TOM MCQ

Chapter 1
Fundamentals of Kinematics
and Mechanisms
Multiple Choice Questions for Online Exam
Introduction to Theory of Machine
Q. 1
Theory of machines is the branch of science, which deals
with the study of _____.
(a) the relative motion between the parts of a machine and
the study of forces which acts on those parts
(b) the relative motion between the parts of a machine
(c) the forces acting on the parts of the machine
(d) none of the above
Ans. : (a)
• Explanation : The theory of machine is an applied science
which is used to understand the relative motion between the
geometry and motions of the parts of a machine or the
mechanisms and the study of the forces which produce these
motions.
Q. 2 Kinematic of machines deals with _____________.
(a) Apparatus for applying mechanical power.
(b) The forces acting on machine parts.
(c) The number of inter related parts.
(d) The relative motion between the parts, neglecting the
consideration of the force.
Ans. : (d)
•
Explanation : Kinematic of machines is to study the
relative motion between the parts of a machine without
considering forces i.e. it is the study of position,
displacement, rotation, speed, velocity and acceleration.Q. 3 Which of the following disciplines provides study of relative
motion between the parts of a machine ?
(a) Theory of machines
(b) Applied mechanics
(c) Kinetics
(d) Kinematics.
Ans. : (d)
Explanation : Refer Q 2
Q. 4 Dynamics of machines deals with _____________.
(a) The number of related parts, each having a definite
motion.
(b) The forces acting on the parts of the machine.
(c) The relative motion between the parts of the machine.
(d) None of the above.
Ans. : (b)
Explanation : It relates to study of forces acting on
various parts of a machine. The subject of dynamics is
further subdivided in two parts as follows :
(a)
Statics : It deals with the forces which act on
parts of a machine which may be assumed
without mass.
(b)
Kinetics : It deals with the inertia forces due to
combined effect of mass and motion of the parts.
Q. 5 Which of the following disciplines provides study of inertia
forces arising from the combined effect of the mass and the
motion of the parts ?
(a) Theory of machines
(b) Applied mechanics
(c) Kinetics
(d) Kinematics.
Ans. : (c)
Explanation : Kinetics deals with the inertia forces due to
combined effect of mass and motion of the parts.
Q. 6 Which of the following disciplines provides study of the
relative motion between the parts of a machine and the
forces acting on the parts ?
(a) Theory of machines
(b) Applied mechanics(c) Kinetics
Ans. : (a)
Explanation : Refer Q.1
Q. 7
Q. 8
(d) Kinematics.
The study of relative motion between the parts of a
machine is called _________.
(a) Statics
(b) Hydrodynamics
(c) Kinematics
(d) Kinetics
Ans. : (c)
Explanation : Refer Q.2
Which of the following is false statement in respect of
differences between machine and structure?
(a) Machines transmit mechanical work, whereas
structures transmit forces
(b) In machines, relative motion exists between its
members, whereas same does not exist in case of
structures
(c) Machines modify movement and work, whereas
structures modify forces
(d) Efficiency of machines as well as structures is below
100%.
Ans. : (d)
Explanation :
Sr.
No. Structure Machine
1 There is no relative motion
between its links. There exists a relative motion
between the links of a machine.
2. It serves to modify and
transmit the forces only and
it cannot transmit the energy
for doing mechanical work. It serves to transfer both forces
and motion. It transforms the
available energy into mechanical
work or some other form of
energy.
Q. 9
The difference between a machine and structure is
_____________.
(a) The machine serves to modify and transmit the forces
and relative motion whereas the structure serves to
modify an transmit the forces only.Q. 10
(b) The machine serves to modify and transmit forces only
whereas structure serves to modify and transmit
mechanical work.
(c) The relative motion exists between the parts of a
structure but it does not exist in machines.
(d) None of the above.
Ans. : (a)
Explanation : Refer Q 8
In engine rectilinear motion of piston is converted into
rotary by _____________.
(a) cross head
(b)
slider crank
(c) connecting rod
(d)
gudgeon pin
Ans. : (c)
Explanation : The reciprocating motion of the piston is
converted into rotary motion of the crank shaft with the
help of connecting rod and crank.
Kinematic Link or Element
Q. 11
The kinematic link is _____________.
(a) Any part of the machine.
(b) Any resistant body which is capable of transmitting
the required motion and
forces with negligible
deformation.
(c) Only the fixed part of the machine.
(d) None of the above.
Ans. : (b)
Explanation : Each part of a machine which has relative motion
to some other part of the machine is known as kinematic link or
an element. A link is not necessary a rigid body but it is a
resistant body. A resistant body is one which is capable of
transmitting the required motion and forces with negligible
deformation.
Q. 12
Link or element is a _____________.
(a) Part of a machine
(b) Stationary part of a machine
(c) Part of a machine which has motion relative to some
other part(d) None of the above.
Ans. : (c)
Explanation : Refer Q.11
Q. 13 Kinematic link or element of mechanism is _____________.
(a) Any part of the machine
(b) Any stationary part of a machine
(c) Any resistant body or assembly of resistant bodies
which go to make part of a machine connecting other
parts which have a motion relative to it.
(d) None of the above.
Ans. : (c)
Explanation : Refer Q.11
Q. 14 In a reciprocating engine _____________.
(a) Piston, gudgeon pin form two kinematic links
(b) Piston, gudgeon pin form one kinematic link
(c) Piston, gudgeon pin and connecting rod form one
kinematic link
(d) None of the above statement is true.
Ans. : (b)
Explanation : A link may consist of number of parts
connected in such a way that they have no relative motion
between them.
Q. 15 In reciprocating engine mechanism _____________.
(a) Piston, gudgeon pin form one kinematic link
(b) Piston, gudgeon pin and connecting rod form one
kinematic link
(c) Piston, gudgeon pin and crank form one kinematic link
(d) None of the above
Ans. : (a)
Explanation : Refer Q.14
Q. 16 In a reciprocating engine _____________.
(a) Crank shaft and flywheel form two kinematic links
(b) Crank shaft and flywheel form one kinematic link
(c) Crank shaft and flywheel do not from kinematic link
(d) Flywheel and crank shaft separately form kinematic
links.Ans. : (b)
Explanation : Refer Q.14
Q. 17 In a reciprocating steam engine, which of the following
forms a kinematic link ?
(a) Crank shaft and flywheel
(b) Cylinder and piston
(c) Piston and connecting rod
(d) Flywheel and engine frame
Ans. : (a)
Explanation : Refer Q.14
Q. 18 In IC engine, which of the following forms a kinematic link
?
(a) cylinder and piston
(b) piston rod and connecting rod
(c) crank shaft and flywheel
(d) flywheel and engine frame
Ans. : (c)
Explanation : Refer Q .14
Q. 19 Which of the following would constitute a link ?
(a) piston, piston rings and gudgeon pin
(b) piston, and piston rod
(c) piston rod and cross head
(d) piston, piston-rod and cross head.
Ans. : (d)
Explanation : Refer Q.14
Q. 20 The function of an element is to _____________.
(a) transmit motion and forces
(b) to serve as a support
(c) to guide other elements
(d) all of these
Ans. : (d)
Explanation : Refer Q.11
Q. 21 The purpose of a link is to _____________.(a) transmit motion
(c) act as a support
Ans. : (d)
Explanation : Refer Q.11
(b) guide other links
(d) all of these
Q. 22 Piston from a single slider crank kinematic chain is a
_________ link.
(a) binary
(b)
ternary
(c) quaternary
(d)
fluid
Ans. : (a)
Explanation : The link which is attached at two different
points in a mechanism is called as binary link.
Q. 23 _______ link can transmit only compressive force.
(a) Rigid
(b)
Flexible
(c) Fluid
(d)
Both (a) and (b)
Ans. : (c)
Explanation : The links which transmit the motion by
fluid pressure as compression are called fluid links e.g.
liquid used in hydraulic press, lift, jacks and brakes of
an automobile etc.
Q. 24 Slotted disc from double slider kinematic chain is a
__________ link.
(a) Binary
(b) Ternary
(c) Quaternary
(d) Flexible
Ans. : (a)
Explanation : Refer Q 22
Q. 25 Track rod of ‘Davis steering gear mechanism’ is a ________
link.
(a) Binary
(b)
Ternary
(c) Quaternary
(d)
Fluid
Ans. : (a)
Explanation : Refer Q 22
Q. 26 Link of a mechanism should be _____________.
(a) rigid
(b)
resistant•
(c) separate part
(d)
all of above
Ans. : (d)
Explanation : A link is not necessary a rigid body but it is a
resistant body. A resistant body is one which is capable of
transmitting the required motion and forces with negligible
deformation.
Q. 27 The links of a structure transmit _____________.
(a) Forces only
(b) Motion only
(c) Forces and motion
(d) None of the above
Ans. : (a)
Explanation : Structure serves to modify and transmit the
forces only and it cannot transmit the energy for doing
mechanical work.
Q. 28 The links of machine may transmit _____________.
(a) Force only
(b) Motion only
(c) Forces and motion
(d) None of the above
Ans. : (c)
Explanation : links of machine serves to transfer both
forces and motion. Machine transforms the available
energy into mechanical work or some other form of energy.
Q. 29 Hydraulic press is _____________.
(a) Rigid link
(b) Flexible link
(c) Fluid link
(d) None of the above
Ans. : (c)
•
Explanation : The links which transmit the motion by
fluid pressure as compression are called fluid links e.g.
liquid used in hydraulic press, lift, jacks and brakes of
an automobile etc.
Types of Constrained MotionQ. 30 The motion of a bar with rectangular cross-section, sliding
in corresponding rectangular slot is an example of
_____________.
(a) successfully constrained motion.
(b) un-constrained motion
(c) incompletely constrained motion
(d) completely constrained motion
Ans. : (d)
Explanation : A completely constrained motion is one in
which the relative motion between the two links can be
place only in a definite direction and it can be predicted.
Q. 31 Piston and Cylinder pair from an IC engine is an example
of _________.
(a) completely constrained motion
(b) un-constrained motion
(c) normally constrained motion
(d) successfully constrained motion
•
Ans. : (d)
Explanation : When the relative motion between the
links is not completely constrained by itself but it is
made so by some other means is called the successfully
constrained motion.
Q. 32 The motion of a piston in the cylinder is an example of
_____________.
(a) completely constrained motion
(b) incompletely constrained motion
(c) successfully constrained motion
(d) none of the above
Ans. : (a)
Explanation : Refer Q.30
Q. 33 The motion of shaft in a circular hole is an example of
_____________.
(a) incompletely constrained motion
(b) completely constrained motion(c) successfully constrained motion
(d) none of the above
•
Q. 34
Ans. : (a)
Explanation : When the two links are connected in a
such a manner that their relative motion can take place
in more than one direction, then the motion is said to be
incompletely constrained motion. In such cases, the
motion of the link cannot be predicted.
A round bar passes through the cylindrical hole. Which one
of the following statements is correct in this regard ?
(a) The two links shown form a kinematic pair
(b) The pair is completely constrained
(c) The pair has incomplete constraint
(d) The pair is successfully constrained
Ans. : (c)
Explanation : Refer Q 33
Q. 35
Consider the following statements out of these statements
which is correct.
1. A round bar in a round hole form a turning pair
2. A square bar in a square hole forms a sliding pair
3. a vertical shaft in a footstep bearing forms a successful
constraint
(a) 1 and 2 are correct
(b)
2 and 3 are correct
(c) 1 and 3 are correct
(d)
1,2 and 3 are correct
Ans. : (d)
Explanation : Refer Q.30, Q.31 and Q.33Q. 36 A circular bar moving in a round hole is an example of
_____________.
(a) incompletely constrained motion
(b) partially constrained motion
(c) completely constrained motion
(d) successfully constrained motion
Ans. : (a)
Explanation : Refer Q 33
Q. 37 If some links are connected such that motion between them
can take place in more than one direction, it is called
_____________.
(a) incompletely constrained motion
(b) partially constrained motion
(c) completely constrained motion
(d) successfully constrained motion
Ans. : (a)
Explanation : Refer Q.33
A completely constrained motion can be transmitted with
_____________.
(a) 1 link with pin joints
(b) 2 links with pin joints
(c) 3 links with pin joints
(d) 4 links with pin joints
Ans. : (d)
Explanation : Refer Q.30
Q. 38
Q. 39 Rectangular bar in a rectangular hole is the following type
of motion.
(a) completely constrained motion
(b) partially constrained motion
(c) incompletely constrained motion
(d) freely constrained motion
Ans. : (a)
Explanation : Refer Q.30
Q. 40 A foot step bearing and rotor of a vertical turbine form
examples of _________.(a) incompletely constrained motion
(b) partially constrained motion
(c) completely constrained motion
(d) successfully constrained motion
Ans. : (d)
•
Explanation : The motion of the shaft in a foot step
bearing as shown in Fig. 1.5.4. The sliding motion of the
shaft is prevented in the upward direction due to load
and only the rotational motion is possible. Therefore, its
motion has been successfully constrained in one
direction only.
Q. 41
Q. 42
Q. 43
Which one of the following is an example of completely
constrained motion ?
(a) Rotor of a vertical turbine
(b) A footstep bearing
(c) A shaft with collars at each end rotating in a round
hole
(d) A circular bar moving in a round hole
Ans. : (c)
Explanation : Refer Q 30
The motion of a rotating shaft in a foot step bearing
constitutes between the elements of kinematic pair
_____________.
(a) Successfully constrained motion
(b) Completely constrained motion
(c) Incompletely constrained motion
(d) None of the above. It is not a kinematic pair.
Ans. : (a)
Explanation : Refer Q.40
The motion of a circular shaft with collars at each end
rotating in a round hole constitutes between the elements
of kinematic pair _____________.
(a) Successfully constrained motion
(b) Completely constrained motion
(c) Incompletely constrained motion
(d) None of the above. It is not kinematic pair
Ans. : (b)Q. 44
Q. 45
Explanation : Refer Q.30
The motion of a piston in the cylinder of a steam engine is
an example of ______.
(a) Completely constrained motion
(b) Incompletely constrained motion
(c) Successfully constrained motion
(d) None of the above
Ans. : (c)
Explanation : Refer Q.31
The motion of a circular shaft in a circular hole constitutes
between the elements of kinematic pair _____________.
(a) Successfully constrained motion
(b) Completely constrained motion
(c) Incompletely constrained motion
(d) None of the above.
Ans. : (c)
Explanation : Refer Q.33
Kinematic Pair
Q. 46 Kinematic pairs are those which have _____________.
(a) point or line contact between the two elements when in
motion
(b) surface contact between the two elements when in
motion
(c) elements of pairs not held together mechanically
(d) two elements that permit relative motion
Ans. : (d)
Explanation : When two kinematic links are connected in
such a way that their motion is either completely or
successfully constrained, these two links are said to form a
kinematic pair.
Q. 47 A kinematic pair is formed by _____________.
(a) connecting rigidly two elements
(b) connecting not more than two elements permitting
relative motion.(c) connecting two or more elements permitting relative
motion
(d) two or more elements without relative motion
Ans. : (c)
Explanation : Refer Q.46
The elements or links which are connected together in such
a way that their relative motion is completely constrained,
form a _____________.
(a) Kinematic pair
(b)
Machine
(c) Mechanism
(d)
Kinematic chain
Ans. : (a)
Explanation : Refer Q.46
Q. 48
Q. 49
Joint of two elements that permits relative motion which is
completely constrained or successfully constrained is called
_____________.
(a) Mechanism
(b)
Machine
(c) Structure
(d)
Kinematic pair
Ans. : (d)
Explanation : Refer Q.46
Spherical pair is a _________ pair.
(a) higher (b) lower (c) open (d) constrained
Q. 50
•
Q. 51
Screw pair has ________ number to degrees of freedom
(DOF).
(a) 1
(b) 2
(c) 3
(d) 4
•
Q. 52
Ans. : (a)
Explanation : When one links forms a spherical shape
and turns inside a fixed element, the resulting
kinematic pair is called a spherical pair.
Ans. : (a)
Explanation : When the nature of contact between the
pair of links is such that one of the links can turn about
the other link by screw threads they form a screw pair.
In this case the motion of the screw will have both
rotational and sliding motion.
Which of the following is a turning pair ?(a)
(b)
(c)
(d)
Piston and cylinder of a reciprocating steam engine
Shaft with collar at both ends fitted in a circular hole
Lead screw of a lathe with nut
Ball and socket joint
Ans. : (b)
•
Q. 53
Explanation : When the two links are so connected that
one link is constrained to turn or revolute relative to the
other link, they are said to form a turning pair.
Sometimes, they are also called as hinged pair.
The lead screw of a lathe with nut forms a _____________.
(a) Sliding pair
(b)
Rolling pair
(c) Helical pair
(d)
Turning pair
Ans. : (c)
Explanation : Refer Q.51
Q. 54
A screw pair with zero pitch can be called as _______.
(a) Cylindrical or Revolute pair
(b) Prismatic pair
(c) Spherical pair
(d) Sliding pair
Ans. : (a)
Explanation : Screw having zero pitch means no threads on
screw, it will act as a circular bar which will turn or
revolute relative to other link, therefore such pair is called
cylindrical pair or revolute pair or turning pair.
Q. 55
Q. 56
Q. 57
The type of constraint in a cylindrical pair is ________.
(a) Incomplete
(b)
Complete
(c) Successful
(d)
Both (b) and (c)
Ans. : (a)
Explanation : Refer Q.33
Two links that permits relative motion which is either
completely constrained or successfully constrained is called
_____________.
(a) Structure
(b)
Mechanism
(c) Machine
(d)
Kinematic pair.
Ans. : (d)
Explanation : Refer Q.46
Classification of kinematic pairs is based on _____________.(a) Nature of contact between the links.
(b) Nature of relative motion between the links.
(c) Nature of mechanical arrangement for constraint
between elements.
(d) All of the above.
Ans. : (d)
Explanation :
Q. 58 A ball and socket joint forms _____________.
(a) Rolling pair
(b)
Spherical pair
(c) Turning pair
(d)
None of these.
Ans. : (b)
Explanation : When one links forms a spherical shape
and turns inside a fixed element, the resulting kinematic
pair is called a spherical pair.
Q. 59 A bolt and nut forms _____________.
(a) Turning pair
(b) Rolling pair
(c) Screw pair
(d) Sliding pair
Ans. : (c)
Explanation : Refer Q 51Q. 60
In case of lower pair the contact between two links while
transmitting motion is ____________.
(a) Surface contact
(b) Line contact
(c) Point contact
(d) Surface and line contact
Ans. : (a)
•
Q. 61
Explanation : When the two links have surface contact
or area contact while transmitting the motion, they form
a lower pair. The relative motion between the links of
lower pairs is purely sliding or turning type.
Choose the higher pairs from _____________.
(a) Toothed gears in mesh
(b) Roller bearings
(c) Cam and follower
(d) All of the above.
Ans. : (d)
•
Explanation : When the two links are connected in such
a way that they have a line or point contact while
transmitting the motion, these links are said to form
higher pair.
Q. 62
A shaft revolving in a bearing forms a _____________.
(a) Higher pair
(b)
Sliding pair
(c) Lower pair
(d)
Turning pair.
Ans. : (c)
Explanation : Refer Q.60
Q. 63
The balls with its bearing forms a _____________.
(a) Turning pair
(b)
Screw pair
(c) Rolling pair
(d)
Spherical pair
Ans. : (c)
•
Explanation : When two links are connected in such a
way that one of the links rolls over the other link, they
are said to form rolling pair.
Q. 64
When the elements of the pair are kept is connected by the
action of external forces. The pair is said to be a
_____________.
(a) Higher pair
(b)
Lower pair
(c) Force closed pair
(d)
Self closed pairAns. : (c)
•
Explanation : When the two links of a pair are not held
together mechanically while transmitting relative
motion are called force closed pairs.
Q. 65 Higher pairs have got _________.
(a) surface contact
(b)
Larger height
(c) line contact
(d)
large weight
Ans. : (c)
Explanation : Refer Q.61
Q. 66 Which one of the following is a higher pair ?
(a) Rolling pair
(b)
Screw pair
(c) Sliding pair
(d)
Spherical pair
Ans. : (a)
Explanation : Refer Q.61
Roller bearings have got _____________.
(a) line contact
(b)
point contact
(c) surface contact
(d)
none of the above
Ans. : (b)
Explanation : Refer Q 63
Kinematic pairs are classified as _____________.
(a) sliding pair
(b)
rolling pair
(c) spherical pair
(d)
all of the above
Ans. : (d)
Explanation : Refer Q.46
Piston and cylinder of a reciprocating steam engine forms a
_____________.
(a) turning pair
(b)
rolling pair
(c) sliding pair
(d)
spherical pair
Ans. : (c)
Q. 67
Q. 68
Q. 69
•
Q. 70
Explanation : A sliding pair or prismatic sliding pair is
formed by two links in such a manner that one link is
constrained to have a sliding motion relative to other
link.
A ball and a socket joint form a _____________.Q. 71
Q. 72
Q. 73
Q. 74
Q. 75
Q. 76
(a) turning pair
(b)
rolling pair
(c) sliding pair
(d)
spherical pair.
Ans. : (d)
Explanation : Refer Q.50
Circular shaft fitted into a circular hole with collars at both
ends forms a ______.
(a) turning pair
(b)
rolling pair
(c) sliding pair
(d)
spherical pair.
Ans. : (a)
Explanation : Refer Q.52
When the two elements of a pair have a surface contact
when in motion and the surface of one element slides over
the surface of the other element, the pair formed is called a
_____________.
(a) higher pair
(b)
lower pair
(c) force-closed pair
(d)
none of the above
Ans. : (b)
Explanation : Refer Q 60
When the two elements of a pair have line or point contact
when in motion, the pair formed is called a _____________.
(a) higher pair
(b)
lower pair
(c) unclosed pair
(d)
none of the above
Ans. : (a)
Explanation : Refer Q.61
The lower pair is _____________.
(a) closed pair
(b)
unclosed pair
(c) surface contact pair
(d)
both (a) and (c)
Ans. : (d)
Explanation : ReferQ.60
The higher pair is a _____________.
(a) closed pair
(b) unclosed pair
(c) point contact pair
(d) both (b) and (c)
Ans. : (d)
Explanation : Refer Q.61
Which one of the following is a lower pair ?
(a) ball and roller bearing (b) automobile steering gear
(c) cam and follower
(d) belt and chain drives.Q. 77
Q. 78
Ans. : (b)
Explanation : Refer Q.60
Which of the following is a higher pair ?
(a) ball and roller bearing (b) belt and chain drives
(c) cam and follower
(d) all of the above
Ans. : (d)
Explanation : Refer Q.61
Choose the wrong statement.
(a) An element needs not be a rigid body, but it must be a
resistant body.
(b) Cam and follower is an example of lower pair.
(c) A sliding pair has a completely constrained motion
(d) The motion of a shaft in a circular hole is an example
of incompletely constrained motion
Ans. : (b)
Explanation : Refer Q.60
Q. 79
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 _____________.
(a) turning pair
(b)
rolling pair
(c) sliding pair
(d)
spherical pair
Ans. : (a)
Q. 80
Explanation : Refer Q.52
Which of the following is a lower pair ?
(a) ball and socket
(b)
piston and cylinder
(c) cam and follower
(d)
both (a) and (b)
Ans. : (d)
Explanation : Refer Q.60
Q. 81
The example of lower pair is _____________.
(a) shaft revolving in a bearing
(b) straight line motion mechanisms
(c) automobile steering gear
(d) all of the above
Ans. : (d)
Explanation : Refer Q.60
Q. 82
Pulley in a belt drive acts as _____________.(a) cylindrical pair
(c) rolling pair
(b)
(d)
turning pair
sliding pair
Ans. : (c)
Explanation : Refer Q.63
Q. 83
The example of rolling pair is _____________.
(a) bolt and nut
(b) lead screw of a lathe
(c) ball and socket joint
(d) ball bearing and roller bearing
Ans. : (d)
Explanation : Refer Q.63
Q. 84 A universal joint is an example of _____________.
(a) higher pair
(b)
lower pair
(c) rolling pair
(d)
sliding pair
Ans. : (b)
Explanation : Refer Q.60
Q. 85 The example of spherical pair is _____________.
(a) bolt and nut
(b) lead screw of a lathe
(c) ball and socket joint
(d) ball bearing and roller bearing
Ans. : (c)
Explanation : Refer Q.50
Q. 86 Cross head and guides form a _____________.
(a) higher pair
(b)
turning pair
(c) rolling pair
(d)
sliding pair
Ans. : (d)
Explanation : Refer Q 69
Q. 87 Kinematic pairs are those which have two elements that
_____________.
(a) have line contact
(b) have surface contact
(c) permit relative motion (d) are held together
Ans. : (c)
Explanation : Refer Q.46
Q. 88 The lower pair is a _____________.
(a) open pair
(b)
closed pair(c) point contact pair
Ans. : (b)
Explanation : Refer Q.64
(d)
does not exist
Q. 89 Automobile steering gear is an example of _____________.
(a) higher pair
(b)
sliding pair
(c) turning pair
(d)
lower pair.
Ans. : (d)
Explanation : Refer Q.60
Q. 90 In higher pair, the relative motion is _____________.
(a) purely turning
(b) purely sliding
(c) purely rotary
(d) combination of sliding and turning.
Ans. : (d)
Explanation : Refer Q.61
Q. 91 Which of the following has sliding motion ?
(a) crank
(b)
connecting rod
(c) crank pin
(d)
cross-head
Ans. : (d)
Explanation : Refer Q.69
Q. 92 Which of the following mechanism is obtained from lower
pair ?
(a) pantograph
(b)
valve and valve gears
(c) generated straight line motions
(d) all of the above.
Ans. : (d)
Explanation : Refer Q.60
Elements of pairs held together mechanically is known as
_____________.
(a) closed pair
(b)
open pair
(c) mechanical pair
(d)
rolling pair
Ans. : (a)
Explanation : Refer Q.64
Shaft revolving in a bearing is the following type of pair.
(a) lower pair
(b)
higher pair
(c) spherical pair
(d)
cylindrical pair
Q. 93
Q. 94Ans. : (a)
Explanation : Refer Q.60
A slider crank chain consists of following numbers of
turning and sliding pairs.
(a) 1, 3
(b) 2, 2
(c) 3, 1
(d) 4, 0
Ans. : (c)
Explanation : Refer Q69 and Q.52
Lower pairs are those which have _____________.
(a) point or line contact between the two elements when in
motion
(b) surface contact between the two elements when in
motion
(c) elements of pairs not held together mechanically
(d) two elements that permit relative motion
Ans. : (b)
Explanation : Refer Q.60
Q. 95
Q. 96
Q. 97
Higher pairs are those which have _____________.
(a) point or line contact between the two elements when in
motion
(b) surface contact between the two elements when in
motion
(c) elements of pairs not held together mechanically
(d) two elements that permit relative motion
Ans. : (a)
Explanation : Refer Q.61
Open pairs are those which have _____________.
(a) point or line contact between the two elements when in
motion
(b) surface contact between the two elements when in
motion
(c) elements of pairs not held together mechanically
(d) two elements that permit relative motion
Ans. : (c)
Q. 98
•
Explanation : When the two links of a pair are not held
together mechanically while transmitting relative
motion are called force closed pairs.Q. 99 The main disadvantage of the sliding pair is that it is
_____________.
(a) bulky
(b) wears rapidly
(c) difficult to manufacture
(d) both (a) and (b)
Ans. : (d)
Explanation : Refer Q.69
Q. 100 A kinematic pair is a joint of _____________.
(a) two links which are fixed
(b) two links having same velocity
(c) two links having relative motion between them
(d) none of above
Ans. : (c)
Explanation : Refer Q46
In a pen stand mechanism a ball and a socket joint forms a
_____________.
(a) turning pair
(b)
rolling pair
(c) sliding pair
(d)
spherical pair
Ans. : (d)
Explanation : Refer Q.50
Balls in a bearing with casing forms a _____________.
(a) turning pair
(b)
rolling pair
(c) sliding pair
(d)
spherical pair
Ans. : (b)
Explanation : Refer Q 63
When the two elements of a pair have a surface contact
when in motion and the surface of an element slides over
the surface of the other elements, the pair formed is called
a _____________.
(a) higher pair
(b) lower pair
(c) forced-closed pair
(d) none of the above
Ans. : (b)
Explanation : Refer Q.60
Kinematic pairs are those which have _____________.
Q. 101
Q. 102
Q. 103
Q. 104
(a) Two elements that permit relative motion
(b) Paired elements held together mechanically
(c) Pairs of element having line contact or surface contact.Q. 105
Q. 106
Q. 107
Q. 108
(d) all of above
Ans. : (d)
Explanation : Refer Q.46
When the two elements have point or line contact motion,
the pair so formed is known as _____________.
(a) Higher pair
(b) Lower pair
(c) Screw pair
(d) Closed pair
Ans. : (a)
Explanation : Refer Q.61
When the elements of the pair are kept is contact by the
action of external forces, the pair is said to be a
_____________.
(a) lower pair
(b)
higher pair
(c) self closed pair
(d)
force closed pair
Ans. : (d)
Explanation : Refer Q.64
Which of the following is a turning pair ?
(a) Piston and cylinder of a reciprocating steam engine
(b) Shaft with collars at both ends fitted in a circular hole
(c) Lead screw of a lathe with nut
(d) Ball and socket joint
Ans. : (b)
Explanation : Refer Q 52
In a kinematic pair, when the elements have surface
contact while in motion, it is a _____________.
(a) higher pair
(b)
closed pair
(c) lower pair
(d)
unclosed pair
Ans. : (c)
Explanation : Refer Q.60
Explanation : Refer section 1.6
Q. 109
Classification of kinematic
between elements may give
(a) Sliding pair
(b)
(c) Rolling pair
(d)
Ans. : (d)
Explanation : Refer Q 57
pairs based on relative motion
_____________.
Turning pair
All of the above.Q. 110
Q. 111
Q. 112
Q. 113
Various kinematic pairs are given below. Choose the lower
pair.
(a) Ball bearings
(b) Tooth gears in mesh
(c) Cam and follower (d) Crank shaft and bearing.
Ans. : (d)
Explanation : Refer Q.60
Various kinematic pairs are given. Choose the higher pair.
(a) Roller bearing
(b)
Tooth gears in mesh
(c) Cam and follower
(d)
All of the above
Ans. : (d)
Explanation : Refer Q.61
Choose the correct statement.
(a) Tooth gears in mesh constitute a higher kinematic pair
(b) Belt on pulley drive constitute a higher kinematic pair
(c) Chain and sprocket drive constitute a higher kinematic
pair
(d) All of the above.
Ans. : (d)
Explanation : Refer Q.61
In a four bar chain or quadric cycle chain _____________.
(a) each of the four pairs is a turning pair
(b) one is a turning pair and three are sliding pairs
(c) three are turning pairs and one is sliding pair
(d) each of the four pairs is a sliding pair.
Ans. : (a)
Explanation : Refer Q.52
Kinematic Chain
Q. 114
A kinematic chain is _____________.
(a) a combination of kinetic pair in which one link is fixed
(b) a combination of kinetic pair in which each link forms
put of two pairs and the relative motion between the
links is completely constrained
(c) the same as mechanism(d) also called inversion.
Ans. : (b)
Explanation :
Kinematic chain is defined as the
combination of kinematic pairs in which each link forms a
part of two kinematic pairs and the relative motion
between the links is either completely constrained or
successfully constrained.
Q. 115
A Kinematic chain _____________.
(a) Comprises a chain of links in space with constrained
motion
(b) Comprises a chain of links with at least one link fixed
and completely constrained motion.
(c) Comprises chain of links with at least one link fixed
and successfully constrained motion
(d) Comprises chain of links with incompletely constrained
motion.
Ans. : (a)
Explanation : Refer Q114
Q. 116
A combination of kinematic pairs, joined in such a way that
the relative motion between the links is completely
constrained, is called a _____________.
(a) Structure
(b)
Mechanism
(c) Kinematic chain
(d)
Inversion
Ans. : (c)
Q. 117
Explanation : Refer Q114
In a kinematic chain having constrained motion, the
minimum number of kinematic pairs required are
_____________.
(a) Two (b) Three (c) Four
(d) Five
Ans. : (c)
Explanation : A kinematic chain with four kinematic pairs
is called a Simple kinematic chain and a kinematic chain
with more than four links is called a compound kinematic
chainQ. 118
The relation between the
number of joints (j) to form
equation _____________.
2
(a) j = (n + 2)
3
n
(c) j = (n + 2) –
2
number of links (n) and the
kinematic chain is given by the
1
(n + 2) + 2
2
3
(d) j = n – 2
2
(b) j =
Ans. : (d)
•
Explanation :
chain are :
The required conditions to form a kinematic
n = 2 p − 4
and,
where,
j +
j = 3
n − 2
2
H
2 = 3
n − 2
2
n =Number of links, p = Number of lower pairs
j = Number of joints ,H=Number of higher pairs
Q. 119
Q. 120
Q. 121
Relationship between the number of links (n) and number
of pairs (P) is _______.
(a) P = 2n – 4
(b)
P = n/2 + 4
n
(c) P = + 2
(d)
P = 2n – 2
2
Ans. : (c)
Explanation : Refer Q118
A kinematic chain requires minimum of _____________.
(a) 3 links an 3 turning pairs
(b) 4 links and 5 turning pairs
(c) 4 links and 4 turning pairs
(d) 4 links and 3 sliding pairs
Ans. : (c)
Explanation : Refer Q 117
The equation between the number of links (n) and number
2
of joints (j) with lower pairs is given as, n =
3
(j + 2). If this relation is applied to chains constituted by
higher pairs then _____________.Q. 122
Q. 123
Q. 124
(a) Each higher pair must be taken equivalent to two
lower pairs and one additional link.
(b) Each higher pair must be then equivalent to two lower
pairs.
(c) Each higher pair must be take as one lower pair and
one additional link.
(d) Each higher pair must be take as one lower pair and
two additional links.
Ans. : (a)
Explanation : If above Equations is to be used to a
kinematic chain using higher pairs, then each higher pair
may be taken as equivalent to two lower pairs and one
additional element or link
A kinematic chain with 7 joint needs the following number
of links.
(a) 4
(b) 6
(c) 9
(d) 12
Ans. : (b)
Explanation:
Given:
j = 7 , n = ?
We know that,
3
3
j = n – 2 ;
7 =
n – 2
2
2
3
9 = n ;
 n = 6
...Ans.
2
If J = Number of binary joints in a chain, H = Number or
unclosed pairs, and n = Number of links, then to determine
whether a chain is a locked, constrained or unconstrained,
the relation used is given by _____________.
3
(a) J + H = [n – 2]
(b)
J + H = 3n – 2
2
H 3
J
3
(c) J + = n – 2
(d)
+ H = n + 2
2 2
2
2
Ans. : (c)
Explanation : Refer Q118
If the right-hand side of the equation (which is used to
determine whether a chain is locked, constrained or
unconstrained) is greater than the left-hand side, then the
chain is _____________.
(a) locked
(b)
constrained(c) unconstrained
Ans. : (c)
(d)
none of the above.
Explanation : In order to determine whether a chain is a
structure or a kinematic chain or unconstrained chain, A W
Klein specified a rule called criteria of constraint which is as
follows :
If
L.H.S > R.H.S, the chain is locked or structure.
L.H.S = R.H.S, the chain is constrained.
L.H.S < R.H.S, the chain is unconstrained.
Q. 125
Q. 126
Q. 127
If R.H.S. = L.H..S of the equation (which is used to
determine whether a chain is locked, constrained or
unconstrained) then the chain is _____________.
(a) locked
(b) constrained
(c) unconstrained
(d) none of the above.
Ans. : (b)
Explanation : Refer Q.124
If R.H.S. is less than L.H.S. of the equation (used for
determining whether a chain is locked, constrained or
unconstrained), then the chain is _____________.
(a) locked
(b) constrained
(c) unconstrained
(d) none of the above
Ans. : (a)
Explanation : Refer Q124
The relation between the number of links ( l ) and the
number of binary joints (j) for a kinematic chain having
3
constrained motion is given by j = l – 2. If the left hand
2
side of this equation is greater than right hand side, then
the chain is _____________.
(a) locked chain
(b) completely constrained chain
(c) successfully constrained chain
(d) incompletely constrained chain
Ans. : (a)
Explanation : Refer Q124Q. 128
The equation for criterion of constraint for kinematic chain
having plane motions is given by 2J + H = 3n – 4. If L. H.
S. of equation is greater than R. H. S. then _____________.
(a) Chain is locked
(b) Chain is completed constrained
(c) Chain is successfully constrained
(d) Chain is incompletely constrained.
Ans. : (a)
Explanation : Refer A124
Q. 129
The equation for criterion of constraint for kinematic chain
having plane motion
is given by 2J + H = 3n – 4. If L. H. S. of equation is less
than R.H.S. then ____________.
(a) Chain is locked
(b) Chain is completed constrained
(c) Chain is successfully constrained
(d) Chain is incompletely constrained.
Ans. : (d)
Explanation : Refer Q124
Q. 130
The equation for criterion of constraint for kinematic chain
having plane motion
is given by 2J + H = 3n – 4. If L. H. S. of equation is equal
to R. H.S. then _____________.
(a) Chain is locked
(b) Chain is completed constrained
(c) Chain is successfully constrained
(d) Chain is incompletely constrained.
Ans. : (b)
Explanation : Refer Q124Q. 131
Q. 132
Ternary joint is a joint at which, three links are joined at
the same connection. It is equivalent to _____________.
(a) one binary joint
(b) two binary joints
(c) three binary joints
(d) combination of (a) and (b).
Ans. : (b)
Explanation : One ternary joint is equivalent to two binary
joints
Quaternary joint is a joint at which four links are joined at
the same connection and it is equivalent to _____________.
(a) one binary joint
(b) two binary joints
(c) three binary joints
(d) combination of (a) and (b)
Ans. : (c)
Explanation : One quaternary joint is equivalent to three
binary joint.
Q. 133
Q. 134
Q. 135
___________ number of pairs are associated with one
ternary joint.
(a) 1
(b) 2
(c) 3
(d) 4
Ans. : (b)
Explanation : Refer Q 131
One ternary joint is equivalent to _____________ binary
joints.
(a) 1
(b) 2
(c) 3
(d) 4
Ans. : (b)
Explanation : Refer Q131
A quaternary joint is equivalent to ___________ joint.
(a) two
(b) three
(c) four
(d) one
Ans. : (b)
Explanation : Refer Q132
Q. 136
_____________ number of
quaternary joint.
(a) 1
(b) 2
(c) 3
Ans. : (c)
Explanation : Refer Q132
pairs are
(d) 4
associated
withQ. 137
Q. 138
Q. 139
If 'n' links are connected at the same joint, the joint is
equivalent to _________.
(a) (n – 1) binary joints
(b) (2n – 1) binary joints
(c) (n – 2) binary joints
(d) (n – 3) binary joints.
Ans. : (a)
Explanation : Refer Q131
In a kinematic chain if ‘n’ links are connected at the same
joint, the joint is equivalent to _____________.
(a) (n – 1)binary joints
(b) (n – 2)binary joints
(c) (n – 3)binary joints
(d) (2n – 1) binary joints
Ans. : (a)
Explanation : Refer Q131
In a translating roller follower number of links n, number
of lower pairs l and number of higher pairs h are given by
_____________.
(a) n = 4; l = 2; h = 1
(b)
n = 3; l = 3; h = 1
(c) n = 3; l = 4 and
(d)
n = 4; l = 3; h = 2
Ans. : (a)
Explanation : Refer Fig Q139
Number of Links: Three (1,2 and3), NumberLower pair:
Two(Pair1-2 and Pair 1-3), Number of Higher Pair:
One(Pair2-3)
Fig Q139
Q. 140
Number of revolute pairs for constrained motion of a simple
six bar mechanism is _____________.
(a) 5
(b) 4
(c) 7
(d) 6
Ans. : (c)
Explanation : Refer section 1.8Q. 141
Fig. 1 is an example of a _____________.
(a) locked chain
(b) unconstrained chain
(c) constrained chain
(d) none of the above
Ans. : (a)
Explanation :
Fig. Q141
number of links, n = 3 ; number of pairs, p = 3
using Equation
n = 2p − 4 ;

