R

Objective

Mechanical Engineering with Study Material

R. Gupta's®

STUDY MATERIAL WITH MULTIPLE CHOICE QUESTIONS

by

RPH EDITORIAL BOARD

Published by O.P. Gupta for Ramesh Publishing House Admin. Office 12-H, New Daryaganj Road, Opp. Officers' Mess, New Delhi-110002  23261567, 23275224 E-mail: [email protected] For Online Shopping: www.rameshpublishinghouse.com Showroom Balaji Market, Nai Sarak, Delhi-6  23253720, 23282525 4457, Nai Sarak, Delhi-6,  23918938 © Reserved with the Publisher No Part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical including photocopying, recording or by any transformation storage and retrieval system without written permission from the Publisher. Indemnification Clause: This book is being sold/distributed subject to the exclusive condition that neither the author nor the publishers, individually or collectively, shall be responsible to indemnify the buyer/user/possessor of this book beyond the selling price of this book for any reason under any circumstances. If you do not agree to it, please do not buy/accept/use/possess this book.

Book Code: R-111 ISBN: 978-93-5012-662-2 25th Edition: February, 2021 HSN Code: 49011010

CONTENTS 1. Applied Mechanics .............................................................................. 3 Parallelogram Law of Forces; Triangle Law of Forces; Law of Moments; The Conditions of Equilibrium of a Body; Centroid; M.I. by Routh’s Rule; Laws of Frictions; Limiting Equilibrium and Limiting Friction; Angle of Repose; Projectile; The Equation of the Trajectory; Simple Harmonic Motion; Periodic Time; Oscillation; Amplitude; Equation of a Particle Moving with S.H.M.; Acceleration of a Particle Moving with S.H.M.; Angular Acceleration; Angular Velocity; Laws of Angular Motion; D’Alembert’s Principle; Motion of Lift Moving Upward; Acceleration of a Body on an Inclined Plane; Motion of Lift Moving Downward; Recoil of a Gun; Motion of Two Bodies Connected by a String; Multiple Choice Questions; Answers

2. Fluid Mechanics ................................................................................. 27 Type of Fluid; Buoyancy and Archimedes’ Principle; Bernoulli’s Theorem; Flow Through Pipes; Specific Energy; Hydraulic Turbines; Reaction Turbines; Impulse Turbines; Pumps; Water Power; Total Pressure and Centre of Pressure; Buoyancy and Flotation; Loss of Energy Due to Friction; Minor Energy Losses; Boundary Layer Flow; Multiple Choice Questions; Answers

3. Strength of Materials ......................................................................... 53 First and Second Law of Beam Diagrams; Relationship between Elastic Constants; Proof Resilience; Resilience; Strain Energy and Complimentary Energy; Multiple Choice Questions; Answers

4. Thermodynamics and Heat Transfer ................................................ 79 First Law of Thermodynamics; Second Law of Thermodynamics; Heat Transfer; Fins; Kirch off's Law; Multiple Choice Questions; Answers

5. Theory of Machines ......................................................................... 107 Kinematic Chain; Flat Belt Drives; Types of Governors; Cams; Governors; Vibrations; Multiple Choice Questions; Answers

6. Engineering Materials ...................................................................... 131 Introduction; Solidification; Crystallographic Structure; Allotropic Forms; Heat Treatment; Effect of Alloying Elements; Cast Irons; Flow Sheet for Production of Iron and Steel; Classification of Materials; Non-Metals; Crystal Structure; Crystal Imperfections; Solid Solutions; Equilibrium Phase Diagrams; Heat Treatment; Multiple Choice Questions; Answers

7. Production Technology .................................................................... 160 Foundry Practices; Vacuum Casting; Mechanics of Metal Cutting; Single Point Cutting Tool; Effect of Various Angles; Cutting Fluids (Lubricants or Coolants); Non-conventional Methods of Machining; Pattern and Mould; Pattern Allowances; Types of Patterns; Casting (iii)

