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WORKSHOP CALCULATION, & SCIENCE, (As Per NSQF), 1st Year, Common for All Engineering Trades under CTS, (For all 1 Year and 2 Year Trades), , DIRECTORATE GENERAL OF TRAINING, MINISTRY OF SKILL DEVELOPMENT & ENTREPRENEURSHIP, GOVERNMENT OF INDIA, , NATIONAL INSTRUCTIONAL, MEDIA INSTITUTE, CHENNAI, Post Box No. 3142, CTI Campus, Guindy, Chennai - 600 032, , Copyright free, under CC BY Licence

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Workshop Calculation & Science (NSQF) - 1st Year, Common for All Engineering Trades Under CTS, (For All 1 year & 2 year Trades), , Copyright © 2018 National Instructional Media Institute, Chennai, First Edition :, , 2019, , Copies :, , Rs. /-, , All rights reserved., No part of this publication can be reproduced or transmitted in any form or by any means, electronic or mechanical,, including photocopy, recording or any information storage and retrieval system, without permission in writing from the, National Instructional Media Institute, Chennai., , Published by:, NATIONAL INSTRUCTIONAL MEDIA INSTITUTE, P. B. No.3142, CTI Campus, Guindy Industrial Estate,, Guindy, Chennai - 600 032., Phone : 044 - 2250 0248, 2250 0657, 2250 2421, Fax : 91 - 44 - 2250 0791, email : [email protected], [email protected], (ii), Website: www.nimi.gov.in, , Copyright free, under CC BY Licence

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FOREWORD, , The Government of India has set an ambitious target of imparting skills to 30 crores people, one out of every, four Indians, by 2020 to help them secure jobs as part of the National Skills Development Policy. Industrial, Training Institutes (ITIs) play a vital role in this process especially in terms of providing skilled manpower., Keeping this in mind, and for providing the current industry relevant skill training to Trainees, ITI syllabus, has been recently updated with the help of comprising various stakeholder's viz. Industries, Entrepreneurs,, Academicians and representatives from ITIs., The National Instructional Media Institute (NIMI), Chennai, has now come up with instructional material to, suit the revised curriculum for Workshop Calculation & Science 1st Year (For All 1 year & 2 year, Trades) NSQF Commom for all engineering trades under CTS will help the trainees to get an international, equivalency standard where their skill proficiency and competency will be duly recognized across the globe, and this will also increase the scope of recognition of prior learning. NSQF trainees will also get the, opportunities to promote life long learning and skill development. I have no doubt that with NSQF the, trainers and trainees of ITIs, and all stakeholders will derive maximum benefits from these IMPs and that, NIMI's effort will go a long way in improving the quality of Vocational training in the country., The Executive Director & Staff of NIMI and members of Media Development Committee deserve appreciation, for their contribution in bringing out this publication., Jai Hind, , RAJESH AGGARWAL, Director General/ Addl. Secretary, Ministry of Skill Development & Entrepreneurship,, Government of India., , New Delhi - 110 001, , (iii), , Copyright free, under CC BY Licence

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PREFACE, The National Instructional Media Institute(NIMI) was set up at Chennai, by the Directorate General of Training,, Ministry of skill Development and Entrepreneurship, Government of India, with the technical assistance, from the Govt of the Federal Republic of Germany with the prime objective of developing and disseminating, instructional Material for various trades as per prescribed syllabus and Craftsman Training Programme(CTS), under NSQF levels., The Instructional materials are developed and produced in the form of Instructional Media Packages (IMPs),, consisting of Trade Theory, Trade Practical, Test and Assignment Book, Instructor Guide, Wall charts,, Transparencies and other supportive materials. The above material will enable to achieve overall improvement, in the standard of training in ITIs., A national multi-skill programme called SKILL INDIA, was launched by the Government of India, through a, Gazette Notification from the Ministry of Finance (Dept of Economic Affairs), Govt of India, dated 27th, December 2013, with a view to create opportunities, space and scope for the development of talents of, Indian Youth, and to develop those sectors under Skill Development., The emphasis is to skill the Youth in such a manner to enable them to get employment and also improve, Entrepreneurship by providing training, support and guidance for all occupation that were of traditional, types. The training programme would be in the lines of International level, so that youths of our Country can, get employed within the Country or Overseas employment. The National Skill Qualification Framework, (NSQF), anchored at the National Skill Development Agency(NSDA), is a Nationally Integrated Education, and competency-based framework, to organize all qualifications according to a series of levels of Knowledge,, Skill and Aptitude. Under NSQF the learner can acquire the Certification for Competency needed at any, level through formal, non-formal or informal learning., The Workshop Calculation & Science 1st Year (For All 1 year & 2 year Trades) (Comon for All Engineering, Trades under CTS) is one of the book developed by the core group members as per the NSQF syllabus., The Workshop Calculation & Science (Common for All Engineering Trades under CTS as per NSQF), 1st Year (For All 1 year & 2 year Trades) is the outcome of the collective efforts of experts from Field, Institutes of DGT, Champion ITI’s for each of the Sectors, and also Media Development Committee (MDC), members and Staff of NIMI. NIMI wishes that the above material will fulfill to satisfy the long needs of the, trainees and instructors and shall help the trainees for their Employability in Vocational Training., NIMI would like to take this opportunity to convey sincere thanks to all the Members and Media Development, Committee (MDC) members., , R. P. DHINGRA, EXECUTIVE DIRECTOR, , Chennai - 600 032, , (iv), , Copyright free, under CC BY Licence

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ACKNOWLEDGEMENT, The National Instructional Media Institute (NIMI) sincerely acknowledge with thanks the co-operation and, contribution of the following Media Developers to bring this IMP for the course Workshop Calculation & Science, (1st Year (For All 1 year & 2 year Trades)) as per NSQF., , MEDIA DEVELOPMENT COMMITTEE MEMBERS, Shri. M. Sangara pandian, , -, , Training Officer (Retd.), CTI, Guindy, Chennai., , Shri. G. Sathiamoorthy, , -, , Jr.Training Officer (Retd.), Govt I.T.I, DET - Tamilnadu., , NIMI CO-ORDINATORS, Shri. Nirmalya nath, , -, , Deputy General Manager,, NIMI, Chennai - 32., , Shri. G. Michael Johny, , -, , Assistant Manager,, NIMI, Chennai - 32., , NIMI records its appreciation of the Data Entry, CAD, DTP Operators for their excellent and devoted services in, the process of development of this IMP., NIMI also acknowledges with thanks, the efforts rendered by all other staff who have contributed for the development of this book., , (v), , Copyright free, under CC BY Licence

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INTRODUCTION, The material has been divided into independent learning units, each consisting of a summary of the topic and an, assignment part. The summary explains in a clear and easily understandable fashion the essence of the mathematical, and scientific principles. This must not be treated as a replacment for the instructor’s explanatory information to be, imparted to the trainees in the classroom, which certainly will be more elaborate. The book should enable the, trainees in grasping the essentials from the elaboration made by the instructor and will help them to solve independently, the assignments of the respective chapters. It will also help them to solve the various problems, they may come, across on the shop floor while doing their practical exercises., The assignments are presented through ‘Graphics’ to ensure communications amongst the trainees. It also assists, the trainees to determine the right approach to solve the problems. The required relevent data to solve the problems, are provided adjacent to the graphics either by means of symbols or by means of words. The description of the, symbols indicated in the problems has its reference in the relevant summaries., At the end of the exercise wherever necessary assignments, problems are included for further practice., Time allotment:, Duration of 1st Year, , : 84 Hrs, , Time allotment for each module has been given below. Common to all 1 year and 2 year Engineering Trades., S.No, , Title, , Exercise No., , 1, , Unit, Fractions, , 1.1.01 - 1.1.07, , 2, , Square root, Ratio and Proportions, Percentage, , 1.2.08 - 1.2.14, , 3, , Material Science, , 1.3.15 - 1.3.19, , 4, , Mass, Weight, Volume and Density, , 1.4.20 & 1.4.21, , 5, , Speed and Velocity, Work, Power and Energy, , 1.5.22 - 1.5.25, , 6, , Heat & Temperature and Pressure, , 1.6.26 - 1.6.32, , 7, , Basic Electricity, , 1.7.33 - 1.7.38, , 8, , Mensuration, , 1.8.39 - 1.8.43, , 9, , Levers and Simple machines, , 1.9.44 & 1.9.45, , 10, , Trigonometry, , 1.10.46 - 1.10.49, , LEARNING / ASSESSABLE OUTCOME, On completion of this book you shall be able to, • Demonstrate basic mathematical concept and principles to perform, practical operations., • Understand and explain basic science in the field of study including, simple machine., , (vi), , Copyright free, under CC BY Licence