3 = 2  3 − 4
 3  2

LHS > RHS
Therefore, three bar chain is called as locked chain or
structure.
Q. 142
Fig 2 is an example of a _____________.
(a) locked chain
(b) unconstrained chain
(c) constrained chain
(d) none of the above.
Ans. : (c)
Explanation :
Fig. Q142
number of links, n = 4 ;
number of pairs, p = 4
using Equation
n = 2p − 4 ;

4 = 2  4 − 4 ;
 4 = 4 ;
i.e.
LHS = RHS
Therefore, four bar chain is called as kinematic chain.Q. 143
Fig. 3 is an example of _____________.
(a) locked chain
(b) unconstrained chain
(c) constrained chain
(d) none of the above.
Ans. : (b)
Explanation :
Fig. 3
Number of links,
n = 5 ;
Number of pairs, p = 5
using Equation
n = 2p − 4

i.e.
5 = 2  5 − 4

5  6
LHS < RHS
Therefore, five bar chain is called chain having unconstrained
motion.
Q. 144 For a kinematic chain, the relationship between number of
pairs (P) and number of links (L) is _____________.
(a) L = P – 2
(b)
L = 2P– 4
(c) L = 4P – 2
(d)
L = P – 4.
Ans. : (b)
Explanation : Refer Q.118
Q. 145 For a kinematic-chain, the relationship between number of
joints (j) and number of links (L) is _____________.
2
2
(a) L = (J + 2)
(b)
L = (J – 2)
3
3
3
3
(c) L = (J + 2)
(d)
L = (J – 2)
2
2
Ans. : (a)
Explanation : Refer Q.118
Q. 146
Choose the correct statement.(a) A chain consisting of three links and three joints is
known as locked chain.
(b) A chain consisting of four links and four joints is
known as kinematic chain
(c) Quaternary joint is equivalent to three binary joints
(d) All of the above
Ans. : (d)
Explanation : Refer Q 132
Q. 147
Choose the wrong statement.
(a) A chain consisting of three links and three joints is
known as locked chain
(b) A chain consisting of four links with four joints is
known as kinematic chain
(c) Quaternary joint is equivalent to three binary joints
(d) Rectangular bar in a rectangular hole is the example of
incompletely constrained motion
Ans. : (d)
Explanation : Rectangular bar in a rectangular hole is the
example completely constrained motion.
Q. 148
Q. 149
Q. 150
A chain comprises of 5 links having 5 joints. Is it kinematic
chain ?
(a) yes
(b)
no
(c) it is a marginal case
(d)
unpredictable.
Ans. : (b)
Explanation : Refer Q 143
For a kinematic chain to be considered as mechanism
_____________.
(a) two links should be fixed
(b) one link should be fixed
(c) none of the links should be fixed
(d) there is no such criterion
Ans. : (b)
Explanation : Refer Q.114
A kinematic chain is known as a mechanism when
_____________.
(a) none of the links are fixed
(b) one of the link is fixed(c) two of the links are fixed
(d) all of the links are fixed
Q. 151
•
•
Ans. : (b)
Explanation : Refer Q.114
A 6-bar chain can be formulated to give constrained motion
by using _________.
(a) 5 turning pairs
(b) 6 turning pairs
(c) 7 turning pairs
(d) 8 turning pairs.
Ans. : (c)
Explanation : Refer section 1.7 and Fig. 4.
In Fig. 1.8.2, the joint B
and C are called as
ternary joint and joint
A, D and E is called
binary joint.
One ternary joint is
equivalent
to
two
binary joints, therefore
equivalent binary joint
for given Fig. 1.8.2 is,
Fig. Q 151
Number of binary joints = 2 (Number of ternary joints) + Number of
binary joints
= 2 (2) + 3 = 7
Q. 152
In a 6-bar kinematic chain for constrained motion there
will be _____________.
(a) 6 binary links
(b) 4 binary and 2 ternary links
(c) 5 binary and 2 ternary links
(d) 3 binary and 2 ternary links
Ans. : (c)
Explanation : Refer section 1.8 and Fig. 4.
Q. 153
A 6-bar kinematic chain can be formed to give constrained
motion by using _____________.
(a) 5 turning pairs
(b)
6 turning pairs(c) 7 turning pairs
(d)
8 turning pairs
Ans. : (c)
Explanation
Fig 4
Q. 154
A kinematic chain requires at least _____________.
(a) 2 links and 3 turning pairs
(b) 3 links and 4 turning pairs
(c) 4 links and 4 turning pairs
(d) 5 links and 4 turning pairs
Ans. : (c)
Explanation :
number of links, n = 4 ;
number of pairs, p = 4
using Equation
n = 2p − 4 ;

4 = 2  4 − 4 ;
 4 = 4 ;
i.e.
LHS = RHS
Therefore, four bar chain is called as kinematic chain.
Q. 155
For a kinematic chain which relation is true.
(a) l = 2p – 4
(b)
l = 2p + 4
(c) l = p – 4
(d)
l = p + 4
Ans. : (a)
Explanation : Refer Q.118Q. 156
The relation between the number of pairs (p) forming a
kinematic chain and the number of links ( l ) is
_____________.
(a) l = 2p – 2
(b)
l = 2p – 3
(c) l = 2p – 4
(d)
l = 2p – 5
Ans. : (c)
Explanation : Refer Q.118
Mechanism & Inversion
Q. 157
Inversion of a mechanism is _____________.
(a) changing of a higher pair to lower pair
(b) obtained by fixing different links in a kinematic chain
(c) turning it upside down
(d) obtained by reversing the input and output motion
Ans. : (b)
Explanation : A “mechanism” is defined as a kinematic chain with
one of its link fixed. Usually, the fixed link of the mechanism is
called the frame.
When the different links of a kinematic chain are used as a frame,
the relative motion of various links are not altered, however, the
absolute value of motion of such mechanism is drastically altered.
The process of choosing different links of a chain for the frame is
called “kinematic inversion”. Therefore, if a mechanism has ‘n’
number of links, then we can obtain ‘n’ number of mechanism by
fixing each link as a frame.
Q. 158
Q. 159
Inversion of a mechanism is _____________.
(a) its mirror image
(b) obtained by fixing other link
(c) obtained by reducing size of links
(d) obtained by increasing size of links.
Ans. : (b)
Explanation : Refer Q.157
Inversions are _____________.Q. 160
(a) Different mechanisms obtained by fixing different links
in a kinematic chain with the object of changing
relative motions of links with respect to one another.
(b) Different mechanisms obtained by fixing different links
in a kinematic chain but keeping relative motions of
links unchanged with respect to one another.
(c) Different mechanisms obtained by fixing different links
in kinematic chain to modify the mechanical
advantage.
(d) None of the above.
Ans. : (b)
Explanation : Refer Q.157
Choose the correct statement.
(a) Mechanism transmits and modifies motion
(b) Mechanism is the skeleton outline of machine to
produce definite motion between various links
(c) Machine modifies mechanical work
(d) All of the above
Ans. : (d)
Explanation :
S.N Mechanism Machine
1. A mechanism is a kinematic chain
with one link fixed used to transmit
and transform the motion. A machine is an assemblage of
mechanism use to transmit both
motion and forces.
2. A mechanism is a skeleton diagram
of a machine to transmit the
definite motions between various
links. A machine may have number of
mechanism which transmits the
available energy into used work.
Examples are : type writer, clock
work etc.
Q. 161
Examples are : I.C. engine, shaper
machine etc.
In a kinematic chain with four lower pairs, if all the four
lower pairs are turning pairs, the mechanism is classified
into the chain known as _____________.
(a) crossed slider crank chain
(b) four bar chain
(c) slider crank chain(d) double slider crank chain
Ans. : (b)
Explanation : : It consists of four turning or revolute pairs
(4R) as shown in Fig. Q161
Q. 162
In a kinematic chain with four lower pairs, if one is sliding
pair and three turning pairs, the mechanism is classified
into the chain known is ___________.
(a) crossed slider crank chain
(b) four bar chain
(c) slider crank chain
(d) double slider crank chain
Ans. : (c)
Explanation : It consists of three turning or revolute pairs
and one sliding or prismatic pair (3R – 1P) as shown in Fig.
Q162
FigQ162Q. 163
In a kinematic chain with four lower pairs, if there are two
sliding pairs and two turning pairs and two similar pairs
are adjacent then the mechanism is classified into the
chain known as _____________.
(a) crossed slider crank chain
(b) four bar chain
(c) slider crank chain
(d) double slider crank chain
Ans. : (d)
Explanation : : It consists of two turning or revolute pairs
and two sliding or prismatic pairs (2R – 2P) as shown in
Fig. Q163
Fig. Q163
Q. 164
In a 2R-2P kinematic chain, we must ground ________ to
get Scotch-Yoke mechanism.
(a) slotted disc (b) link connecting the two sliders
(c) slider
(d) None of the above
Ans. : (c)
Explanation :
•
This mechanism is obtained by fixing link 1 of double
slider crank chain.
•
This mechanism is used to convert rotary motion into
reciprocating motion. In this mechanism link 1 is fixed
as shown in Fig. 1.11.23(b).
•
The driving crank 2 rotates about point ‘O’. The slider 3
which is attached to driving crank will also reciprocates
in guide block in the form of link 4 which caused to
reciprocate link 4.Q. 165
Fig. Q.164 : First inversion of double slider crank chain
In a single slider chain, if crank is grounded, we get
___________.
(a) oscillating cylinder engine
(b) IC engine mechanism
(c) door closer mechanism
(d) hand pump
Ans. : (c)
Explanation : This is an example of second inversion of
single slider crank chain mechanism. In this mechanism
the link-2 is fixed, link-3 is driving crank, link-4 is
piston and link-1 is the cylinder.Q. 166
Q. 167
•
_______ mechanism can be obtained from a double slider
simple kinematic chain.
(a) Two (b) Three (c) Four
(d) Five
Ans. : (b)
Explanation : The double slider kinematic chain is a four
bar kinematic chain consisting of two turning and two
sliding pairs. Different inversion of such kinematic chain
are Scotch Yoke mechanism, Oldham’s Coupling and
Elliptical trammel :
Equivalent Linkage of a ‘Cam Follower Pair’ is a
_____________.
(a) Crank-Rocker mechanism
(b) Double slider mechanism
(c) Four bar mechanism
(d) Slider crank mechanism
Ans. : (c)
Explanation :
The example of higher pair replaced by two turning pairs is cam and
follower which is shown in Fig. Q167
Fig. Q167
Q. 168
Equivalent Linkage of a radial cam driving reciprocating
follower is a _______.
(a) Crank-Rocker mechanism(b) Double slider mechanism
(c) Four bar mechanism
(d) Slider crank mechanism
Ans. : (d)
Explanation : In radial cam driving reciprocating follower
mechanism, radial cam is acting like crank and
reciprocating follower is acting like piston. Therefore
Equivalent Linkage of a radial cam driving reciprocating
follower is a Slider crank mechanism.
Q. 169
Fig Q 159
Equivalent linkage of a spur gear pair is a __________.
(a) crank – rocker mechanism
(b) double slider mechanism
(c) four bar mechanism
(d) double crank mechanism
Ans. : (c)
Explanation :Fig. P. 1.8.4
Soln. : Since the gears will have point contact, therefore, they form a
higher pair. It is equivalent to two lower pairs and one additional
link. It implies that mechanism is equivalent to four links and four
lower pairs. Therefore, the mechanism represents a kinematic link
since it satisfies the equation of
n = 2 p − 4