(iv) processes; Dry Sand Mould Casting; Shell Mould Casting; Investment Casting; Gravity Die Casting; Die Casting; Centrifugal Casting; Slush casting; Defects in Castings; Fundamentals of Hot and Cold Working Processes (Metal Working) ; Multiple Choice Questions; Answers

8. Referigeration and Air-Conditioning .............................................. 191 Air Refrigerator Working on Reversed Carnot Cycle; Refrigeration; Psychrometric Processes; Stefan Boltzman Law; Psychrometry; Grey Body; Emissivity; Psychrometric Relations; Air Conditioning; Refrigerants; Multiple Choice Questions; Answers

9. Power Plant Engineering ................................................................. 211 Calorific Value (CV); Steam Generators; Classification of Boilers; Boiler Mountings; Classification of Steam Engines; Steam Turbines; Flow Through Nozzles; Stagnation Properties; Advantages of super-critical boilers over critical; Performance of Boilers; Reaction Turbine; Proton; Neutron; Isotopes; Isobars; Isotones; Nuclear Fission; Nuclear Fusion; Thermal Neutrons; Work Function; Essential Components of a Nuclear Reactor; Types of Nuclear Reactors; Nuclear Power Plant; Nuclear Reactors; Multiple Choice Questions; Answers

10. I.C. Engines ...................................................................................... 240 Introduction; Scavenging of I.C. Engines; Ignition System of Petrol Engines; Governing of I.C. Engines; Spark Plug; Carburettor; Supercharging of I.C. Engines; Rating of C.I. Engine Fuels-Cetane Number; Thermodynamic Tests for I.C. Engines; Internal Combustion Engine; Classification of Internal Combustion Engine; Application of I.C. Engines; Major Terms Connected with I.C. Engine; Multiple Choice Questions; Answers

11. Industrial Engineering and Management ....................................... 267 Principles of Organisation; PPC; Kind of Inspection; Definition of Time-Study; Time Study Equipments; Stop-Watches and Other Time Measuring Devices; Work Sampling; Job Evaluation Methods; Grading Method or Classification Method; Factors Comparison Method (Qualitative Approach); Point Method (Quantitative Approach); Merit Rating; Merit Rating Plans Classification; Break-even Analysis; Cost Accounting; Gantt Chart (Bar Chart); Multiple Choice Questions; Answers

12. Machine Design ................................................................................ 290 Introduction; Pressure Vessels; Pipes and Pipe Joints; Sliding Contact Bearings; Wedge Film Journal Bearings; Bearing Characteristic Number; Heat Generated in a Bearing; Rolling Contact Bearings; Spur Gears; Law of Gearing; Lewis Equation; Helical Gears; Bevel Gears; Worm Gears; Multiple Choice Questions; Answers

2

Mechanical Engineering

Applied Mechanics

Parallelogram Law of Forces It states, “If two forces acting simultaneously on a particle, be represented in magnitude and direction by the two adjacent sides of a parallelogram, then the resultant may be represented in magnitude and direction by the diagonal of the parallelogram which passes through their point of intersection”.

Resultant of a Number of Forces Acting at a Point Let, H = algebraic sum of resolved parts of the forces along X-axis. V = Algebraic sum of resolved parts of the forces along Y-axis. The resultant

Let, P and Q be two forces (Fig. 1)

R =

P 2 Q 2 2PQ. cos Q sin  tan  = P  Q cos  R =

and

3

and tan  =

( H ) 2  ( V ) 2 V H

Law of Moments It states “If a point remains in equilibrium under the effect of a number of coplanar forces, the sum of clockwise moments must be equal to the sum of anti-clockwise moments about any point in the same plane.”

R

P

Fig. 1

Lami’s Theorem (Fig. 3) If three coplanar forces acting on a point be in equilibrium, then each force is proportional to the sine of the angle between the other two. Mathematically Q R P = = sin  sin  sin 

Fig. 2 : Resolution of a force where is angle of inclination and R is the resultant of the forces. Triangle Law of Forces It states “If two forces acting simultaneously on a particle be represented in magnitude and direction by the two sides of a triangle, taken in order, their resultant may be represented in magnitude and direction, by the third side of the triangle, taken in opposite order.”