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CONTENTS, Exercise No., , Title of the Exercise, , Page No., , Unit, Fractions, 1.1.01, , Unit - Classification of unit system, , 1, , 1.1.02, , Unit - Fundamental and Derived units F.P.S, C.G.S, M.K.S and SI units, , 2, , 1.1.03, , Unit - Measurement units and conversion, , 4, , 1.1.04, , Factors, HCF, LCM and problems, , 14, , 1.1.05, , Fractions - Addition, substraction, multiplication & division, , 15, , 1.1.06, , Decimal fractions - Addition, subtraction, multilipication & division, , 17, , 1.1.07, , Solving problems by using calculator, , 20, , Square root, Ratio and Proportions, Percentage, 1.2.08, , Square root - Square and suare root, , 27, , 1.2.09, , Square root - Simple problems using calculator, , 28, , 1.2.10, , Square root - Applications of pythagoras theorem and related problems, , 29, , 1.2.11, , Ratio and proportion, , 30, , 1.2.12, , Ratio and proportion - Direct and indirect proportions, , 32, , 1.2.13, , Percentage, , 36, , 1.2.14, , Precentage - Changing percentage to decimal and fraction, , 39, , Material Science, 1.3.15, , Material science - Types metals, types of ferrous and non ferrous metals, , 40, , 1.3.16, , Material science - Physical and mechanical properties of metals, , 42, , 1.3.17, , Material science - Introduction of iron and cast iron, , 45, , 1.3.18, , Material science - Difference between iron & steel, alloy steel and carbon steel, , 48, , 1.3.19, , Material science - Properties and uses of rubber, timber and insulating materials, , 50, , Mass, Weight, Volume and Density, 1.4.20, , Mass, volume, density, weight and specific gravity, , 53, , 1.4.21, , Related problems for mass, volume, density, weight and specific gravity, , 55, , Speed and Velocity, Work, Power and Energy, 1.5.22, , Speed and velocity - Rest, motion, speed, velocity, difference between speed, and velocity, acceleration and retardation, , 61, , 1.5.23, , Speed and velocity - Related problems on speed & velocity, , 65, , 1.5.24, , Work, power, energy, HP, IHP, BHP and efficiency, , 69, , 1.5.25, , Potential energy, kinetic energy and related problems with assignment, , 72, , Heat & Temperature and Pressure, 1.6.26, , 1.6.27, , Heat & Temperature - Concept of heat and temperature, effects of heat,, difference between heat and temperature, boiling point & melting point of, different metals and non-metals, , 74, , Heat & Temperature - Scales of temperature, celsius, fahrenheit, kelvin and, conversion between scales of temperature, , 76, , (vii), , Copyright free, under CC BY Licence

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Exercise No., 1.6.28, , Title of the Exercise, , Page No., , Heat &Temperature - Temperature measuring instruments, types of, thermometer, pyrometer and transmission of heat - Conduction, convection, and radiation, , 78, , Heat & Temperature - Co-efficient of linear expansion and related problems, with assignments, , 80, , 1.6.30, , Heat & Temperature - Problem of heat loss and heat gain with assignments, , 82, , 1.6.31, , Heat & Temperature - Thermal conductivity and insulators, , 86, , 1.6.32, , Concept of pressure - Units of pressure, atmospheric pressure, absolute, pressure, gauge pressure and gauges used for measuring pressure, , 88, , 1.6.29, , Basic Electricity, 1.7.33, , Basic electricity - Introduction and uses of electricity, molecule, atom, how, electricity is produced, electric current AC,DC their comparison, voltage,, resistance and their units, , 98, , 1.7.34, , Basic electricity - Conductor, insulator, types of connections - series and parallel, , 102, , 1.7.35, , Basic electricity - Ohm’s law, relation between V.I.R & related problems, , 108, , 1.7.36, , Basic electricity - Electrical power, energy and their units, calculation with, assignments, , 112, , Basic electricity - Magnetic induction, self and mutual inductance and EMF, generation, , 115, , Basic electricity - Electrical power, HP, energy and units of electrical energy, , 118, , 1.7.37, 1.7.38, , Mensuration, 1.8.39, , Mensuration - Area and perimeter of square, rectangle and parallelogram, , 121, , 1.8.40, , Mensuration - Area and perimeter of Triangles, , 125, , 1.8.41, , Mensuration - Area and perimeter of circle, semi-circle, circular ring, sector, of circle, hexagon and ellipse, , 130, , Mensuration - Surface area and volume of solids - cube, cuboid, cylinder,, sphere and hollow cylinder, , 138, , Mensuration - Finding the lateral surface area, total surface area and capacity, in litres of hexagonal, conical and cylindrical shaped vessels, , 145, , 1.8.42, 1.8.43, , Levers and Simple machines, 1.9.44, , 1.9.45, , Simple machines - Effort and load, mechanical advantage, velocity ratio,, efficiency of machine, relationship between efficiency, velocity ratio and, mechanical advantage, , 148, , Lever & Simple machines - Lever and its types, , 150, , Trigonometry, 1.10.46, , Trigonometry - Measurement of angles, , 154, , 1.10.47, , Trigonometry - Trigonometrical ratios, , 156, , 1.10.48, , Trigonometry - Trigonometrical tables, , 162, , 1.10.49, , Trigonometry - Application in calculating height and distance, (Simple applications), , 173, , (viii), , Copyright free, under CC BY Licence

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SYLLABUS, 1st Year, , Common for All Engineering Trades under CTS, , S.no., I, , Duration: One Year, , Syllabus, Unit, Fractions, , Time, , Marks, , 4, , 3, , 7, , 4, , 10, , 5, , 10, , 5, , 5, , 3, , 1 Classification of Unit System, 2 Fundamental and Derived Units F.P.S, C.G.S, M.K.S and SI Units, 3 Measurement Units and Conversion, 4 Factors, HCF, LCM and Problems, 5 Fractions – Addition, Subtraction, Multiplication and Division, 6 Decimal Fractions - – Addition, Subtraction, Multiplication and Division, 7 Solving Problems by using calculator, II, , Square Root: Ratio and Proportions, Percentage, 1 Square and Square Root, 2 Simple problems using calculator, 3 Application of Pythagoras Theorem and related problems, 4 Ratio and Proportions, 5 Direct and Indirect proportion, 6 Percentage, 7 Changing percentage to decimal, , III, , Material Science, 1 Types of metals, 2 Physical and Mechanical Properties of metals, 3 Types of ferrous and non-ferrous metals, 4 Introduction of iron and cast iron, 5 Difference between iron and steel, alloy steel and carbon steel, 6 Properties and uses of rubber, timber and insulating materials, , IV, , Mass, Weight, Volume, and Density, 1 Mass, volume, density, weight & specific gravity, 2 Related problems for mass, volume, density, weight & specific gravity, , V, , Speed and Velocity, Work Power and Energy, 1 Rest, motion, speed, velocity, difference between speed and velocity, acceleration, and retardation, 2 Related problems on speed and velocity, 3 Potential energy, Kinetic Energy and related problems with related problems, 4 Work, power, energy, HP, IHP, BHP and efficiency, , (ix), , Copyright free, under CC BY Licence

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S.no., VI, , Syllabus, Heat & Temperature and Pressure, , Time, , Marks, , 15, , 9, , 12, , 7, , 9, , 4, , 5, , 3, , 5, , 5, , 84, , 50, , 1 Concept of heat and temperature, effects of heat, difference between heat and, temperature, 2 Scales of temperature, Celsius, Fahenhiet, Kelvin and Conversion between scales, of temperature, 3 Temperature measuring instruments, types of thermometer, pyrometer and, transmission of heat - Conduction, convection and radiation, 4 Co-efficient of linear expansion and related problems with assignments, 5 Problem of Heat loss and heat gain with assignments, 6 Thermal conductivity and insulators, 7 Boiling point and melting point of different metals and Non metals, 8 Concept of pressure and its units in different system, VII, , Basic Electricity, 1 Introduction and uses of electricity, molecule, atom, how electricity is produced,, electric current AC, DC and their comparison, voltage , resistance and their units, 2 Conductor, Insulator, types of connections- Series and Parallel, Ohm’s Law,, relation between VIR & related problems, 3 Electrical power, energy and their units, calculation with assignments, 4 Magnetic induction, self and mutual inductance and EMF generation, 5 Electrical Power, HP, Energy and units of electrical energy, , VIII, , Mensuration, 1 Area and perimeter of square, rectangle and parallelogram, 2 Area an Perimeter of Triangle, 3 Area and Perimeter of Circle, Semi-circle , circular ring, sector of circle, hexagon, and ellipse, 4 Surface area and Volume of solids- cube, cuboids, cylinder, sphere and, hollow cylinder, 5 Finding lateral surface area , total surface area and capacity in liters of hexagonal,, conical and cylindrical shaped vessels, , IX, , Levers and Simple Machines, 1 Simple machines, Effort and load, mechanical advantage, velocity ratio, efficiency, of machine, relation between efficiency, velocity ratio and mechanical advantage, 2 Lever and its types, , X, , Trigonometry, 1 Measurement of Angle, Trigonometrical Ratios, Trigonometric Table, 2 Trigonometry-Application in calculating height and distance (Simple Applications), Total, , (x), , Copyright free, under CC BY Licence