3 + 1 = 2  4 − 4

4 = 2  4 − 4

4 = 4
i.e.
L.H.S
=
R.H.S, Hence it is constrained
chain or four bar mechanism
Q. 170 Choose the correct statement.
(a) A sliding pair has incompletely constrained motion
(b) A pair of friction discs constitutes a lower pair
(c) Rectilinear motion of a piston is converted into rotary
motion by slider crank mechanism
(d) Automobile steering gear is an example of higher pair
Ans. : (c)
Explanation : Refer Q.162
Q. 171 The mechanism used in a petrol engine is _____________.
(a) Crank mechanism
(b) double slider mechanism
(c) Single slider crank mechanism
(d) None of the above.
Ans. : (c)
Q. 172
Explanation : Refer Q.162
Which of the following are the inversion of single slider
kinematic chain ?
(a) Reciprocating engine
(b) Quick return motion of slotted lever type,
(c) Whitworth mechanism
(d) All of the above.Ans. : (d)
Explanation :
Q. 173 The Watt’s engine Indicator is based on the following
kinematic chain.
(a) Four bar
(b)
Single slider
(c) Double slider
(d)
None of these.
Ans. : (a)
Explanation :
• Fig. Q173 shows Watt’s engine indicator mechanism
which is based on four bar chain mechanism.Fig. Q173 : Watt’s engine indicator
•
Link 1 is fixed, link 2 is ABC and link 3 is CDP, both
acts as levers. Link BE and Link ED form link 4 since
these two links do not have relative motion between
them.
Q. 174
•
Kinematic chain is called mechanism when _____________.
(a) One of the links is fixed
(b) Two of the links are fixed
(b) None of the links is fixed
(d) All of the links are fixed
Ans. : (a)
Explanation : A “mechanism” is defined as a kinematic chain
with one of its link fixed. Usually, the fixed link of the
mechanism is called the frame.
Q. 175
Which of the following is an inversion of double slider
crank chain ?
(a) Beam engine
(b)
Watt's indicator
(c) Oscillating cylinder
(d)
Elliptical trammels
Ans. : (d)
Explanation :Q. 176
The kinematic chain having N links will have
_____________.
(a) (n – 1) inversions
(b) (n – 2) inversions
(c) (n – 3) inversions
(d) N inversions
Ans. : (d)
Explanation : The process of choosing different links of a
chain for the frame is called “kinematic inversion”.
Therefore, if a mechanism has ‘n’ number of links, then we
can obtain ‘n’ number of mechanism by fixing each link as a
frame.Q. 177
If a kinematic chain having constrained motion has ‘n’
links then number of mechanism obtained are
_____________.
(a) (n – 1)
(b) (n – 2)
(c) (n + 1)
(d) n
Ans. : (d)
Explanation : Refer Q.176
Q. 178
For a kinematic chain of n-links, total number of possible
inversions are _____.
(a) n + 1 (b) n (n – 1)/2
(c) n
Ans. : (c)
Explanation : Refer Q.176
Q. 179
(d) n 2
If there are ‘n’ number of links in a mechanism then
number of possible inversions is equal to _____________.
(a) n + 1 (b) n – 1 (c) n
(d) n + 2
Ans. : (c)
Explanation : Refer Q176
Q. 180 Oldham's coupling is the inversion of _____________.
(a) Single slider crank chain
(b) Double slider crank chain
(c) Four bar chain
(d)
None of the above
Ans. : (b)
Explanation : Refer Q 175
Q. 181 The mechanism used in a shaping machine is
_____________.
(a) A closed 4 bar chain having 4 revolute pairs.
(b) A closed 6 bar chain having 6 revolute pairs.
(c) A closed 4 bar chain having 2 revolute and 2 sliding
pairs
(d) An inversion of the single slider crank chain
Ans. : (d)
Explanation : Refer Q172
For the planer mechanism
shown in Fig. 8 select the
most appropriate choice for
Q. 182the motion of link 2 when
link 4 is moved upward
__________.
(a) Link 2 rotates clockwise
(b) Link 2 rotates counter-clockwise
(c) Link 2 does not move
(d) Link 2 motion cannot be determined
Ans. : (b)
Explanation: If link 4 Move upward then link 2 will rotate
in counterclockwise direction about pivot 1 through link 3.
Q. 183 Any point on a link connecting double slider crank chain
will trace a _________.
(a) straight line
(b)
circle
(c) ellipse
(d)
parabola
Ans. : (c)
Explanation : Any point on the extension of link-2 of
elliptical trammel will trace the path of an ellipse.
Q. 184 A fourth inversion of double slider crank chain traces a
__________.
(a) circle
(b)
parabola
(c) hyperbola
(d)
ellipse.
Ans. : (d)
Explanation : Refer Q183
Choose the correct statement.
(a) If a kinematic chain has ‘ l ’ links, then ‘ l ’ different
mechanism are obtained by fixing each of the links in
turn
(b) A machine serves to modify and transmit energy (or
force and motion) while the structure modifies and
transmits force only
(c) When one of the links of a kinematic chain is fixed, the
chain is known as a mechanism
(d) All of the above
Q. 185
Ans. : (d)
Explanation : Refer Q160Q. 186
Q. 187
Q. 188
Oldham’s coupling and elliptic trammels are the inversion
of _____________.
(a) double slider crank chain
(b) single slider crank chain
(c) four bar chain
(d) crossed slider crank chain
Ans. : (a)
Explanation : Refer Q175
Choose the correct statement.
(a) The ordinary reciprocating steam engine is an
inversion of double slider crank chain.
(b) Scotch Yoke mechanism is an inversion of slider crank
chain.
(c) Quick return mechanism is an inversion of slider crank
chain.
(d) The number of inversions in a mechanism having
number of links equal to l , will be ( l + 1).
Ans. : (c)
Explanation : Refer Q172
Quick return mechanism is an inversion of _____________.
(a) four bar chain
(b) single slider crank chain
(c) double slider crank chain
(d) crossed slider crank chain
Ans. : (b)
Explanation : Refer Q172
Q. 189
Which of the following are inversions of a double slider
crank chain ?
(a) Whitworth return motion
(b) Scotch yoke
(c) Oldham’s coupling
(d) Rotary engine
Ans. : (c)
Explanation : Refer Q175Q. 190
Q. 191
•
The number of inversions for a slider crank mechanism is
_____________.
(a) 6
(b) 5
(c) 4
(d) 3
Ans. : (c)
Explanation : Refer Q172
Which of the following statements is INCORRECT ?
(a) Grashof’s rule states that for a planar crank-rocker
four bar mechanism, the sum of the shortest and
longest link lengths cannot be greater than the sum of
the remaining two link lengths
(b) Inversions of a mechanism are created by fixing two
different links one at a time.
(c) Geneva mechanism is an intermittent motion device
(d) Gruebler’s criterion assumes mobility of a planar
mechanism to be one
Ans. : (b)
Explanation : The process of choosing different links of a
chain for the frame is called “kinematic inversion”. Therefore,
if a mechanism has ‘n’ number of links, then we can obtain ‘n’
number of mechanism by fixing each link as a frame.
Q. 192
Oldham's coupling is the _____________.
(a) second inversion of double slider crank chain
(b) third inversion of double slider crank chain
(c) second inversion of single slider crank chain
(d) third inversion of slider crank chain
Ans. : (a)
Explanation : Refer Q175
Q. 193
•
A mechanism is an assemblage of _____________.
(a) two links
(b)
three links
(c) four links or more than four links
(d) all of the above
Ans. : (c)
Explanation : A “mechanism” is a kinematic chain with any
one link fixed which is used to transmit the required motion.•
A mechanism with four links is called a simple mechanism
and the mechanism with more than four links is called a
compound mechanism.
Q. 194
•
Pantograph is a _____________.
(a) graph of fine functions
(b) Mechanism for reproducing the path to enlarged or
reduced scale
(c) Mechanism with high pairs
(d) Mechanism for generating straight line
Ans. : (b)
Explanation : A Pantograph is a mechanism used to produce paths on
an enlarged or reduced scale as exactly as possible the path described by
a given point. It is based on four bar kinematic chain.
Q. 195
The number of links in pantograph mechanism is equal to
_____________.
(a) 2
(b) 3
(c) 4
(d) 5
Ans. : (c)
Explanation :
•
•
A Pantograph is a mechanism used to produce paths on an enlarged or
reduced scale as exactly as possible the path described by a given point.
It is based on four bar kinematic chain.
It consists of four links joined in such a way that it forms a
parallelogram
ABCD
as
shown
in
Fig. Q195
Fig. Q195 : PantographQ. 196
•
Q. 197
It is required to connect two parallel shafts, the distance
between whose axes is small and variable. The shafts are
coupled by _____________.
(a) universal joint
(b)
knuckle joint
(c) oldham's coupling
(d)
flexible coupling
Ans. : (c)
Explanation : This oldham's coupling is used to connect
two parallel shafts when they are not co-axial and the
distance between their centre lines is small.
In elliptical trammels _____________.
(a) all four pairs are turning
(b) three pairs turning and one pair sliding
(c) two pairs turning and two pairs sliding
(d) one pair turning and three pairs sliding
Ans. : (c)
•
Q. 198
Explanation : By fixing cylinder (link 4) of double slider
crank chain the fourth inversion is obtained called
elliptical trammel. In this mechanism two pairs turning
and two pairs sliding.
Typewriter constitutes _____________.
(a) machine
(b)
(c) unconstrained mechanism
(d)
Ans. : (a)
structure
inversion
Explanation : A machine is an assemblage of mechanism use to
transmit both motion and forces. A machine may have number of
mechanism which transmits the available energy into used work.
Therefore typewriter is called as machine.
Q. 199 If the opposite links of a four bar linkage are equal, the
links will always form a _____________.
(a) triangle
(b)
rectangle
(c) parallelogram
(d)
pentagon
Ans. : (c)
• Explanation : When two opposite links are parallel andequal in length, then any one of the link can be made
fixed. The two links which are adjacent to the fixed link
will become cranks and the link opposite to fixed link
becomes a coupler. Such mechanism is called as parallel
crank mechanism or double crank mechanism or crank-
crank
mechanism
or
rotary-rotary
converter
mechanism.
Q. 200
Q. 201
Q. 202
Q. 203
Whitworth quick return mechanism is an inversion of
_____________.
(a) double slider crank chain
(b) single slider crank chain
(c) four bar chain
(d) none of the above
Ans. : (b)
Explanation : Refer Q175
In its simplest form, a cam mechanism consists of following
number of links.
(a) 1
(b) 2
(c) 3
(d) 4
Ans. : (c)
Explanation : Refer Q167
In a mechanism, usually one link is fixed. If the fixed link
is changed in a kinematic chain, then relative motion of
other links _____________.
(a) will remain same
(b) will change
(c) could change or remain unaltered depending on which
link is fixed
(d) will not occur
Ans. : (a)
Explanation : Refer Q157
The following is the inversion of slider crank chain
mechanism.
(a) whit worth quick return mechanism
(b) hand pump
(c) oscillating cylinder engine
(d) all of the aboveQ. 204
Ans. : (d)
Explanation : Refer Q172
A typewriter mechanism has 7 number of binary joints, six
links and none of higher pairs. The mechanism is
_____________.
(a) kinematically sound
(b) not sound
(c) soundness would depend upon which link is kept fixed
(d) data is not sufficient to determine same
Ans. : (a)
Explanation:
Given:
j = 7 , n = 6
We know that,
3
j = n – 2
2
3
7 = 6 – 2
2
7 = 7
 L.H.S = R.H.S, hence it is chain having constrained
motion so this mechanism is kinematically sound
Q. 205
Q. 206
Oldham’s coupling and elliptic trammels are the inversions
of _____________.
(a) double slider crank chain
(b) single slider crank chain
(c) four bar chain
(d) none of the above
Ans. : (a)
Explanation : Refer Q175
In a mechanism which one is true _____________.
(a) All links fixed
(b) more than one link fixed
(c) All links free
(d) None of the above
Ans. : (d)
Explanation : Refer Q157Q. 207
Q. 208
Q. 209
Q. 210
Q. 211
Q. 212
In a slider crank mechanism there are _____________.
(a) Three links
(b) Four links
(c) Five links
(d) None of the above
Ans. : (b)
Explanation : Refer Q162
Which of the following is the inversion of double slider
crank chain ?
(a) Beam engine
(b) Elliptical trammel
(c) Watt's indicator mechanism
(d) Quick return motion mechanism.
Ans. : (b)
Explanation : Refer Q175
Which of the following mechanism is used to enlarge or
reduce the size of a drawing ?
(a) Pantograph
(b)
Graphometer
(c) Oscillograph
(d)
Clinograph.
Ans. : (a)
Explanation : Refer Q195
A slider crank mechanism is a special case of a
_____________.
(a) 3-bar mechanism
(b) 2-bar mechanism
(c) 6-bar mechanism
(d) 4-bar mechanism
Ans. : (d)
Explanation : Refer Q162
Using four revolute pairs, one can construct _____________.
(a) Only a planar mechanism
(b) Only a spherical mechanism
(c) Only a spatial mechanism
(d) All of the above
Ans. : (a)
Explanation : Refer Q161
Pantograph is an instrument used to _____________.
(a) an equal scale
(b) no scale
(c) infinite scale
(d) an enlarged or a reduced scale
Ans. : (d)
Explanation : Refer Q195Q. 213
Q. 214
Q. 215
Q. 216
In a pantograph, all the pairs are _____________.
(a) turning pairs
(b)
sliding pairs
(c) spherical pairs (d)
self-closed pairs
Ans. : (a)
Explanation : Refer Q195
Which of the following is an inversion of single slider crank
chain?
(a) Beam engine
(b) Watt’s indicator mechanism
(c) Elliptical trammels
(d) Whitworth quick return motion mechanism
Ans. : (d)
Explanation : Refer Q162
Which of the following is an inversion of double slider
crank chain ?
(a) Coupling rod of a locomotive
(b) Pendulum pump
(c) Elliptical trammels
(d) Oscillating cylinder engine
Ans. : (c)
Explanation : Refer Q163
Which of the following is an inversion of single-slider crank
chain?
(a) Elliptical trammel
(b)
Hand pump
(c) Scotch yoke
(d)
Oldham’s coupling
Ans. : (b)
Explanation : Refer Q162Q. 217
Q. 218
Q. 219
Q. 220
Which of the following is an inversion of double-slider
crank chain?
(a) Whitworth quick return mechanism
(b) Reciprocating compressor
(c) Scotch yoke mechanism
(d) Rotary engine
Ans. : (c)
Explanation : Refer Q163
Oldham’s coupling is used to connect two shafts which are
_____________.
(a) intersecting
(b)
parallel
(c) perpendicular
(d)
co-axial
Ans. : (b)
Explanation : Refer Q196
A pantograph consists of _____________.
(a) 4 links
(b)
6 links
(c) 8 links
(d)
10 links
Ans. : (a)
Explanation : Refer Q195
A whitworth mechanism can be obtained as an inversion of
slider crank chain by fixing _____________.
(a) slotted link
(b)
slider
(c) connecting rod
(d) link connected to slotted link and crank
Ans. : (d)
Explanation :(a) Kinematic inversion
(b) Whitworth quick return mechanisms
Fig. Q220 : Second inversion of single slider crank chain
Q. 221
Q. 222
Q. 223
Q. 224
In two inversions obtained for the same parent chain, the
relative velocity between two links _____________.
(a) change in proportion to the lengths of the two frame
links.
(b) change in proportion to the lengths of links opposite to
the frame links
(c) does not change
(d) none of the above
Ans. : (c)
Explanation : Refer Q162
A pantograph mechanism is a _____________.
(a) double crank mechanism
(b) double rocker mechanism
(c) crank-rocker mechanism
(d)
none of these.
Ans. : (b)
Explanation : Refer Q195
A hand-pump mechanism is the inversion of a slider crank
mechanism by fixing _____________.
(a) crank
(b)
connecting rod
(c) slotted link
(d)
slider
Ans. : (c)
Explanation : Refer Q162
Match the following :
Type of mechanism
P. Scott-Russel mechanism
Q. Geneva-mechanism
R. Off-set slider crank mechanism
S. Scoten yoke mechanism
(a) P → 2,
(b) P → 1,
(c) P → 3,
(d) P → 4,
Ans. (d)
Q → 1,
Q → 2,
Q → 2,
Q → 1,
R → 4,
R → 3,
R → 1,
R → 2,
Motion achieved
(1) Intermittent motion
(2) Quick return motion
(3) Simple return motion
(4) Straight line motion
S → 3
S → 4
S → 4
S → 3Q. 225
Match the pairs by selecting most appropriate alternative
from column (X) for the elements from column (Y) and
write down the correctly formed pair.
Column (X) Column (Y)
(Mechanisms) (Its application)
(1) Scotch yoke
(A) Steam pump
(2) Crank
lever
(B) Slotting machine
(3) Double crank (C) Coupled wheel
locomotive
(4) Elliptic trammel (D) To draw ellipse
(a) 1 → B,
(b) 1 → A,
(c) 1 → C,
(d) 1 → D,
Ans. : (b)
Q. 226
slotted
2 → A,
2 → B,
2 → D,
2 → C,
3 → D,
3 → C,
3 → A,
3 → B,
of
4 → C
4 → D
4 → B
4 → A
Match List-I with List-II and select the correct answer
using the codes given below the lists.
(A)
(B)
(C)
(D)
(a)
(b)
(c)
(d)
Q. 227
and
List-I
Quadric cycle chain
Single cycle crank
chain
Double slider crank
chain
Crossed slider crank
chain
A → 5,
A → 3,
A → 5,
A → 3,
B → 4,
B → 1,
B → 3,
B → 5,
C → 2,
C → 5,
C → 4,
C → 1,
1.
2. List-II
Ellipatic trammel
Rapsons slide
3. Ackerman steering
4. Eccentric mechanism
5. Pendulum pump
D → 1
D → 4
D → 2
D → 2
Ans. : (d)
Match List-I with List-II and select the correct answer
using the codes given below the lists.
List-I
(A) Crank shaft
List-II
1.
Support the revolving partsand transmits toque
Q. 228
(B) Wire shaft 2. Transmits motion between
shafts where it is not
possible to effect a rigid
coupling between them
(C) Axle 3. Converts linear motion into
rotary motion
(D) Plain shaft 4. Supports only the revolving
parts
(a) A → 3, B → 2, C → 1, D → 4
(b) A → 4, B → 2, C → 3, D → 1
(c) A → 3, B → 2, C → 4, D → 1
(d) A → 1, B → 4, C → 2, D → 3
Ans. : (a)
Match the items in Columns I and II.
Column I
P. Higher kinematic pair
Q. Lower kinematic pair
R. Quick return mechanism
S. Mobility of a linkage
(a)
(b)
(c)
(d)
P → 2,
P → 6,
P → 6,
P → 2,
Q → 6,
Q → 2,
Q → 2,
Q → 6,
R → 4,
R → 4,
R → 5,
R → 5,
Column II
1. Grubler’s equation
2. Line contact
3. Euler’s equation
4. Planer
5. Shaper
6. Surface contact
S → 3
S → 1
S → 3
S → 1
Ans. : (d)
Explanation :
Q. 229
(1)
(2)
(3)
Match the pairs by selecting most appropriate alternative
from column Q for the elements from column P and write
down the correctly formed pairs.
Column P
Wrist watch
Telescope
Splined shaft
A.
B.
C.
Column Q
Binary link
Cylindrical pair
MachineColumn P
Column Q
and splined hub
D.
Mechanism
(4)
Car glass wiper
E.
Crank
(5)
Grounded link in
F.
Prismatic pair
Whitworth mechanism
(6)
Piston
(a) 1. →C, 2. → A, 3. → E, 4. → F, 5. → B, 6. → D
(b) 1. → E, 2. → F, 3. → D, 4. → C, 5. → B, 6. → A
(c) 1. → D, 2. → B, 3. → F, 4. → C, 5. → E, 6. → A
(d) 1. → B, 2. → C, 3. → A, 4. → D, 5. → F, 6. → E
Ans. : (c)
Q. 230 Two shafts connected by Oldham's coupling have a centre
distance of 20 mm and the driving shaft rotates at angular
speed of 20 radians per second, the speed of driven shaft
in radians/s is _____________.
(a) 15
(b) 18
(c) 20
(d) 25
Ans. : (c)
Explanation : In case of Oldham’s coupling the driving
shaft and driven shaft will have the same angular speed.
Q. 231 Two shafts connected by Oldham's coupling have a centre
distance of 20 mm and the driving shaft rotates at angular
speed of 20 radians per second, the maximum velocity of
sliding of the intermediate piece in the slots of flanges is
_____________.
(a) 0.2 m/s
(b)
0.4 m/s
(c) 0.8 m/s
(d)
1.0 m/s.
Ans. : (b)
Explanation :
Given :
r = 20 mm = 0.02 m
 = 20 rad/sec
 maximum velocity of sliding is,
V s =   r = 20  0.02 = 0.4 m/s
Q. 232
...Ans.
Fig. 9(a) shows a quick return mechanism the crank on
rotates clockwise informally OA = 2 cm, OO = 4 cm, The
ratio of time for forward motion to that for return motion.
(a) 0.5
(b) 2.0
(c)
2
(d) 1(a)
(b)
Fig. 9
Ans. : (b)
Explanation :
Refer Fig. 9(b) cos =
2
= 0.5 ,
3
 = 60
and (360 – 2) = 240
Forward stroke
(360 – 2)
240
=
=
= 2.0
Return stroke
120
2 
 2  = 120 ,
...Ans.
Grashof’s Law
Q. 233
The necessary condition for Drag Link Quick Return
mechanism is that ______.
(a) Shortest link is fixed link. Sum of the shortest link and
the longest link is less than the sum of other two links.
(b) Longest link is a fixed link. Sum of the shortest link
and the longest link is greater than the sum of other
two links.
(c) Shortest link is fixed link. Sum of the shortest link and
the longest link is greater than the sum of other two
links.
(d) Longest link is fixed link Sum of the shortest link and
the longest link is less than the sum of other two links.
Ans. : (a)
Explanation : When shortest link s is fixed, then link l and
q rotate through full revolution i.e. l and q become cranksQ. 234
and link p also makes one complete revolution relative to
the fixed link s i.e. p become coupler. Such mechanism is
called
as
double
crank
mechanism
or
crank-crank mechanism or drag crank mechanism or
rotary-rotary converter mechanism.
In a 4 revolute class I kinematic chain, if link adjacent to
the shortest link is grounded, we get _____________.
(a) double rocker mechanism
(b) crank and rocker mechanism
(c) double crank mechanism
(d) structure
Ans. : (b)
Explanation : When any of the adjacent links of shortest
link s i.e. link l or q is fixed, then link s rotate through full
revolution i.e. link s become crank and a link opposite to it
i.e. link p become rocker or lever which can oscillates such
mechanism is called as crank and rocker mechanism or
crank and lever mechanism or rotary oscillating converter
mechanism.
Q. 235
Q. 236
In a four bar 'Grashoffian Linkage', if shortest link is
grounded, we get ______.
(a) Double crank mechanism
(b) Crack-rocker mechanism
(c) Double rocker mechanism
(d) Deltoid mechanism
Ans. : (a)
Explanation : Refer Q233
In a 4 revolute ‘Grashoffian chain’, if the opposite to
shortest link is grounded, we get _____________.
(a) double rocker mechanism
(b) crank rocker mechanism
(c) double crank mechanism
(d) structure
Ans. : (a)
Explanation : When the link opposite to shortest link s i.e.
link p is fixed, then link s become coupler and links l and q
became rocker which oscillates. Such mechanism is called
as double rocker mechanism or rocker-rocker mechanism orQ. 237
Q. 238
double lever mechanism or oscillating-oscillating converter
mechanism.
The length of links of a 4 bar linkage with revolute pairs
only are p, q, r and s units. Given that p < q < r < s. which
of these links should be the fixed one for obtaining a
“double crank” mechanism ?
(a) Link of length p
(b)
Link of length q
(c) Link of length r
(d)
Link of length s
Ans. : (d)
Explanation : Refer Q233
In a class two four-bar mechanism, if the shorter link is
fixed, the type of mechanism obtained is _____________.
(a) rocker-rocker mechanism
(b) crank-crank mechanism
(c) double-lever mechanism
(d) none of these
Ans. : (a)
Explanation : The different mechanisms obtained from class -
II four bar linkage by fixing of any of the links always result in
double rocker mechanism or rocker-rocker mechanism or
double lever mechanism.
In such mechanism the link opposite to fixed link always
became coupler and remaining two links became rocker
which oscillates
Q. 239 In a four-bar linkage, S denotes the shortest link length, L
is the longest link length, P and Q are lengths of other two
links. Atleast one of the three moving links will rotate by
360 if _____________.
(a) S + L  P + Q
(b) S + L > P + Q
(c) S + P  L + Q
(d) S + P > L + Q
Ans. : (a)
•
•
Explanation :
According to Grashof’s law, one of the link,
particular the shortest link will rotate continuously relative to
the other three links if,
s + l

p + qQ. 240 In a darg link quick return mechanism, the shortest link is
always fixed. The sum of the shortest and longest link is
_____________.
(a) equal to sum of other two
(b) greater than sum of other two
(c) less than sum of other two
(d) there is no such relationship
Ans. : (c)
Explanation : Refer Q233
Q. 241 In a four-link mechanism, the sum of the shortest and the
longest link is less than the sum of the other two links. It
will act as a double crank mechanism if _____________.
(a) the longest link is fixed
(b) the shortest link is fixed
(c) any link adjacent to shortest link is fixed.
(d) the link opposite to the shortest link is fixed
Ans. : (b)
Explanation : Refer Q233
Q. 242 In a four-link mechanism, the sum of the shortest and the
longest link is less than the sum of the other two links. It
will act as a crank-rocker mechanism if _____________.
(a) the link opposite to the shortest link is fixed
(b) the shortest link is fixed
(c) any link adjacent to shortest link is fixed.
(d) the longest link is fixed
Ans. : (c)
Explanation : Refer Q234
Q. 243 In a four-link mechanism, the sum of the shortest and the
longest link is less than the sum of the other two links. It
will act as a rocker-rocker mechanism if _____________.
(a) the link opposite to the shortest link is fixed
(b) the shortest link is fixed.
(c) any link adjacent to shortest link is fixed.
(d) the longest link is fixed
Ans. : (a)
Explanation : Refer Q236Q. 244
Q. 245
Q. 246
Q. 247
Q. 248
For a four bar mechanism to act as a double crank
mechanism _____________.
(a) the largest link should be follower
(b) the shortest link should be crank
(c) the shortest link should be frame
(d) none of the above
Ans. : (b)
Explanation : Refer section 1.12.1
In a four bar chain, if l is length of the longest link, s is
length of the shortest link and p and q are lengths of the
remaining two links. Grashof’s law states that
_____________.
(a) l + s < p + q
(b)
l + s > p + q
(c) l – s < p – q
(d)
l – s > p + q
Ans. : (a)
Explanation : Refer Q233
When the shortest link is coupler link, it results into
_____________.
(a) crank-crank mechanism
(b) crank-rocker mechanism
(c) double rocker mechanism
(d) none of the above
Ans. : (c)
Explanation : Refer Q238
When the shortest link is crank and any other link is
frame, it results into _____________
(a) crank-crank mechanism
(b) crank-rocker mechanism
(c) double rocker mechanism
(d) none of the above
Ans. : (b)
Explanation : Refer Q234
A mechanism in which two links of equal length are placed
adjacent and the longest link is fixed, is called
_____________
(a) parallelogram mechanism
(b) deltoid mechanism
(c) galloway mechanism(d) rocker-rocker mechanism
Ans. : (b)
Explanation :
When two links which are equal in
lengths are adjacent to each other then mechanism is
called as deltoid four bar linkage.
Q. 249
Q. 250
A mechanism in which two links of equal length are not
adjacent is called _____.
(a) parallelogram mechanism
(b) deltoid mechanism
(c) galloway mechanism
(d) crank-rocker mechanism
Ans. : (a)
Explanation : Refer section 1.12.3
A mechanism in which two links of equal length are placed
adjacent and the shorter link is fixed is called
_____________.
(a) parallelogram mechanism
(b) double rocker mechanism
(c) double crank mechanism
(d) crank-rocker mechanism
Ans. : (c)
• Explanation : When two opposite links are parallel and
equal in length, then any one of the link can be made
fixed. The two links which are adjacent to the fixed link
will become cranks and the link opposite to fixed link
becomes a coupler. Such mechanism is called as parallel
crank mechanism or double crank mechanism or crank-
crank
mechanism
or
rotary-rotary
converter
mechanism.
Q. 251 In a four bar chain, lengths of three links are 20 mm,
130 mm and 80 mm. If length of the fourth link is _______
the kinematic chain will be a ‘Grashoffian Chain’.
(a) 30 mm
(b) 50 mm
(c) 60 mm
(d) 75 mm
Ans. : (d)Explanation :
Given : s = 40 mm, l =130mm, p = 80 mm, q = ?,
For Grashoffian Chain mechanism,
s + l < p + q
Q. 252
20 + 130 < 80 + q
This condition will satisfy when we choose q = 75 mm from
above four option, therefore length of fourth link is, q = 75
mm
...Ans.
In a four-bar chain it is required to give an oscillatory
motion to the follower for a continuous rotation of the
crank. For the lengths of 40 mm of crank and 80 mm of the
follower, determine theoretical maximum length of coupler.
The distance between fixed pivots of crank and followers is
100 mm.
(a) 140 mm
(b) slightly less than 140 mm
(c) slightly more than 140 mm
(d) none of the above.
Ans. : (b)
Explanation :
Given : s = 40 mm, p = 80 mm, q = 100 mm,
Q. 253
For class I four bar chain mechanism,
s + l < p + q ;
40 + l < 80 + 100
l < 140
... Ans.
Four links of a mechanism have lengths of 3,7,5,4 units
respectively. By fixing link of length 3 units resulting
mechanism will be _____________.
(a) crank-crank mechanism
(b) crank rocker mechanism
(c) rocker-rocker mechanism
(d) a mechanism with change points
Ans. : (c)
Explanation :
Given : s = 3, l = 7, p = 5, q = 4
since s + l > p + q,
3 + 7 > 5 + 4,
It belongs to class II four bar chain mechanism and
shortest link is fixed, so it is double rocker mechanism.Q. 254
Fig 10. shows four bar chain mechanism.
The type of mechanism is _____________.
(a) Double crank mechanism
(b) Crank and rocker mechanism
(c) Double rocker mechanism
(d) All of above
Ans. : (c)
Explanation :
Fig. 10
Length of shortest link, s = 6
Length of longest link, l = 11
Length of other two links, p and q = 8 and 10
Since, 6 + 11 < 8 + 10, it belongs to class-I four bar chain
mechanism and the link opposite to shortest link is fixed, so it
is a double rocker mechanism.
Q. 255
Fig. 11 shows four bar chain mechanism.
The type of mechanism is _____________.
(a) Double crank mechanism
(b) Crank and rocker mechanism
(c) Double rocker mechanism
(d) All of above
Ans. : (a)
Fig. 11
Explanation :
Length of shortest link, s = 4
Length of longest link, l = 10
Length of other two links, p and q = 8 and 8
Since, 4 + 10 < 8 + 8, it belongs to class - I four bar chain
mechanism and the shortest link is fixed, so it is a double
crank mechanism.
Q. 256
Fig. 12 shows four bar chain mechanism. The type of
mechanism is ________.
(a) Double crank mechanism
(b) Crank and rocker mechanism
(c) Double rocker mechanism
(d) All of above
Ans. : (c)
Explanation :
Fig. 12Length of shortest link, s = 3
Length of longest link, l = 10
Length of other two links, p and q = 8 and 4
Since 3 + 10 > 8 + 4, it belongs to class-II four bar chain
mechanism and the link adjacent to shortest link is
fixed, so it is double rocker mechanism.
Q. 257 Fig. 13 shows four bar chain mechanism. The type of
mechanism is _________.
(a) Double crank mechanism
(b) Crank and rocker mechanism
(c) Double rocker mechanism
(d) All of above
Ans. : (c)
Fig. 13
Explanation :
Length of shortest link, s = 4
Length of longest link l = 10
Length of other two links, p and q = 6 + 5
Since 4 + 10 > 6 + 5, it belongs to class-II four bar chain
mechanism and the shortest link is fixed, so it is double rocker
mechanism.
Degree of Freedom
Q. 258 In a kinematic chain, if the specification of one co-ordinate,
or dimension or position of a single link is sufficient to
define the position of all other links, then the chain is
called a kinematic chain of _____________.
(a) two degrees of freedom
(b) one degree of freedom
(c) three degrees of freedom
(d) none of the above
Ans. : (b)
Explanation : DOF is the number of independent co-
ordinates required to define the position of each link, in a
mechanism.
Q. 259 The Grubler’s criterion for determining the degrees of
freedom (f) of a mechanism having n links and P 1 pairs is
given by _____________.(a) f = 3(n – 1) – 2 P 1 (b) f = 6 (n – 1) – 2 P 1
(c) f = 5n – 2 P 1 (d) f = 3 (n + 1) – 2 P 1
Ans. : (a)
Explanation : In planer mechanism having lower and higher pairs, the
degree of freedom is,
f = 3 (n – 1) – 2 p 1 – 1p 2
where, n =Number of links in mechanism
p 1 = Number of lower pairs (pairs having 1 DOF)
p 2 = Number of higher pairs (pairs having 2 DOF)
The above equation is known as Grubler’s criterion for planer
mechanism.
Q. 260 The Grublers criterion for determining the degrees of
freedom of a mechanism having plane motion is
_____________.
(a) f = 3 + (n + 1) + 2 p 1 – 1p 2
(b) f = 3 (n –1) – 2 p 1 – 1p 2
(c) f = 2 (n –1) – 1 p 1 – p 2
(d) f = 2 (n –1) – 1 p 1 + 1p 2
Ans. : (b)
Explanation : Refer Q.260
Q. 261
Q. 262
The Grubler’s criterion for determining the degrees of
freedom (n) of a mechanism having plane motion is
_____________.
(a) n = ( l – 1) – j
(b)
n = 2 ( l – 1) – 2j
(c) n = 3 ( l – 1) – 2j
(d)
n = 4 ( l – 1) – 3j
Where l = Number of links,
and j = Number of binary joints.
Ans. : (c)
Explanation : Refer Q260
The Grubler’s criterion for obtaining the dof of a planar
mechanism with n number of links and j number of binary
joints, is given by _____________.
(a) f = 3 (n – 1) – j
(b)
f = 3 (n – 1) – 2j
(c) f = 2 (n – 1) – j
(d)
f = 2 (n – 1) – 2j
Ans. : (b)Explanation : Refer Q260
Q. 263
Grubler’s Equation to determine degrees of freedom is
written as _________.
(n = number of links, p 1 = number of lower pairs,
p 2 = number of roll-slide pairs, f = degree of freedom)
(a) f = 3 (n – 1) – 2 (p 1 ) – 1 (p 2 )
(b) f = 3 (n – 1) – 2 (p 1 ) – 2 (p 2 )
(c) f = 3 (n – 1) – 1 (p 1 ) – 2 (p 2 )
(d) f = 2 (n – 1) – 1 (p 1 ) – 2 (p 2 )
Ans. : (a)
Explanation : Refer Q260
Q. 264
The Grubler’s Equation for two spur gears of involute tooth
profile in mesh can be written as f = 3 (n – 1) – 2 (p 1 ) – 1
(p 2 ) where _____________.
(a) n = 3, p 1 = 2 and p 2 = 1
(b) n = 4, p 1 = 4 and p 2 = 0
(c) f = + 1, n = 4 and p 1 = 4
Q. 265
(d) All of the above
Ans. : (d)
Explanation : Refer Q260
Degree of freedom of a chain having ‘n’ number of links and
p 1 number of lower pairs the d.o.f. is given by
_____________.
(a) f = 3n – p 1
(b)
f = 3n – 2p 1
(c) f = 3(n – 1) – 2 p 1
Q. 266
(d)
f = 3n – 3p 1
Ans. : (c)
Explanation : Refer Q260
A cylindrical pair has ________ DOF (degrees of freedom).
(a) 1
(b) 2
(c) 3
(d) 4
Ans. : (b)
Explanation :
The motion of a round bar in a round hole of a fixed block is
the case of incompletely constrained motion since the bar
can rotate or reciprocate or both motions of rotation and
sliding of bar can take place simultaneously. Therefore
such pair has two DOF (degrees of freedom).Q. 267
In a spatial mechanism, number of degree of freedom is
________.
(a) 1
(b) 2
(c) 3
(d) 6
Ans. : (d)
Explanation : In this type of mechanisms, the complete
motions cannot be represented in a single plane. That is, to
describe the motion of such mechanisms, more than one
plane would be required. They have three dimensional
motion paths, So spatial mechanism, number of degree of
freedom is six
Q. 268
In a planar pair, number of degrees of freedom is
_____________.
(a) 1
(b) 2
(c) 3
(d) 6
Ans. : (c)
Explanation :
•
The complete motion paths of the mechanism could be
represented on a single plane. Or in other words, the
entire mechanism could be represented to a scale on a
sheet of paper.
•
Every link of the mechanism can have two translatory
motions in the X and Y axis. In addition they can have
one rotation about the Z-axis.
•
Hence the DOF for a planar mechanism will reduce to
only three DOF. Due to this less number of DOF, the
analysis of planar mechanism will be much simpler as
compared to a spatial mechanism.
Q. 269
A globular (spherical) pair has _________ degrees of
freedom.
(a) one
(b) two
(c) three
(d) four
Ans. : (c)
Explanation :
•
When one links forms a spherical shape and turns inside
a fixed element, the resulting kinematic pair is called a
spherical pair and such pair has three DOF.
Q. 270
Unconstrained rigid body in space will have ______ Degrees
of freedom.
(a) three
(b) six
(c) nine
(d) fourAns. (b)
Explanation : Refer section 1.13
Q. 271 When three links are joined by a single pin, we must count
_____ number of pairs at the connections.
(a) one (b) two
(c) three
(d) four
Ans. : (b)
Explanation :
•
This rigid body have possesses following independent motion.
1)Translation motions along any three perpendicular axis x, y
and z.
2)Three rotational motions about the axis x, y and z.
•
Thus degree of freedom of rigid body has six degree of freedom in free
space.
Q. 272 A Bolt and Nut pair has __________ DOF (degrees of
freedom).
(a) 1
(b) 2
(c) 3
(d) 4
Ans. : (a)
Explanation :
The motion of bolt in a fixed nut is the example of screw
pair. It can be noted that in this case the bolt has both
rotational and sliding motion but for a specific amount of
rotation of bolt, it results in a strictly proportionate amount
of axial displacement (sliding) relative to nut. Thus Bolt
and Nut pair has 1 DOF
Q. 273 The mechanism called a structure when the number of
degrees of freedom is equal to _____________.
(a) 0
(b) 1
(c) 2
(d) 3
Ans. : (a)
Explanation : Structure is an assemblage of resistant
bodies without any relative motion between the links, thus
DOF of Structure is zero.
Q. 274 If the binary link, is constrained to move with planer
motion. It possesses _____.
(a) 2 degrees of freedom(b) 3 degrees of freedom
(c) 4 degrees of freedom
(d) 6 degrees of freedom
Ans. : (b)
Explanation : Refer Q268
Q. 275
A negative degree of freedom for a linkage means
_____________.
(a) Constrained motion of the linkage
(b) Unconstrained motion of the linkage
(c) Any of the linkage
(d) A statically indeterminate structure
Ans. : (d)
Explanation : Refer Fig Q275
Number of links, n = 6
Number of binary joint = 2 (Number of ternary joint) = 2 (4)
Number of binary = 8
Number of binary joint = Number of lower pairs = 8
Number of lower pairs, p 1 = 8
Number of higher pairs, p 2 = 0
Step 2 :
Apply Grubler’s criteria.
The degree of freedom for a given mechanism is,
f =3 (n − 1) − 2 p 1 − 1 p 2
Fig. Q275
 f = 3 (6 − 1) − 2  8 − 0
f = − 1
DOF is negative in this case. This means there are redundant
constraints in the kinematic chain. Or it is statically indeterminate
structure. Thus a mechanism with DOF zero or less than zero is not
a mechanism but a structure or redundant structure.
Refer section 1.16
Q. 276
Zero dof for a mechanism means _____________.
(a) statically indeterminate structure
(b) statically determinate structure
(c) constrained motion mechanism(d) unconstrained motion mechanism
Ans. : (b)
Explanation : Refer Q273
Q. 277
A spherical pair allows _____________.
(a) 2 d.o.f.
(b)
4 d.o.f.
(c) 1 d.o.f
(d)
3 d.o.f
Ans. : (d)
Explanation : Refer Q269
Q. 278
A cylindrical pair allows _____________.
(a) 2 d.o.f.
(b)
3 d.o.f.
(c) 4 d.o.f
(d)
1 d.o.f
Ans. : (a)
Explanation : Refer Q266
Q. 279
A screw pair allows _____________.
(a) 2 d.o.f.
(b)
3 d.o.f.
(c) 5 d.o.f
(d)
1 d.o.f
Ans. : (d)
Explanation : Refer Q272
Q. 280
The mechanism forms a structure, when the number of
degrees of freedom (n) is equal to _____________.
(a) 0
(b) 1
(c) 2
(d) – 1
Ans. : (a)
Explanation : Refer Q273
Q. 281
Q. 282
Degree of freedom of a constrained mechanism is
_____________.
(a) less than one
(b)
greater than one
(c) equal to one
(d)
equal to zero
Ans. : (c)
Explanation :
The mechanism has only 1 DOF. That means a single input
a sufficient to impart a constrained motion to all other
links. Thus a mechanism with single DOF is said to have a
constrained motion or the mechanism is said to be
constrained.
The dof of a rigid link placed on a table is _____________.
(a) 0
(b) 1
(c) 2
(d) 3Ans. : (d)
Explanation : Refer Q268
Q. 283
Q. 284
Unconstrained rigid link in a plane has _____________.
(a) One degree of freedom
(b) Two degrees of freedom
(c) Three degrees of freedom
(d) Zero degree of freedom
Ans. : (c)
Explanation : Refer Q268
In a nine link mechanism of d. o. f. = 2, number of binary
joints are ________.
(a) 11
(b) 10
(c) 13
(d) 9
Ans. : (a)
Explanation :
Given :
f = 2, n = 9
We know that, Grubler’s criterion is,
f = 3(n – 1) – 2p 1 or
Q. 285
f = 3(n – 1) – 2j
...[ Number of lower pairs = number of binary joints]
2 = 3(9 – 1) – 2j
j = 11
..Ans.
A mechanism consisting of 8 kinematic links and 10
kinematic pairs will have _________ DOF.
(a) 1
(b) 2
(c) 3
(d) 4
Ans. : (a)
Explanation :
Given :
Number of links, n = 8,
Number of lower pairs, p 1 = 10,
Number of higher pairs, p 2 = 0
Applying Grublers criterion,
f = 3 (n –1) – 2 p 1 – p 2
Q. 286
 f = 3 (8 –1) – 2  10 – 0
 f = 1
...Ans.
A mechanism consisting of 8 kinematic links and 9
kinematic pairs will have _________ DOF.
(a) 1
(b) 2
(c) 3
(d) 4Ans. : (c)
Explanation :
Given :
Number of links, n = 8,
Number of lower pairs, p 1 = 9,
Number of higher pairs, p 2 = 0
Applying Grublers criterion,
f = 3 (n –1) – 2 p 1 – 2p 2
Q. 287
 f = 3 (8 –1) – 2  9 – 2  0
 f = 21 – 18
 f = 3
..Ans.
A planar mechanism has 8 links and 10 rotary joints. The
number of degrees of freedom of the mechanism, using
Grubler’s criterion, is ___________.
(a) 0
(b) 1
(c) 2
(d) 3
Ans. : (b)
Explanation :
Given :
Number of links, n = 8,
Number of lower pairs, p 1 = 10,
Number of higher pairs, p 2 =
0
Applying Grublers criterion,
f = 3 (n –1) – 2 p 1 – p 2
 f = 3 (8 –1) – 2  10 – 0
 f = 1
Q. 288
..Ans.
The number of degrees of freedom for mechanism shown in
Fig. 14 is ________.
(a) 0
(b) 1
(c) 2
(d) 3
Ans. : (a)
Explanation :
Given :
Number of links, n = 3,
Number of lower pairs,
Fig. 14
P 1 = 3, Number of higher pairs, P 2 = 0Applying Grublers Criteria,
f = 3 (n –1) – 2 p 1 – 1p 2
Q. 289