Fig. 3. Lami’s theorem 3

4

Mechanical Engineering

The Conditions of Equilibrium of a Body For the equilibrium of a body H = 0, V = 0, and M = 0

Solution

Centroid The point in a plane figure at which whole area of the figure is assumed to act, is called centroid.

Moment of Inertia of Important Figures (Fig. 4) 1. Rectangular section :

Centroid of Composite Figures If a1, a2, a3......an be the areas of individual figures. x¯ 1, x¯ 2, x¯ 3........x¯ n be the distances of centroids from any given axis.

X , the distance of the composite figure. then, an x n ax a x a x X = 1 1 2 2 3 3 a1 a2 a3 an Similarly, Y

=

X

=

Y

=

a1 y1 a2 y2

a3 y3

an y n

a1 a2 a3 an The equations for the centre of gravity of composite bodies are:

v1x1 v2 x2 v3 x3 ... v1 v2 v3 ...... v1 y1

v2 y2 v1

v2

vn xn vn

v3 y3 v3

vn yn vn

where letters carry their usual meanings. M.I. By Routh’s Rule I =

A S for rectangular or square sections 3

A S = for circular or elliptical sections 4 VS = for spherical bodies 5 A = area; V = volume. S = sum of the squares of remaining two semi-axes. Illustration: Find the moment of inertia of a circular section by Routh’s rule.

Ixx =

Ixx =

r 4 A. S r 2  ( r 2  O 2 ) = = 4 4 4

bd 3 ; 12

bd 3 12 2. Hollow rectangular section : 1 Iyy = (DB 3 – db 3); 12 1 Ixx = (BD 3 – bd 3) 12 3. A triangular section : Iyy =

bh 3 ; 36 bh 3 Ibase = 12 4. Circular section : d 4 r 4 Ixx = Iyy = = 4 64 where d = diameter of circle r = radius of circle r 4 d 4 Izz = = 32 2 where ZZ is an axis perpendicular to both XX and YY axes 5. Hollow circular section : Ixx =

(D4 – d4) 64 (R4 – r4)

Ixx = Iyy = =

4

Laws of Friction A. Laws of static friction : Following are the laws of static friction : 1. The force of friction acts in a direction, opposite to that in which, the body tends to move. 2. The magnitude of the force of friction, is always equal to the applied force.

Applied Mechanics Table 1. C.G. for centroid of common figures SL. No.

Figure

Area/Volume

Distance of C.G.

¯x = 1.

A=

A=

1 2

3

¯y =

¯y =

B.H.

2

2 B.H. 3

x=

3 B 8

y=

3 H 8

x=

y=

3

3

¯y =

(A+B)×H

H

H

D 2 8

A=

3.

5.

1

A=

2.

4.

B.H. 2

A=

B

2D 3

A 2  AB  B 2 3 ( A  B)

H 3

×

(2 A B) ( A B)

6.

Vol. =

2 3

3

y=

3r 8

7.

Vol. =

1 2 r h 3

y=

h 4

5

6

Mechanical Engineering 4. The force of friction depends upon the roughness of the surface. 5. The force of friction is independent of the area of contact between the bodies. B. Laws of dynamic friction : Following are the laws of dynamic friction: 1. The force of friction always acts in a direction opposite to that on which the body is moving. 2. The magnitude of the dynamic friction bears a constant ratio to the normal reaction between the surface. It is slightly less than that in case of static friction. 3. For moderate speeds, the force of friction remains constant and decreases with a slow rate with the increase of speed. Limiting Equilibrium and Limiting Friction The stage when the body acted upon by an external force, is just on the point of moving, is called limiting equilibrium of the body. The force of friction which is offered by the rough surface at the stage of limiting equilibrium, is called limiting friction. When a body starts moving, the force of friction offered by the surface, is called the dynamic friction. Angle of Friction When a body is at the point of limiting eqilibrium, the force of friction is maximum. The angle which the resultant of the maximum force of friction and the normal reaction makes with the normal reaction, is called the angle of friction. It is denoted by  Maximum force of friction tan = Normal reaction Coefficient of Friction The ratio of the limiting friction and the normal reaction is called coefficient of friction.