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Unit - Classification of unit system, , Exercise 1.1.01, , Necessity, , Fundamental units and derived units are the two classifications of units., , All physical quantities are to be measured in terms of, standard quantities., Unit, , Length, mass and time are the fundamental units in all the, systems (i.e) F.P.S, C.G.S, M.K.S and S.I. systems., , A unit is defined as a standard or fixed quantity of one kind, used to measure other quantities of the same kind., , Example, Length: What is the length of copper wire in the roll , if the, roll of copper wire weighs 8kg, the dia of wire is 0.9cm and, the density is 8.9 gm/cm3?, , Classification, Fundamental units and derived units are the two classifications., , Solution, , Fundamental units, , mass of copper wire in the roll = 8kg (or)8000grams, Dia of copper wire in the roll = 0.9cm, Density of copper wire = 8.9 gm/cm3, , Units of basic quantities of length, mass and time., Derived units, , Area of cross section of copper wire, , Units which are derived from basic units and bear a, constant relationship with the fundamental units.E.g. area,, volume, pressure, force etc., , =, ,  d2, , , ,   (0.9 2 ), , 4, 4, Volume of copper wire, , Systems of units, – F.P.S system is the British system in which the basic, units of length, mass and time are foot, pound and, second respectively., , , ,  0.636cm 2, , Mass of copper wire, 8000grams, ,  898.88cm3, Density of copper wire 8.9 gm/cm3, , Length of copper wire, , – C.G.S system is the metric system in which the basic, units of length, mass and time are centimeter, gram and, seconds respectively., , =, , – M.K.S system is another metric system in which the, basic units of length, mass and time are metre, kilogram and second respectively., , Volume of copper wire, 898.88cm 3, , Area of cross section of copper wire, 0.636cm 2, = 1413.33 cm, , Length of copper wire =1413cm., Time: The S.I. unit of time, the second, is another base, units of S.I., it is defined as the time interval occupied by, a number of cycles of radiation from the calcium atom. The, second is the same quantity in the S.I. in the British and in, the U.S. systems of units., , – S.I. units are referred to as Systems International units, which is again of metric and the basic units, their, names and symbols are as follows., , Fundamental units of F.P.S, C.G.S, M.K.S and S.I, S.No., , Basic quantity, , British units, F.P.S, , Metric units, , Symbol, , C.G.S, , Symbol, , International units, , M.K.S, , Symbol, , S.I Units, , Symbol, , 1, , Length, , Foot, , ft, , Centimetre cm, , Metre, , m, , Metre, , m, , 2, , Mass, , Pound, , lb, , Gram, , g, , Kilogram, , kg, , Kilogram, , Kg, , 3, , Time, , Second, , s, , Second, , s, , Second, , s, , Second, , s, , 4, , Current, , Ampere, , A, , Ampere, , A, , Ampere, , A, , Ampere, , A, , 5, , Temperature, , Fahrenheit, , °F, , Centigrade, , °C, , Centigrade, , °C, , Kelvin, , K, , 6, , Light intensity, , Candela, , Cd, , Candela, , Cd, , Candela, , Cd, , Candela, , Cd, , 1, , Copyright free, under CC BY Licence

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2, , Copyright free, under CC BY Licence, , Physical quantity, , Area, , Volume, , Density, , Speed, , Velocity (linear), , Acceleration, , Retardation, , Angular velocity, , Mass, , Weight, , Force, , Power, , Pressure,Stress, , Energy, Work, , Heat, , Torque, , Temperature, , Specific heat, , S.No, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 13, , 14, , 15, , 16, , 17, , 18, , 2, , °F, , lbf.ft, , Btu, , ft.lb, , lb/in, , hp, , ft.lb/sec, , lbf, , lb, , lb, , Deg/sec, , ft/s2, , ft/s2, , ft/s, , ft/s, , lb/ft3, , ft3, , ft2, , Symbol, , BTU per pound degree Btu/lb°F, fahrenheit, , Degree Fahrenheit, , Pound force foot, , British thermal unit, , Foot.pound, , Pound per square inch, , Horse power, , Foot pound per, second, , Pounds, , Pound, , Pound (slug), , Degree per second, , Foot per square, Second, , Foot per square, second, , Foot per second, , Foot per second, , Pound per cubic, foot, , Cubic foot, , Square foot, , FPS, , British units, , g/cm3, , cm3, , cm2, , Symbol, , Kilogram per cubic, metre, , Cubic metre, , Square metre, , MKS, , Calorie per gram, degree Celsius, , Degree Centigrade, , Newton millimetre, , Kelvin, , Kilogram metre, , joule, , Cal/g°C Joule per kilogram, kelvin, , °C, , N mm, , Cal, , g.cm, , Kilogram per, square metre, Kilogram metre, , g/cm, , kilogram metre per, second, , Kilogram force, , Kilogram weight, , Kilogram, , Gram per square, centimetre, Gram centimetre, , 2, , g.cm/, sec, , dyn, , g, , g, , watt, , calorie, , m/sec2, , m/sec2, , m/sec, , m/sec, , kg/m3, , m3, , m2, , Symbol, , 2, , J/(kgK), , K, , kg.m, , J, , kg.m, , kg/m, , W, , kg.m/, sec, , kgf, , kg, , kg, , rad/sec Radian per second rad/sec, , cm/sec2 Metre per square, second, , cm/sec2 Metre per square, second, , Erg per second, , Gram.centimetre/sec, , dyne, , Gram, , Gram, , Radian per second, , Centimetre per, square second, , Centimetre per, square second, , Centimetre per second cm/sec Metre per second, , Centimetre per second cm/sec Metre per second, , Gram per cubic, centimetre, , Cubic centimetre, , Square centimetre, , CGS, , Metric units, , Derived units of F.P.S, C.G.S, M.K.S and SI system, , Unit - Fundamental and Derived units F.P.S, C.G.S, M.K.S and SI units, , -, , Joule per, kilogram kelvin, , Kelvin, , Newton metre, , joule, , Newton per square, metre, joule, , watt, , Newton, , Newton, , Kilogram, , Radian per second, , Metre square second, , Metre per square, second, , Metre per second, , Metre per second, , Kilogram per cubic, metre, , Cubic metre, , Square metre, , SI Units, , International units, , J/(kgK), , K, , Nm, , J(Nm), , J(Nm), , N/m2, , W(J/sec), , -, , N(kgm/sec2), , N, , kg, , rad/sec, , m/sec2, , m/sec2, , m/sec, , m/sec, , Kg/m3, , m3, , m2, , Symbol, , Exercise 1.1.02

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Angle, , Specific volume, , Specific resistance Ohm foot, , Specific weight, , Fuel consumption Miles per gallon, , Dynamic viscosity Pound force per, square foot, , Surface tension, , Entropy, , Electric current, , Electric voltage, , Electric resistance Ohm, , Electric, conductance, , Light intensity, , Specific gravity, , 23, , 24, , 25, , 26, , 27, , 28, , 29, , 30, , 31, , 32, , 33, , 34, , 35, , 36, , No unit, , Candela, , Mho, Siemens, , Volt, , Columb per second, , British thermal unit, per degree Fahrenheit, , Poundal per foot, , Pound per cubic foot, , Cubic foot per pound, , degree, , Pounds foot, , Moment of force, , Pound second, , Pound force foot, square second, , 22, , Moment of inertia, , 20, , Cycle per second, , Momentum, , Frequency, , 19, , FPS, , British units, , 21, , Physical quantity, , S.No, , 2, , Gram per cubic, centimetre, Centimetre per cubic, centimetre, Centi poise, dyne per centimetre, Calorie per degree, centigrade, , Biot, Volt, Ohm, Mho, , lbf/ft3, m/gal, lbf/ft2, pdl/ft, Btu/ F, C/s, V, , s, , -, , Cd, , No unit, , Candela, , Cubic centimetre per, gram, Ohm centimetre, , ft3/lbs, , 0, , degree, , deg, ,  ft, , Gram centimetre, , lbs.ft, , 2, , Kilogram per cubic kg/m3, metre, , g/cm3, , Ampere, Volt, Ohm, Siemens, , Bi, V, , s, , -, , Cd, , No unit, , Candela, , Joule per kelvin, , Cal/ c, , -, , Cd, , s, , , , V, , A, , J/K, , No unit, , Candela, , Siemens, , Ohm, , Volt, , Ampere, , Joule per kelvin, , Newton per metre, , Newton per metre N/m, , dyn/cm, 0, , pascal second, , Pa.s, , pascal second, , CP, , Newton per cubic, metre, , Cubic metre per, kilogram, Ohm meter, , Metre per cubic metre, , m, , m3/kg, , Radian, , cm/cm3 Kilometre per litre km/l, ,  cm, , Cubic metre per, kilogram, Ohm meter, , Cm3/g, , deg, , degree, , deg, , Newton metre, , kg.m, , Kilogram square, metre, , Hertz, , Kilogram metre, , kg.m, , Hz, , g.cm, , Kilogram, square metre, , Hertz, , SI Units, , kg.m/sec Kilogram metre, per second, , 2, , Symbol, , International units, , g.cm/sec Kilogram metre, per second, , g.cm, , Gram square, centimetre, Gram centimetre, per second, , Hz, , Symbol MKS, , Hertz, , CGS, , Metric units, , lb.s, , lbf.ft.s, , 1/s, , Symbol, , , , Copyright free, under CC BY Licence, , , Workshop Calculation & Science : (NSQF) Exercise 1.1.02, , 3, , -, , Cd, , s, , (V/A), , V, , A, , J/K, , N/m, , Pa.s, , m/m3, , N/m3, , m, , m3/kg, , rad, , Nm, , Kg.m/, sec, , Kg.m2, , Hz, , Symbol

