gradually increased. It is found that the block starts slipping when the plank makes an angle of 18° with the horizontal. However, once started the block can continue with uniform speed if the inclination is reduced to 15°. Find the coefficient of static friction between the block and the plank.
a) tan 180
b) tan 150
c) tan 330
d) tan 30
Answer: a
Clarification: The coefficient of static friction is
μs = tan 180.

14. A wooden block is kept on a polished wooden plank and the inclination of the plank is gradually increased. It is found that the block starts slipping when the plank makes an angle of 18° with the horizontal. However, once started the block can continue with uniform speed if the inclination is reduced to 15°. Find the coefficient of kinetic friction between the block and the plank.
a) tan 180
b) tan 150
c) tan 330
d) tan 30
Answer: b
Clarification: The coefficient of kinetic friction is
μk = tan 150.

250+ TOP MCQs on Pulleys and Answers

Machine Kinematics Multiple Choice Questions on “Pulleys”.

1. In a wheel and differential axle, the diameter of the effort wheel is 400 mm. The radii of the load axles are 150mm and 100 mm respectively. The diameter of the rope is 10 mm. Find the load which can be lifted by an effort of 100 N, assuming an efficiency of the machine to be 75%.
a) 800 N
b) 725 N
c) 615 N
d) none of the mentioned

Answer: c
Clarification: Diameter of effort wheel, D = 400 mm
Diameter of longer axle, d1 = 2 x 150 = 300 mm
Diameter of the smaller axle, d2 = 2 x 100 = 200 mm
Diameter of the rope, dr = 10 mm

therefore, V.R. = 2(D + dr )/(d1 + dr ) – (d2 + dr )
= 2(400 + 10)/ (300 + 10) – (200 + 10)
= 820/100 = 8.2
Effort, P = 100 N
ȵ = 75%
Let W = load which can be lifted by the machine

ȵ = M.A./V.R.
0.75 = W/P x 8.2
W = 0.75 x 100 x 8.2 = 615 N.

2. Four movable pulleys are arranged as in the first system. If the weight of each pulley is 5 N, calculate the effort which can lift a load of 10 kN.
a) 629.7 N
b) 615 N
c) 625 N
d) none of the mentioned

Answer: a
Clarification: We know that M.A. = 2nW/W + w(2n – 1)
where W = load to be lifted
w = weight of each pulley
n = no. of movable pulleys

therefore, M.A. = 24 x 10000/10000 + 5(24 – 1) = 10000/P
P = 10000 + 5(24 – 1)/ 24 = 629.7 N.

3. A person weighing 600 N platform attached to the lower block of a system of 5 pulleys arranged in the second system. The platform and the lower block weigh 100N. The man himself supports by exerting a downward pull at the free end of the rope. Neglecting friction, the minimum pull of the man will be
a) 1000 N
b) 200 N
c) 116.7 N
d) none of the mentioned

Answer: c
Clarification: n = total number of pulleys in the system = 5
W = 600 N
Weight of the lower block and platform = 100 N
Total weight = 600 + 100 = 700 N
Let the pull exerted by the man be P newton.
Due to this pull the effective load on the lower platform will reduce to (700 – P)
nP = effective load = 700 – P
therefore, 5P = 700 -P
5P = 700
P = 116.7 N.

4. Five pulleys are arranged in the second system of pulleys. When not loaded the effort required to raise the movable block is 35N. Further wastage in friction increases the pull at the rate of 3% of the load lifted. What is the effort required to raise a load of 2kN?
a) 500 N
b) 400 N
c) 495 N
d) none of the mentioned

Answer: c
Clarification: n = no. of pulleys = 5
Frictional effort at zero loading = 35N
Frictional effort at 2 kN loading = 35 + 2000 x 3/100 = 95 N
When the system is considered frictionless nP = W
5P = 2000
P = 400 N
Hence total effort = 400 + 95 = 495 N.

5. Five pulleys are arranged in the second system of pulleys. When not loadwd the effort required to raise the movable block is 35N. Further wastage in friction increases the pull at the rate of 3% of the load lifted. What is the efficiency of the system at 2kN?
a) 80%
b) 80.81%
c) 80.50%
d) none of the mentioned

Answer: b
Clarification: n = no. of pulleys = 5
Frictional effort at zero loading = 35N
Frictional effort at 2 kN loading = 35 + 2000 x 3/100 = 95 N
When the system is considered frictionless nP = W
5P = 2000
P = 400 N
Hence total effort = 400 + 95 = 495 N
Efficiency at this load = effort without friction/effort with friction
= 400/495 x 100
= 80.81%.

6. In a weston differential pulley block, the number of recesses in the smaller wheel is 9/10 of that of the larger wheel. If the efficiency of the machine is 50%, find the load lifted by an effort of 300N.
a) 2000N
b) 3000N
c) 4000N
d) none of the mentioned

Answer: b
Clarification: Let the recesses in the larger wheel, n1 = 10
Recesses in the smaller wheel, n2 = 9/10 x 10 = 9

V.R. = 2n1 /n1 – n2 = 2×10/10 – 9 = 20
and mechanical advantage M.A. = W/P
= W/300
efficiency = M.A./V.R.
0.5 = W/300×20
W = 3000N.