Fig. 4 3. The magnitude of the limiting friction, bears a constant ratio to the normal reaction between F the body and the surface, i.e., = a constant. R

Angle of Repose The angle which an inclined rough surface makes with the horizontal when a body placed on it, is on the point of moving down, is called Angle of Repose. The angle of repose is always equal to the angle of friction.

Applied Mechanics

7

Maximum height attained by the projectile h =

u2 sin 2

2g For maximum range, angle of projection  = 45° The velocity and direction of motion of a projectile at a given height h, above the point of projection Fig. 5

v =

u 2  2 gh

2gh u 2 sin 2 u cos Time of flight of a projectile up an inclined plane

tan  =

2u sin 2    g cos  where,  = angle of projection  = angle of inclined plane. Time of flight of a projectile down an inclined plane 2u sin    t = g cos  The range of projectile on an inclined plane

t =

Fig. 6 Projectile The path travelled by a projectile in air (neglecting air resistance) is a parabola as shown in (Fig. 7).

R =

Fig. 7 The Equation of the Trajectory The equation of the path traversed by a projectile is gx 2 y = x. tan  – . 2u 2 cos 2  The velocity of projectile after t-seconds v =

u 2  g 2 t 2  2u gt sin  ,

u sin   gt u cos  Time of flight of a projectile usin  = g where  =

2u 2 sin   cos 

g cos 2  For maximum range down an inclined plane    = – 4 2 For maximum range up an inclined plane    = + 4 2 SIMPLE HARMONIC MOTION The to and fro motion of the foot of perpendicular of a point moving along the circumference of any diameter is called simple harmonic motion. Periodic Time The time taken by the body for one complete oscillation is called periodic time.  T =  T = Periodic time  = Angular velocity in radians/sec.

8

Mechanical Engineering

Oscillation One complete vibration of a body moving in simple harmonic motion is called oscillation. Amplitude The maximum displacement of a body from its mean position is called amplitude. Equation of a Particle Moving with S.H.M. The equation for velocity of a particle moving with simple harmonic motion is given as v =  r 2  y2 r = the amplitude y = distance from mean position. Acceleration of a Particle M oving w ith S.H.M. The acceleration of a particle moving with simple harmonic motion f = – 2y It is maximum when the projectile is at its extreme position. Angular Acceleration The rate of change of angular velocity is called angular acceleration. It is expressed in radians/ sec/sec. Angular Velocity A body making N r.p.m. is said to possess angular velocity N  = rad./sec. 60 Laws of If, w 0 w t

Angular Motion = initial angular velocity = final angular velocity = time taken by the particle to change its velocity from 0 +  = constant angular acceleration in rad./sec 2.  = total angular displacem ent in radians. Then, w = w0 + t = w0t + ½ t2 w 2 = w02 + 2  The relationship between linear and angular velocities is v = rw.

where r is radius of the circular path of the moving body. D’Alembert’s Principle If a rigid body is acted upon by a system of forces, the system may be reduced to a single resultant force whose magnitude, direction and line of action may be found out by the method of graphic statics, i.e., P = mf = O. Motion of Lift Moving Upward A lift of weight W kg is moving upward with a uniform acceleration F. The tension in the cable supporting the lift is given by

FG H

T = W 1

IJ K

f kgf. g

NEWTON’S LAWS OF MOTION There are three laws of motion : 1. First law: Every body continues in its state of rest or of uniform motion in a straight line, unless it is acted upon by some external force. 2. Second law: The rate of change of momentum is directly proportional to the impressed force and takes place in the same direction in which the force acts, i.e., P = mf. 3. Third law: To every action, there is always an equal and opposite reaction. Acceleration of a Body on an Inclined Plane The acceleration f with which a load W kg slides along the inclined smooth surface, making an angle with horizontal is given by f = g sin .