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Unit - Measurement units and conversion, , Exercise 1.1.03, , Units and abbreviations, Quantity, , Units, , Abbreviation of unit, , Calorific value, , kilojoules per kilogram, , kJ/kg, , Specific fuel, , kilogram per hour per newton, , kg/hr/N, , Length, , millimetre, metre, kilometre, , mm, m, km, , Mass, , kilogram, gram, , kg, g, , Time, , seconds, minutes, hours, , s, min, h, , Speed, , centimetre per second,, metre per second, kilometre per hour, miles, per hour, , cm/s, m/s, , consumption, , km/h, mph, , Acceleration, , metre-per-square second, , m/s2, , Force, , newtons, kilonewtons, , N,kN, , Moment, , newton-metres, , Nm, , Work, , joules, , J, , Power, , horsepower, watts, kilowatts, , Hp, W, kW, , Pressure, , newton per square metre, kilonewton per square metre, , N/m2, kN/m2, , Angle, , radian, , rad, , Angular speed, , radians per second, radians-per-square second, revolutions per minute, revolutions per second, , rad/s, rad/s2, Rpm, rev/s, , Decimal multiples and parts of unit, Decimal power, , Value, , Prefixes, , Symbol, , Stands for, , 1012, , 1000000000000, , tera, , T, , billion times, , 10, , 1000000000, , giga, , G, , thousand milliotimes, , 106, , 1000000, , mega, , M, , million times, , 10, , 1000, , kilo, , K, , thousand times, , 102, , 100, , hecto, , h, , hundred times, , 101, , 10.101, , deca, , da, , ten times, , 10, , 0.1 10, , deci, , d, , tenth, , 10-2, , 0.01, , centi, , c, , hundredth, , 10, , 0.001, , milli, , m, , thousandth, , 10-6, , 0.000001, , micro, , , , millionth, , 10-9, , 0.000000001, , nano, , n, , thousand millionth, , 10, , 0.000000000001, , pico, , p, , billionth, , 9, , 3, , -1, , -3, , -12, , -1, , 4, , Copyright free, under CC BY Licence

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SI units and the British units:, Quantity, , SI unit  British unit, , British unit  SI unit, , Length, , 1 m = 3.281 ft, 1 km = 0.621 mile, , 1 ft = 0.3048 m, 1 mile = 1.609 km, , Speed, , 1 m/s = 3.281 ft/s, 1 km/h = 0.621 mph, , 1 ft/s = 0.305 m/s, 1 mph = 1.61 km/h, , Acceleration, , 1 m/s2 = 3.281 ft/s2, , 1 ft/s2 = 0.305 m/s2, , Mass, , 1 kg = 2.205 lb, , 1 lb = 0.454 kg, , Force, , 1 N = 0.225 lbf, , 1 lbf = 4.448 N, , 1 MN, , 1 million newtons, , Torque, , 1 Nm = 0.738 lbf ft, , 1 lbf ft = 1.355 Nm, , Pressure, , 1 N/m2 = 0.000145 lbf/in2, 1 Pa = 1 N/m2, 1 bar = 14.5038 lbf/in2, , 1 lbf/in2 = 6.896 kN/m2, , 1 J = 0.738 ft lbf, 1 J = 0.239 calorie, 1 kJ = 0.948 Btu, (1 therm = 100 000 Btu), 1 kJ = 0.526 CHU, , 1 ft lbf = 1.355 J, 1 calorie = 4.186 J, 1 Btu = 1.055 kJ, , Power, , 1 kW = 1.34 hp, , 1 hp = 0.7457 kW, , Fuel consumption, , 1km/L = 2.82 mile/gallon, , 1 mpg = 0.354 km/L, , Specific fuel, consumption, , 1 kg/kWh = 1.65 lb/bhp h, 1 litre/kWh=1.575 pt/bhp h, , 1 lb/bhp h = 0.606 kg/kWh, 1 pt/bhp h = 0.631 litre/kWh, , Calorific value, , 1 kJ/kg = 0.43 Btu/lb, 1 kJ/kg = 0.239 CHU/lb, , 1 Btu/lb = 2.326 kJ/kg, 1 CHU/lb = 4.188 kJ/kg, , Energy, work, , 1 lbf/in2 = 6.895 kN/m2, , 1 CHU = 1.9 kJ, , Workshop Calculation & Science : (NSQF) Exercise 1.1.03, , Copyright free, under CC BY Licence, , 5

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Units in measuring practice with definitions, Quantity, , Unit, , Force, , F, , Newton, , N, , 1 Newton is equal to the force which imports an, acceleration of 1m/s2 to a body of mass 1 kg, 1N = 1 kg m/s2, , Pressure, , P, , Newton, per square, metre, , N, m2, , 1 Newton per square metre (1 pascal), is equal to the pressure with which, the force of 1 N is exercised perpendicular, to the area of 1 m2, , Pascal, , Pa, , 1Pa = 1 N/m2. 1 Bar (bar) is the special name, for 100 000 Pa., , Newton per, square, metre, , N, m2, , Joule, , J, , 1 Joule is equal to the work that is done when, the point of application of the force of 1 N is, shifted by 1 m in the direction of the force., 1 J = 1 Nm = 1 Ws = 1 kgm2/s2, 3600 000 J = 1 kWh, , Newton, metre, joule, , Nm, J, , 1 Newton is equal to the moment of a force, which results from the product of the force, of 1 N and the lever arm of 1 m., 1 Nm = 1 J = 1 Ws = 1 kgm2/s2, , Watt, , W, , Normal stress, tensile or, compressive, stress, Shear stress, , Heat Energy, Quantity of heat, , Moment of a force, (torque), , W, , M, , Power, Energy flow, , P, , Heat flow, , ø, , 1 Newton per square metre (1 pascal), to the mechanical stress with which the, force of 1 n is exercised on the area of 1 m2., In many branches of engineering the mechanical stress and strength are specified in N/m2., 1 N/m2 = 1000 000 Pa = 1 MPa, , 1 Watt is equal to the power with which the, energy of 1 J is converted during the time of 1s., The unit watt is also called volt ampere in, the specification of apparent electric power, 1 W = 1 J/s = 1 Nm s = 1 VA, , H, , Joule per, kilogram, , J, kg, , 1 Joule per kilogram is equal to the quantity of, heat which on complete burning of the mass of, 1 kg releases the energy of 1 J, , Fuel, consumption, , P, , gram per, kilowatthour, , g, kwh, , 1 gram per kilowatt-hour is equal to the fuel, consumption of the mass of 1 g for the work, of 1 kWh., , Temperature, , T, , Kelvin, , K, , The kelvin is defined as the fraction 273.16 of, the thermodynamic temperature of the triple, point of water., , Electric current, , I, , Ampere, , A, , 1 Ampere is the strength of a current which, would bring about an electrodynamic force of, 0.2.10 N per 1 m length between two parallel, conductors placed at a distance of 1 m., , Specific, heat value, , 6, , Explanation, , 1, , Workshop Calculation & Science : (NSQF) Exercise 1.1.03, , Copyright free, under CC BY Licence