f = 3 (3 –1) – 2  3 – 0  f = 0
...Ans.
The number of degrees of freedom for mechanism shown in
Fig. 15 is ________.
(a) 0
(b)
1
(c) 2
(d)
3
Ans. : (b)
Explanation :
Given :
Number of links, n = 4,
Number of lower pairs, p 1 = 4,
Fig. 15
Number of higher pairs, p 2 = 0
Applying Grublers criterion,
f = 3 (n –1) – 2 p 1 – 1p 2
Q. 290
 f = 3 (4 –1) – 2  4 – 0
 f = 1
...Ans.
The minimum number of links in a single degree-of-
freedom planner mechanism with four lower kinematic
pairs is _____________.
(a) 2
(b) 3
(c) 4
(d) 5
Ans. : (c)
Explanation :
Given : Number of links, n = ?
Number of lower pairs, p 1 = 4,
Number of higher pairs, p 2 = 0
Degree of freedom, f = 1
Applying Grublers criterion,
f = 3 (n –1) – 2 p 1 – 1p 2
Q. 291
 1 = 3 (5 –1) – 2  4 – 0
 n = 4
The number of degrees of
freedom of a five link plane
mechanism with five revolute
pair as shown in Fig. 16 is
__________.
..Ans.(a) 0
(c) 2
(b)
(d)
1
3
Ans. : (c)
Explanation :
Given : Number of links, n = 5,
Number of lower pairs, p 1 = 5,
Number of higher pairs, p 2 =
Fig. 16
0
Applying Grublers criterion,
f = 3 (n –1) – 2 p 1 – 1p 2
Q. 292
 f = 3 (5 –1) – 2  5 – 0  f = 2
...Ans.
The number of degrees of freedom of a planer linkage with
8 links and 9 simple revolute joints is __________.
(a) 1
(b) 2
(c) 3
(d) 4
Ans. : (c)
Explanation :
Given :
Number of links, n = 8, Number of lower
pairs p 1 = 9
(··· 9 revolute joints = 9 lower pairs) , Number of higher
pairs, p 2 = 0
Applying Grublers criterion,
f = 3 (n –1) – 2 p 1 – 1 p 2
Q. 293
 f = 3 (8 –1) – 2  9 – 0
 f = 3
For the given mechanisms as shown
in Fig. 17 find the degree of freedom.
(a) 1
(b) 2
(c) 3
(d) 4
Ans. : (b)
...Ans.
Fig. 17
Explanation :
Number of links, n = 7
Number of lower pairs, p 1 = 8
Number of higher pairs, p 2 = 0
Apply Grubler’s criteria;f = 3 (n – 1) – 2 p 1 – 1 p 2
f = 3 (7 – 1) – 2 8
f = 3  6 – 16 = 2
DOF of mechanism is 2
Q. 294
...Ans.
Fig. 17(a)
For the given mechanisms as shown in Fig. 18 find the
degree of freedom.
Fig. 18
(a) 1
(b) 2
Ans. : (a)
Explanation :
(c) 3
(d) 4
Number of links, n = 10
Number of lower pair, p 1 = 13
Number of higher pair, p 2 = 0
Apply Grubler’s criteria,
f = 3 (n – 1) – 2p 1 – 1 p 2
f = 3 (10 – 1) – 2  13 – 1  0 = 3  9 – 13  2
f = 27 – 26 = 1
DOF of mechanism is 1.
..Ans.Fig. 18(a)
Q. 295
For the given mechanisms
as shown in Fig. 19 find the
degree of freedom.
(a) 1
(b) 2
(c) 3
(d) 4
Ans. : (a)
Explanation :
Fig. 19
Number of links,
n = 4
Number of lower pairs, p 1 = 4
Number of higher pairs, p 2 = 0
Applying Grubler’s criteria,
f = 3 (n – 1) – 2 p 1 – p 2
f = 3 (4 – 1) – 2 (4) – 0
f = 9 – 8
f = 1
DOF of mechanism is 1
...Ans.
Fig. 19(a)
Q. 296
For the given mechanisms as shown in Fig. 20 find the
degree of freedom.(a) 1
(b) 2
Ans. (b)
Explanation :
Fig. 20
(c) 3
(d) 4
Number of links,
n = 7
Number of lower pairs, p 1 = 8
Number of higher pairs, p 2 = 0
Applying Grubler’s criteria,
f = 3 (n – 1) – 2 p 1 – p 2
f = 3 (7 – 1) – 2  8 – 0
f = 18 – 16
f = 2
DOF of mechanism is 2
...Ans.
Fig. 20(a)
Q. 297
For the given mechanisms as shown in Fig. 21 find the
degree of freedom.Fig. 21
(a) 1
(b) 2
(c) 3
(d) 4
Ans. : (a)
Explanation :
Number of links,
n = 3
Number of lower pairs, p 1 = 2
Number of higher pairs, p 2 = 1
Applying Grubler’s criteria,
f = 3 (n – 1) – 2 p 1 – p 2
f = 3 (3 – 1) – 2  (2) – 1 = 6 – 4 – 1
f = 1
DOF of mechanism is 1
...Ans.
Fig. 21(a)
Q. 298
For the given mechanisms as shown in Fig. 22 find the
degree of freedom.
Fig. 22
Explanation :
Number of links,
n = 6
Number of lower pairs, p 1 = 7
Number of higher pairs, p 2 = 0
Applying Grubler’s criteria,f = 3 (n – 1) – 2 p 1 – p 2
f = 3 (6 – 1) – 2  7 – 0
f = 1
DOF of mechanism is 1
...Ans.
Fig. 22(a)
Q. 299
Find the degree of freedom of the following mechanism.
Fig. 23
(a) 1
(b) 2
(c) 3
(d) 4
Ans. : (b)
Explanation :
Number of links, n = 7
Number of lower pair, p 1 = 8
Number of higher pair, p 2 = 0
Applying Grubler’s criteria,
f = 3 (n – 1) – 2 p 1 – 1 p 2
= 3 (7 – 1) – 2  8 – 0 = 3  6 – 16
f = 2
 DOF of mechanism is 2.
...Ans.Fig. 23(a)
Q. 300
Find the degree of freedom of the following mechanism.
Fig. 24
(c) 3
(a) 1
(b) 2
(d) 4
Ans. : (a)
Explanation :
Number of links, n = 6
Number of lower pair, p 1 = 7
Number of higher pair, p 2 = 0
Applying Grubler’s criteria,
f = 3 (n – 1) – 2 p 1 – 1 p 2
= 3 (6 – 1) – 2  7 = 15 – 14
f = 1
DOF of mechanism is 1.
...Ans.Fig. 24(a)
Q. 301
Find the degree of freedom of the following mechanism.
Fig. 25
(a) 4
(b) 5
Ans. : (d)
Explanation :
(c) 6
(d) 7
Number of links, n = 7
Number of lower pair, p 1 = 5
Number of higher pair, p 2 = 1
Applying Grubler’s criteria,
f = 3 (n – 1) – 2 p 1 – 1 p 2
Fig. 25(a)
= 3 (7 – 1) – 2  5 – 1 = 18 – 10 – 1
f = 7
Q. 302
DOF of mechanism is 7.
...Ans.
Find the degree of freedom of the following mechanism.
(a) 1
(b) 2
Ans. : (b)
Explanation :
Fig. 26
(c) 3
Number of links, n = 5
Number of lower pair, p 1 = 4
Number of higher pair, p 2 = 2
(d) 4Applying Grubler’s criteria,
f = 3 (n – 1) – 2 p 1 – 1 p 2
= 3 (5 – 1) – 2  4 – 2 = 12 – 10 = 2
DOF of mechanism is 2
...Ans.
Fig. 26(a)
Q. 303
Find the degree of freedom of the following mechanism.
Fig. 27
(a) 1
(b) 2
(c) 3
(4) 4
Ans. : (a)
Explanation :
Number of links, n = 6
Number of lower pairs, p 1 = 7
Number of higher pairs, p 2 = 0
Applying Grubler’s criteria,
f = 3 (n – 1) – 2 p 1 – 1p 2
= 3 (5) – 2  7 = 15 – 14
f = 1
DOF of mechanism is 1
...Ans.Fig. 27(a)
Q. 304
Find the degree of freedom of the following mechanism.
Fig. 28
(a) 1
(b) – 1
Ans. : (b)
Explanation :
(c) 2
(d) – 2
Number of links, n = 12
Number of lower links, p 1 = 17
Number of higher links, p 2 = 0
Applying Grubler’s criteria,
f = 3 (n – 1) – 2 p 1 – 1 p 2
= 3 (12 – 1) – 2  17 – 0 = 33 – 34 = – 1
DOF of mechanism is – 1
Fig. 28(a)
...Ans.Q. 305
Q. 306
Find the degree of freedom of the
following mechanism.
(a) 1
(b) 2
(c) 3
(d) 4
Ans. : (a)
Explanation :
Number of links, n = 3
Number of lower pairs, p 1 = 2
Number of higher pairs, p 2 = 1
Apply Grubler’s criteria,
f = 3 (n – 1) – 2 p 1 – 1 p 2
f = 3 (3 – 1) – 2  2 – 1
= 3  2 – 2  2 – 1
= 6 – 4 – 1 = 6 – 5 = 1
DOF of mechanism is 1
Find the degree of freedom of the
following mechanism.
(a) 0
(b) 1
(c) 2
(d) 3
Ans. : (a)
Explanation :
Number of links, n = 4
Number of lower pairs, p 1 = 4
Fig. 29
Fig. 29(a)
...Ans.
Fig. 30
Number of higher pairs, p 2 = 1
Apply Grubler’s criteria,
f = 3 (n – 1) – 2 p 1 – 1 p 2
f = 3 (4 – 1) – 2  4 – 1
= 3  3 – 8 – 1 = 9 – 9 = 0
DOF of mechanism is zero.
Fig. 30(a)
...Ans.Q. 307
Find the degree of freedom of the following mechanism.
Fig. 31
(a) 1
(b) 2
Ans. : (b)
Explanation :
(c) 3
(d) 4
Number of links, n = 4
Number of lower pairs, p 1 = 3
Number of higher pairs, p 2 = 1
Apply Grubler’s criteria,
f = 3 (n – 1) – 2 p 1 – 1 p 2
Fig. 31(a)
f = 3 (4 – 1) – 2  3 – 1  1
Q. 308
= 9 – 6 – 1 = 9 – 7 = 2
DOF of mechanism is 2.
...Ans.
Find the degree of freedom of the following mechanism.
(a) 1
(b) 2
Ans. (a)
Explanation :
Fig. 32
(c) 3
(d) 4
Total number of links, n = 3
Number of lower pairs, p 1 = 2
Number of higher pairs, p 2 = 1
Apply Grubler’s criteria,
Fig. 32(a)
f = 3 (n – 1) – 2 p 1 – 1 p 2
f = 3 (3 – 1) – 2  2 – 1 = 3  2 – 2  2 – 1Q. 309
= 6 – 4 – 1 = 6 – 5 = 1
DOF of mechanism is 1
...Ans.
Find the degree of freedom of the following mechanism.
Fig. 33
(a) 0
(b) 1
Ans. : (a)
Explanation :
(c) 2
(d) 3
Total number of links, n = 5
Number of lower pairs, p 1 = 6
Number of higher pairs, p 2 = 0
Apply Grubler’s criteria,
= 3 (n – 1) – 2 p 1 – 1 p 2
= 3 (5 – 1) – 2  6 = 3  4 – 12 = 0
DOF of mechanism is zero
Fig. 33(a)
...Ans.Straight Line Generating Mechanism
Q. 310
Q. 311
The approximate straight line mechanism is a __________.
(a) four bar linkage
(b) 6 bar linkage
(c) 8 bar linkage
(d) 3 bar linkage
Ans. : (a)
Explanation : To obtain approximate straight line the
mechanism must be four bar linkage.
Hart mechanism has _____________.
(a) eight links
(b) six links
(c) four links
(d) twelve links
Ans. : (b)
Explanation :
This is also an exact straight line
mechanism. It has six links.
Fig Q311
Q. 312
Which
of
the
following
mechanisms
mathematically an exact straight line motion ?
(a) Grasshopper mechanism
(b) Watt mechanism
(c) Peaucellier's mechanism
(d) Tchabichiff mechanism
Ans. : (c)
Explanation :
producesQ. 313
Peaucellier's mechanism has _____________.
(a) Sliding pairs
(b)
Turning pairs
(c) Spherical pairs
(d)
None of the above
Ans. : (b)
Explanation : In Peaucellier's mechanism consist of eight
links andall links are having pin joints so pairs form in this
mechanism is turning pair.
Fig. Q313 : Actual Peaucellier mechanism
Q. 314
A Peaucellier mechanism has _____________.
(a) 4 links
(b)
6 links
(c) 8 links
(d)
10 links
Ans. : (c)
Q. 315
Explanation : Refer Q313
Which of the following mechanism is made up of turning
pairs?
(a) Scott Russel’s mechanism
(b) Peaucellier’s mechanism(c) Hart’s mechanism
Ans. : (d)
Q. 316
Q. 317
Q. 318
(d)
Both (b) and (c)
Explanation :
Peaucellier’s mechanism and Hart’s mechanism is made up
all turning pairs.
Which of these mechanisms gives an approximately
straight line ?
(a) Hart
(b)
Watt
(c) Paucellier
(d)
Kempe
Ans. : (b)
Explanation :
Fig:Q316
Which of these mechanisms has six links?
(a) Tchebicheff
(b)
Hart
(c) Paucellier
(d)
Watt
Ans. : (b)
Explanation : Refer Q311
Indicate exact straight line motion mechanism out of
_____________.
(a) Hart’s mechanism
(b) Watt’s mechanism
(c) Grasshopper mechanism
(d) Robert’s mechanism
Ans. : (a)
Explanation : Refer Q312Q. 319
Q. 320
Q. 321
Indicate approximate straight line motion mechanism out
of _____________.
(a) Hart’s mechanism
(b) Peaucellier mechanism
(c) Scott Russel mechanism
(d) Tchiebicheff’s mechanism
Ans. : (d)
Explanation : Refer Q316
Exact straight line motion mechanism using sliding pair is
_____________.
(a) Peaucellier mechanism
(b) Hart’s mechanism
(c) Scott Russel mechanism
(d) Robert’s mechanism
Ans. : (c)
Explanation : Refer Q312
The Scott-Russell mechanism consists of _____________.
(a) sliding and turning pairs
(b) sliding and rotary pairs
(c) turning and rotary pairs
(d) sliding pairs only
Ans. : (a)
Explanation : In Scott-Russell mechanism when link OQ is
given a rotary motion the point P will move along a
straight line perpendicular to OS.
Fig. Q321 : Scott Russel mechanismSteering Gear Mechanism
Q. 322
Which of the following statements is correct for Ackermann
steering gear for automobiles?
(a) It has only turning pairs
(b) It is preferred over Davis-Steering gear
(c) It fulfills the fundamental equation of correct gearing
in the extreme position.
(d) All of the above
Ans. : (a)
Q. 323 Explanation : Ackermann steering gear for automobiles
consist of four turning pairs.
Q. 324 The Ackermann steering gear is the inversion of a
_____________.
(a) slider crank chain
(b) four bar chain
(c) double slider crank chain
(d) crossed slider crank chain.
Ans. : (b)
Q. 325
Explanation : Ackermann gear mechanism which is based
on four bar kinematic chain represented by quadrilateral.
The Ackermann steering mechanism is preferred to the
Davis type in automobiles because _____________.
(a) the former is mathematically accurate
(b) the former is having turning pair
(c) the former is most economical
(d) the former is most rigid
Ans. : (b)
Q. 326
Explanation : Ackermann steering gear for automobiles
consist of all turning pairs. Due to non sliding pairs friction
is less and hence wear is less, hence it is preferred.
Davis steering gear consists of _____________.
(a) sliding pairs
(b)
turning pairsQ. 327
Q. 328
(c) rolling pairs
(d)
both (a) and(b)
Ans. : (d)
Explanation : It consists of two turning and two sliding
pairs
Ackermann steering gear consists of _____________.
(a) sliding pairs
(b)
turning pairs
(c) rolling pairs
(d)
higher pairs
Ans. : (b)
Explanation : Refer Q322
The condition for correct steering of a Davis steering gear is
_____________.
b
l
(a) tan  =
(b)
cot  =
2 l
2b
(c) cos  =
l
l
(d)
tan  =
b
b
where b = wheel base
l = distance the pivots of the front axle
 = angle made by the links to the vertical.
Ans. : (a)
Explanation : The condition for correct steering of a Davis
b
steering gear is tan  =
2 l
Q. 329
The condition for correct steering of a Davis steering gear is
_____________.
(a) sin  = b/ l
(b)
cos  = l /b
(c) tan  = b/2 l
(d)
cot  = l /2b
Where  = Angle of inclination of the links to the vertical,
b = Wheel base, and
l = Distance between the pivots of the front axle.
Ans. : (c)
Q. 330
Explanation : Refer Q328
Which statement is false ?
(a) The Ackermann steering gear consists of turning pairs.
(b) The Davis steering gear consists of sliding pairs.
(c) Whole of the mechanism in the Davis steering gear is
on the front of the front wheels(d) Davis steering gear is preferred to Ackermann steering
gear mechanism.
Ans. : (d)
Explanation :
Sr.
No. Davis Steering Gear Mechanism 1. It consists of two turning and two sliding
pairs It consists of four turning pairs
2. This mechanism
correct steering of This mechanism does not meet
exactly the condition of correct
steering
3. The mechanism lie front side of front
wheels The mechanism lie back side of front
wheels
4. Due to presence of sliding pairs, the
friction is more and they wear out
rapidly Due to non sliding pairs friction is
less and hence wear is less
Q. 331
Q. 332
Q. 333
gives
condition
Ackermann Steering Gear
Mechanism
The Ackerman steering gear mechanism is preferred to the
Davis steering gear mechanism because _____________.
(a) whole of the mechanism in the Ackerman steering gear
is on the back of the front wheels.
(b) the Ackerman steering gear consists of turning pairs
(c) the Ackerman steering gear is most economical
(d) both (a) and (b)
Ans. : (d)
Explanation : Refer Q330
Davis steering gear is not used because _____________.
(a) it has turning pairs
(b) it is costly
(c) it has sliding pairs
(d) it does not fulfill the condition of correct gearing
Ans. : (c)
Explanation : Refer Q330
For correct steering of a four wheeled vehicle the condition
to be satisfied is _____________.(a) cot  – cot  = W
H (b) tan  – tan  =
W
H
(c) cos  – cos  = W
H (d) cot  – cot  W
=
H

where
 = Angle through which the axis of inner forward wheel
turns
 = Angle through which the axis of outer forward wheel
turns.
W = Distance between points of front wheels
H = Wheel base.
Ans. : (a)
Explanation :
Equation (a) represents the fundamental equation for
correct steering. If this condition is satisfied then the
motion of wheels will be of pure rolling and no skidding will
take place. The mechanisms used for automatically
adjusting the values of  and  for correct steering are
called steering gears.
Q. 334
In a Davis steering gear the distance between the pivots of
the front axle is 120 cm and the wheel base is 260 cm.
When the vehicle is moving on a straight path, The
inclinations of track arms to the vertical.
(a) 11.99
(b) 12.99
(c) 10.29
(d) 10.40
Ans. : (b)
Explanation :
Given :
Distance between the pivot of the front axle, b = 120 cm
wheel base, l = 260 cm
b
We know that, for correct steering,
tan  =
2 l
120

tan  =
= 0.23075
2  260
Therefore, inclination of track arms with the vertical.
 = 12.99
...Ans.
Q. 335 The ratio between the width of the front axel and that of
wheel base of a steering mechanism is 0.44. At the instantwhen the front inner wheel is turned by 18 what should be
the angle turn by the outer front wheel for perfect steering.
(a) 18.9
(b) 12.45
(c) 15.9
(d) 10.87
Ans. : (c)
Explanation :
Given :
Ratio of the front axle to wheel base, b / l = 0.44
Angle made by inner wheel,  = 18
We know that, for correct steering,
b
cot  − cot  =
 cot  − cot 18 = .44
l
cot  − 3.078 = .44
 cot  = 3.518
 = 15.9
...Ans.
❑❑❑Chapter 2
Static and Dynamic Force
Analysis
Multiple Choice Questions for Online Exam
Radius of Gyration of Rigid Bodies
Q. 1
Inertia force acts __________.
(a) perpendicular to the accelerate force
(b) along the direction of accelerate force
(c) opposite to the direction of accelerate force
(d) in any direction w.r.t. accelerate force depending on
the magnet two
Ans. : (c)
Explanation :
•

Inertia force is an imaginary force, which when acted
upon a rigid body, brings it to an equilibrium position.
It’s magnitude is equal to the accelerating force, but
opposite in direction. Mathematically,
Inertia force, F I = − Acceleration force
F I = − mf
Q. 2
where, m = mass of the rigid body
f = linear acceleration
A rigid body is said to be in equilibrium if __________.
(a)  F x = 0
(b)
 F y = 0
(c)  M z = 0
(d)
all of the above
Ans. : (d)
Explanation :D’Alembert’s principle states that, “The resultant force or
resultant torque acting on the body together with inertia
force or inertia torque respectively will keep the body in
equilibrium position.”
Q. 3
In simple harmonic motion __________.
(a) the acceleration is inversely proportional
displacement
(b) the acceleration is always equal to zero
(c) the acceleration is directly proportional
displacement
(d) none of the above.
to
to
Ans. : (c)
Explanation :
The Equation of acceleration of simple harmonic motion is
given by, Acceleration =  2  Displacement
Hence in simple harmonic motion the acceleration is
directly proportional to displacement.
Q. 4
In simple Harmonic Motion the acceleration is ___-
_______.
(a) directly proportional to displacement
(b) inversely proportional to displacement
(c) directly proportional to velocity
(d) inversely proportional to velocity.
Ans. : (a)
Explanation : Refer Q3
Q. 5
In a simple pendulum the time of oscillation depends
upon __________.
(a) mass of the pendulum and the length of the
pendulum
(b) mass of the pendulum and the local acceleration due
to gravity
(c) total acceleration due to gravity and length of
pendulum
(d) mass of pendulum, the length of pendulum and the
local acceleration due to gravity.Ans. : (c)
Explanation :
The periodic time of oscillation of simple pendulum is given by,
t p =
Q. 6
1
= 2
f n
l
g
sec/cycle
If the radius of gyration of a compound pendulum about
an axis through C. G. is more, then its frequency of
oscillation will be __________.
(a) less (b) more
(c) same (d) data are insufficient to determine same
Ans. : (a)
Explanation :
Therefore the frequency of oscillation of compound pendulum
is given by,
Q. 7
•
•
f n = 1
2 
f n = 1
2 
Angular acceleration
Angular displacement
g l
k 2 + l 2

cycle/sec or Hz
The Bifilar suspension method is used to determine ___-
_______.
(a) natural frequency of vibration
(b) position of balancing weights
(c) moment of inertia
(d) centripetal acceleration
Ans. : (c)
Explanation :
We know that the mass moment of inertia of rigid body is
given by,
I =
m k 2
where, m =
mass of rigid body and
k =radius of gyration of rigid body
For symmetrical and simple geometrical shapes, we can find
the mass moment of inertia by using simple geometrical
relations. But for components having complicated geometry•
Q. 8
•
Q. 9
like connecting rod, it is not possible to find the mass
moment of inertia by geometrical relationship. In such cases,
the radius of gyration of complicated components is
determined by experimentally.
Following experimental methods are used for finding out the
radius of gyration of components which have complicated
geometry.
1. Compound pendulum
2. Bifilar suspension
3. Trifilar suspension.
The simple method to find mass moment of inertia of
connecting is __________.
(a) Compound pendulum (b)
Bifilar suspension
(c) Trifilar suspension
(d)
All of above
Ans. : (a)
Explanation : If a rigid body is suspended vertically, and it
oscillates with small amplitude under the action of the force
of gravity, then the body is known as compound pendulum.
The mass moment of inertia of flywheel can be determine
by __________.
(a) Compound pendulum (b)
Bifilar suspension
(c) Trifilar suspension
(d)
All of above
Ans. : (c)
Explanation :
•
The mass moment of inertia of a rigid body can also be
determined experimentally by an apparatus called Trifilar
suspension or Torsional pendulum.
In this method, the rigid body whose mass moment of
inertia is to be determined (say disc or flywheel) is
suspended by three long parallel strings.
Q. 10
The period of oscillation of bifilar suspension system is
given by __________.
(a) t p =
2k
xy
l
g
(b)
t p = 2k
l
gxy(c) t p =
2k
xy
g
l
(d)
t p =
l
2k
g
xy
Ans. : (b)
Explanation :
Therefore the frequency of oscillation of bar AB is given by,
f n = 1
2 
= 1
2 

f n =
Angular acceleration
Angular displacement
g  x  y
l k 2
g  x  y
l
1
2  k
cycle/sec or Hz
The periodic time of oscillation of bar AB is given by,
t P =
1
= 2  k
f n
1
g xy
sec/cycle
When wires are attached at equal distance from the centre of
gravity of rigid body (i.e. x = y) in such case periodic time of
oscillation of rigid body is given by,
t P =
Q. 11
2  k
x
l
g
sec/cycle
In bifilar suspension system if two wires are attached at
equal distance from C.G. of rigid body then period of
oscillation is given by __________.
(a) t p = 2k
(c) t p = 2
g
l
g
l k
(b) t p = 2
(d) t p = 2k
l k
g
l
g
Ans. : (d)
Explanation : Refer Q10
Q. 12
The period of oscillation of trifilar suspension system is
given by __________.
(a) t p = 2k
l
gx
(b)
t p =
2k
g
l
x(c) t p =
l
2k
x
(d)
g
t p =
2k
x
g
l
Ans. : (c)
Explanation :
The frequency of oscillation of disc becomes,
f n =
g  x 2
Angular acceleration
1
=
Angular displacement 2 
1
2 

f n =
k 2  l
g
l
x
2  k
cycle/sec or Hz
The periodic time of oscillation of disc is given by,
t P =
Q. 13
l
1 2  k
=
f n
x
sec/cycles.
g
If k = radius of gyration of compound pendulum about
axis through C.G., l = distance of the C.G. from axis of
suspension, then the period of oscillation of the compound
pendulum will be __________.
(a) 2
(c) 2
l
g
gh
l + k
2
2
(b) 2
(d) 2
gh
l + k 2
2
l 2 + k 2
g l
Ans. : (d)
Explanation : Refer section 2.5.1
Therefore the frequency of oscillation of compound pendulum
is given by,
f n = 1
2 
f n = 1
2 
Angular acceleration
Angular displacement
g l
k 2 + l 2

cycle/sec or Hz
The periodic time of oscillation of connecting rod is given by,
t p
=
1
= 2 
f n
k 2 + l 2
sec/cycle
g lQ. 14
•
Q. 15
Q. 16
Q. 17
A compound pendulum can be treated for analysis as
simple pendulum if __________.
(a) length of compound pendulum is same as that of
simple pendulum
(b) mass of compound pendulum is same as that of
simple pendulum.
(c) time period of oscillation of compound pendulum is
same as that of simple pendulum
(d) compound pendulum and simple pendulum give time
period of oscillation of only 1 sec.
Ans. : (b)
Explanation : If we convert compound pendulum system of
actual connecting rod into equivalent simple pendulum by
keeping mass ‘m’ constant, the length of simple pendulum is
going to change.
When l = k, the time period of oscillation of compound
pendulum is __________.
(a) maximum
(b)
minimum
(c) zero
(d)
1 sec.
Ans. : (b)
Explanation : Refer Q13
The minimum time period of oscillation of compound
pendulum is __________.
(a) 2 2k
g (b) 2 k
g
(c) 2 k
g (d) 2 k
2g
Ans. : (a)
Explanation : Refer Q13
A connecting rod of mass 5 kg is placed on a platform
whose mass is 3 kg. It is suspended by 3 equal wires each
1.5 m long from a rigid support. The wires are equally
spaced around the circumference of a circle 150 mm
radius. When the C.G. of connecting rod coincides with
axis of circle and platform makes 15 oscillations in 40 sec,
find the M.I. of the system.
(a) 0.3162
(b) 0.2107
(c) 0.4108 (d) 0.1078
Ans. : (b)Explanation :
Given :
Mass of connecting rod, m 1 = 5 kg
Mass of platform, m 2 = 3 kg
Length of each wires, l = 1.5 m
Distance between string and C.G. of platform, x = 150 mm
= 0.15 m
Periodic time of oscillation of whole system is,
40
t P =
= 2.66 sec/cycle
15
The period of oscillation of whole system when suspended
by three wires by, trifilar suspension theory is given by
the equation,
2  k
l
t P =
x
g
2  k
1.5
=
0.15
9.81
Radius of gyration, k = 0.1623 m
The mass moment of inertia of whole system about the
C.G. of platform is,
I G = (m 1 + m 2 ) k 2 = (5 + 3) (0.1623) 2
2.66 =
Q. 18
= 0.2107 kg m 2
...Ans.
A connecting rod with mass 3 kg oscillates 50 times in one
minute when suspended at a small end. Find its mass
moment of inertia about the axis passing through its
centre of gravity which is located at 300 mm from small
end.
(a) 0.012 (b) 0.023 (c) 0.042
(d) 0.052
Ans. : (d)Explanation :
Given : Mass of connecting rod, m = 3 kg
Distance of C.G. from small end, l = 300 mm = 0.3 m
Periodic time for one oscillation is,
60
t P =
= 1.2 sec/cycle
50
t P = 2 
Period of oscillation,
k 2 + l 2
g l
2

1.2 = 2 
2
k + (0.3)
9.81  0.3
 Radius of gyration, k = 0.1317 m
Moment of inertia of connecting rod about the centre of
gravity is,
I G = m k 2 = 3 (0.1317) 2

Q. 19
I G = 0.052 kg  m 2
...Ans.
A connecting rod of mass 3.5 kg is suspended by two wires
each of 2 m length. The wires are attached to the rod at
points 150 mm on either side of the centre of gravity. If
the connecting rod makes 35 oscillations in 60 seconds.
Find the radius of gyration and the mass moment of
inertia of the connecting rod about its centre of gravity.
(a) 0.0287
(b)
0.0187
(c) 0.0365
(d)
0.0485
Ans. : (a)
Explanation :
Given :
Mass of the connecting rod, m = 3.5 kg
Length of wires, l = 2m
Distance of wire from centre of gravity on either side,
x = y = 150 mm

x = y = 0.15 m
Periodic time of one oscillation of connecting rod is,
60
t P =
= 1.7142 sec/cycle
35The period of oscillation of connecting rod when suspended
by two strings is given by, bifilar suspension theory is,
t P = 2  k

1.7142 = 2  k

l
g xy
2
9.81  (0.15) (0.15)
k = 0.09064 m
Mass moment of inertia of connecting rod about the centre
of gravity is,
I G = m  k 2 = 3.5 (0.09064) 2
= 0.0287 kg m 2
...Ans.
Dynamically Equivalent System and Correction Couple
Q. 20
A rigid body can be replaced by two masses, connected
rigidly together. The system of two masses will be
dynamically equivalent to the rigid body if __________.
(a) the mass of the two systems are the same
(b) the centre of gravity of the two mass system coincides
with the centre of gravity of the rigid body
(c) the total mass moment of inertia of the two mass
system about an axis through the centre of gravity is
equal to that of rigid body about the same axis
(d) all of the above
Ans. : (d)
Explanation :
•
A two mass dynamically equivalent system must satisfy the
following three conditions :
(1) The sum of the two masses must be equal to the mass
of the rigid body.

m 1 + m 2 = m
...(i)
(2) The centre of gravity of the two masses must coincide
with that of the rigid body.
...(ii)

m 1 l 1 = m 2 l 2
(3) The moment of inertia of the two mass systems must
be the equal to moment of inertia of the rigid body,I G 1 + I G 2 = I G