Fig. 8

Applied Mechanics

9

Motion of Lift Moving Downward A lift of weight W kg is moving downward with a uniform acceleration f. The tension in the cable supporting the lift is given by f T = W 1 kgf. g

FG H

IJ K

Recoil of a Gun The velocity of recoil V of a gun of mass M from which a bullet of mass m is fired with velocity v is related by MV = mv. Motion of Two Bodies Connected by a String (1) For the system of two bodies connected by a string and passing over a smooth pulley as shown in Fig.10. Acceleration ( m1  m2 ) g f = metre/sec2 2 m1 + m2 Tension 2 m1 m2 T = g Newtons m1 + m2 (2) The acceleration developed in a body of mass m1 hanging freely and connected by a string to other body of mass W lying on a smooth horizontal plane, (Fig. 9). m1 g f = metre/sec2. m1 + W

Fig. 10 (4) A body of mass m1 is hanging freely and is connected by a string passing over a pulley to another body of mass W lying on a rough horizontal plane. The tension T in the string m1W (1 +  ) T = × g Newtons. m1 + W (5) A body of mass m1 is hanging freely and is connected by a string passing over a pulley to another body of mass m2 lying on a rough horizontal plane. The acceleration of the system g (m1  m2 ) f = metre/sec 2. m1 + m2 (6) A body of mass m1 is hanging freely and is connected by a string passing over a pulley, to another body of mass m2, lying on smooth inclined plane (Fig. 11). The acceleration of the system. f =

Fig. 9 (3) A body of mass m1 is hanging freely and is connected by a string passing over a pulley to another body of mass W lying on a smooth horizontal plane. Tension T in the string is given by m1W T = Newtons. m1 + W

g (m1 m2 sin ) m/sec2. m1 + m2

Fig. 11 The tension in the string T =

m1 . m2 (1 sin  ) Newtons. m1 + m2

10

Mechanical Engineering

(7) A body of mass m 1 is hanging freely and another body of mass m 2 is lying on a rough inclined plane. and being coefficient of friction and angle of inclination respectively.

f =

g (m1  m2 sin  m2 cos  )

T = m 1 m2

( m1 + m2 ) 2

m/sec2.

g (1  sin   cos  ) kg.f. m1 m2

EXERCISE 1. The resolved part of the resultant of two forces inclined at an angle  in a given direction is (a) algebraic sum of the resolved parts of the forces in the direction (b) arithmetical sum of the resolved parts of the forces in the direction (c) difference of the forces multiplied by cosine ° (d) sum of the forces multiplied by the sine (e) sum of the forces multiplied by the tangent ° 2. A circular section of radius r is cut out of a uniform disc of radius R. The centre of hole being R/2 from the centre of original disc. The centre of gravity of the resulting body will lie on the line joining the centre of hole to the centre of disc at a distance of (a)

(b)

(c)

r2R

from centre of disc and away 2 (R 2  r 2 ) from the hole

r2R 2 (R 2  r 2 ) the hole r2R

from centre of disc towards

from centre of hole towards 2 (R 2  r 2 ) the centre of disc (d) none of the above 3. The following is in unstable equilibrium (a) a uniform solid cone resting on a generator on a smooth horizontal plane (b) a uniform solid cone resting on its base on a horizontal plane (c) a solid cube resting on one edge (d) a satellite encircling the earth

4. A point subjected to a number of forces will be in equilibrium if (a) sum of resolved parts in any tw o directions at right angles are both zero (b) algebraic sum of the forces is zero (c) two resolved parts in any two directions at right angles are equal (d) algebraic sum of the moments of the forces about the point is zero (e) none of these 5. Minimum potential energy of a system will be in the position of (a) stable equilibrium (b) unstable equilibrium (c) neutral equilibrium (d) all of the above 6. The forces which meet at one point and have their lines of action in different planes are called (a) coplaner non-concurrent forces (b) non-coplaner concurrent forces (c) non-coplaner non-current forces (d) intersecting forces (e) none of these 7. A rigid body is in a stable equilibrium if the application of any force (a) can raise the CG of the body but cannot lower it (b) tends to lower the CG of the body (c) neither raises nor lower the CG of the body (d) none of the above 8. According to Law of Triangle of Forces (a) three forces acting at a point can be represented by the sides of a triangle each side being in proportion to the force

Applied Mechanics

9.