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Electric voltage, , V, , Volt, , V, , 1 Volt is equal to the electric voltage between, two points of a metallic conductor in which a, power of 1 W is expended for a current of 1 A, strength., , Electric resistance, , R, , Ohm, , , , 1 Ohm is equal to the electric resistance between two points of a metallic conductor in, which an electric current of 1 A flows at a, voltage of 1 V., , Electric conductance, , G, , Siemens, , S, , 1 Siemens is equal to the electric conductance, of a conductor of electric resistance of 1ohm, , Quantity, of electricity, , Q, , Coulomb, ampere-second, , C, As, , 1 Coulomb is equal to the quantity, of electricity which flows through the conductor, cross-section during the time of 1 s at an, electric current of 1A., , Prefixes for decimal multiples and submultiples, , Units of physical quantities, , Use, 1 Megapascal = 1 MPa = 1000000 Pa, 1 Kilowatt, , = 1 kW, , = 1000 W, , 1 Hectolitre, , = 1 hL= 100 L, , Decanewton, , = 1 daN = 10 N, , Decimetre, , = 1 dm = 0.1 m, , 1 Centimetre, , = 1 cm, , 1 Millimetre, , = 1 mm = 0.001 m, , 1 Micrometre, , = 1 um, , = 0.01 m, , = 0.000001 m, , Conversion factors, 1 inch, 1 mm, 1 metre, 1 micron, 1 kilometre, , =, =, =, =, =, , 25.4 mm, 0.03937 inch, 39.37 inch, 0.00003937", 0.621 miles, , 1 pound, 1 kg, 1 metric ton, , = 453.6 gr, = 2.205 lbs, = 0.98 ton, , Units of length, Micron, Millimetre, Centimetre, Decimetre, Metre, Kilometre, Inch, Foot, Yard, Nautical mile, Geographical mile, , 1, 1 mm, 1 cm, 1 dm, 1m, 1 km, 1", 1", 1 Yd, 1 NM, 1, , Workshop Calculation & Science : (NSQF) Exercise 1.1.03, , Copyright free, under CC BY Licence, , =, =, =, =, =, =, =, =, =, =, =, , 0.001 mm, 1000 , 10 mm, 10 cm, 10 dm, 1000 m, 25.4 mm, 0.305 m, 0.914 m, 1852 m, 1855.4 m, , 7

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TIME, T, , Time or time interval, , s, , Second, , s, , nu, , Rotational frequency, , l/min, , Reciprocal second, , l/s, , u,v,w,c, , Velocity speed, , m/min, , Metre per second, , m/s, , , , Angular velocity, , rad/s, , Radian per second, , rad/s, , g, , Acceleration of freefall, , m/s2, , Metre per second square, , m/s2, , a, , Acceleration, , m/s2, , Metre per second square, , m/s2, , Retardation, , m/s2, , Metre per second square, , m/s2, , FORCE AND PRESSURE, F, , Force, , kgf, , Newton (1kgf = 9.80665N) N, , G(P,W), , Weight, , kgf, , Newton, , N, , , , Specific weight, , kgf/m3, , Newton per cubic metre, , N/m3, , M, , Moment of force, (force x distance), , kgf.m, , Newton metre, , N.m, , p, , Pressure (force/ area), , kgf/cm2, , pascal, Newton per, square metre, , Pa,N/m2, , p, , Normal stress, , kgf/mm2, , bar (1 bar = 10 N/m), ,  p, , Shear stress, , kgf/mm2, , bar, , E, , Modulus of elasticity, , kgf/mm2, , Newton per square metre, , N/m2, , G, , Shear modulus, , kgf/mm2, , Newton per square metre, , N/m2, , , , Co-efficient of friction, , No Unit, , TEMPERATURE, Scale, , Freezing point, , Boiling point, , Centigrade (°C), , 0°C, , 100°C, , Fahrenheit (°F), , 32°F, , 212°F, , Kelvin (K), , 273K, , 373K, , Reaumur (°R), , 0°R, , 80°R, , °C, °R, = 100 = K- 273 = °F- 32, , 100, 180, , 10, , Workshop Calculation & Science : (NSQF) Exercise 1.1.03, , Copyright free, under CC BY Licence

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HEAT, WORK, ENERGY,FORCE, A,W, , Work, , kgfm, , Joule (1 Joule=1 N.m), , J (Nm), , P, , Power, , kgfm/s, , Watt, , W (J/s), , E,W, , Energy, , kgfm, , Joule, , J (Nm), , , , Efficiency, , -, , -, , -, , W,A,E,Q, , Quantity of heat, , kcal, , Joule, , J, , C, , Specific heat, , kcal/kgf°C, , Joule per newton per, , J/N.°K, , degree Kelvin, Thermal conductivity, , kcal/mh°C, , Joule per metre per, , J/ms°K, , second per degree, Kelvin, Force In C.G.S. System : Force (Dyne), , = Mass (gm)XAcceleration (cm/sec2), , In F.P.S. System : Force (Poundal) = Mass (Ib) X Acceleration (ft./sec2), In M.K.S System : Force (Newton) = Mass (Kg) x Acceleration (mtr./sec2), 1 Dyne, , = 1 gm x1 cm/sec2, , 1 Poundal, , = 1 Ib x 1 ft/sec2, , 1 Newton, , = 1 kg x 1 mtr/sec2 = 105 dynes, , 1gm weight, , = 981 Dynes, , 1 Ib weight, , = 32 Poundals, , 1 kg weight, , = 9.81 Newtons, , ELECTRICAL QUANTITIES, V, , Electric potential, , V, , Volt, , V(W/A), , E, , Electromotive force, , V, , Volt, , V(W/A), , I, , Electric current, , A, , Ampere, , A, , R, , Electric resistance, , , , Ohm, , V/A), , e, , Specific resistance, , m, , Ohm metre, , Vm/A, , G, , Conductance, ,  -1, , Siemens, , S, , Workshop Calculation & Science : (NSQF) Exercise 1.1.03, , Copyright free, under CC BY Licence, , 11

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Assignment - Conversions of length, mass, force, work, power amd energy, 1 Convert the following as indicated, , 9 Convert the following S.I. units as required., , a 5 yards into metres, , ______, , a Length, , b 15 miles into kilometres, , ______, , i, , 3.4 m, , =, , ______ mm, , c 7 metres into yards, , ______, , ii 1.2 m, , =, , ______ cm, , d 320 kilometres into miles, , ______, , iii 0.8 m, , =, , ______ mm, , iv 0.02 km =, , ______ cm, , 2 Convert, a 5 pounds into kilograms, , ______, , v 10.2 km =, , ______ mile, , b 8.5 kilograms into pounds, , ______, , vi 6 m, , =, , ______ km, , c 5 ounces into grams, , ______, , vii 18 m, , =, , ______ mm, , d 16 tons into kilograms, , ______, , viii 450 m, , =, , ______ km, , ix 85 cm, , =, , ______ km, , 3 Convert, a 40 inches into centimetres, , ______, , b 12 feet into metres, , ______, , c 5 metres into inches, , ______, , d 8 metres into feet, , ______, , x 0.06 km =, , ______ mm, , b Mass, i, , 4 Convert, , 650 g, , =, , ______ kg, , ii 300 cg, , =, , ______ g, , iii 8 g, , =, , ______ dg, , a 234 cubic metres into gallons, , ______, , b 2 cubic feet into litres, , ______, , iv 120 mg, , =, , ______ g, , c 2.5 gallons into litres, , ______, , v 8 dag, , =, , ______ mg, , d 5 litres into gallons, , ______, , vi 2.5 g, , =, , ______ mg, , vii 2.5 g, , =, , ______ kg, , viii 20 cg, , =, , ______ mg, , ix 0.05 Mt, , =, , ______ kg, , 1.2 N, , =, , ______ kg, , ii 2.6 N, , =, , ______ kg, , The same car travels a distance of 120 kilometer. what, is the consumption of fuel in litres., , iii 800 N, , =, , ______ KN, , iv 14.5 kg, , =, , ______ N, , 7 Write equivalent British units for the given metric units, , v 25 kg, , =, , ______ N, , 5 Answer the following questions, a 120°C =, , ______ °F., , b 8 mm = ______ inches, 12 mm = ______ inches, , c Force, , 6 Convert and find out, , i, , A car consumes fuel at the rate of one gallon for a travel, of 40 miles., , a Seconds, minutes, Hours, , d Work, energy, amount of heat, , b Grams, Kilograms, , i, , c Litres, Cubic meters, d Square centimeter, Square kilometer, 8 Expand the abbreviations of the following, a km/l, b N/m2, , 12, , 2 Nm, , =, , ______ Ncm, , ii 50 Ncm, , =, , ______ Nm, , iii 120 KJ, , =, , ______ J, , iv 40 J, , =, , ______ KJ, , v 300 wh, , =, , ______ kwh, , e, , Power, , c KW, , i, , 200 mW, , = ______ W, , d m/s2, , ii, , 0.2 kW, , = ______ W, , e RPM, , iii, , 300 kW, , = ______ mW, , Workshop Calculation & Science (NSQF) - 1st Semester, , Copyright free, under CC BY Licence