•
Q. 21
2
2
m 1 l 1 + m 2 l 2 = m k 2
...(iii)
If all three conditions are satisfied then the system is called
as two point mass dynamically equivalent system.
A rigid body, under the action of external forces, can be
replaced by two masses placed at a fixed distance apart.
The two masses form an equivalent dynamical system, if
__________.
(a) the sum of two masses is equal to the total mass of
the body
(b) the centre of gravity of the two masses coincides with
that of the body
(c) the sum of mass moment of inertia of the masses
about their centre of gravity is equal to the mass
moment of inertia of the body
(d) all of the above
Ans. : (d)
Q. 22
Explanation : Refer Q20
Any distributed mass can be replaced by two point
masses to have the same dynamical properties if ___-
_______.
(a) the sum of the two masses is equal to the total mass
(b) the combined centre of mass coincides with that of
the mass.
(c) the moment of inertia of two point masses about
perpendicular axis through their combined centre of
mass is equal to that of the rod.
(d) all of above.
Ans. : (d)
Q. 23
Explanation : Refer Q20
The rigid body is replaced by two concentrated masses
rigidly connected together. The system will be
kinematically equivalent of the body if __________.
(a) the mass of two masses is equal to the mass of rigid
body
(b) the centre of gravity of two mass system coincides
with the centre of gravity of rigid body(c) the mass moment of inertia of the two mass system
and the rigid body about centre of gravity are equal
(d) all of the above.
Ans. : (d)
Explanation : Refer Q20
Q. 24
Which one of the following conditions is satisfied for a
system to be dynamically equivalent ?
(a) l 1  l 2 = k 2 (b)
(c) l 1 – l 2 = k 2 (d)
l 1 + l 2 = k 2
l 2  l 2 = k
Where l 1 and l 2 = distance of two masses from C.G. of the
body, and k = radius of gyration of the body.
Ans. : (a)
•
Explanation :
A two mass dynamically equivalent system must satisfy the
following three conditions :
(1) The sum of the two masses must be equal to the mass
of the rigid body.

m 1 + m 2 = m
...(i)
(2) The centre of gravity of the two masses must coincide
with that of the rigid body.
...(ii)

m 1 l 1 = m 2 l 2
(3) The moment of inertia of the two mass systems must
be the equal to moment of inertia of the rigid body,
I G 1 + I G 2 = I G

•
•
2
2
m 1 l 1 + m 2 l 2 = m k 2
...(iii)
From above three conditions if only first two conditions are
satisfied, the system is said to be two point mass statically
equivalent systems.
If all three conditions are satisfied then the system is called
as two point mass dynamically equivalent system.
Substituting the value of m 2 from Equation (ii) in Equation
(i), we get,
m 1 l 1
m 1 +
= m m 1 (l 1 + l 2 ) = m l 2
l 2
Similarly we get,
m 1 = m l 2
(l 1 + l 2 ) ...(1)
m 2 = m l 1
(l 1 + l 2 ) ...(2)
On substituting the value of m 1 and m 2 in Equation (iii) we
get,
m l 2
m l 1
(l ) 2 +
(l ) 2
(l 1 + l 2 ) 1
(l 1 + l 2 ) 2
or
(l 1 + l 2 ) l 1 l 2
(l 1 + l 2 )
or,
•
•
Q. 25
=
m k 2
= k 2
k 2 = l 1 l 2
...(3)
If the distance of one of the mass (i.e. either l 1 or l 2 ) is
selected arbitrarily then the other distance can be obtained
from Equation (3)
Therefore, the Equations (1), (2) and (3) represents the
essential condition-of placing the two masses, such that the
system becomes dynamically and kinematically equivalent
system.
Fig. 1 shows a rigid
body of mass m having
radius of gyration K
about its centre of
gravity. It is to be
replaced
by
in
equivalent dynamical
system of two masses
placed at A and B. The
mass at A should be
__________.
a  m
b  m
(a)
(b)
(a + b)
(a + b)
m
m
(c)
(d)
2
3
Ans. : (a)
Explanation : Refer Q24
Fig. 1Q. 26
Fig. 1 shows a rigid body of mass m having radius of
gyration K about its centre of gravity. It is to be replaced
by in equivalent dynamical system of two masses placed
at A and B. The mass at B should be __________.
a  m
b  m
(a)
(b)
(a + b)
(a + b)
m
m
(c)
(d)
2
3
Ans. : (b)
Q. 27
Explanation : Refer Q24
Refer Fig. 2. The mass
of rigid body is m and
radius of gyration is k.
The
kinetically
equivalent
system
shows masses m 1 and
m 2 . The mass m 1 should
be __________.
Fig. 2(a) m 1 = m l 2
l 1 + l 2 (b) m 1 =
m l 1
l 1 + l 2
(c) m 1 = m (l 1 l 2 )
l 1 + l 2 (d) m 1 = 
l 2 
m
 l 1 + l 2  2
Ans. : (a)
Q. 28
Explanation : Refer Q24
Refer Fig. 2. The mass of rigid body is m and radius of
gyration is k. The kinetically equivalent system shows
masses m 1 and m 2 . The mass m 2 should be __________.
(a) m 2 = m l 2
l 1 + l 2 (b) m 2 =
m l 1
l 1 + l 2
(c) m 2 = m (l 1 l 2 )
l 1 + l 2 (d) m 1 = 

l 2 
m
l 1 + l 2  2
Ans. : (b)
Explanation : Refer Q24
Q. 29
Two systems shall be dynamically equivalent when ___-
_______.
(a) the mass of two are same
(b) example of two coincides
(c) M.I. of two about an axis through example is equal
(d) all of the above
Ans. : (d)
Q. 30
Explanation : Refer Q24
The essential condition of placing the two masses, so that
the system becomes dynamically equivalent is __________.
2
(a) l 1  l 2 = k G
(b)
l 1  l 2 = k G
(c) l 1 = k G
(d)
l 2 = k G
where l 1 and l 2 = Distance of two masses from the centre
of gravity of the body, and k G = Radius of gyration of the
body.
Ans. : (a)
Explanation : Refer Q24Q. 31
For a link to be dynamically equivalent, which condition
needs to be satisfied ?
(a) k 2 = l 1 / l 2
k 2 = l 2 / l 1
(b)
2
k 2 = l 1 + l 2
(c) k = l 1 l 2
(d)
Ans. : (c)
Explanation : Refer Q24
Q. 32
In
a
dynamically-equivalent
system,
a
uniformly
distributed mass is divided into _______point masses.
(a) two
(b)
three
(c) four
(d)
five
Ans. : (a)
Q. 33
Explanation : Refer Q24
If m is the mass of connecting rod, k is the radius of
gyration of connecting rod, k 1 is the new radius of
gyration of connecting rod and  is the angular
acceleration of connecting rod then correction couple is
given by __________.
[
= m [ k
(a) T C = m k 2 – k 21
(c) T C
2
1
k 2
]
]  (b) T C = m k 21 – k 2
[
m
=
k
 [
 (d) T C
2
1
] 
– k ]
2
Ans. : (b)
Explanation :
Torque required to accelerate the connecting rod or
dynamically equivalent system is,
T = I  
Torque
required
to
T
accelerate
= m (k) 2  
the
non
...(i)
dynamically
equivalent system is,
T 1 = I 1  
T 1
= m (k 1 ) 2  
...(ii)
The correction couple is the difference between the torque
required to accelerate non dynamically equivalent system and
torque required to accelerate the dynamically equivalent system.
T c = T 1 − T
On substituting value of T 1 and T in above equation we get,
T C = m (k 1 ) 2  − m (k) 2 

2
T C = m [k 1 − k 2 ] ...(1)
2
On substituting value of k 1 = l 1 l 3 and k 2 = l 1 l 2 in above
equation we get,
T C = m [l 1 l 3 − l 1 l 2 ] 

•
T C = m l 1 [l 3 − l 2 ] ...(2)
This correction couple must be applied, when the masses are
placed arbitrarily to make the system dynamically
equivalent.
Static and Dynamic Force Analysis
Q. 34
Crank effort is the net force applied at the crankpin
________to the crank which gives the required turning
moment on the crankshaft.
(a) Parallel
(b)
Perpendicular
(c) at 45
(d)
135
Ans. : (b)
Explanation :
F T = Tangential force acting at crank pin or force in
direction perpendicular to the crank,Q. 35
State which of the following statement is true ?
(a) The inertia force is equal in magnitude and opposite
in direction to accelerating force
(b) The magnitude of inertia force is given by the
expression
cos 2 
F I = m r  2 r  cos  +
where symbols have their
2n 

usual meanings
(c) D-Alembert’s principle is used to reduce a dynamic
problem into an equivalent static problem
(d) All are true statements
Ans. : (d)
Explanation : Refer Q1,Q2 and
We know that acceleration of reciprocating part is given by,
f P
=  2 r  cos  +

cos 2  
n 
Therefore axial force acting on piston or inertia force due to
mass at point P is,
F P = F I = equivalent mass acting at point P  acceleration of
reciprocating parts


Q. 36
F P = F I = m e  f P
F P = F I = (m R + m 1 )  2 r  cos  +

cos 2  
n 
In horizontal engine piston effort is given by __________.
(a) F p = F g  F I – F F (b) F p = F g + F I ∓ F F
(c) F p = F F + F g ∓ F I (d) F p = F g ∓ F I + F F
Where, F g = Net gas force; F I = Inertia force;
F F = Friction force
Ans. : (b)
Explanation :
Piston effort,
−
F P =Net gas force − Inertia force + frictional force
−
 F P = F g − F I + F FIf F p is the piston effort and  is the obliquity angle of
Q. 37
connecting rod, the piston side thus F N is __________.
(a) F p tan 
(c) F p cos 
(b)
(d)
F p sin 
F p cot 
Ans. : (a)
Explanation :
Consider triangles OBC and PBC in Fig. 2.8.2 we can write,
CB = l  sin = r  sin
r
sin 
 ··· n = l 

sin  = sin  =
l
n
r 

cos  =
But,
1 − sin 2 
On substituting value of sin 2  ,we can write above equation
as,
cos 

1 −
=
cos  =
1
n
sin 2 
=
n 2
n 2 − sin 2 
n 2
n 2 − sin 2 
Consider triangle of
forces PCB shown in Fig. Q37
On resolution of forces,
We can write,
F N = F P  tan
Q. 38
Fig Q37
If F p is the piston effort and  is the obliquity angle of
connecting rod, the force along connecting rod F Q is ___-
_______.
F p
(a)
tan 
F p
(c)
sin 
(b) F p
cos 
(d) F p cos 
Ans. : (b)
Explanation :
Form Fig. Q37 the force on connecting rod,F P
F Q =
cos 
On substituting value of cos  from above Equation we get,
F P
F Q =
1
n 2 − sin 2 
n
F Q =
n F P
n − sin 2 
If F Q is force acting on connecting rod,  is the obliquity
angle of connecting rod and Q is the angle made by crank,
the tangential force acting on crank pin F T is __________.
Q. 39
(a) F Q sin ( + )
(c) F Q sin ( – )
2
(b) F Q cos ( + )
(d) F Q cos ( – )
Ans. (a)
Explanation :
From Fig. Q39 the tangential force,
F T = F Q sin ( + )
or
F T =F Q  (sin  cos  + cos  sin )
Fig. Q39
Q. 40
If F Q is force acting on connecting rod,  is the obliquity
angle of connecting rod and Q is the angle made by crank,
the radial force acting along the crank shaft F R is ___-
_______.
(a) F Q sin ( + )
(b)
F Q sin ( – )
(c) F Q cos ( + )
(d)
F Q cos ( – )
Ans. : (c)
Explanation :
Refer Fig. Q.39. The component of force F Q in radial direction
i.e. the force acting along the crank is given by
F R = F Q cos ( + )Q. 41
The turning moment of crank shaft is __________.
F T
(a)
F T  r
(b)
r
(c) F T r
(d)
F T
r
Where, F T = Crank effort
r = radius of crank
Ans. : (c)
Explanation : The torque acting on crank shaft called
turning moment is the product of the crank effort (F T ) and the
radius of crank, (r).
Turning moment of crankshaft,
Q. 42
T = F T  r
If P is the piston effort and  is the obliquity angle of
connecting rod, the expression for thrust in connecting
rod is __________.
(a) P/cos 
(b)
P  cos 
(c) P tan 
(d)
P/sin 
Ans. : (a)
Explanation : Refer Q38
Q. 43
If piston effort of 100 kN is acting and crank is made 45 
from TDC and obliquity ratio is 2, then the thrust in
connecting rod is __________.
(a) 110 kN
(b)
106.90 kN
(c) 95.20 kN
(d)
109.70 kN
Ans. : (b)
Explanation :
Given :
F P = 100 kN,
 = 45 ,
 sin  =
n = 2
sin 
sin 45 
=
= 0.3535
n
2
  = 20.70
Thrust in connecting rod is,
F Q =
F p
100
=
cos 
cos 20.70 
= 106.90 kN
...Ans.Q. 44
If piston effort of 100 kN is acting and crank is made 45 
from TDC and obliquity ratio is 2, then the piston side
thrust is __________.
(a) 38.40 (b) 35.40 (c) 36.58
(d) 37.78
Ans. : (d)
Explanation :
Given :
F p = 100 kN,  = 45  ,
sin  =
n = 2
sin  sin 45 
=
= 0.3535
n
2
 = 20.70
Piston side thrust is,
F N = F p tan 
= 100  tan (20.70  )
= 37.78 kN
Q. 45
...Ans.
If piston effort of 100 kN is acting and crank is made 45 
from TDC and obliquity ratio is 2, then tangential force
acting on crank shaft is __________.
(a) 95.42 (b) 97.42
(c) 93.42
(d) 92.38
Ans. : (b)
Explanation :
Given :
F p = 100 kN,  = 45  ,
n = 2
sin  sin 45 
sin  =
=
= 0.3535
n
2
F p
 = 20.70 and F Q =
cos 
=
100
= 106.90 kN
cos 20.70 
Tangential force acting on crank shaft is,
F T = F Q sin ( + )
= 106.90 sin (45  + 20.70  )
F T = 97.42 kN
...Ans.Q. 46
If piston effort of 100 kN is acting and crank is made 45 
from TDC and obliquity ratio is 2, then radial force or
load on main bearing is __________.
(a) 44.99 (b) 42.99
(c) 43.99
(d) 41.99
Ans. : (c)
Explanation :
F p = 100 kN,  = 45  , n = 2
Given :
sin  sin 45 
=
= 0.3535
n
2
F p
 = 20.70 and F Q =
cos 
sin  =
=
100
= 106.90 kN
cos 20.70 
Radial force acting on main bearing is,
F R = F Q cos ( + )
= 106.90 cos (45 + 20.70)
Q. 47
= 43.99 kN
...Ans.
The tangential force acting on engine is 100 kN and
radius of crank is 500 mm, then torque acting on crank is
__________.
(a) 60 kN m
(b)
50 kN m
(c) 45 kN m
(d)
55 kN m
Ans. : (b)
Explanation :
Given :
F T = 100 kN,
r = 500 mm = 0.5 m
The torque acting on crank shaft is,
T = F T  r = 100  0.5 = 50 kN  m
...Ans.
Friction
Q. 48
According to the law of dry friction, the force of friction
__________.
(a) Is independent of area of contact.
(b) Is directly proportional to the normal pressure
between the surface of contact.(c) Is independent of velocity of sliding.
(d) Is dependent on material of bodies in contact.
(e) All of the above.
Ans. : (e)
Explanation :
Laws of solid or dry friction are given as follows :
• The force of friction is directly proportional to the normal
reaction between the surfaces in contact.
• The force of friction depends upon the nature of material of
the surfaces in contact.
• The force of friction is independent of the area of contact
between the surfaces.
• The force of friction is independent of the velocity of sliding of
one surface relative to the other surface.
Q. 49
Which of the following statements is true when ___-
_______.
(a) tan  =  body will move over the surface
(b) tan  >  the motion of body over the surface is not
possible
(c) tan  cannot be smaller than 
(d) tan  <  the motion of body over the surface is not
possible
Ans. : (a)
Explanation :
According to first law of dry friction,
Friction force is directly proportional to the normal reaction
between the surfaces in contact i.e.
F r  R n


F r =  R n
F r
 =
R n
...(ii)
where,  =is constant of proportionality and called
as coefficient of friction and it’s value dependsupon material of two surfaces and surface
roughness
Also from Fig. 2.14.1(b) we can write,
tan  = F r
Friction Force
=
Normal Reaction R n
 tan  =  R n
...[ ··· F r =  R n ]
R n
 tan  = 
or
 = tan − 1 () ...(iii)
This angle  is known as limiting angle of friction or angle of
friction.
Q. 50
Angle of inclination of plane at which body starts moving
downwards is called __________.
(a) Angle of projection
(b)
Horizontal angle
(c) Angle of friction
(d)
Angle of repose.
Ans. : (d)
Explanation :
(1)
A body will not move over the surface until the angle of reaction R
with normal reaction R n becomes equal to limiting angle  i.e. when
tan  = , the body will move over the surface irrespective of the
magnitude of the external force F o .
(2) If tan  < , then motion of body over the surface is not possible.
(3) tan  cannot be greater than , or  cannot be greater than tan − 1 .
Q. 51
If the angle of friction is tan , then which of the following
cannot be the value of tan  ?
(a) 0.1  (b) 0.5 
(c) 
(d) 1.5 
Ans. : (d)
Explanation : Refer Q50Q. 52
Shaft of radius r is revolving inside a bearing then radius
of friction angle is given by __________.
(a) r sin 
(b) r tan 
(c) r cos 
(d) r cot 
Ans. : (a)
Explanation :
Let,
Fig. Q52 shows journal bearing in which shaft is stationary.
W = Vertical (radial) load on shaft ;
r = Radius of shaft (Journal)
R n = Normal reaction exerted by bearing surface on shaft


R

=
=
=
=
Coefficient of friction between shaft and bearing
Friction angle
Resultant reaction
Angular speed of shaft
(a) When shaft is stationary
(b) When shaft is rotating
Fig. Q52 Friction circle
•
Consider now a shaft is rotating in clockwise direction as
shown in Fig. 2.16.1(b).
•
The lubrication is used to reduce friction which forms a thin
layer between shaft and bearing. It gives rise to boundary
friction.
•
When shaft rotates, a frictional force F =  R n acts at the
circumference of the shaft which opposes the motion of shaft
and also shift the point of contact from point A to B as shown
in
Fig. 2.16.1(b). Therefore, the normal reaction does not act
vertically upwards, but it acts at point B. The resultantreaction R produced by bearing will act at an angle  with
R n .
•
Under equilibrium, R must be equal to W, the two forces W
and R are parallel but not in same line. Therefore, these
forces form a couple which opposes the torque required to
produce the motion. This couple is known as friction couple
or frictional torque which is given by,
T = W  OC = W  OB sin  = W  r sin 
 T = W  r tan 
...[ ··· sin  = tan 
when  is very small]
 T = W  r  
...[ ···  = tan ]
 T =  W r
Power lost in friction is given by,
P = T  
where,  is angular speed of shaft.
Therefore, if a circle is drawn with O as centre and radius
OC = r sin  or r , then the circle is called the friction circle of a
bearing.
Q. 53
Boundary film lubrication is __________.
(a) thin film lubrication
(b)
thick film lubrication
(c) solid lubrication
(d)
none of the above
Ans. : (a)
Explanation : When friction occurs between two rubbing
surfaces having a very thin layer of lubricant between the
surfaces, it is known as boundary friction or non-viscous
friction.
Q. 54
When a journal is rotating in its bearing, the resultant of
the normal force (R N ) and frictional force (R N ) ___-
_______.
(a) passes through the centre of the journal
(b) is tangent to the circle of the journal at the point of
contact
(c) is tangent to a small circle of radius r  sin  where
 = tan –1 (d) none of the above
Ans. : (c)
Explanation : Refer Q52
Q. 55
Friction circle in case of journal bearing is __________.
(a) the circle in which resultant force F is tangent
(b) the circle which is having radius equal to half of the
radius of the journal
(c) the circle which is having radius equal to   R where
R is the radius of the journal
(d) both (a) and (c)
Ans. : (d)
Q. 56
Explanation : Refer Q52
The radius of friction circle increases __________.
(a) with the increase of radius of journal
(b) with the increase of speed
(c) with the increase of co-efficient of sliding friction
(d) (a) and (c) only
Ans. : (d)
Q. 57
Explanation : Refer Q52
A friction circle is a circle drawn when the journal rotates
in a bearing. Its radius depends on the coefficient of
friction and __________.
(a) magnitude of the forces on the journal
(b) angular velocity of the journal
(c) clearance between the journal and the bearing
(d) radius of the journal
Ans. : (d)
Q. 58
Explanation : Refer Q52
The tendency of a body to resist change from rest or
motion is known as __________.
(a) mass
(b)
friction
(c) inertia
(d)
resisting force
Ans. : (c)
Explanation : Inertia is the property of body by virtue of which it
remains in its state or uniform in motion until a force is
applied on it.Q. 59
The friction force, which comes into play, when the sliding
motion of the body is just to commence, is called
__________.
(a) static friction
(b)
dynamic friction
(c) limiting friction
(d)
none of these
Ans. : (c)
Q. 60
Explanation : Refer Q49
The kinematic friction is the friction experienced by a
body, when the body __________.
(a) is at rest
(b)
just beings to slide
(c) is in motion
(d)
none of these.
Ans. : (c)
• Explanation : It is impossible to produce a perfectly smooth
surface. If a block is placed on a surface, interlocking of
projecting particles takes place due to surface roughness of
the surfaces.
• When block tends to move with respect to the mating surface,
an opposing force acts in a direction opposite to motion of the
block which called as force of friction or simply friction.
Q. 61
The radius of a friction circle for a shaft of radius r,
rotating slide a bearing with coefficient, of friction  = tan
, is __________.
(a) (r/2) tan 
(b)
r  sin 
r
(c) r  cos 
(d)
sin 
2
Ans. : (b)
Q. 62
Q. 63
Explanation : Refer Q52
The friction force is independent of __________.
(a) material
(b)
area of contact
(c) normal reaction (d)
surface finish
Ans. : (b)
Explanation : Refer Q48
Choose the correct statement.
(a) The force of friction always opposes the relative
motion of one surface over another.(b) The force of friction is directly proportional to normal
load between surfaces in contact.
(c) The force of friction is independent of are as of
contact between the surfaces for a given load.
(d) All of the above.
Ans. : (d)
Explanation : Refer Q48
Q. 64
The radius of the friction circle __________.
(a) increases as the load increases
(b) increases as the radius of journal increases
(c) increases as the speed of the shaft in journal
increases
(d) increases as the co-efficient of friction increases.
Ans. : (d)
Explanation : Refer Q52
Q. 65
The radius of friction circle for a shaft rotating inside a
bearing of radius r is equal to __________.
(a) r sin 
(b) r cos 
(c) r
(d) r/cos 
Ans. : (a)
Explanation : Refer Q52
Q. 66
When journal is rotating, the resultant of the normal
force R n and tangential force R n __________.
(a) passes approximately through the centre of journal
(b) approximately tangential to the circle of the journal
of radius r at contact point
(c) is approximately tangential to the small circle of
radius = r
(d) none of the above.
Ans. : (c)
Explanation : Refer Q52
❑❑❑Chapter 3
Friction Clutches
Multiple Choice Questions for Online Exam
Bearings
Q. 1
The uniform wear conditions are used __________.
(a) for maximum power to be transmitted
(b) for safe design of clutches
(c) for safe design of bearings
(d) for overcoming the frictional torque.
Ans. : (b)
Explanation :
For given dimensions, the frictional torque determined by uniform pressure
theory is more than the torque calculated by uniform wear theory. For safe
design of bearings, the uniform pressure theory should be used since it will
give us maximum power that is lost in overcoming the friction so that net
power available is minimum for safe design.While in case of clutches, friction
is used for transmission of power. It would be safe to design a clutch by using
uniform wear theory .
Q. 2
Q. 3
The uniform pressure conditions are used __________.
(a) for maximum power to be transmitted
(b) for safe design of clutches
(c) for safe design of bearings
(d) for overcoming the frictional torque.
Ans. : (c)
Explanation : Refer Q1
For safe design of bearing, we assume __________.
(a) maximum power to be transmittedQ. 4
(b) safe design of clutches
(c) safe design of bearings
(d) overcoming the frictional torque.
Ans. : (a)
Explanation : Refer Q1
For safe design of clutches, we assume__________.
(a) Uniform pressure conditions
(b) Uniform wear conditions
(c) Neither uniform pressure nor uniform wear
(d) All of the above.
Ans. : (b)
Explanation : Refer Q1
Q. 5
According to uniform wear theory __________.
(a) P  = constant
(b)
P r = constant
(c) P W = constant
(d)
P T = constant
Where  = Angular velocity of shaft,
r = radius of shaft
W = Axial lead on shaft,
T = Fractional torque
Ans. : (b)
Explanation : Assuming that wear which takes place is
uniform over the entire bearing surface i.e.
p  r = Constant = C
Q. 6
In case of foot step bearing, the frictional force under the
conditions of uniform pressure may be assumed to be
acting at __________.
1
3
2
(a) r
(b)
r
(c)
r
(d)
r
2
2
3
Ans. : (d)
Explanation :
The torque required to overcome the friction between shaft and
bearing surface for flat pivot bearing considering uniform pressure
2
theory is given by T=
 W R
3
.Q. 7
In case of foot step bearing, the frictional force under the
conditions of uniform wear may be assumed to be acting
at __________.
1
3
2
(a) r
(b)
r
(c)
r (d)
r
2
2
3
Ans. : (b)
Explanation :
The torque required to overcome the friction between shaft and
bearing surface for flat pivot bearing considering uniform wear
1
theory is given by, T =  W R
2
.
Q. 8
The ratio of friction torque for uniform pressure to
uniform wear for the same dimensions and load is ___-
_______.
1
(a) > 1 (b) = 1
(c) < 1
(d) =
2
Ans. : (a)
Explanation : Friction torque transmitted for uniform
2
pressure is WR and frictional torque transmitted for
3
1
uniform wear is WR, Therefore
2
2
WR
3
4
Ratio =
=
= 1.33 > 1
1
3
WR
2
Q. 9
...Ans.
Torque required to overcome the friction between shaft
and bearing surface in case of multiple flat collar bearing
is __________.
(a) Independent of no of collars
(b) Dependent on radius of collar
(c) Dependent of no of collars
(d) None of above.
Ans. : (a)Q. 10
Sr.
Explanation : Thus torque required to overcome the
friction between shaft and bearing surface is independent
on the number of collars used. The number of collars used
will affect the intensity of pressure only
According to uniform wear theory frictional torque
transmitted by conical pivot bearing is __________.
2
1
(a)
 W R cosec 
(b)
 W R cosec 
3
2
1
(c)
 W R cos 
(d)
 W R cosec 
2
Ans. : (b)
Explanation :
Type of bearing
Uniform pressure Uniform wear
theory theory
p = Constant p  r = Constant
No.
1.
Flat pivot
T =
bearing
2.
Single and
multiple flat
2
 W R
3
T =
1
 W ( R 1 + R 2
2
 R 1 3 – R 2 3 
2
T =  W  2
2 
3
 R 1 – R 2  T = 2
 W R cosec 
3 T = 1
 W R cosec 
2
 R 1 3 – R 2 3 
2
 W  2
2 
3
 R 1 – R 2  T = 1
 W ( R 1 + R 2
2
)
collar bearing
3.
Conical pivot
T =
bearing
4.
Truncated
conical or
1
 W R
2
T =
trapezoidal
) cosec 
cosec 
bearing
Q. 11
Frictional torque
__________.
(a) P (R o + R i )
(c)
(R o3 + R i3 )
1
P
2
(R o2 + R i2 )
as
per
uniform
wear
(b) 1
P (R o + R i )
2
(d) P(R o3 – R i3 )
2(R o2 – R i2 )
theory
isQ. 12
Ans. : (b)
Explanation : Refer Q10
For a flat collar, frictional torque as per uniform pressure
theory is __________.
1
(a) P (R o – R i )
(b)
P (R o – R i )
2
(c)
Q. 13
•
(R o3 – R i3 )
2
P
3
(R o2 – R i2 )
(d)
P(R o3 – R i3 )
2(R o2 – R i2 )
Ans. : (c)
Explanation : Refer Q10
In case of flat pivot bearing, the rubbing velocity is ___-
_______.
(a) maximum at the centre of the contact area
(b) zero at the centre and maximum at the outer radius
(c) uniform throughout the contact area
(d) zero at the outer radius
Ans. : (b)
Explanation : When the bearing becomes old, the bearing
surface will wear out. The wear of the surface at different
points of contact depends upon the product of the intensity of
pressure and the velocity of rubbing at the point of contact.
•
The velocity of rubbing is also different since it depends upon
the radii from the axis of rotation for the given uniform
angular speed of the shaft. It follows that the wear may be
different at different radii and this causes the redistribution
of intensity of pressure. Under these conditions, for uniform
wear, we can assume,
p.r = Constant
Q. 14
The rate of wearing in case of flat pivot bearing is ___-
_______.
(a) inversely proportional to the pressure intensity
(b) directly proportional to the rubbing velocity
(c) directly proportional to the pressure intensity
(d) both (b) and (c)Q. 15
Q. 16
Q. 17
Q. 18
Ans. : (d)
Explanation : Refer Q13
For calculating the friction power for bearing, the
assumptions made in practice are __________.
(a) the rate of wearing at the surface is uniform
(b) the rubbing velocity is constant
(c) the pressure intensity is uniform
(d) (a) and (c) only
Ans. : (d)
Explanation : Refer Q1
In case of flat pivot bearing, the frictional torque
transmitted for uniform pressure as compared to the
frictional torque transmitted
for uniform wear
__________.
(a) is more
(b)
is less
(c) is constant
(d)
may be more or less
Ans. : (a)
Explanation : Refer Q1
The frictional torque, transmitted in case of flat pivot
bearing for uniform pressure, is equal to __________.
2
1
1
(a) WR
(b) WR
(c) WR (d) WR
3
3
2
Where W = Total axial load carried by pivot,
 = Co-efficient of friction, and
R = Radius of bearing surface
Ans. : (b)
Explanation : Refer Q10
The frictional torque, transmitted in case of flat pivot
bearing for uniform wear, is equal to __________.
2
1
1
(a) WR
(b) WR
(c) WR
(d) WR
3
3
2
Where W = Total axial load carried by pivot,
 = Co-efficient of friction, and
R = Radius of bearing surface
Ans. : (d)
Explanation : Refer Q10Q. 19 The ratio of frictional torque transmitted for uniform
pressure to the frictional torque for uniform wear in case
of flat pivot, is equal to __________.
(a) 1/3 (b) 2/3
(c) 1.0
(d) 4/3
Ans. : (d)
Explanation : For flat pivot bearing friction torque
transmitted for uniform pressure is 2/3 WR and
1
frictional torque transmitted for uniform wear is WR,
2
Therefore
2
WR
3
4
Ratio =
=
...Ans.
1
3
WR
2
Q. 20 The frictional torque transmitted in case of a flat collar
bearing for uniform pressure, is equal to __________.
 R 31 – R 32 
(a) W  2
2 
 R 1 – R 2 
(b)
 R 1 – R 2 
2
W  2
2 
3
R – R

3 3
1 2

W
W
(R 1 + R 2 )
(d)
(R 1 – R 2 )
2
2
Where R 1 = External radius of the collar,
(c)
R 2 = Internal radius of the collar, and
Q. 21
W = Total axial load carried by the shaft
Ans. : (b)
Explanation : Refer Q10
The frictional torque transmitted for uniform wear, in
case of flat collar bearing, is equal to __________.
 R 31 – R 32 
(a) W  2
2 
 R 1 – R 2  (b)
W
(R 1 + R 2 )
2 (d)
(c)
Q. 22
 R 1 – R 2 
2
W  2
2 
3
R – R