10.

11.

12.

(b) three forces acting along the sides of a triangle are always in equilibrium (c) if three forces acting on a point can be represented in magnitude and direction, by the sides of a triangle taken in order, these will be in equilibrium (d) if three forces acting at a point are in equilibrium each force is proportional to the sine of the angle between the other two (e) if the forces acting on a particle be represented in magnitude and direction by the two sides of a triangle taken in order, their resultant will be represented in magnitude and direction by the third side of the triangle, taken in opposite order A heavy ladder resting on a floor and against a vertical wall may not be in equilibrium if (a) floor is smooth and the wall is rough (b) floor is rough and the wall is smooth (c) floor and wall both are smooth surfaces (d) floor and wall both are rough surfaces A uniform rod 9 m long weighing 40 kg is pivoted at a point 2 m from one end where a weight of 120 kg is suspended. The required force acting at the end in a direction perpendicular to rod to keep it in equilibrium at an inclination of 60º with horizontal is (a) 40 kg (b) 60 kg (d) 100 kg (c) 10 kg The centre of gravity of a plane lamina will not be at its geometrical centre if it is a (a) circle (b) equilateral triangle (c) rectangle (d) square (e) right angled triangle The mass moment of inertia of a rectangular plate of mass M and sides a and b about an axis perpendicular to plate through its centre is (a)

M (a2 b2 ) 3

(b)

(c)

M 2 (a + b 2) 12

(d)

M 2 (a + b2) 4 Ma2 Mb2 + 4 12

11

13. Varigon’s theorem of moments states that the (a) arithmetical sum of the moments of two forces about any point is equal to the moments of their resultant about that point (b) algebraic sum of the moments of two forces about any point, is equal to the moment of their resultant about that point (c) arithmetical sum of the moments of the forces about any point in their plane, is equal to the moment of their resultant about that point (d) algebraic sum of the moments of the forces about any point in their plane, is equal to the moment of their resultant about that point 14. The necessary condition of equilibrium of a body is : (a) algebraic sum of horizontal components of all the forces must be zero (b) algebraic sum of vertical components of all the forces must be zero (c) algebraic sum of the moments of the forces about a point must be zero (d) all (a), (b) and (c) 15. For rotation about a central axis, a solid cylinder of mass M and radius R is equivalent to a hoop of same mass and radius equal to (a)

R 2

(c) 2 R

(b) (d)

2R

R 2

16. There are three solids viz., a hoop, a cylinder and a sphere, all having the same mass M and radius R. The rotational inertia about the axis passing through the centre of cross-section is (i) maximum for hoop (ii) minimum for sphere (iii) maximum for hoop and minimum for cylinder (iv) same for hoop, sphere and cylinder The correct answer is (a) only (iii) (b) only (iv) (c) (i) and (ii)

About the Book This concise book on Objective Mechanical Engineering is specially published for the aspirants appearing in various competitive examinations in engineering organised by SSC, UPSC, PSC, DDA, FCI, JKSSB, DSSSB, DMRC, JMRC, RRB, NTPC, BSNL etc. and all other recruitment exams organised by various technical institutions. The book contains ample Study and Practice Material with numerous Multiple Choice Question-Answers on the subject, important from the point of view of the exam. All the practice questions in the book have been modelled on previous exam-questions and solved by learned subject-expert with due diligence. The book is highly recommended for the aspirants to Sharpen their Problem Solving Skills with thorough practice of actual exam-style questions and hundreds of practice questions provided in the book, and prepare themselves to face the exam with Confidence, Successfully. While the specialised study and practice material of this book Paves the Way for your Success, your own diligent study and practice with it will ensure you a Successful Career.

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