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Factors, HCF, LCM and problems, Prime Numbers and whole Numbers, , Exercise 1.1.04, Factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2, , Factor, , 2 128, 2 64, 2 32, 2 16, 2, 8, 2, 4, 2, , A factor is a small number which divides exactly into a, bigger number.e.g., To find the factors of 24, 72, 100 numbers, 24 = 2 x 2 x 2 x 3, 72 = 2 x 2 x 2 x 3 x 3, 100 = 2 x 2 x 5 x 5, The numbers 2,3,5 are called factors., , •, , 3,5,7,11,13,17,19,23,29, , Definition of a prime factor, Prime factor is a number which divides a prime number into, factors.e.g., 57 = 3 x 19, , Select prime numbers from 3 to 29, , •, , Find the HCF of the following group of numbers HCF of, 78, 128, 196, 78 = 2 x 3 x 13, , The numbers 3 and 19 are prime factors., , 2, 3, , They are called as such, since 3 & 19 also belong to prime, number category., , 78, 39, 13, , 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2, , Definition of H.C.F, The Highest Common Factor, , 2, 2, 2, 2, 2, 2, , The H.C.F of a given group of numbers is the highest, number which will exactly divide all the numbers of that, group.e.g., To find the H.C.F of the numbers 24, 72, 100, 24 = 2 x 2 x 2 x 3, 72 = 2 x 2 x 2 x 3 x 3, 100 = 2 x 2 x 5 x 5, , 128, 64, 32, 16, 8, 4, 2, , 196 = 2 x 2 x 49, 2 196, 2 98, 49, , The factors common to all the three numbers are, 2 x 2 = 4. So HCF = 4., Definition of L.C.M, , HCF = 2, , Lowest common multiple, , •, , LCM =, , The lowest common multiple of a group of numbers is the, smallest number that will contain each number of the given, group without a remainder.e.g., •, , Factorise the following numbers, 7,17 - These two belong to Prime numbers. Hence no, factor except unity and itself., , 2 20, 2 10, 5, Factors of 66 = 2 x 3 x 11, 2 66, 3 33, 11, , 2, 2, 3, , 84, 92, 76, 42, 46, 38, 21, 23, 19, 7, 23, 19, , LCM = 2 x 2 x 3 x 7 x 23 x 19 = 36708, , 7,17,20,66,128, , Factors of 20 = 2 x 2 x 5, , Find LCM of 84,92,76, , •, , To find out the LCM of 36, 108, 60, 2 36, 108, 60, 2 18, 54, 30, 3 9, 27, 15, 3 3, 9, 5, 1, 3, 5, LCM of the number, , 36, 108, 60 = 2 x 2 x 3 x 3 x 3 x 5 = 540, The necessity of finding LCM and HCF arises in subtraction, and addition of fractions., , 14, , Copyright free, under CC BY Licence

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Fractions - Addition, substraction, multiplication & division, , Exercise 1.1.05, , Description, , Division, , A minimal quantity that is not a whole number. For e.g. ., , When a fraction is divided by another fraction the dividend, is multiplied by the reciprocal of the divisor. (Fig 4), , 1, , A vulgur fraction consists of a numerator and denomi5, nator., Numerator/Denominator, The number above the line in a vulgar fraction showing how, many of the parts indicated by the denominator are taken, is the numerator. The total number of parts into which the, whole quantity is divided and written below the line in a, vulgar fraction is the denominator. e.g., 1 3 7, , ,, 4 4 12, , 1,3,7 - numerators, , 4,12 - denominators, , Addition and Subtraction, The denominators of the fractions should be the same, when adding or subtracting the fractions. Unequal, denominators must first be formed into a common, denominator. It is the lowest common denominator and it, is equal to the product of the most common prime numbers, of the denominators of the fractions in question.(Fig 5), , Fraction: Concept, Every number can be represented as a fraction.e.g., 1, , 1, 4, , , , 5, 4, , , A full number can be represented as an apparent, , fraction.e.g. (Fig 1), , Examples, •, , Multiply, 3, , The value of a fraction remains the same if the numerator, and denominator of the fraction are multiplied or divided by, the same number.(Fig 2), , •, , 3, , Divide, , 8, , 3, , , , 8, , •, , Add, , 3, , 3, 4, , When fractions are to be multiplied, multiply all the, numerators to get the numerator of the product and multiply, all the denominators to form the denominator of the, product. (Fig 3), , •, , sub, 17, 32, , 7, 16, , , , , , 2, 3, by, 3, 4, , 2, 3, , 16, , , , ,, , 3, , 6, , , , 12, , 3, 4, , 2, , , , 1, 2, , ,, 3, , , , x, , 8, , 4, , , , 3, , 1, 2, , ,, , 3, , , from, , 7, , 2, , by, , and, , 4, , Multiplication, , 4, , , , 4, , Fraction: Value, , 3, , 9, 12, , , , 8, 12, , , , 17, 12, , 1, , 5, 12, , 17, 32, , 17, 32, , , , 14, 32, , , , 17  14  , 32, , 3, 32, , Types of fractions, •, , Proper fractions are less than unity. Improper fractions, have their numerators greater than the denominators., , •, , A mixed number has a full number and a fraction., 15, , Copyright free, under CC BY Licence

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Addition of fraction, Add, , 1, , 1, , , , 2, , , , 8, , Solution, Step 1: Add whole numbers = 3 + 6 - 4 = 5, , 5, 12, , 3 7 5, 9,  , , 4 8 16 32, , Step 2: Add fractions =, , To add these fractions we have to find out L.C.M of, denominators 2,8,12., , L.C.M of 4,8,16,32 is 32, , Find L.C.M of 2,8,12, 24  28  10  9, , Step 1 L.C.M, , 32, , 2 2,8,12, 2 1,4,6, 1,2,3, , , , , Factors are 2,2,2,3, Hence L.C.M = 2 x 2 x 2 x 3 = 24, , , , 2, , , 1, 8, , , , 5, 12, , , , 12  3  10, 24, , 12, 24, , , , , 25, 24, , 3, 24, , , , 1, , 24, , 24, , 32, , 1, , 3, , 6, , 32, , 3, 32, , Common fractions, , ., , •, , 15, 9, 9, 15, 9, from17, or(17, ), 32, 16, 16, 32, , Multiply, , •, , Step 2: L.C.M of 16,32 = 32, Since number 16 divides the number 32, , 3, 32, , 3, , a, , Step 1: Subtract whole number first 17 - 9 = 8, , Subtracting fractions =, , 32, , Examples, , Subtraction of fraction, subtract 9, , 1, , we get 5  1, , 10, , 1, , 32, 33, , Step 3: Adding again with the whole number, , Step 2, 1, , 52  19, , •, , 4, , by, , 8, , 3, , , , 7, , 8, , x, , 4, , , , 7, , 5, 5, 5 32, , , x, 2, 16 32 16 5, , b, , 4, , Addition, , L..C.M  2,4,8  8, , 2, 1, , solve 3, , 3, 4, , 6, , 7, 8, , 4, , 5, 16, , , , 9, , 2, , •, , 32, , , , 1, 4, 1, 4, , 1, , , , 2 add all + Numbers, 3 Add all - Numbers, 4 Find L.C.M of all denominators, , 1, , , , 8, 1, , , , 4, , 3, ,  2, , 1 Add all whole numbers, , 16, , , , 8, , , , 4  2 1, 8, , , , 7, 8, , Subraction, , 5, , Rule to be followed, , 2, 3, , x, , 3, 4, , x, , 5, 8, , , , 5, 16, , 2, 1 14 22 14 7, 49, 16, 3 , , , x, , 1, 3, 7, 3, 7, 3 22 33, 33, , 3, 3, we get 8 , 8, 32, 32, , Example, , 14, , a, , 1, , Problems with plus and minus sign, , b, , Division, , Adding with whole number from Step 1, , Common fractions, , 3, , , , 9, 4, 6, 4, , , , , 1, 4, , 3, 4, , , 3, 4, 3, 2, , , ,  53, 3, 4, , 2, , 1, 4, , 1, 4, , , , , , 3, 4, , 3, 4, , 93, , 1, , 4, 1, 2, , Workshop Calculation & Science : (NSQF) Exercise 1.1.05, , Copyright free, under CC BY Licence