3 3
1 2

W
(R 1 – R 2 )
2
Ans. : (c)
Explanation : Refer Q10
When the axial load carried by the shaft in case of flat
collar bearing, is large ; a number of collars are provided
on the shaft. This is done __________.Q. 23
Q. 24
(a) to reduce the frictional torque
(b) to increase the frictional torque
(c) to reduce the intensity of pressure
(d) to increase the intensity of pressure.
Ans. : (c)
Explanation : Refer Q9
Choose the correct statement regarding the safe design of
a bearing.
(a) When horse power lost in friction is to be determined,
uniform rate of wear is assumed
(b) When horse power transmitted is to be determined,
uniform pressure is assumed
(c) When horse power lost in friction is to be determined,
uniform pressure is assumed and when horse power
transmitted is to be determined, uniform wear is
assumed
(d) None of the above
Ans. : (c)
Explanation : Refer Q1
The frictional torque transmitted for uniform pressure, in
case of a conical pivot bearing having semi-angle of the
cone as , is equal to __________.
WR
WR
(a)
(b)
sin 
2 sin 
(c)
Q. 25
2 WR
sin 
(d)
2WR
3 sin 
Where W = Total axial load and R = Radius of shaft
Ans. : (d)
Explanation : Refer Q10
The frictional torque transmitted for uniform wear, in
case of a conical pivot bearing having semi-angle of the
cone as , is equal to __________.
WR
WR
(a)
(b)
sin 
2 sin 
(c)
2 WR
sin 
(d)
Ans. : (b)
Explanation : Refer Q10
2WR
3 sin Q. 26
For uniform pressure the ratio of the frictional torque for
a flat collar bearing to the frictional torque for a conical
collar bearing is equal to __________.
(a) 1.33
(b) 1.0
(c) sin 
(d) cos 
Where  = Semi-angle of the cone
Ans. : (c)
Explanation : For uniform pressure frictional torque for a
2
flat collar bearing is WR and frictional torque for
3
2
2 WR
conical collar or bearing is WR cosec or
.
3
3 sin 
2
WR
3
 Ratio =
= sin 
2 WR
3 sin 
Q. 27
...Ans.
For uniform wear, the ratio of the frictional torque for a
flat collar bearing to the frictional torque for a conical
collar bearing, is equal to __________.
(a) 1.33
(b) 1.0
(c) sin 
(d) cos 
Ans. : (c)
Explanation : For uniform pressure frictional torque for a
flat collar bearing is 1/2 WR and frictional torque for
WR
conical collar or bearing is 1/2 WR cosec or 1/2
sin 
1
WR
2
 Ratio =
= sin 
1 WR
2 sin 
Q. 28
...Ans.
The ratio of frictional torque for conical collar bearing to
that for flat collar bearing is proportional to __________.
(a) sin 
(b) cos 
(c) tan 
(d) 1/sin 
Ans. : (d)
Explanation : For uniform pressure frictional torque for
WR
conical collar or bearing is 1/2 WR cosec or 1/2
sin 
and a frictional torque for flat collar bearing is 1/2 WR1 WR
2 sin 
 Ratio =
= 1/sin 
1
WR
2
Q. 29
...Ans.
For flat and conical pivots, ratio of friction torque with
uniform wear to friction torque with uniform pressure is
__________.
(a) 2/3
(b) 3/2
(c) 4/3
(d) 3/4
Ans. : (d)
Explanation : For flat and conical pivots friction torque
with uniform wear is 1/2 and with uniform pressure is
2/3, Therefore
Ratio =
Q. 30
1/2 3
=
2/3 4
...Ans.
The ratio of frictional torque produced for uniform wear to
frictional torque produced for uniform pressure is
__________.
3
4
(b)
4
3
Ans. : (a)
(a)
(c)
2
3
(d) 1.
Explanation : The friction torque with uniform wear is 1/2
and with uniform pressure is 2/3, Therefore
Ratio =
1/2 3
=
2/3 4
...Ans.
Q. 31 The frictional torque for the same diameter in a conical
bearing is ____________than in a flat bearing.
(a) more
(b)
less
(c) equal
(d)
may be more or less
Ans. : (a)
Explanation : Refer Q10
Q. 32 The frictional torque transmitted by a conical collar
bearing is __________.
(a) less than that of flat collar bearing
(b) equal to that of flat collar bearing
(c) more than that of flat collar bearingQ. 33
Q. 34
Q. 35
(d) none of the above
Ans. : (c)
Explanation : Refer Q10
The frictional torque produced due to friction for same
shaft diameter, in conical bearing is __________.
(a) less than that for flat bearing
(b) equal to that for flat bearing
(c) more than that for flat bearing
(d) may be more or less depending upon cone angle.
Ans. : (c)
Explanation : Refer Q10
Choose the wrong statement.
(a) The frictional torque transmitted by a shaft fitted
with a large number of collars is the same as
transmitted by a shaft with one collar
(b) The frictional force for uniform pressure, in case of a
flat pivot bearing, can be assumed to be acting at a
distance of two-thirds of the radius of the shaft from
the axis of shaft.
(c) When horse power lost in friction is to be determined,
uniform pressure is assumed for safe design of
bearing
(d) The frictional force for uniform pressured, in case of a
flat collar pivot bearing, can be assumed to be acting
at a distance of half of the sum of internal and
external radii of the shaft
Ans. : (d)
The frictional torque transmitted in a conical pivot
bearing, considering uniform wear, is __________.
1
2
(a)
   W. R. cosec 
(b)
   W. R. cosec 
2
3
3
(c)
   W. R. cosec 
(d)  W. R. cosec 
4
Where
R = Radius of the shaft, and
 = Semi-angle of the cone.
Ans. : (a)
Explanation : Refer Q10Q. 36
For a pivot bearing with r 1 and r 2 as the outer and inner
radii and W as the axial load, the friction torque,
assuming uniform pressure, is given by __________.
1
1
(a)
 W (r 1 + r 2 )
(b)
 W (r 1 – r 2 )
2
2
 r 1 – r 2 
2
(c)
 W  2
2 
3
 r 1 – r 2 
3
3
(d)
2
 W 
3
 r 21 – r 22 
 2
2 
 r 1 + r 2 
Ans. : (c)
Explanation : Refer Q10
Q. 37
For collared bearing with r 2 and r 1 as inner and outer
radii and W as axial load, assuming uniform wear, is
given by __________.
1
1
(a)
 W (r 1 + r 2 )
(b)
 W (r 1 – r 2 )
2
2
 r 1 – r 2 
1
(c)
 W   2
2 
2
r – r
3

Q. 38
1
3

2
 r 1 – r 2 
2
 W  2
2 
3
r + r
3
(d)

1
3
2

Ans. : (a)
Explanation : Refer Q10
The ratio of frictional torque produced for uniform wear to
that for uniform pressure is __________.
2
4
3
(a) 1
(b)
(c)
(d)
3
3
4
Ans. : (d)
Explanation : For friction torque with uniform wear is 1/2
and with uniform pressure is 2/3, Therefore
1/2 3
Ratio =
=
...Ans.
2/3 4
Q. 39
In case of pivot bearing the rate of wear is __________.
(a) maximum at the centre of contact area
(b) zero at the centre of contact area
(c) uniform throughout the contact area
(d) zero at the maximum radius of the contact area.
Ans. : (b)
Explanation :•
The velocity of rubbing is also different since it depends upon
the radii from the axis of rotation for the given uniform
angular speed of the shaft. It follows that the wear may be
different at different radii and this causes the redistribution
of intensity of pressure. Under these conditions, for uniform
wear, we can assume,
p.r = Constant
Q. 40
Q. 41
The number of collars are provided to carry a fixed axial
force in a flat collar bearing __________.
(a) to increase torque
(b) to decrease torque
(c) to decrease intensity of pressure
(d) to increase intensity of pressure
Ans. : (c)
Explanation : Refer Q9
In flat pivot bearing the total moment of friction force
__________.
(a) for uniform wear is greater than that for uniform
pressure
(b) for uniform wear is less than that for uniform
pressure
(c) for uniform wear is equal to that for uniform pressure
(d) for uniform wear may be more or less and cannot be
predicted.
Ans. : (b)
Explanation : Refer Q1
Q. 42
In flat pivot bearing the total power lost in friction
__________.
(a) for uniform wear is greater than that for uniform
pressure
(b) for uniform wear is less than that for uniform
pressure
(c) for uniform wear is same as that for uniform pressure
(d) for uniform wear may be more or less pressure and
cannot be predicted.Ans. : (b)
Explanation : Refer Q1
Q. 43
The frictional torque produced in case of flat pivot bearing
for uniform wear and uniform pressure respectively is
(W = total axial load,  = co-efficient of friction, r = radius
of pivot)
1
2
2
1
(a)
 W r and W  r
(b)
 W r and  W r
2
3
3
2
1
1
1
1
(c)
 W r and  W r
(d)
 W r and  W r
3
2
2
3
Ans. : (a)
Explanation : Refer Q10
Q. 44
Q. 45
Q. 46
To be on the safe side in design, the uniform pressure and
uniform wear is assumed respectively __________.
(a) for power transmitted by friction and power lost due
to friction
(b) for power lost due to friction and power transmitted
by friction
(c) for torque transmitted by friction and frictional
torque overcome
(d) none of the above
Ans. : (b)
Explanation : Refer Q1
Choose the correct statement for safe design.
(a) for power lost in friction uniform wear is assumed
(b) for power lost in friction uniform pressure is assumed
(c) for power transmitted due to friction uniform
pressure is assumed
(d) for power transmitted due to friction or lost in friction
uniform wear is assumed.
Ans. : (b)
Explanation : Refer Q1
Choose the correct statement for safe design.
(a) for power transmitted by friction uniform wear is
assumed
(b) for power transmitted by friction uniform pressure is
assumedQ. 47
Q. 48
(c) for power lost in friction and power transmitted
uniform wear is assumed
(d) power lost in friction and power transmitted uniform
pressure is assumed
Ans. : (a)
Explanation : Refer Q1
The purpose of providing multiple collars on a flat collar
pivot bearing is to __________.
(a) Establish self sustaining bearing conditions.
(b) Increase the frictional resistance.
(c) Distribute the axial load because of limiting bearing
pressures on the bearing.
(d) Increase the power absorbed by the bearing.
Ans. : (c)
Explanation : Refer Q9
Frictional torque produced for uniform pressure for
conical pivot bearing having semi angle of cone =  is
equal to __________.
  W r 1
2  W r 1
(a)
(b)
2 sin 
3 sin 
(c)
Q. 49
 W r 1
3 sin 
(d)
3  W r 1
4 sin 
Ans. : (b)
Explanation : Refer Q10
Frictional torque developed due to uniform wear for
conical bearing with semi-cone angle =  is __________.
 W r 1
2  W r 1
(a)
(b)
2 sin 
3 sin 
(c)
 W r 1
3 sin 
(d)
3  W r 1
4 sin 
Ans. : (a)
Explanation : Refer Q10
Q. 50
In a multi-collared thrust bearing consisting of ‘n’ collars
and subjected to axial load W, load sheared by each collar
is __________.
(a) W
(b)
W/n
(c) W/ (n – 1)
(d)
none of theseAns. : (b)
Explanation :
For ‘n’ number of collar, the intensity of pressure is given by
P =
Load
Number of collars  Bearing surface area

Q. 51
3
2
(c)
3
n 
2
 r 1 3 – r 32 
 2 2 
 r 2 – r 2 
( R 21 – R 22 )
(b) r 1 + r 2
4
(d) r 1 + r 2
2
Ans. : (c)
Explanation : Refer Q10
In case of flat collar pivot bearing, the frictional force in
case of uniform wear may be assumed to be acting at
__________.
2
2
2 r 1 – r 2
(a)
3 r 1 – r 2
2
(c)
3
Q. 53
W
=
In case of flat collar pivot bearing, the frictional force in
case of uniform pressure may be assumed to be acting at
__________.
2 r 1 – r 2
(a)
3 r 1 – r 2
Q. 52
p
 r 1 3 – r 32 
 2 2 
 r 2 – r 2 
(b) r 1 + r 2
4
(d) r 1 + r 2
2
Ans. : (d)
Explanation : Refer Q10
A flat foot step bearing 20 cm in diameter supports a load
of 2500 kgf. The shaft is running at 420 r.p.m. and the co-
efficient of friction is 0.09. The frictional torque, for
uniform pressure, will be equal to __________.
(a) 11.25 N-m
(b)
15 N-m
(c) 18.50 N-m
(d)
20.0 N-m
Ans. : (b)
Explanation :
Given : D = 20 cm, R = 10 cm = 0.10 m, W = 2500 N,N = 420 rpm,
2  420
  =
= 43.98 rad,  = 0.09
60
Frictional torque for uniform pressure is,
2
2
T = WR =  0.09  2500  0.10
3
3
Q. 54
Q. 55
= 15 N-m
...Ans.
A flat foot step bearing 20 cm in diameter supports a load
of 2500 kgf. The shaft is running at 420 r.p.m. and the co-
efficient of friction is 0.09, the frictional torque, for
uniform wear, will be equal to __________.
(a) 11.25 N-m
(b)
15 N-m
(c) 18.50 N-m
(d)
20.0 N-m
Ans. : (a)
Explanation :
Given : D = 20 cm, R = 10 cm = 0.10 m, W = 2500 N,
N = 420 rpm,
2  420
  =
= 43.98 rad,  = 0.09
60
Frictional torque for uniform wear is,
1
1
T = WR =  0.09  2500  0.1
2
2
= 11.25 N-m
...Ans.
A flat foot step bearing 20 cm in diameter supports a load
of 2500 kgf. The shaft is running at 420 r.p.m. and the co-
efficient of friction is 0.09, the power lost in overcoming
friction at the bearing, will be equal to __________.
(a) 650 Watt
(b)
880 Watt
(c) 659.7 Watt
(d)
680.20 Watt
Ans. : (c)
Explanation :
Given : D = 20 cm, R = 10 cm = 0.10 m, W = 2500 N,
N = 420 rpm,
2  420
  =
= 43.98 rad,  = 0.09
60
Frictional torque for uniform pressure is,
2
2
T = WR =  0.09  2500  0.1 = 15 N.m
3
3Q. 56
Power lost in friction is,
P = T = 43.98  15 = 659.7 Watt
...Ans.
A flat foot step bearing 20 cm in diameter supports a load
of 2500 kgf. The shaft is running at 420 r.p.m. and the co-
efficient of friction is 0.09, the power transmitted by the
friction between the surface, will be equal to __________.
(a) 493.875 Watt
(b)
880 Watt
(c) 659.7 Watt
(d)
680.20 Watt
Ans. : (a)
Explanation :
Given : D = 20 cm, R = 10 cm = 0.10 m, W = 2500 N,
N = 420 rpm,
2  420
  =
= 43.98 rad,  = 0.09
60
Frictional torque for uniform wear is,
1
1
T = WR =  0.09  2500  0.1 = 11.25 N-m
2
2
Power transmitted by the friction is,
P = T = 43.90  11.25
= 493.875 Watt
Q. 57
...Ans.
A thrust bearing, having collars of external and internal
diameters as 40 cm and 20 cm respectively, carries an
axial load of 3000 N. The co-efficient of friction is 0.09 and
shaft runs at 420 r.p.m. The power lost in friction is to be
determined. The corresponding value of friction torque
would be __________.
(a) 40.5 N-m (b) 42 N-m
(c) 45.0 N-m (d) 48.5 N-m
Ans. : (b)
Explanation :
Given :
D 1 = 40 cm  R 1 = 20 cm = 0.2 m
D 2 = 20 cm  R 2 = 10 cm = 0.1 m
W = 3000 N,
 = 0.09, N = 420 rpm
2  420
  =
= 43.98 rad/s
60
Frictional torque for uniform pressure is,T =
=
 R 31 – R 32 
2
W  2
2 
3
 R 1 – R 2 
2
0.2 3 – 0.1 3 
 0.09  3000  2
3
 0.2 – 0.1 2 
T = 42 Nm
Q. 58
...Ans.
A thrust bearing, having collars of external and internal
diameters as 40 cm and 20 cm respectively, carries an
axial load of 3000 N. The co-efficient of friction is 0.09 and
shaft runs at 420 r.p.m, the power lost in friction would
be equal to __________.
(a) 1.94 kW (b) 1.84 kW (c) 2 kW (d) 2.5 kW
Ans. : (b)Explanation :
Given :
D 1 = 40 cm  R 1 = 20 cm = 0.2 m
D 2 = 20 cm  R 2 = 10 cm = 0.1 m
W = 3000 N,  = 0.09, N = 420 rpm
2  420
  =
= 43.98 rad/s
60
Frictional torque for uniform pressure is,
T =
=
 R 31 – R 32 
2
W  2
2 
3
 R 1 – R 2 
2
0.2 3 – 0.1 3 
 0.09  3000  2
3
 0.2 – 0.1 2 
T = 42 Nm
Power lost in friction is
P = WT = 43.98  42 = 1847.16 Watt
Q. 59
= 1.84 kW
...Ans.
A flat footstep bearing 10 cm radius supports a load of 3
kN. The shaft runs at 600 / r.p.m. and co-efficient of
friction is 0.1. The frictional torque and frictional power
in watts with uniform pressure are respectively
__________.
(a) 10 N-m and 400 watts
(b) 20 N-m and 400 watts
(c) 30 N-m and 600 watts
(d) not possible to find for insufficient data.
Ans. : (b)
Explanation :
Given : R = 10 cm = 0.1 m, W = 3 kN = 3000 N
N = 600/ rpm
2  600
  =
= 20 rad/sec
60  
 = 0.1
Frictional torque for uniform pressure is,
2
2
T =  WR =
 0.1  3000  0.1
3
3= 20 Nm
...Ans.
Frictional power is,
Q. 60
P = T = 20  20 = 400 Watt
...Ans.
A flat footstep bearing 10 cm radius supports a load of
3 kN. The shaft runs at 600/ r.p.m. and co-efficient of
friction is 0.1. The frictional torque and frictional power
in watts with uniform wear are respectively __________.
(a) 30 N-m and 600 watts
(b) 15 N-m and 300 watts
(c) 15 N-m and 600 watts
(d) not possible to find for insufficient data
Ans. : (b)
Explanation :
Given : R = 10 cm = 0.1 m, W = 3 kN = 3000 N
2  600
N = 600/ rpm  W =
= 20 rad/sec
60  
 = 0.1
Frictional torque for uniform wear is,
1
1
T =
WR =  0.1  3000  0.1
2
2
= 15 Nm
...Ans.
Frictional power is,
Q. 61
P = T = 20  15 = 300 Watt
...Ans.
A flat foot step bearing 225 mm in diameter supports a
load of 7500 N. If the co-efficient of friction is 0.09 and the
shaft rotates at 60 r.p.m., calculate the power lost in
friction. The friction torque and frictional power in watts
with uniform pressure is __________.
(a) 50.62 Nm and 317.92 Watt
(b) 60.62 Nm and 320.92 Watt
(c) 57.68 Nm and 321.92 Watt
(d) 55.50 Nm and 410.12 Watt
Ans. : (a)
Explanation :
Diameter of bearing, D = 225 mm = 0.225 m Radius of bearing, R =
D 0.225
=
= 0.1125 m
2
2
Axial load, W = 7500 N
Coefficient of friction,  = 0.09
Speed of shaft, N = 60 rpm
2  N 2  60
 Angular speed of shaft,  =
=
60
60
= 6.28 rad/sec
For bearing we are considering uniform pressure
theory,
 Torque required to overcome the friction for flat foot
step bearing considering uniform pressure theory is,
2
2
T =
 W R =
0.09  7500  0.1125
3
3
 T = 50.62 Nm
...Ans.
Power lost in friction is,
P = T = 6.28  50.62 = 317.92 Watt
Q. 62
 P = 0.31792 kW
...Ans.
Following are the details pertaining to thrust bearing :
i) External and internal dia of contacting surfaces
– 320 mm and 200 mm resp.
ii) Axial load – 80 kN
iii) Pr. Intensity − 350 kN/m 2
iv) Shaft speed – 400 rpm
v) Coeff of friction – 0.06
Calculate power lost in overcoming friction. The
frictional torque and power lost in friction
(a) 600.35 Nm and 26.60 kW
(b) 630.30 Nm and 28.40 kW
(c) 635.08 Nm and 26.60 kW
(d) 650.52 Nm and 27.35 kW
Ans. : (c)Explanation :
Given : Thrust bearing
External diameter of contacting surface,
D 1 = 320 mm = 0.32 m
R 1 = 0.16 m
Internal diameter of contacting surface,
D 2 = 200 mm = 0.2 m
R 2 = 0.1 m
Axial load, W = 80 kN = 80  10 3 N
Pressure intensity, P = 350 kN/m 2 = 3.5  10 6 N/m 2
Shaft speed, N = 400 r.p.m.
Coefficient of friction,  = 0.06
Power lost in overcoming friction = p = ?
2   N 2  400
 =
=
60
60
= 41.89 rad/sec.
Torque required to overcome the friction for thrust bearing
considering uniform pressure theory is
 R 31 − R 32 
2
T = W  2

3
R − R 2

1

2
2
0.16 3 − 0.1 3 
=  0.06  80  10 3   

3
 0.16 2 − 0.1 2 
= 635.08 Nm.
...Ans.
Power lost in overcoming friction,
P =  T = 41.89  635.08 = 26603.5 W
Q. 63
P = 26.603 kW
...Ans.
A conical pivot supports a load of 20 kN, and its semicone
angle is 41.81. The friction coefficient is 0.06 and shaft
runs at 200 rpm. The torque required to over come
friction and power lost in friction is __________.
(a) 135.20 Nm and 2.93 kW
(b) 125.40 Nm and 2.93 kW
(c) 130.35 Nm and 2.53 kW
(d) 135.36 Nm and 2.83 kWAns. : (d)
Explanation :
Given : W = 20 kN = 20  10 3 N,  = 0.06,  = 41.81
N = 200 rpm
2  200
 W =
= 20.93 rad / s
60
Torque required to overcome the friction in conical
pivot bearing considering uniform pressure theory is,
2
T =
 W R cosec 
3
2
 T =
 0.06  20  10 3  0.1128  cosec 41.81
3

T = 135.36 Nm
Power lost in friction is,
...Ans.
p =  T = 20.93  135.36
= 2833.53 Watt

P = 2.83 kW
...Ans.
Clutches
Q. 64
•
The clutch is a device used __________.
(a) to absorb frictional torque
(b) to absorb frictional power
(c) to transmit power from one shaft to another
(d) all of the above
Ans. : (c)
Explanation :
The clutch is a mechanical device used to connect or
disconnect the driven shaft from the driving shaft at the will
of the operator while power is transmitted from driving to
driven shaft.
•
Clutch is mounted between driving and driven shaft and
power is transmitted from one shaft to the another shaft
which is required to be started and stopped frequently.Q. 65
Advantage of positive clutch is __________.
(a) They can be engaged at high speed
(b) They can use for high speed applications
(c) Shock and noise is there during engagement
(d) Due to no slip between contact surfaces no heat is
generated
Ans. : (d)
Explanation : Advantages of positive clutches :
(i) Positive clutches do not slip
(ii) No
heat
is
generated
at
clutch
surface
during
the
engagement.
Q. 66
Disadvantage of positive clutch is __________.
(a) They can be engaged at high speed
(b) They don't provide positive engaged
(c) Shock and noise is there during engagement
(d) Due to slip between contact surfaces heat is
generated
Ans. : (c)
Explanation :
Disadvantages of positive clutches :
(i) Shock and noise is there when engaged in motion.
(ii) They cannot be engaged at high speeds and sometimes
cannot be engaged at rest unless the jaws are aligned.
Q. 67
Torque transmitted by friction clutch __________.
(a) is more on the assumption of uniform pressure as
compared to uniform wear
(b) is less on the assumption of uniform pressure as
compared to uniform wear
(c) is same on the assumption of uniform pressure and
uniform wear(d) is more or less on the assumption of uniform pressure
as compared to uniform wear depending upon
material.
Ans. : (a)
Explanation : Refer Q1
Q. 68
•
Q. 69
For multiple plate clutch having ‘n’ is the effective
number of surfaces for transmitting power then no. of
plates are __________.
(a) (n + 1)
(b)
(n – 1)
(c) n
(d)
n – 2
(e) (n + 2)
Ans. : (a)
Explanation : If the plates are one side effective, then the
total number of plates are required to transmit the torque
is ( n + 1 ). For single plate clutch normally the both sides
of the plate are effective. Therefore, single plate clutch has
two pairs of contacting surface i.e. n = 2.
Frictional torque transmissions thorough plate clutch is
equivalent to __________.
(a) Conical pivot bearing
(b) Flat pivot bearing
(c) Trapezondial pivot bearing
(d) Flat roller bearing
Ans. : (d)
Explanation : Since equation for torque transmitting
capacity of Flat roller bearing and plate clutch is same.
Q. 70
In case of cone clutch, power transmission is same as that
of __________.
(a) Flat pivot bearing
(b) Flat collar bearing
(c) Trapezoidal pivot bearing
(d) Conical pivot bearing
Ans. : (c)
Explanation : Since equation for torque transmitting
capacity of Trapezoidal pivot bearing and cone clutch is
same.Q. 71
Multiplate clutch is a kind of __________.
(a) Positive clutch
(b)
Cone clutch
(c) Centrifugal clutch
(d)
Disc clutch
Ans. : (d)
•
Q. 72
Explanation : A multiplate clutch is consists of two sets of
plates the driving plates and driven plates arranged
alternately, driving shaft, spring, splined driven shaft etc. It
consists of more than one driving as well as driven plates. So
the number of pair of contacting surfaces is more than two.
Among which clutch heat dissipation is a serious
problem?
(a) Multiplate clutch
(b)
Positive clutch
(c) Cone clutch
(d)
Centrifugal clutch
Ans. : (a)
• Explanation : For given torque capacity radial size of
multiplate clutch is smaller than that of single plate clutch.
Therefore, it is compact in construction. The heat dissipation
is a serious problem because of compact arrangement.
• It is necessary to use cooling oil in case of multiplate clutch.
Because of this multiplate clutch is sometime called as wet
clutch.
Q. 73
In case of clutch, uniform pressure theory is applicable for
__________.
(a) Old clutch
(b)
New clutch
(c) Rusted clutch (d)
Clutch made of hard neutral
Ans. : (b)
Explanation :
In case of new clutch, the intensity of pressure is uniform along the
surface of clutch but in case of old or worn out clutch uniform wear
theory is more correct. In actual practice, for computation purpose the
assumption which gives result on safer side are preferred i.e. usually
the theory of uniform wear is used in the analysis of clutches.Q. 74
Centrifugal clutches are used in __________.
(a) Truck
(b)
Racing car
(c) Hack saw
(d)
Mopeds
Ans. : (d)
Explanation :
•
Q. 75
Q. 76
Q. 77
Centrifugal clutch works on the principle of centrifugal force
i.e. centrifugal force is increases with the increase in speed.
It is used when it is required to engage the driven member
automatically after the driving member has attained certain
speed.
A friction clutch should be designed on the assumption of
__________.
(a) uniform wear
(b)
uniform pressure
(c) constant friction
(d)
any one
Ans. : (a)
Explanation : Refer Q73
Considering the safe design, the friction clutch should be
designed __________.
(a) assuming uniform pressure
(b) assuming uniform wear
(c) assuming any criteria, either uniform pressure or
uniform wear
(d) assuming uniform pressure for high torque and
uniform wear for low torque.
Ans. : (b)
Explanation : Refer Q73
Frictional torque transmissions through multi plate
clutch is equivalent to __________.
(a) Conical pivot bearing
(b) Flat pivot bearing
(c) Trapezoidal pivot bearing
(d) Flat roller bearing
Ans. : (d)
Explanation :
Since equation for torque transmitting capacity of Flat
roller bearing and multi plate clutch is same.Q. 78
Q. 79
In case of multiplate clutch, power transmission is same
as that of __________.
(a) Flat pivot bearing
(b) Flat collar bearing
(c) Trapezoidal pivot bearing
(d) Conical pivot bearing
Ans. : (b)
Explanation : Refer Q77
Single plate clutch is a kind of __________.
(a) Positive clutch
(b)
Come clutch.
(c) Centrifugal clutch
(d)
Disc clutch
Ans. : (d)
Explanation :
•
A single plate consists of various elements, such as pressure
plate, friction plate (clutch plate), driving shaft, splined
driven shaft, splined hub, brass bush etc.
Q. 80
Which of the following is an example of friction clutch ?
(a) disc clutch
(b) cone clutch
(c) centrifugal clutch
(d) all of the above
Ans. : (d)
Explanation :Q. 81
Q. 82
The frictional torque transmitted by a disc clutch is same
as that of __________.
(a) flat pivot bearing
(b) flat collar bearing
(c) conical pivot bearing (d) trapezoidal pivot bearing
Ans. : (b)
Explanation : Refer Q69
The frictional torque transmitted by a cone clutch is same
as that of __________.
(a) flat pivot bearing
(b) flat collar bearing
(c) conical pivot bearing
(d) trapezoidal pivot bearing
Ans. : (d)
Explanation : Refer Q70
Q. 83
Q. 84
For a safe design, a friction clutch is designed assuming
__________.
(a) uniform pressure theory
(b) uniform wear theory
(c) any one of the two
(d) none of the above
Ans. : (b)
Explanation : Refer Q73
In a multiple friction clutch, the number of active friction
surfaces is __________.
(a) 2n
(b) n
(c) 2 (n – 1)
(d) n – 1
Ans. : (d)
Explanation : Refer Q62
Q. 85
Friction torque transmitted by a disc or plate clutch is
same as that of __________.
(a) flat pivot bearing
(b) flat collared bearing
(c) conical pivot bearing
(d) tructed conical pivot bearing.
Ans. : (b)
Explanation : Refer Q69Q. 86
For multiple plate friction clutch with n plates, the
number of active surfaces n a for transmitting power is
__________.
(a) n a = 2n
(b)
n a = (2n – 1)
(c) n a = (n – 1)
Q. 87
(d)
n a = 2(n – 1)
Ans. : (c)
Explanation : Refer Q62
A single plate clutch is transmitting power, the inner and
other radii of the plate are 10 cm and 22.5 cm
respectively. Using the uniform wear theory, the
maximum pressure intensity occurs at a radius of
__________.
(a) 10 cm
(b) 15 cm
(c) 16.25 cm
(d) 22.5 cm
Ans. : (a)
Explanation :
Given : R 1 = 22.5 cm, R 2 = 10 cm
Maximum pressure intensity is,
C
C
P max =
=
R min R 2
Therefore maximum pressure intensity occurs at radius of
R 2 = 10 cm
...Ans.
Q. 88
Torque transmitted by single plate clutch with axial
spring load = 1 kN, and inner and outer radii of 10 cm
and 15 cm and  = 0.5 with assumption of uniform wear is
__________.
(a) 250 N-m
(b) 125 N-m
(c) 62.5 N-m
(d) not possible to find with given data.
Ans. : (c)
Explanation :
Given : W = 1 kN = 1000 N, R 1 = 15 cm = 0.15 m,
R 2 = 10 cm = 0.1 m,  = 0.5
The torque transmitted by uniform wear theory is,
1
1
T =
W [R 1 + R 2 ] =  0.5  1000 [0.5 + 0.1]
2
2
T = 62.5 Nm
...Ans.Q. 89
A multiple clutch with 5 plates transmits torque of 400 N-
m. If  = 0.5 and inner radius = 10 cm and outer
radius = 15 cm and four springs exert-axial load, the load
per spring is __________.
(a) 200 N
(b) 300 N (c) 400 N
(d) 1280 N.
Ans. : (c)
Explanation :
Given : number of plate = 5
 number of pairs in contact, n = (5 – 1) = 4
T = 400 Nm,  = 0.5
R 1 = 15 cm = 0.15 m, R 2 = 10 cm = 0.1 cm
Torque transmitted by considering uniform wear is,
1
T = nW (R 1 + R 2 )
2
1
 400 =  4  0.5  W (0.15 + 0.1)
2
W = 1600 N per four springs
Q. 90
Therefore axial load exert by one spring is,
= 1600/4 = 400 N per spring
...Ans.
Determine the axial force required to engage a cone clutch
transmitting 25 kW power at 750 rpm. Average friction
diameter of the cone is 400 mm, semicone angle 10and
coefficient of friction 0.25.
(a) 325.30 Nm and 1100.25 N
(b) 318.31 Nm and 1105.47 N
(c) 320.25 Nm and 1008.20 N
(d) 325.89 Nm and 1120.30 N
Ans. : (b)
Explanation :
Given :
Power transmitted, P = 25 kW = 25  10 3 Watt
Speed of clutch, N = 750 r.p.m.
Average diameter, D m = 400 mm = 0.4 m
R 1 + R 2 D m 0.4
=
=
2
2
2
= 0.2 m
Coefficient of friction,  = 0.25
 Mean radius, R m =Semi-cone angle,  = 10
Power transmitted by cone clutch is,
2NT
P =
60
2  750  T
60
T = 318.31 N-m
...Ans.
Torque transmitted by cone clutch considering uniform
wear theory is,
1
T =
W ( R 1 + R 2 ) cosec 
2
 25  10 3 =
Q. 91
318.31 = 0.25  W  0.2  cosec 10
W = 1105.47 N
...Ans.
Determine the axial force required to engage a cone clutch
transmitting 25 kW power at 750 rpm. Average friction
diameter of the cone is 400 mm, semi cone angle 60 and
coefficient of friction 0.25.
(a) 325.30 N-m and 1100.25 N
(b) 318.31 N-m and 1105.47 N
(c) 320.25 N-m and 1008.20 N
(d) 325.89 N-m and 1120.30 N
Ans. : (b)
Explanation :
Given :
Power transmitted,
P = 25 kW = 25  10 3 Watt
Speed of clutch,
N = 750 r.p.m.
Average diameter, D m = 400 mm = 0.4 m
 Mean radius, R m =
R 1 + R 2 D m 0.4
=
=
= 0.2 m
2
2
2
Coefficient of friction,  = 0.25
Semi-cone angle,  = 10
Power transmitted by cone clutch is,
2NT
P =
60
25  10 3 =
2  750  T
60
T = 318.31 N-m
...Ans.Torque transmitted by cone clutch considering uniform wear
theory is,
1
T =
W ( R 1 + R 2 ) cosec 
2
318.31
W
= 0.25  W  0.2  cosec 10
= 1105.47 N
...Ans.
❑❑❑Chapter 4
Brakes and Dynamometer
Multiple Choice Questions for Online Exam
Brakes
Q. 1
Q. 2
Brake is a device used for bringing a moving body
__________.
(a) To rest
(b) To retard the motion
(c) To keep it in a state of the rest against the external
forces.
(d) All of the above
Ans. : (d)
Explanation :
The brake is a device by means of which an artificial
frictional resistance is applied to a moving body in order
to retard or stop the motion of a body.
In case of band and block brake, having ‘n’ blocks and
each block subtending a semi-cone angle ‘’ with as
coefficient of friction, the ratio of tension on two sides will
be __________.
(a) T n  1 +  tan  n
=
T o  1 –  tan  
(c) T n  n +  tan  n
=
T o  n –  tan  
(e) T n  1 + n tan  n
=
T o  1 – n tan  
Ans. : (a)
(b)
(d)
T n  1 –  tan 
=
T o  1 +  tan  
T n   + n tan  n
=
T o   – n tan  Explanation : For ‘n th ’ block ratio of tensions will be given
by,
T n
=
T n – 1
1 +  tan 
1 –  tan 
Hence the ratio of the tensions to the tight and slack sides of
complete band and block brake can be obtained by multiplying
above equations as follows.
T n T 1 T 2 T 3
T n
=