7. If the velocity ratio for an open belt drive is 8 and the speed of driving pulley is 800 r.p.m, then considering an elastic creep of 2% the speed of the driven pulley is
a) 104.04 r.p.m
b) 102.04 r.p.m
c) 100.04 r.p.m
d) 98.04 r.p.m

Answer: d
Clarification: Velocity Ratio = Velocity of belt on driver/Velocity of belt on driven
Velocity of belt on driven = 800/8 = 100 r.p.m
Elastic creep = velocity of belt at driven pulley – Velocity of driven pulley
0.02 × Vp = [100-Vp]
Vp = 100/1.02 = 98.04r.p.m.

8. If the angle of wrap on smaller pulley of diameter 250 mm is 1200 and diameter of larger pulley is twice the diameter of smaller pulley, then the centre distance between the pulleys for an open belt drive is
a) 1000 mm
b) 750 mm
c) 500 mm
d) 250 mm

Answer: d
Clarification: sin α = (D -d)/2c
Angle of wrap on smaller pulley = п – 2α
2п/3 = п – 2sin-1(D -d)/2c
c = 250 mm.

9. In short open-belt drives, an idler pulley is used in order to decrease the angle of contact on the smaller pulley for higher power transmission.
a) True
b) False

Answer: b
Clarification: In short open-belt drives, an idler pulley is used in order to increase the angle of contact on the smaller pulley for higher power transmission.

10. In design of arms of a pulley, in belt drive, the cross-section of the arm is elliptical with minor axis placed along the plane of rotation.
a) True
b) False

Answer: b
Clarification: Arms of a pulley in belt drive are subjected to complete reversal of stresses and is designed for bending in the plane of rotation.

250+ TOP MCQs on Bevel Gears – 2 and Answers

Machine Kinematics Questions and Answers for Aptitude test on “Bevel Gears – 2”.

1. The mathematical form of the bevel tooth profile which closely resembles a spherical involute but is fundamentally different is
a) crown
b) back cone
c) octoid
d) none of the mentioned
Answer: c
Clarification: Octoid is the mathematical form of the bevel tooth profile which closely resembles a spherical involute but is fundamentally different is
The sharp corner orming the outside diameter is crown.

2. The angle formed between an element of the pitch cone and the bevel gear axis is
a) shaft angle
b) pitch angle
c) spiral angle
d) none of the mentioned
Answer: b
Clarification: The angle between the tooth trace and an element of the pitch cone, corresponding to helix angle in helical gears is spiral angle.
The angle formed between an element of the pitch cone and the bevel gear axis is pitch angle.

3. The angle between the tooth trace and an element of the pitch cone, corresponding to helix angle in helical gears is
a) shaft angle
b) pitch angle
c) spiral angle
d) none of the mentioned
Answer: c
Clarification: The angle between the tooth trace and an element of the pitch cone, corresponding to helix angle in helical gears is spiral angle.
The angle formed between an element of the pitch cone and the bevel gear axis is pitch angle.

4. The diameter and plane of rotation surface or shaft centre which is used for locating the gear blank during fabrication of the gear teeth is known as
a) crown
b) back cone
c) generating mounting surface
d) none of the mentioned
Answer: c
Clarification: The diameter and plane of rotation surface or shaft centre which is used for locating the gear blank during fabrication of the gear teeth is known as generating mounting surface.

5. The sharp corner orming the outside diameter is
a) crown
b) back cone
c) octoid
d) none of the mentioned
Answer: a
Clarification: The sharp corner orming the outside diameter is crown.
The length of teeth along the cone distance is face width.

6. The angle between elements of the face cone and pitch cone is
a) addendum angle
b) pitch angle
c) spiral angle
d) none of the mentioned
Answer: a
Clarification: The angle between mating bevel gear axes, also the sum of the two pitch angles is spiral angle.
The angle between elements of the face cone and pitch cone is addendum angle.

7. The angle between mating bevel gear axes, also the sum of the two pitch angles is
a) shaft angle
b) pitch angle
c) spiral angle
d) none of the mentioned
Answer: c
Clarification: The angle between mating bevel gear axes, also the sum of the two pitch angles is spiral angle.
The angle between elements of the face cone and pitch cone is addendum angle.

8. The length of teeth along the cone distance is
a) crown
b) face width
c) octoid
d) none of the mentioned
Answer: b
Clarification: The length of teeth along the cone distance is face width.
The sharp corner orming the outside diameter is crown.

9. The angle of a cone whose elements are tangent to a sphere containing a trace of the pitch circle is
a) crown
b) back cone
c) octoid
d) none of the mentioned
Answer: b
Clarification: The angle between elements of the root cone and pitch cone is dedendum angle.
The angle of a cone whose elements are tangent to a sphere containing a trace of the pitch circle is back cone.

10. The angle between elements of the root cone and pitch cone is
a) addendum angle
b) dedendum angle
c) spiral angle
d) none of the mentioned
Answer: b
Clarification: The angle between elements of the root cone and pitch cone is dedendum angle.
The angle of a cone whose elements are tangent to a sphere containing a trace of the pitch circle is back cone.

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