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Decimal fractions - Addition, subtraction, multilipication & division, Exercise 1.1.06, Divide 0.75 by 0.25, , Description, Decimal fraction is a fraction whose denominator is 10 or, powers of 10 or multiples of 10 (i.e.) 10, 100, 1000, 10000, etc. Meaning of a decimal number:12.3256 means, (1 X 10)  (2 X 1) , , 3, 2, 5, 6, , , , 10 100 1000, 10000, , Representation, The denominator is omitted. A decimal point is placed, at different positions of the number corresponding to the, magnitude of the denominator, Ex., , 5, 35, 127, 3648,  0.5,,  0.35,  0.0127,,  3.648, 10, 100, 10000, 1000, , Addition and subtraction, Arrange the decimal fractions in a vertical order, placing the, decimal point of each fraction to be added or subtracted, in, succession one below the other, so that all the decimal, points are arranged in a straight line. Add or subtract as, you would do for a whole number and place the decimal, point in the answer below the column of decimal points., Decimal fractions less than 1 are written with a zero before, the decimal point. Example: 45/100 = 0.45 (and not simply, .45), , 0.25 0.75, 0.75 100 75, x, , 0.25 100 25, 25 75  3, Move the decimal point in the multiplicand to, the right to one place if the multiplier is 10, and, to two places if the multiplier is 100 and so on., When dividing by 10 move the decimal point, one place to the left, and, if it is by 100, move, them point by two places and so on., Example, Allowing 3 mm for cutting off each pin how many pins, can, be made from a 900 mm long bar? How much material will, be left out?, Length of pin, = 2.25 + 55.36 + 12.18, = 69.79 mm, , Add 0.375 + 3.686, 0.375, 3.686, 4.061, Subtract 18.72 from 22.61, 22.61, 18.72, , Length of the bar = 900 mm, Step 1, , 3.89, , Let the number of pins to be made = x, , Multiplication, , Length of x number of pins = x x 69.79 mm, , Ignore the decimal points and multiply as whole numbers., Find the total number of digits to the right of the decimal, point. Insert the decimal point in the answer such that the, number of digits to the right of the decimal point equals to, the sum of the digits found to the right of the decimal points, in the problem., Multiply 2.5 by 1.25, = 25 x 125 = 3125. The sum of the figures to the right, of decimal point is 3. Hence the answer is 3.125., , Step 2, , Division, Move the decimal point of the divisor to the right to make, it a full number. Move the decimal point in the dividend to, the same number of places, adding zeroes if necessary., Then divide., , x = 900  72.79, , Waste for each pin = 3 mm, Waste for x number of pins = 3 x x mm=3x mm, Adding step (1) + step (2) and equating to length of bar, 69.79x mm + 3x mm = 900 mm, , x (69.79mm + 3mm) = 900mm, x (72.79mm) = 900mm, Hence Number of pins to be made = 12, , 17, , Copyright free, under CC BY Licence

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Secondly, •, , Left out materials, = Total length of bar - Length for 12 pins+ wastage of, cutting, , Convert, 0.875, 8 7000, , = 900mm - (12 x 69.79 + 12 x 3)mm, , 64, , = 900 - (837.48 + 36)mm, , 60, 56, , = 900 - 873.48mm, =26.52mm, , 40, , Left out material 26.52 mm, , 40, 0, , Conversion of Decimals into fractions and vice-versa, •, , to a decimal, , Convert decimal into fractions, , 7, , Example, , 8, , Convert 0.375 to a fraction, Now place 1 under the decimal point followed by as many, zeros as there are numbers, ,  0.875, , Recurring decimals, While converting from fraction to decimals, some fractions, can be divided exactly into a decimal. In some fractions the, quotient will not stop. It will continue and keep recurring., These are called recurring decimals., Examples, •, , •, , convert, , 1 2 1, , ,, 3 3 7, , Convert fraction into decimal, , Example, • Convert, , a, to a decimal, , Proceed to divide, , in the normal way of division but put, , b, , zeros (as required) after the number 9 (Numerator), c, , 1, 3, 2, 3, , , , , , 10000, 3, 20000, 3, ,  0.3333  Recurring, ,  0.666  Recurring, ,  1 10000, ,  0.14285714 2  Recurring ,  , 7, 7, , , , These are written as below with a dot over the figure, ., 0.3333 ——> 0.3, ., 0.6666 ——> 0.6, ., ., 0.14857142 —> 0.14857, Note the dots marked over numbers., , = 0.5625, , We normally carry the decimal points upto 4 places in, Engineering calculations., Approximations in Measured Value calculations, In Measured Value calculations 4 places of decimals are, sufficient and in many dimensions of parts even 3 decimal, places are near enough to complete the maintenance job, operations., , 18, , Workshop Calculation & Science : (NSQF) Exercise 1.1.06, , Copyright free, under CC BY Licence

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Method of writing approximations in decimals,  (4  100)  (5  10)  (3  1) , , 1.73556 = 1.7356 Correct to 4 decimal places, 5.7343 = 5.734, Correct to 3 decimal places, 0.9345 = 0.94, Correct to 2 decimal places, Multiplication and division by 10,100,1000, Multiplying decimals by 10, ,  453, , •, , A decimal fraction can be multiplied by 10,100,1000 and so, on by moving the decimal point to the right by as many, places as there are zeros in the multiplier., •, •, •, , 4.645 x 10, 4.645 x 100, 4.645 x 1000, , = 46.45 (one place), = 464.5 (two places), = 4645 (three places), , A decimal fraction can be divided by 10,100,1000 and so, on, by moving the decimal point to the left by as many, places as required in the divisor by putting zeros, , 7, 100, , , , 3, 100, , 273, 1000, , Write the representation of decimal places in the given, number 0.386, 3 - Ist decimal place, 8 - IInd decimal place, 6 - IIIrd decimal place, , •, , Write approximations in the following decimals to 3, places., , b 8.7456 ——> 8.746, • Convert fraction to decimal, , Examples, , 21, , 3.732 ÷ 10, 3.732 ÷ 100, 3.732 ÷ 1000, , 10, , , , a 6.9453 ——> 6.945, , Dividing decimals by 10, , •, •, •, , 2, , = 0.3732 (one place), = 0.03732 (two places), = 0.003732 (three places), , Examples, , 24, , 7, , , , 8, ,  0.875, , • Convert decimal to fraction, 0.0625 , , • Rewrite the following number as a fraction, , 625, 5, 1, , , 10000 80 16, , 453.273, 453.273, , ASSIGNMENT, 1 Write down the following decimal numbers in the, expanded form., , 4 Convert the following fraction into decimals, a, , a 514.726, b 902.524, , b, , 2 Write the following decimal numbers from the expansion., a, , 500  70  5 , , b, , 200  9 , , 1, 10, , , , 3, 10, , , , 3, 100, , 2, 100, , , , , , 9, , 5, 10, 4, , 24, , 1000, , 5, 1000, , 3 Convert the following decimals into fractions in the, simplest form., a 0.72, , c, , 3, , d, , 54, 1000, , 12, 25, , e, , 8, 25, , f, , 1, , 3, 25, , b 5.45, c 3.64, d 2.05, , Workshop Calculation & Science : (NSQF) Exercise 1.1.06, , Copyright free, under CC BY Licence, , 19

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Solving problems by using calculator, , Exercise 1.1.07, , A pocket calculator allows to spend less time in doing, tideous calculations. A simple pocket calculator enables, to do the arithmatical calculations of addition, substraction,, multiplication and division, while a scientific type of calculator, can be used for scientific and technical calculations also., No special training is required to use a calculator. But it is, suggested that a careful study of the operation manual of, the type of the calculator is essential to become familiar, with its capabalities. A calculator does not think and do., It is left to the operator to understand the problem, interpret, the information and key it into the calculator correctly., Constructional Details (Fig 1), , •, , +, , Addition key, , -, , Subtraction key, , x, , Multiplication key, , ÷, , Division key, , =, , Equals key to display the result, , Function keys, , , , Pi key, , x, , Square root key, , %, , Percentage key, , +/- Sign change key, x2, The key board is divided into five clear and easily recogniz, able areas and the display, •, , 1, X, , Data entry keys, , The entry keys are from, , 0, , and a key for the decimal point, •, , ..............to, , ., , 9, , •, , Square key, , Reciprocal key, , Memory keys, , M Store the display number, , Clearing keys, , These keys have the letter ‘C’, C, CE, , CLR, , Clear totally, , Clear entry only, , CM , MC, , M+, , The displayed value is added to the memory, , M-, , The displayed value is subtracted from the, memory, , MR RCL The stored value is recalled on to the, display, , Clear memory, , 20, , Copyright free, under CC BY Licence

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Further functional keys included in Scientific calculators, are as shown in Fig 2., , Rules and Examples:, •, , Addition: Example 18.2 + 5.7, Sequence, , Input of the 1st term, of the sum, , Input, 1, , Press + key, , +, , Input 2nd term of the, sum. the first term, goes into the register, , 5, , Press the = key, , =, , •, , (, , Cos Tan, , ) For trigonometric functions and, for brackets, , Exp Exponent key, Some of the keys have coloured lettering above or, INV below them. To use a function in coloured lettering,, press INV key. INV will appear on the display., Then press the key that the coloured lettering, identifies. INV will disappear from the display., log , INV, , 10, , x, , to obtain the logarithm of the dis-, , played number and the antilogarithm of the displayed value., INV, , R–P to convert displayed rectangular coordinates, , into polar coordinates., INV, , P–R to convert displayed polar coordinates into, , The display shows the input data, interim results and, answers to the calculations., The arrangement of the areas can differ from, one make to another. Keying in of the numbers, is done via. an internationally agreed upon set, of ten keys in the order that the numbers are, written., , ., , 7, , 5.7, 23.9, , Display, 8, , -, , Enter the minnend., The subtrahend goes, into the register, Press the = key, , 128.8, 128.8, , 9 2 ., , 9, , =, , 92.9, 35.9, , Multiplication: Example 0.47 x 2.47, Sequence, , Input, , Enter multiplicand, , . 4, , Press x key, Enter multiplier,, multiplicand goes, to register, , Display, 7, , x, , 0.47, 0.47, , 2 . 4, , 7, , =, , 2.47, 1.1609, , Division: Example 18.5/2.5, Sequence, , The display, , 18.2, 18.2, , 1 2 8 ., , Press - key, , •, , 2, , Input, , Press = key, , rectangular coordinates., •, , Enter the subtrahend, , •, , ., , Subtraction: Example 128.8 - 92.9, Sequence, , Sin, , 8, , Display, , Enter the dividend, Press ÷ Key, Enter the divisor, goes to the register, Press = key, , Workshop Calculation & Science : (NSQF) Exercise 1.1.07, , Copyright free, under CC BY Licence, , Input, 1, , 8 ., , Display, 5, , ÷, 2 ., =, , 18.5, 18.5, , 5, , 2.5, 7.4, , 21