 ... 
T 0 T 0 T 1 T 2
T n – 1

T n
T 0
=
 1 +  tan    1 +  tan    1 +  tan  
 1 +  tan  
 1 –  tan     1 –  tan     1 –  tan    ....   1 –  tan  

 
 




Q. 3
T n
=
T 0
Material used
__________.
for
brake
 1 +  tan  
 1 –  tan  


lining
n
should
not
have
(a) High heat dissipation capacity
(b) Low heat resistance.
(c) High strength
(d) High coefficient of friction
Ans. : (b)
•
Explanation : The material used for the brake lining should
have the following characteristics :
(1) It should have high coefficient of friction with
minimum heating i.e. coefficient of friction should
remain constant with change in temperature.
(2) It should have low wear rate.
(3) High heat resistance.
(4) It should have high heat dissipation capacity.
(5) It should have adequate mechanical strength.
(6) It should not be affected by moisture and oil.Q. 4
Pivoted block brake is a kind of __________.
(a) Single block brake
(b) Shoe brake
(c) Differential band brake
(d) Internal expanding shoe brake
Ans. : (b)
Explanation :
Fig. Q4 : Pivoted block or shoe brake
Q. 5
•
Equivalent coefficient of friction in case of pivoted block
brakes is __________.
4 sin 
4 sin 
(a)
(b)
2 + sin 2 
2 + sin 
2 sin 
2  + sin 2
(c)
(d)
2 + cos 
4 sin 
Ans. : (a)
Explanation :
pivoted block brakes have more life and provide higher
braking torque. Braking torque for pivoted block or shoe is
given by,
T B = F t  r =  R n  r
where  =Equivalent coefficient of friction=
4  sin 
2  + sin 2 
 =Actual coefficient of friction.
Q. 6
Which type of breaks are used in electric crane ?
(a) Internal expanding shoe brake
(b) Differential band brake
(c) Single block brake
(d) Double block brake
Ans. : (d)
Explanation :
Fig.6 : Double block or shoe brake
•
Q. 7
When force P is applied to bell crank lever, then spring gets
compressed and brake is released. When there is no force P
on bell crank lever, the brake is engaged automatically. In
this case braking torque become two times. Therefore double
block or shoe brake is used in electric cranes.
In a band brake ‘a’ is perpendicular distance between
slack side and pivot point and ‘b’ is perpendicular
distance between tight side and pivot point. If applied
force is downwards then __________.
(a) a > b
(b)
a = b(c) a < b
Ans. : (a)
Explanation :
(d)
a = b 2
Fig. Q 7 : Differential band brake
Q. 8
Brakes used in Railway engine are __________.
(a) Band brake
(c) Shoe brake
Ans. : (c)
Explanation :
(b)
(d)
Internal expending brake
Band and brake brake.
• Shoe brakes consist of blocks which are pressed against the
surface of a rotating drum.
• The block which is rigidly attached or pivoted to the lever is
lined with friction material. The friction between friction
lining on the block and drum retards the rotation of the
drum.
• The block or shoe is made up of softer material than the rim
of the drum. The material of the block for light and slow
vehicles is wood and rubber and for heavy and fast vehicles
it is cast steel. This type of brake is commonly used in
bicycles and railway trains.Q. 9
Q. 10
•
The brake commonly used in motor cars __________.
(a) shoe brake
(b) band brake
(c) bank and block brake
(d) internal expanding brake
Ans. : (d)
Explanation : Internal expanding shoe brake is used in
motor cars and light trucks.
When will the distance travelled by 4-Wheel be less after
braking ?
(a) When brakes applied to rear wheel
(b) When brakes applied to front wheel
(c) When brakes applied to all 4-wheels
(d) When brakes applied to rear wheel gradually.
Ans. : (c)
Explanation : When the brakes are applied to all 4-wheels of
moving vehicle, it is retarded and its speed decreases suddenly
and hence the distance travelled by 4-Wheel be less .
Q. 11 Brakes are self-energizing when __________.
Q. 12 (a) frictional torque assists the applied torque
(b) frictional torque opposes the applied torque
(c) frictional torque vanishes
(d) none of the above
Ans. : (a)
Explanation : Brake is said to be self-energizing when
frictional force helps to apply the brake
In a differential
band brake, as
shown in Fig. 1
the length OA is
greater than OB.
In order to apply
the
brake,
the
force at C should
__________.
(a) be zero
(b) act in upward direction(c) act in downward direction
(d) none of the above
Ans. : (c)
Explanation : Refer Q7
Q. 13
For the brake to be self locking, the force P at C as shown
in Fig. 1 should ____.
(a) be zero
(b) act in upward direction
(c) act in downward direction
Ans. : (a)
Explanation : We know that when P is negative or zero
the brake is called self locking brake.
Q. 14
When brakes are applied to all the four wheels of a
moving car, the distance travelled by car before it is
brought to rest, will be __________.
(a) Maximum
(c) Both (a) and (b)
Ans. : (b)
Explanation : Refer Q10
Q. 15
(b)
(d)
Minimum
None of above
Which of the following brakes is commonly used in two
wheelers ?
(a) disc brake
(b) band brake
(c) internal expanding shoe brake
(d) (a) and (c)
Ans. : (d)
Explanation : Refer Q9
Q. 16
Brakes commonly used in trains are ________brakes.
(a) band
(b)
(c) band and block
(d)
Ans. : (b)
Explanation : Refer Q8
Q. 17
shoe
internal expanding shoe
In a self-locking brake, the force required to apply the
brake is __________.(a) minimum
(b)
(c) maximum
(d)
Ans. : (b)
Explanation : Refer Q13
Q. 18
When the frictional force helps the applied force in
applying the brake, the brake is __________.
(a) self-locking
(c) self-energising
Ans. : (c)
Explanation : Refer Q11
Q. 19
zero
none of these
(b) automatic
(d) (a) and (c)
A motor car uses __________.
(a) a band brake
(b) a band and block brake
(c) an external shoe brake
(d) an internally expanding shoe-brake
Ans. : (d)
Explanation : Refer Q9
Q. 20
•
For a simple band brake to be effective __________.
(a) tight side is connected to free end of lever and effort
is applied at end so as to move this end away from
drum
(b) same as (a) above but effort is so applied at end as to
move this end towards the drum
(c) slack side is connected to the free-end of the lever and
the effort is applied at end so to move it away from
drum.
(d) same as (c) but effort is applied at end so as to move
this end towards the drum.
Ans. : (d)
Explanation :
A simple band brake is shown in FigQ20. In this one end of
the band is attached at the fulcrum of the lever while the
other end is at a distance ‘b’ from fulcrum.(a) Clockwise rotation of drum
(b) Anticlockwise rotation of
drum
Fig. Q 20 : Simple band brake
•
Q. 21
When the force P is applied at the free end of the lever, it
turns about the fulcrum O. It tightens the band on the drum
and brakes are applied. The braking force is provided by the
friction between the band and the drum.
For the differential band brake shown in Fig. Q. 21, for
c.c.w. direction of drum, rotation with tension T 1 in band
at A and T 2 at B, the effort required at end is __________.
(a) zero
(c) P =
(b)
P =
(T 1 a – T 2 b)
l
downwards
(T 1 a – T 2 b)
upwards
l
(d) P = T 1 (a/ l ) downwards
Fig. 2
Ans. : (b)
Explanation :
When drum rotates in anticlockwise direction, the end of the
band
attached
at
point
A will be tight side with tension T 1 and end of band attached
at point B will be slack side with tension T 2 .
Taking moment about fulcrum ‘O’.
 M o
= 0
 P  l – T 1  a + T 2  b = 0 P  l = T 1 a – T 2  b
 P =
Q. 22
Q. 23
T 1  a – T 2  b
l
For c. c. w direction of drum rotation in Fig. 2 brake
applies if __________.
(a) a > b and P acts downwards
(b) a < b and P acts downwards
(c) a = b and P acts downwards
(d) a > b and P acts upwards
Ans. : (a)
Explanation : Refer Q7
The band brake in Fig. 2 will be self-energizing when
__________.
(a) drum rotates c.c.w. and ‘P’ acts upwards
(b) drum rotates c.c.w. and ‘P’ acts downwards
(c) drum rotates c.w. and ‘P’ acts upwards
(d) drum rotates c.w. and ‘P’ acts downwards.
Ans. : (d)
Explanation :
Brake is said to be self-energizing when frictional force
helps to apply the brake. In the case of differential band
T 1  a – T 2  b
brake from Equation P =
the moment T 1 b
l
and T 2  b helps in applying brake for clockwise and
anticlockwise rotation of drum respectively.
Q. 24
Self-locking is not possible in the case of __________.
(a) simple band brake
(b) differential band brake
(c) simple block brake
(d) internal expanding brake
Ans. : (a)
Explanation : for self locking break condition is a > b
but in simple band break one end of band is attached at
pivot point so distance ‘a’ is zero Self-locking is not
possible in this case.Q. 25
When moment due to friction force aids in applying the
brake, it is called _____.
(a) simple block brake
(c) self-locking brake
(b)
(d)
pivoted block brake
self-energized brake
Ans. : (d)
Explanation : Refer Q11
Q. 26
Refer Fig. 3 Pivot point O 2 , force at A acting vertically up,
then __________.
Fig. 3
Q. 27
(a) brake is equally effective in both the directions
(b) brake is more effective in counter-clockwise direction
(c) brake is more effective in clockwise direction
(d) brake is not effective in any direction.
Ans. : (a)
Explanation : Brake is equally effective in both the
directions since c = d.
For block brakes  = angle of lap,  = co-efficient of
friction; if  > 45, the equivalent co-efficient of friction is
__________.

4 sin
2
4
sin


(a)  =  
(b)
 = 
 
 sin  +  
sin +
2 2



 4 sin  
2 

(c)  = 
 sin  +  
(d)
Ans. : (c)
Explanation : Refer section 4.3
 = 



 sin  
 sin  +  Q. 28
For block brakes,  = angle of lap, r is the radius of drum,
if  > 45, the equivalent brake radius is __________.

4 r sin
2
4 r sin  
(a) r = 
(b)
r =
 
 sin  +  
sin +
2 2



 4 r sin  
2 

(c) r =
 sin  +  
(d)



r sin  
 sin  +  
r = 
Ans. : (c)
Explanation : Refer Q.5
Q. 29
The ratio of tension on the tight side to that on the slack
side for block brakes is (n = number of blocks,  = semi-
angle subtended by each block and  = co-efficient of
friction) __________.
n – 1 T n
1 +  tan  
(b)
= 
T o  1 –  tan  
n T n
1 +  tan  
(d)
= 
T o  1 +  tan  
T n
1 –  tan  
(a)
= 
T o  1 +  tan  
T n
1 +  tan  
(c)
= 
T o  1 –  tan  
Q. 30
n – 1
n
Ans. : (c)
Explanation : Refer section 4.5
The stopping distance for a vehicle by applying brakes
when all the four wheels are sliding as compared to when
all the four wheels are at impending slide is __________.
(a) more
(b) less
(c) same
(d) not predictable.
Ans. : (a)
Explanation : Refer Q10
Q. 31
Find the brake power of the engine running at 755 rpm.
The length of lever from center of pulley is 1.43 m. The
weight in pan is 40 kg.
(a) 40 kW (b) 42.44 kW
Ans. : (c)
Explanation :
(c) 44.36 kW (d) 45.41 kW
Given : Speed of engine, N = 755 rpm
Length of lever, L = 1.43 mWeight in pan, W = 40  9.81 = 392.4 N
Brake
power
of
the
engine
for
prony
brake
dynamometer is,
P =
W  L  2 N
60
=
392.4  1.43  2   755
60
 P = 44365.01 Watt
 P = 44.365 kW
Q. 32
...Ans.
In a band brake the ratio of tight side band tension to the
tension on the slack side is 3. If the angle of overlap of
band on the drum is 180 the coefficient of friction
required between drum and band is __________.
(a) 0.20
(b) 0.25
Ans. : (d)
Explanation :
(c) 0.30
(d) 0.35
T 1
= 3
T 2
Given :
 = 180 =  radius
T 1
= e 
T 2
but,

3 = e 
 l n (3) =  
1
  =
l n (3) = 0.35

Q. 33
...Ans.
A single block brake as shown in Fig. 6 has a brake drum
diameter of 1 m and the angle of contact is 30. It takes
280 Nm torque at 300 rpm. The coefficient of friction is
0.35. Determine the required force ‘P’ to be applied when
the rotation of drum is clockwise.Fig. 6
(a) 378.99 N
(c) 435.45 N
Ans. : (b)
Explanation :
Given :
(b) 420.99 N
(d) 450.38 N
D = 1 m, r=
D 1
= = 0.5 m
2 2
2  = 30
 = 0.35
Braking torque is, T B =  R n r
 280 = 0.35  R n  0.5
 R n = 1600 N
Fig. 6(a)
For clockwise rotation of drum,
For equilibrium taking moment about fulcrum 0 we get,
R n  200 + F t  30 – P  800 = 0
R n  200 +  R n  30 – P  800 = 0
 1600  200 + 0.35  1600  30 – P  800 = 0
Q. 34
P = 420.99 N
...Ans.
For above question determine the required force ‘P’ to be
applied when the rotation of drum is anti-clockwise.
(a) 378.99 N
(c) 435.45 N
Ans. : (a)
Explanation :
Given :
(b)
(d)
420.99 N
450.38 ND = 1 m,
r =
D 1
= = 0.5 m
2 2
2  = 30
 = 0.35
Braking torque is, T B =  R n r
 280 = 0.35  R n  0.5
 R n = 1600 N
Fig. 7
For anticlockwise rotation of drum
For equilibrium taking moment about fulcrum O we get,
R n  200 – F t  30 – P  800 = 0
R n  200 –  R n  30 – P  800 = 0
 1600  200 – 0.35  1600  30 – P  800 = 0
 P = 378.99 N
Q. 35
...Ans.
A bicycle and rider is mass 125 kg are travelling at a
speed of 10 km/hr. The rider applies a break to the near
wheel of 1 m diameter. How far the bicycle will travel
before it comes to rest ? The pressure applied to the brake
is 120 N and  = 0.05. Also find the number of revolutions
made by bicycle before coming to rest.
(a) 24.58
(c) 23.79
Ans. : (b)
(b)
(d)
25.58
35.48
Explanation :
Given :
m = 125 kg.
V = 10 km/hr. =
D = 1 m.
 = 0.05.
10  1000
= 2.7778 m/sec.
3600
R n = 120 N.When bicycle comes to rest the total kinetic energy of the
bicycle and rider will be absorbed by the brakes.
Work done against friction = K.E. of bicycle and rider
1
 R n  S =
mV 2
2
 S
=
125  2.7778 2
= 80.377 m
0.05  120  2
Total distance travelled by bicycle before it comes to rest
S
Distance travelled in one turn
= 80.377 m
= Circumference of wheel
=   1
Number of revolution of wheel,
Total distance Travelled
N =
Distance travelled in one turn
=
Q. 36
80.377
  1
N = 25.58 rotations
...Ans.
In a simple band brake applied to a shaft carrying a
flywheel of mass 250 kg and radius of gyration of 350
mm, one end of the band is attached to the fulcrum and
the other at a distance 80 mm left from the fulcrum. The
force is applied to the brake lever at a distance 300 mm
from the fulcrum. The angle embraced by the band is 225
and the brake drum diameter is 220 mm coefficient of
friction 0.25 and the shaft speed is 300 rpm clockwise.
Determine the brake torque when a force of 150 N is
applied.
(a) 102.99 N
Ans. : (c)
(b) 105.24 N
(c) 103.26 N
(d) 195.38 N
Explanation :
Given :
m = 250 kg,  = 225 = 225 
D = 220 mm = 0.22 m ,
 = 0.25 ,P=150 N

= 3.9269 radians
180
D 0.22
r = =
= 0.11 m,
2
2Fig. 8
Limiting ratio of tension is,
T 1
= e   = e 0.25 (3.9269) = 2.669
T 2
 T 1
= 2.669 T 2
Taking moment about fulcrum O,
T 2  80
= P  300 = 150  300
 T 2 = 562.5 N
 T 1 = 2.669 T 2 = 2.669  562.5
 T 1 = 1501.3125 N
Braking torque applied is, T B
= ( T 1 – T 2 ) r
= (1501.3125 – 562.5)  0.11
 T B
= 103.269 Nm
...Ans.
Dynamometers
Q. 37
A dynamometer is a device used for measuring ___-
_______.
(a) Power developed by a machine
(b) Torque developed by a machine
(c) Power absorbed by a machine
(d) All of the above
Ans. : (d)
Explanation :
Dynamometer is a device which is used to measure the
frictional resistance. By knowing frictional resistance we can
determine the torque transmitted and hence the power of the
engine.Q. 38
Dynamometer is a device for measuring __________.
(a) torque exerted by the machine
(b) power developed by the machine
(c) power absorbed by the machine
(d) all of the above.
Ans. : (d)
Q. 39
Explanation : Refer Q37
Which of the following is classified a absorption type of
dynamometer ?
(a) Prony broke dynamometer
(b) Rope broke dynamometer
(c) Hydraulic dynamometer
(d) All of the above
Ans. : (d)
Explanation :
Q. 40
Which of the following is classified as transmission type
of dynamometer ?
(a) Prony brake dynameter
(b) Rope brake dynameter
(c) Hydraulic dynamometer
(d) Epicyclic dynamometer
Ans. : (d)
Q. 41
Explanation : Refer Q39
Which of the following
dynamometer ?
(a) prony brake dynamometer
(b) rope brake dynamometer
is
an
absorption
type(c) epicyclic-train dynamometer
(d) both (a) and (b)
Ans. : (d)
Explanation : Refer Q39
Q. 42
In transmission type of dynamometers, the available
power is __________.
(a) Absorbed in overcoming the frictional work.
(b) Transmitted except that the small power is absorbed
due to friction.
(c) Fully transmitted.
(d) None of the above.
Ans. : (b)
Explanation :
•
Q. 43
In transmission type of dynamometers work done by the
engine is not absorbed by frictional resistance but it is used
after the measurement for doing work.
An example of an absorption type dynamometer is
__________.
(a)
(b)
(c)
(d)
prony brake dynamometer
epicyclic train dynamometer
torsion dynamometer
belt transmission dynamometer
Ans. : (a)
Q. 44
Explanation : Refer Q
An example of a transmission dynamometer is ___-
_______.
(a)
(b)
(c)
(d)
prony brake dynamometer
rope brake dynamometer
froude dynamometer
torsion dynamometer
Ans. : (d)
Q. 45
Explanation : Refer section 4.13.2
Which of the following is classified as absorption
dynamometer?
(a) Prony brake dynamometer(b) Rope brake dynamometer
(c) Froude’s hydraulic dynamometer
(d) All of the above.
Ans. : (d)
Q. 46
Explanation : Refer Q39
Which of the following is classified as transmission
dynamometer ?
(a) torsion dynamometer
(b) Froude’s hydraulic dynamometer
(c) belt dynamometer or Tatham dynamometer
(d) prony brake dynamometer
Ans. : (c)
Explanation : Refer Q39
Q. 47
Cooling water is used in __________.
(a)
(b)
(c)
(d)
Prony brake dynamometre
Epicyclic gear train dynamometre
Torsion dynamometre
Rope brake dynamometer
Ans. : (d)
Explanation : In case of Rope brake dynamometer If high
power is produced, then heat is generated due to friction
between rope and pulley or drum. To prevent this, pulley
is provided with internal flanges on the rim, forming
channel for flow of water to cool the pulley or drum.
Q. 48
Tathem dynamometre is name given to __________.
(a)
(b)
(c)
(d)
Rope brake dynamometer
Belt transmission dynamometre
Absorption dynamometer
Epicyclic Train dynamometer
Ans. : (b)
Q. 49
In case of Glibson flash light torsion dynamometre angle
of twist can be measured upto __________.
(a)
1
20
th
and degree
(b)
1
10
th
of degree(c)
1
70
th
of degree
(d)
1
100
th
of degree
Ans. : (d)
Explanation :
The maximum angle of twist can be
1 th
measured up to
of a degree In case of Glibson flash
100
light torsion dynamometer.
Q. 50
The following data refer to a laboratory experiment with
rope brake __________.Diameter of the flywheel = 1 m
Diameter of the rope = 10 mm
Dead weight on the brake = 50 kg
Speed of the engine = 200 rpm
Spring balance reading = 120 N
Find brake power of the engine.
(a) 3.91 kW
(b) 2.90 kW
(c) 4.90 kW
(d) 3.80 kW
Ans. : (a)
Explanation :
Given : Diameter of drum, D = 1 m
Diameter of rope, d = 10 mm = 0.01 m
Weight on the brake, W = 50 kg = 50  9.81 = 490.5 N
Speed of engine, N = 200 rpm
Spring balance reading, s = 120 N
Brake power of the engine for rope brake dynamometer
is given by
 P =
=
( W – s )  ( D + d ) N
60
( 490.5 – 120 )  ( 1 + 0.01 )  200
60
 P = 3918.66 W
 P = 3.9186 kW
Q. 51
...Ans.
A torsion dynamometer is fitted on turbine shaft to
measure the angle of twist. The shaft twist 1.6 for a
length of 8 meters at 600 r.p.m. The diameter of shaft is250 mm. Find the power transmitted by the turbine. Take
G = 80 GPa.
(a) 68.22 kW
(c) 67.22 kW
(b)
(d)
67.22 kW
66.32 kW
Ans. : (b)
Explanation :
Given : Torsion dynamometer :
Angle of twist  = 1.6 = 1.6 

180
= 0.0279 radians.
Length of the shaft L = 8 m.
Speed of shaft N = 600 rpm.
Diameter of shaft d = 250 mm = 0.25 m.
Modules of rigidity G = 80 GPa = 80  10 9 N/m 2 .
Power transmitted, P = ?
Polar moment of inertia, J =
 4 
d =
 (0.25) 4
32
32
= 3.835  10 − 4 m 4
Torque applied to the shaft,
T
G
=
J
L
T =
3.835  10 − 4  80  10 9  0.0279
8
= 1069.965 Nm
Power transmitted by the turbine,
P =
2  NT
2  600  1069.965
=
60
60
= 67.227  10 3 W
P = 67.227 kW
...Ans.
❑❑❑Chapter 5
Kinematic Analysis of
Mechanisms : Analytical Methods
Multiple Choice Questions for Online Exam
Analytical Method for IC Engine
Q. 1
The displacement of the piston of a reciprocating engine is
equal to
(a) r (1 – cos ) + r ( n –
n 2 – sin 2  )
(b) r (1 – cos  – n)
(c) r (1 – cos  – n 2 – sin 2 )
(d) r (1 + cos  – n)
Ans. : (a)
Explanation :
When crank moves from I.D.C. to O.D.C. through angle ‘’ from
I.D.C., then displacement of piston is given by,
 x P =P 1 P = OP 1 − OP =
(l + r) − (r cos  + l cos )
 x P =l + r − r cos  − l cos  = r (1 − cos ) + l (1 − cos )
But


CN = r sin  = l sin 
r
sin  = sin 
l
sin 
l
sin  =
...  ··· n = 
n
r 

We know that,sin 2  + cos 2  =1  cos  = 1 − sin 2 
On substituting the value of sin  from Equation (i) we get,
...(i).

cos  =

cos  =
1 −
sin 2 
n 2
...(ii)
n 2 − sin 2 
n
...(iii)
On substituting the value of cos  in Equation (5.5.1),


x P = r (1 − cos ) + l


x P = r (1 − cos ) + n  r 1 −

Q. 2

 1 −
n 2 − sin 2  
n

n 2 − sin 2  
· ·
n
 ...[ · l = nr]
x P = r (1 − cos ) + r ( n −
n 2 − sin 2  )
The displacement of the piston in a reciprocating steam engine
is given by
cos 2
(a)   r  sin  + n  (b) r (1 – cos ) + l (1 – cos )


(c)  2  r  sin  +

cos 2 
n  (d) none of the above
Ans. : (b)
Q. 3
Explanation : Refer Q1
The velocity of the piston in a reciprocating steam engine is
given by
sin 2
cos 2
(a)   r  sin  + 2n  (b)  2  r  cos  + n 




sin 2
(c)  2  r  sin  + 2n 


Ans. : (a)
(d) none of above
Explanation :
Since, velocity of piston is rate of change of displacement with
respect to time t,
d (x P ) d d

V P =
=

(x )
dt
d dt P
d
d

V P =  
(x )...  · ·  = 
dt 
d P  ·
d

V P =  
r (1 − cos ) + r ( n − n 2 − sin 2  )
d
d

V P
=   r
(1 − cos ) + ( n − n 2 − sin 2  )
d
[
]
[
].
 V P =  r

1
 (− 2 sin  cos )  
 (0 − (− sin ) + ( 0 − 
2
2
 2 n − sin  




sin 2
 V P =   r  sin  +

2
2
2 n − sin  

Since, l >r
 n > 1
2
Therefore n is large value as compared to sin 2 , so neglecting
2
sin  we get,
sin 2
V P =   r  sin  + 2n 


Q. 4
The acceleration of the piston in a reciprocating steam engine
is given by
sin 2
cos 2
(a)   r  sin  + 2n 
(b)  2  r  cos  + n 




sin 2
(c)  2  r  sin  + 2n 


Ans. : (b)
cos 2
(d)  2  r  cos  – 2 n 


Explanation :
Since acceleration is rate of change of velocity with respect to time t,
d (V P )
d
d d

f P = dt (V P ) =
 dt  (V P ) =  
d
d

f P = 
d 
sin 2
cos 2
  r   sin  + 2n   =  2  r  cos  + 2n  2 
d 

 


cos 2 
n 

The velocity of the piston of reciprocating engine is equal to
cos 2
sin 2
(a)  2 r  cos  + 2 n 
(b)  2 r  sin  + 2 n 




f P =  2  r  cos  +
Q. 5
sin 2
(c) r  sin  + 2 n 


(d)
cos 2
r  cos  + 2 n 


Where  = Angular velocity of crank,  = Angular turned by
l
crank from inner dead centre r = Radius of crank, n = r = Ratio
of length of connected rod to crank radius.
Ans. : (c)
Explanation : Refer Q3.
Q. 6
The angular velocity of the connecting rod of a reciprocating
engine is equal to
 sin 
 cos 
(a)
(b)
2
2
n + sin 
n 2 – sin 2 
 cos 
 sin 
(c)
(d)
n 2 – cos 2 
n 2 – sin 2 
Ans. : (b)
Explanation :
sin 
We know that,
sin  =
n
differentiating above equation with respect to time t,
d
d  sin  
We get,
(sin ) =
dt
dt  n 
d d
d d  sin  


 sin  =

d dt
d dt  n 
d
cos  d

cos  
=

dt
n
dt
d
Put
=  = angular velocity of crank
dt
d
and
=  pc = angular velocity of connecting
dt
rod.

cos 
cos    pc = n  

 pc
=
 cos 
n cos 
Put value of cos  from Equation (iii) in above equation

 pc =
Since l > r
n
  cos 
  cos 
=
2
2
n − sin  / n
n 2 − sin 2 
 n > 1
Therefore n is large value as compared to sin 2 , so neglecting
sin 2  we get
2
 pc =
Q. 7
 cos 
n
When the crank is at the inner dead centre, in a horizontal
reciprocating steam engine, then the velocity of the piston will
be.
Q. 8
(a) zero
(b) minimum
(c) maximum (d) (a) and (b) both
Ans. : (a)
Explanation : Refer Q3
The acceleration of the piston in a reciprocating stem engine is
given by
sin 2
cos 2
(a) .r  sin  + n 
(b) .r  cos  + n 




(c)  2  r  sin  +

Q. 9


cos 2 
n 
cos 2 
(d)
4 
r 2  cos  +

cos 2 
n 
Ans. : (d)
Explanation : Refer Q4
The velocity of piston in a reciprocating steam engine is _______
cos 2 
cos 2 
(a) r 2   cos  + n  (b) r  2  cos  + 2n 




(c) r   sin  +

Q. 11
 2  r  cos  +
where
 = Angular velocity of the crank,
r = Radius of the crank,
 = Angle turned by the crank from inner dead centre, and
n = Ratio of length of connecting rod to crank radius.
Ans. : (d)
Explanation : Refer Q4
Acceleration of piston of a reciprocating engine is ________
sin 2
cos 2
(a) r 2  sin  + n  (b) r  cos  + n 




(c) r 2  cos  +
Q. 10
sin 2 
n  (d)
sin 2  
2n 
(d)
r  2  cos  +

cos 2  
n 
Ans. : (c)
Explanation : Refer Q3
When the crank is at the inner dead centre, in a reciprocating
steam engine, the acceleration of the piston will be
n + 1
n – 1
(a) r  2  n 
(b) r  2  n 




n
n
(c) r  2  n + 1 
(d) r  2  n – 1 




Ans. : (a).
Q. 12
Q. 13
Explanation : When crank is at inner dead centre i.e.  = 0.
When the crank is at inner dead centre, in a reciprocating
steam engine, the velocity of the piston will be
n + 1
n + 1
(a) r  2 (b) r   n  (c) r 
(d) r  2  n 




Ans. : (c)
Explanation :
When crank is at inner dead centre
i.e.  = 0.
The displacement of the piston is given by
(a) r (1 – cos ) + l (1 – cos )
sin 2
(b) r  sin  + n 


cos 2 
n 
(d) none of the above
Ans. : (a)
Explanation : Refer Q1
The velocity of piston of a reciprocating engine is given by ( =
angular velocity of crank shaft;  = angle turned by crank from
top dead centre, n = l/r)
sin 2
(a) v p = r  sin  + 2n 


(c)  2 r  cos  +

Q. 14
sin 2
(b) v p =  2 r  sin  + 2n 


(c) v p = r  cos  +

cos 2 
2n 
(d) v p =  2 r  cos  +

cos 2 
2n 
Ans. : (a)
Q. 15
Explanation : Refer Q3
The linear acceleration of piston for reciprocating engine is
given by
cos 2
(a) f p =  2 r  cos  + 2n 


cos 2 
n 

sin 2
(c) f p =  2 r  sin  + 2n 


(b) f p =  2 r  cos  +.
(d) f p =  2 r  sin  +

sin 2 
n 
Ans. : (b)
Explanation : Refer Q4
Angular velocity of the connecting rod of a reciprocating
engine is approximately given by
 cos 
 sin 
(a)
(b)
n
n
Q. 16
 cos 
2n
Ans. : (a)
(c)
(d)
 sin 
2n
Explanation : Refer Q6
Angular acceleration of the connecting rod of a reciprocating
engine is approximately given by
Q. 17
(a)
–  2 cos 
n
 2 cos 
2n
Ans. : (b)
(c)
(b) –  2 sin 
n
(d)  2 sin 
2n
Explanation :
Since angular acceleration is rate of change of angular velocity of
connecting rod with respect to time t,
Therefore, pc
  pc = 
  pc =
=
d
d d
d   cos   d
( ) =

( ) =
dt pc d dt pc d  n  dt
d   cos  
d

...  · · ·
 dt =  
d  n 
 2
−  2
(− sin ) =
sin 
n
n
Negative sign indicates that acceleration of connecting rod is such
that it tends to reduce the angle  or the direction of angular
velocity of crank ‘’ and angular acceleration of connecting rod
have same direction.
Q. 18
The acceleration of the piston at inner dead centre and outer
dead centre is ( = angular velocity of crank,
r = crank radius, l = connecting rod length).
(a) zero and zero respectively
r
r
(b)  2 r  1 + l  and  2 r  l – 1  respectively




r
r
(c)  2 r  1 + 2l  and  2 r  2l – 1  respectively




2r
2r
(d)  2 r  1 + l  and  2 r  l – 1  respectively




Ans. : (b)
Q. 19
Explanation : Refer Q4
In a slider crank mechanism the stroke of slider is
200 mm and the obliquity ratio is 4. The crank rotates at 360
rpm. While crank is approaching the i.d.c. position, then
velocity and acceleration of piston is
(a) 10 m/s and 156 m/s 2
(b) 0 m/s and 177.56 m/s 2
(c) 15.20 m/s and 135.38 m/s 2
(d) 180 m/s and 250.59 m/s 2
Ans. : (b)
Explanation : When crank is approaching the I.D.C.
i.e.  = 0 as shown in Fig. 1.
Fig. 1
Velocity of piston at  = 0
sin 0
V P =  r  sin 0 + 2n  = 0


...Ans.
Acceleration of piston a p at  = 0
cos  
cos 0 
2 
n  =  r  cos 0 + n 
1
1
=  2 r  1 + n  = (37.69) 2  0.1  1 + 4 




f P =  2 · r  cos  +

f P
= 177.56 m/sec 2
...Ans..
Q. 20
In a slider crank mechanism, the stroke of the slider is 200 mm
and the obliquity ratio is 4. The crank rotates uniformly at 360
r.p.m. While crank is perpendicular to crank, then velocity and
Acceleration of piston is
(a) 10 m/s and 156 m/s 2
(b) 0 m/s and 177.56 m/s 2
(c) 3.87 m/s and 3.13 m/s 2
(d) 2.67 m/s and 4.13 m/s 2
Ans. : (c)
Explanation : When connecting rod is perpendicular to crank
is shown in Fig. 2.
Fig. 2
l
tan  = r = n = 4
  = tan − 1 4 = 75.96
Velocity of piston at  = 75.96
sin 2
V P =  r  sin  + 2n 


sin (151.92) 
 V P = (37.69) 0.1  sin (75.96) +
2  4


= 3.87 m/sec.
...Ans.
Acceleration of piston at
 = 75.96
f P =  2 r  cos  +


Q. 21
cos 2 
n 
f P = (37.69) 2  0.1  cos (75.96) +

cos (152.92) 
4


f P = 3.13 m/sec 2
...Ans.
Crank radius and connecting rod length for an IC engine
mechanism are 10 cm and 40 cm respectively. The crank is
rotating uniformly at 1050 rpm clockwise. Using analytical.
method, find out the acceleration of piston as well as the
angular acceleration of connecting rod when the crank is at
20 past the bottom dead center.
(a) 904.59 m/s 2 and 1030.40 rad/s 2
(b) – 904.57 m/s 2 and 1033.67 rad/s 2
(c) 836.38 m/s 2 and 1132.40 rad/s 2
(d) – 836.38 m/s 2 and 1130.40 rad/s 2
Ans. : (b)
Explanation :
Radius or crank,
r = 10 mm
Length of connecting rod,
l = 40 cm
l 40
 obliquity ratio,
n = r = 10 = 4
Speed of crank,
N = 1050 r.p.m.
2   N 2   1050

 =
60 =
60
= 109.95 rad/sec.
Crank angle,

 = 20 from B.D.C.
 = 180 + 20
= 200 from T. D. C.
Acceleration of piston is,
f p =  2 r  cos  +

cos 2 
n 
= 109.95 2  4  cos 200 +

= − 904.57 m/sec 2 .
cos 2  200 
4

...Ans.
Acceleration of connecting rod is,
−  2 sin 
 pc =
n
 pc =
Q. 22
− 109.95 2 sin 200
4
= 1033.67 rad/sec 2
...Ans.
The stroke of steam engine is 15 cm and connecting rod is 30
cm in length. The crank rotates at 600 rpm. When crank has
made 45 from i.d.c. The velocity of piston is _______.
(a) 390.12 cm/s
(b) 392.12 cm/s
(c) 209.36 cm/s
(d) 203.43 cm/s.
Ans. : (b)
Explanation :
Given :
Length of stroke, S =
 Radius of crank, r =
2r = 15 cm
15
2 = 7.5 cm
Length of connecting rod, l = 30 cm
 Obliquity ratio, l / r = 30 / 7.5 = 4
Speed of crank, N = 600 rpm
2   N 2   600

 =
= 20  rad/s
60 =
60
sin 2
Velocity of piston is, V p =   r  sin  + 2n 


sin 90 
or, V p = (20 ) 7.5  sin 45 +
2  4 

Q. 23
= 392.12 cm/s
...Ans.
The stroke of a steam engine is 15 cm and the connecting rod
is 30 cm in length. The crank rotates at 600 rpm. When crank
has made 45 from i.d.c. The acceleration of piston is ______.
(a) 390.12 m/s 2
(b) 392.12 m/s 2
(c) 209.36 m/s 2
(d) 203.43 m/s 2
Ans. : (c)
Explanation :
Given : Length of stroke,
S = 2r = 15 cm
15
 Radius of crank, r= 2 = 7.5 cm
Length of connecting rod, l = 30 cm
 Obliquity ratio, l / r = 30 / 7.5 = 4
Speed of crank, N = 600 rpm
2   N 2   600

 =
= 20  rad/s
60 =
60
Acceleration of the piston is,
cos 2
f p =  2  r  cos  + n 


cos 90
= (20  2 )  7.5  cos 45 + 4 

.

f p = 20936.6 cm/s 2 = 209.366 m/s 2
...Ans..
Q. 24
The stroke of a steam engine is 15 cm and the connecting rod
is 30 cm in length. The crank rotates at 600 rpm. When crank
has made 45 from i.d.c. The angular velocity of connecting rod
is _______.
(a) 11.10 rad/s
(b) 10.11 rad/s
(c) 18.20 rad/s
(d) 16.43 rad/s
Ans. : (a)
Explanation :
Given :
Length of stroke, S = 2r = 15 cm
15
 Radius of crank, r = 2 = 7.5 cm
Length of connecting rod, l = 30 cm
 Obliquity ratio, l / r = 30 / 7.5 = 4
Speed of crank, N = 600 rpm
2   N 2   600

 =
= 20  rad/s
60 =
60
Angular velocity of connecting rod is,
  cos  (20 ) cos 45
 pc =
=
n
4
Q. 25
= 11.107 rad/s 2
...Ans.
The stroke of a steam engine is 15 cm and the connecting rod
is 30 cm in length. The crank rotates at 600 rpm. When crank
has made 45 from i.d.c. The angular acceleration of
connecting rod ______.
(a) 607.89 rad/s 2
(b) 610.89 rad/s 2
(c) 697.89 rad/s 2
(d) – 697.89 rad/s 2
Ans. : (d)
Explanation :
Given :
Length of stroke, S = 2r = 15 cm
15
 Radius of crank, r = 2 = 7.5 cm
Length of connecting rod, l = 30 cm
 Obliquity ratio, l / r = 30 / 7.5 = 4
Speed of crank, N = 600 rpm
2   N 2   600

 =
= 20  rad/s
60 =
60.
Angular acceleration of the connecting rod is,
 2
(20 ) 2
 pc = − n sin  = − 4  sin 45
= − 697.89 rad/s 2
...Ans.
Negative sign indicate that acceleration of connecting rod is
such that it tends to reduce the angle .
Hooke’s Joint
Q. 26
In automobiles the power is transmitted from gear box to
differential through
(a) Oldham's coupling.
(b) Universal joint
(c) Hooke's joint
(d) Knuckle joint
Ans. : (c)
Explanation :
• A Hooke’s joint, commonly known as Universal coupling, is used
to connect two non-parallel and intersecting shafts. It is also used
for transmission of motion and power four shafts with angular
misalignment where flexible coupling does not serve the purpose.
• A common application of this joint is for transmission of power
from engine gear box to the rear axle. The engine shaft rotates at a
uniform speed whereas the driven shaft rotates at varying angular
speeds continuously.
• This joint is also used for transmission of power to different
spindles in a multiple drilling machine, knee joint in a milling
machine, in rolling mills etc.
Q. 27 Transmission of power from the engine to the rear axle of an
automobile is by means of
(a) Compound gears
(b) worm and wheel method
(c) Hooke's joint
(d) crown gear
Ans. : (c)
Explanation : Refer Q26
Q. 28 A Hooke’s joint is used to join two shafts which are
(a) aligned
(b) intersecting.
Q. 29
Q. 30
•
Q. 31
Q. 32
(c) parallel
(d) Perpendicular
Ans. : (b)
Explanation : Refer Q26
The type of coupling used to join two shafts whose axes are
neither in same straight line nor parallel, but intersect is
(a) flexible coupling
(b) universal coupling
(c) chain coupling
(d) oldham's coupling
Ans. : (b)
Explanation : Refer Q26
The Hooke's joint consists of :
(a) two forks
(b) one fork
(c) three forks
(d) four forks
Ans. : (a)
Explanation :
Hooke's joint consists of two U-shaped forks each connected to
driving and driven shafts. The four ends of the two forks are
connected by a centre piece called ‘cross’, the arms of which rests
in the bearings provided at the fork ends.
Tick the false statement
(a) In double Hooke's joint the axes of the driving and driven
shaft are in the same plane, and
(b) Both driving and driven shafts make equal angles with the
intermediate shafts
(c) In single Hooke's joint at four points the speed of the
driving and driven shafts is same
(d) None of the above is correct.
Ans. : (d)
Explanation : Refer Q26
The shafts (i.e. driving and driven shafts) are connected by a
Hooke’s joint. The relation between the angular displacements
of the driver () and driven () shafts and the inclination ()
between the axes of the driver and driven shafts is given by
(a) tan  = tan   tan  (b) tan  = tan   cos 
(c) tan  = cos   tan 
(d) tan  = sin   cos 
Ans. : (c)
Explanation :.
•
Consider two shafts inclined at angle  as shown in Fig. 5.8.3.
These are connected by a joint having two forks AB and CD as
shown in their top and front views. Fork AB of the driving shaft is
in the horizontal plane and the fork CD of the driven shaft is in
vertical plane..
Fig. 5.8.3
•
Consider that the driving shaft, hence, its fork AB rotates through
an angle  in a circle as represented by A 1 B 1 in front view.
Whereas, the fork CD will also move through an angle  of the
same size however, the projection of CD in the plane of paper will
have the trace of an ellipse since it is not in the same plane as AB.
Therefore, CD takes the position C 1 D 1 at an angle  on the ellipse.
In order to find its true angle turned, project C 1 to top view which
cuts the horizontal axis at E and the CD at F 1 . Rotate OF 1 to OF on
horizontal axis. Project F in the front view which cuts the circle at



C 1 . Join OC 1 . Then  COC 1 represents the true angle turned by fork
•
CD which is marked as  .
Therefore, when the driving shaft turns an angle , the driven shaft
turns through an angle . Value of angle  may be greater or less
than angle  or it may be equal to  at a particular position..

Consider triangles OC 1 E and OC 1 F, where,
 OC 1 E = 

Therefore,
But
 OC 1 F = 
and tan  =
tan  = OE
EC 1
tan  OF / FC 1
tan 

FC 1

and
OF

FC 1
...(i)

=
OE / EC 1
= EC 1
tan 
OF OF 1
=
...(ii)
OE
OE
tan 
OE
From OF 1 E,
cos  =
OF 1
On substituting this value in Equation (ii), we get,
tan 
1
=
...(iii)
tan 
cos 
or
tan  = cos  · tan  ...(1)

=
d
dt
d
 2 = Angular speed of driven shaft =
dt
On differentiating Equation (5.8.1) on both sides with respect to
time ‘t’, we get,
d
d
sec 2  ·
= cos  · sec 2  ·
dt
dt
2
2
or
 1 sec  = cos  · sec  ·  2
 2
sec 2 
or
=
...(iv)
 1
cos  · sec 2 
Also we know that,sec 2  = 1 + tan 2 
tan 2 

sec 2  = 1 +
[Using Equation (iii)]
cos 2 
sin 2 
1
or
sec 2  = 1 +

=
2
2
cos

cos 
cos 2  · cos 2  + sin 2 
cos 2  · cos 2 
cos 2  (1 − sin 2 ) + sin 2 
sec 2  =
cos 2  · cos 2 
Let,
 1 = Angular speed of driving shaft =.
− cos 2  · sin 2  + cos 2  + sin 2 
cos 2  · cos 2 
1 − cos 2  · sin 2 
sec 2  =
...(v)
cos 2  · cos 2 
On substituting the value of sec 2  from Equation (v) in Equation
(iv) we get,
 2
sec 2 
=
 1
1 − cos 2  · sin 2  
cos  
 cos 2  · cos 2  
 1 2 
 2
 cos  
or
=
2
 1
1
−
cos
 · sin 2  

2
 cos  · cos 2  
sec 2  =
or
Q. 33
Q. 34
 2
 1
=
cos 
1 − cos 2  · sin 2 
Hooke’s joint connects two shafts.
Angular displacement of driver = , angular displacement of
driven = , angle of inclination of shafts = , then,
(a) tan  = tan  cos 
(b) tan  = tan  tan 
(c) tan  = tan  cos 
(d) tan  = tan  tan .
Ans. : (a)
Explanation : Refer Q32
If the inclination () between the axes of the driver and driven
shafts of a Hooke’s joint is constant, then the velocity ratio of
the driven shaft to the driving shaft is given by
cos 
cos 
(a)
(b)
2
2
1 + cos  sin 
1 – cos 2  cos 2 
(c)
cos 
1 – cos 2  sin 2 
(d)
sin 
1 – cos 2  sin 2 
Where  = angular displacement of the driving shaft
Q. 35
Ans. : (c)
Explanation : Refer Q32
The velocity of the driven shat of a Hooke’s joint is minimum
when  is equal to
(a) 90 (b) 180 (c) 270 (d) (a) and (c).
Ans. : (d)
Explanation :
•
The diagram
showing the
variation of
speed
of
driven shaft
 2
with
respect
to
driven shaft
( 1 ) is called
polar
diagram as
shown in Fig.
5.8.4.
•
During
one
rotation
of
driving shaft,
( 2 ) max
occurs
at
points 1 and
3. At  = 0
and
180
( 2 ) min
occurs
at
points 2 and
4 at  = 90
and
270..
=  2 )
occurs at
( 1
points 5, 6, 7
and 8.
Q. 36
Q. 37
Fig. 5.8.4 : Polar diagram
The velocity of the driven shaft of a Hooke’s joint is maximum
when the value of angular displacement () of the driving shaft
is equal to
(a) 0
(b) 90
(c) 180 (d) (a) and (c)
Ans. : (d)
Explanation : Refer Q35
The value of angular displacement () of the driving shaft, for
which is velocity of the driven shaft is equal to the driver shaft,
is given by the relation
(a) tan  = cos 
(b) tan  =  cos 
(c) tan  =  sin 
(d) tan  =  cos   sin 
Where  = Inclination between the two shafts.
Ans. : (b)
Explanation :
 2
 1
= 1, therefore, from Equation (5.8.2) we have
1 − cos 2  · sin 2  = cos 
or
or
or
cos 2  = 1 − cos 
sin 2 
cos 2  = 1 − cos 
1 − cos 
=
2
1 − cos  (1 − cos ) (1 + cos )
cos 2  = 1
...(5.8.8)
1 + cos 
cos  = 
Also from Equation (1), we have,
1
...(1)
1 + cos .
1 − sin 2  =
or
1
1 + cos 
sin 2  = 1 −
1
cos 
=
1 + cos  1 + cos 
On dividing this equation by Equation (5.8.9), we get
tan 2  = cos 
or
tan  = 
cos 
There are two values each of  corresponding to positive and
negative sign respectively. Hence, there are four values of  at
which  2 =  1 .
Q. 38
If the driving and driven shafts connected by Hooke's joint
have equal speeds, then
(a) cos  = tan 
(b) tan  = 
cos 
(c) cot  = tan 
(d) sin  =  cot 
where
 = Angle through which the driving shaft turns
 = Angle of misalignment between the shafts.
Ans. : (b)
Q. 39
Explanation : Refer Q37
The co-efficient of fluctuation of speed of the driven shaft of a
Hooke’s joint, when the angle between the two shafts is small
and is measured in radian
(a) is proportional to the angle between the two shafts
(b) is proportional to the square of the angle between the two
shafts
(c) is inversely proportional to the angle between the two
shafts
(d) is inversely proportional to the square of the angle
between the two shafts
Ans. : (b)
Explanation :
•
Maximum fluctuation of speed of driven shaft is difference
between maximum and minimum speed of driven shaft,
mathematically it is given by,.
q = ( 2 ) max − ( 2 ) min
or
q =
q =  1
 1
cos 
−  1 cos 
 1 − cos  
 cos 

 1 − cos  
 cos  
2
q =  1
q =  1
 sin   ...(i)
 cos  
2
q =  1 tan   sin ...(5.8.11)
or
Since  is small angle, therefore substituting cos  = 1 and sin  =
 in Equation (i) we get,
q =  1  2
Therefore, maximum fluctuation of the speed of the driven shaft is
approximately varies as the square of the angle between the two
shafts.
Q. 40
The acceleration on the driven shaft of a Hooke’s joint is
maximum for a value of . This approximate value of  is given
by,
2 sin 2 
sin 2 
(a) cos 2 =
(b)
cos
2
=
2 + sin 2 
1 –2 sin 2 
(c) cos 2 =
2 sin 2 
1 – sin 2 
(d)
cos 2 =
2 sin 2 
2 – sin 2 
Ans. : (d)
 2
=
Explanation :
 1 cos 
1 − cos 2  · sin 2 
...From Equation (5.8.3)
Since, angular acceleration of driven shaft, say  1 is,
 2 =
d 2 d 2 d d 2
=

=
·  1
dt
dt d
d.
 1 cos 
d 



d  1 − cos 2  · sin 2  
− sin 2  · 2 cos  · sin  
2
=  1 · cos   

 (1 − cos 2  · sin 2 ) 2 
i.e.,  2 =  1 ·
or  2
2
 2 =
−  1 cos  · sin 2  · sin 2
...(5.8.13)
(1 − cos 2  · sin 2 ) 2
Condition for maximum acceleration :
d 2
•
Condition for maximum acceleration is
= 0. The approximate
d
result obtained is,  2 is maximum when,
or
cos 2 ~
–
Q. 41
Q. 42
•
2 sin 2 
2 − sin 2 
With single Hooke's joint it is possible to connect two shafts,
the axes of which have an angular misalignment up to
(a) 10° (b) 20°
(c) 30°
(d) 40°
Ans. : (d)
Maximum velocity of the driven shaft of a Hooke’s joint is
(a)  1 cos 
(b)  1 / cos 
(c)  1 sin 
(d)  1 / sin 
Ans. : (b)
Explanation :
For a given angle  between the driving and driven shaft, the
velocity ratio from Equation (5.8.4) is given as,
 1 · cos 
 2 =
(1)
1 − cos 2  · sin 2 
(a)
 2 will be maximum when denominator is minimum in
above Equation (1) or cos 2  is maximum for a given angle 
i.e., when for a given angle  i.e., when
cos  = 1 i.e.,  = 0, , 2 , .....
 1 cos 
 1 cos 

( 2 ) max =
=
1 − sin 2 
cos 2 
 1
or
( 2 ) max =
cos .
(b)
 2 will be minimum when denominator is maximum in
Equation (1) or cos 2  is minimum for a given angle  i.e., when
 3 
cos  = 0. i.e.,  = 2, 2 , .....

Q. 43
Q. 44
Q. 45
Q. 46
( 2 ) min =
1 − 0
or
( 2 ) min =  1 · cos 
Minimum velocity of the driven shaft of a Hooke’s joint is
(a)  1 cos 
(b)  1 / cos 
(c)  1 sin 
(d)  1 / sin 
Ans. : (a)
Explanation : Refer Q42
Maximum velocity of the driven shaft of a Hooke’s joint is at 
equal to
(a) 0 and 180
(b) 90 and 270
(c) 90 and 180
(d) 180 and 270
Ans. : (a)
Explanation : Refer Q35
It is good practice for a Hooke’s joint, used in machinary, that
shaft angle of inclination be
(a) around 60
(b) around 45
(c) around 30 and (d) less than or equal to 15
Ans. : (d)
Maximum fluctuation of speed of driven shaft in a single
Hooke’s joint is given by
1 – sin 2  
1 + cos 2  
(a)  
(b)  
2
 sin  
 sin 2  
1 – cos 2  
 2 cos 2 
(d)
(1 – cos 2 )
 cos 2  
Ans. : (c)
Explanation : Refer Q32
Maximum angular acceleration of driven shaft in a single
Hooke’s joint occurs at  given by
2 sin 
2 sin 2 
(a) cos 2  
(b) cos 2  
2
1 – sin 
2 – sin  
(c)  
Q. 47
 1 cos .
(c) cos 2  
Q. 48
Q. 49
Q. 50
Q. 51
Q. 52
2 cos 2 
2 – cos 2 
(d)
none of these
Ans. : (b)
Explanation : Refer Q40
The value of  for which acceleration in driven shaft is
maximum is approximately equal to
(a) cos 2 = 2 sin 2 
2 + sin 2  (b) cos 2 = sin 2 
2 – sin 2 
(c) cos 2 = 2 sin 2 
2 – sin 2  (d) cos 2 = 2 sin 2 
1 – sin 2 
Ans. : (c)
Explanation : Refer Q40
Minimum velocity of the driven shaft of Hooke’s joint is at 
equal to
(a) 0 and 180
(b) 90 and 270
(c) 90 and 180
(d) 180 and 250
Ans. : (b)
Explanation : Refer Q35
Maximum velocity of the driven shaft of Hooke’s joint is at 
equal to
(a) 0 and 180
(b) 90 and 270
(c) 90 and 180
(d) 180 and 250
Ans. : (a)
Explanation : Refer Q35
The ratio of angular speed of driving shaft to the angular speed
of driven shaft for Hooke’s joints
(a)  1 – cos 2  sin 2 
=
(b)
 1
cos  
cos 
=
 1 1 – cos 2  sin 2 
(c)  1 – cos 2  cos 2 
=
(d)
 1
cos   1 – cos 2  sin 2 
=
 1
sin 
Ans. : (a)
Explanation : Refer Q32
The angular velocity ratio of Hooke’s joint is minimum, when 
is equal to
(a)  = 
(b)  = 90
(c)  = 180
(d) none of the above.
Ans. : (c).
Q. 53
Q. 54
Explanation : Refer Q35
The angular velocity ratio of Hooke’s joint is maximum when 
is equal to
(a)  = 
(b)  = 90
(c)  = 180
(d) none of the above.
Ans. : (b)
Explanation : Refer Q35
The angular velocity ratio of Hooke’s joint is unity when
(a) tan  = 
Q. 55
Q. 56
cos 
(b)
tan  = 
cos 
(c) tan  =  tan 
(d) tan  =  tan 
Ans. : (a)
Explanation : Refer Q37
The variation of velocity ration of the two shafts connected by
Hooke’s joint for constant speed of driving shaft is
1
1
(a)
to cos 
(b)
to tan 
cos 
tan 
1
(c)
to sin 
(d) none of the above.
sin 
Ans. : (a)
Explanation : Refer Q37
The driving shaft of a single Hooke’s joint rotates at
1000 r.p.m. The angle between driving and driven shafts is 20.
The maximum and minimum speed of driven shaft in r.p.m. is
(a) 1064.18 and 939.69
(b) 1068.64 and 948.50
(c) 1120.22 and 951.32
(d) 1032.68 and 968.38
Ans. : (a)
Explanation :
Given :
N = 1000 rpm,  = 20
N 1
1000
(N 2 ) max =
=
cos  cos 20
= 1064.18 rpm
(N 2 ) min =
...Ans.
N 1 cos  = 1000 cos 20
= 939.69 rpm
Complex and Vector Algebra
...Ans..
Q. 57
•
Q. 58
The complex algebra method also called as ________
(a) complex flexible method
(b) complex variable method
(c) complex real method
(d) complex imaginary method
Ans. : (b)
Explanation : The complex algebra is analytical method used for
velocity and acceleration analysis of planer mechanism. It is also
known as complex variable method.
→
In complex algebra method Vector R can be represented in
polar form as______
→
(a) R =R  
→
•
(b)
→
→
→
R = R  
→
(c) R =  R
(d) R =   R
Ans. : (a)
Explanation : Complex numbers are not vectors, but they can be
used to represent vectors in a plane. For its representation an
origin, real axis and imaginary axis is to be selected along x axis
and y axis as shown in Fig. 5.6.1.
→
Consider a planer vector R as shown in Fig. 5.6.1 having
magnitude R and its direction  with respect to real axis.
→
Vector R can be represented in polar form as,
→
R = R  
It’s component along the real axis and imaginary axis are
R cos  (R x ) and R sin  (iR y ).
Q. 59
•
Q. 60
Q. 61
For loop closure equation is ________
(a) The sum of the relative position vectors for the links
forming a closed loop in the given mechanism is always
one
b) The sum of the relative position vectors for the links
forming a closed loop in the given mechanism is always
zero
(c) The sum of the relative position vectors for the links
forming a closed loop in the given mechanism is always
less than one
(d) The sum of the relative position vectors for the links
forming a closed loop in the given mechanism is always
more than one
Ans. : (b)
Explanation : The sum of the relative position vectors for the links
forming a closed loop in the given mechanism is always zero. This
is called as loop closure equation.
The loop closure equation is based on fact that the sum of the
relative position vectors for the links forming a closed loop in
the given mechanism is always_____
(a) Zero
(b) one
(c) more than one (d) less than one
Ans. : (a)
Explanation : Refer Q59
The loop closure equation for four bar mechanism having links
AD, AB ,BC and CD is
–→ –→ –→ –→
(a) AB + BC + CD + DA = 0
–→ –→ –→ –→
(b) AC + BD + CA + DA = 0
–→ –→ –→ –→
(c) BA + CB + DC + AD = 0
–→ –→ –→ –→
(d) AD + BC + DC + CA = 0
Ans. : (a)
Explanation :
Fig. Q61 shows four bar mechanism ABCD.
Consider vector  OAB, we get.
Relative position of B w.r.t.
–→ → →
A = AB = B − A ...(i)
Similarly, consider vector  OBC, ODC and OAD we get,
Relative position of C w.r.t. –→ → →
B = BC = C − B
Relative position of D w.r.t. –→ → →
C = CD = D − C ...(iii)
Relative position of A w.r.t. –→ → →
D = DA = A − D ...(iv)
...(ii)
Adding all the equation (i) to (iv) we get,
–→ –→ –→ –→
→ → → → → → → →
AB + BC + CD + DA = B − A + C − B + D − C + A − D
–→ –→ –→ –→
AB + BC + CD + DA = 0
The above Equation represents the loop closure equation for
four bar mechanism.
Fig Q 61 : Four bar mechanism
Q. 62
The loop closure equation for slider crank mechanism having
crank AB ,and connecting rod BC is
–→ –→ –→
–→ –→ –→
(a) AB + BC + CD = 0
(b) AC + BD + CA = 0.
–→ –→ –→
(c) BA + CB + DC = 0
Ans. : (a)
Explanation :
(d)
–→ –→ –→
AD + BC + DC = 0
Fig. Q62 shows slider crank mechanism ABC
Fig. Q62 : Slider crank mechanism
–→ → →
Relative position of B w.r.t. A = AB = B − A ...(i)
–→ → →
Relative position of C w.r.t. B = BC = C − B ...(ii)
–→ → →
Relative position of A w.r.t C = CA = A − C ...(iii)
Adding all equation (i) to (iii), we get
–→ –→ –→
→ → → → → →
AB + BC + CA = B − A + C − B + A − C
–→ –→ –→

AB + BC + CA = 0
The above Equation represents the loop closure equation
for slider crank mechanism.
Q. 63
In engine, crank radius is 50 mm and connecting rod length is
200 mm. The crank is rotating at 100 rad/s clockwise. At a
particular instant the crank is at 40 from TDC position. For
this position of the mechanism,the velocity of piston using
complex algebra method is
(a) 3840 mm/sec
(b) 3837 mm/sec
(c) 3845 mm/sec
(d) 3850 mm/sec
Ans. : (b).
Explanation:
Given :
Crank radius,
→
r = R 2 =50 mm
→
Length of connecting rods, l = R 3 = 200 mm
Angular velocity of crank,
Crank angle,
 2 = 100 rad/sec
 2 = 40 from TDC
Fig. 3
→
→ →
R 1 = R 2 + R 3
 x = r e i 2 + le i 3
 sin  3 =
− 50 sin 40
200
  3 = − 9.24 or 350.76 0
dx
V = dt
  3 =
− r  2 cos  2
l cos  3
= − 19.40
V p = − r  2 sin  2 − l  3 sin  3
V p = − 50x100 sin 40 − 200 (− 19.40) sin (350.76 0 )
V p = 3837 mm/sec.
... Ans.
❑❑❑