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•, , Multiplication & Division:, Example : 2.5 x 7.2 / 4.8 x 1.25, , Sequence, Enter 2.5, Press x key, Enter 7.2, Press ÷ key, Enter 4.8, Press x key, Remember:Before, input of thefirst value, under the fraction, line, the x keymust, be operated, , •, , Input, 2 ., , Display, , 5, , 2.5, , x, 7 ., , 2, , 7.2, , Store the first result in, , Enter 1500, , 1 5 0, , Press x key, , x, , 12, , 4 . 8, , 4.8, , Press INV %, , INV %, , 12, , x, , 3.75, , 1 ., , 2, , 5, , 1.25, , Press = key, , •, , 3.0, , =, , =, , Square root: Example, Sequence, , Input, , Display, , Display, , 2, , 2, , a, , key, , Press a, , Input, , 180, , 2  35, , Enter 2, , 2, , Press + key, , +, , ., , Press bracket key, , (, , ., , 3, , 3, , 2, , 2, , Enter 3, , +, , 2, , Press, , 6, , 6, , =, STO , M, , a, , a key, , ., , Press x key, , x, , ., , 8, , Enter 5, , 5, , 5, , 8, , Press, , a, , ., , Press bracket close key, , ), , ., , =, , 5.2871969, , ) =, , 5.2871969, , a key, , 4, , 4, , Press = key, , +, , 4, , 2, , 3, , 3, , =, , 7, , x, , 7, , RCL or MR, , 8, , =, , 56, , + ( 3, , x 5, , 2  35  5, , •, , Common logaritham: Example log 1.23, Sequence, , 1, , •, , ., , 2, , Input, 3, , log, , =, , Display, 0.0899051, , Power: Example 123 + 302, Sequence, , 1 2 3 + 3 0 INV X2, , 22, , 1500, 2, , Workout for the, 2nd bracket, , Press = key, , 1500, , 1, , or M+, , Recall memory, , 0, , Enter 12, , x, , Press x key, , Display, , 18, , Store in memory Example (2+6) (4+3), , Workout for the first, bracket, , Input, , ÷, , Press = key, , Sequence, , Sequence, , 2.5, , Enter 1.25, , •, , Percentage: Example 12% of 1500, , Workshop Calculation & Science : (NSQF) Exercise 1.1.07, , Copyright free, under CC BY Licence, , Input, , Display, , =, , 1023

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•, , Before starting the calculations be sure to, press the ‘ON’ key and confirm that ‘0’ is, shown on the display., , •, , Do not touch the inside portion of the calculator. Avoid hard knocks and unduly hard, pressing of the keys., , •, , •, , Never use volatile fluids such as lacquer,, thinner, benzine while cleaning the unit., , •, , Take special care not to damage the unit by, bending or dropping., , •, , Do not carry the calculator in your hip pocket., , b, , (389  12.2)x(842  0.05  2.6),  _________, (3.89  0.021)x(28.1  17.04), , Maintain and use the calculator in between, the two extreme temperatures of 0° and 40°, C., , Assignment, 771 Using calculator solve the following, a 625 + 3467 + 20 + 341 + 6278 = ______, b 367.4 + 805 + 0.7 + 7.86 + 13.49 = ______, , 7, , 2a = 450 mm(major axis), , c 0.043 + 1.065 + 13.0 + 34.76 + 42.1 = ______, , 2b = 250mm(minor axis), , d 47160 + 1368.4 + 0.1 + 1.6901 + 134.267 =, _______, , Perimeter of the ellipse, c = _____metre, , 2 Using calculator simplify the following, , Hint C =, , a 24367 - 4385 = ______, ,  2(a 2  b 2 ), , b 9.643 - 0.7983 = ______, c 4382.01 - 381.3401 = ______, d 693.42 - 0.0254 = ______, , 8, , ø = 782 mm,  = 136°, , 3 Using calculator find the values of the following, a, , Area of the sector, , 23 x 87 = ______, , A = ______, , b 1376 x 0.81 = ______, c 678 x 243 = ______, Hint A , , d 0.75 x 0.24 = ______, 4 Using calculator solve the following, , 9, , a 22434 ÷ 3 = ______, , π x d2, α, x, 4, 360, , d = 1.25 metre, V = ______ dm3, , b 4131 ÷ 243 = ______, , Hint V , , c 469890 ÷ 230 = ______, d 3.026 ÷ 0.89 = ______, 5 Solve the following, a, , b, , 1170x537.5, 13x215, , 10, = ________, , L = 1.2 metres, B = 0.6 metre, H = 0.5 metre, , 28.2 x18 x 3500 ________, , 1000 x 3 x 0.8, , 'r' of steel, = 7.85 kg/dm3', m = ______ kg, , 6 Solve the following, a, , 4 3, πr, 3, , (634  128)x(384  0.52),  ________, 8x0.3, , Workshop Calculation & Science : (NSQF) Exercise 1.1.07, , Copyright free, under CC BY Licence, , (mass ‘m = V x r ), , 23

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1 Convert the following into improper fractions., a 1, , b, , 4, , c, , 3, , d, , e, , f, , g, , 5, , 3, , 5, , 7, , 2, ,  __________, , a, , 3,  __________, 5, , b, , 7, , 3, ,  __________, , 5, , 7, , c, , 8, , a, , 1, ,  _________, , 3, , b, , 3, ,  ______, , 4, , c, , 3, ,  __________, , 7, , 1, 74, , d, ,  __________, , 2 Convert the following into mixed numbers., , b, , c, , d, , e, , f, , g, , h, , 24, , 12, 11, 36, 14, 18, 10, 25, 3, , 84, 13, 32, 21, 18, 16, 75, 4, ,  __________, ,  __________, ,  __________, , 11 x, , __________, 13 91, , 3, , 42, , , , 5, , 9, , x, , x, , , , 14, , __________, , 98, , __________, , 4 Simplify., ,  __________, , h 182, , a, , 3 Place the missing numbers., , e, , f, , 45, ,  _________, , 60, 8, ,  __________, , 12, 12, ,  __________, , 14, 56, ,  __________, , 72, , 6, ,  __________, , 14, 3, 4, , x, , 5, 7, , x, , 11 2 14, x x,  __________, 3 4 6, , 5 Multiply., 2,  __________, 3, , a, , 5x, , b, , 3, x 2  __________, 4, , c, , 3 5, x  __________, 4 6, , d, , 3, , 1, x 3  __________, 4, , e, , 2, , 1, 1, x 3  __________, 4, 4, , f, , 5x6, ,  __________, ,  __________, ,  __________, ,  __________, , 1,  __________, 4, ,  __________, , Workshop Calculation & Science : (NSQF) Exercise 1.1.07, , Copyright free, under CC BY Licence

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6 Divide, a, , b, , c, , 1, 4, , , , 6 , , 3, 4, , , 1, , 3, , 3, , 2, , f, , 8 3, , 1, 2, , 2, 1, 4, , 1, 7, , , , ________, , _________, , 7 Place the missing numbers., a, , 2, 3, , , , 6, , 1, 12, , a, , 4 2,   _________, 5 5, , b, , 5 3,   _________, 6 4, , x_________, , 10 Simplify, a, , 2, , b, , 2, , c, , 3, , 5, , c, , 7, 1, , x_________, 8 12, , b, , 3, , 2, 1, , x_________, 36 12, , c, , 1, , g, , h, , 52, 1, , x_________, 36 12, , 11, 1, 3, , x_________, 24 12, , 3, 1, , x_________, 4 12, 7, 1, , x_________, 6 12, , a, , b, , 4, 7, 8, , , , , , 7, 12, 3, 4, , , , 8, 5, 6, 3, 5, , , , 1, 3, , 1, , 1, 16, ,  _________, ,  8  ________, , 1, , 3, 4, , 2, , 1, 2, ,  _________, , 3, 4, 5, 64, 5, , 12, , 12 Reduce to mixed number or whole number, a, , b, , c, , a, ,  _________, ,  _________, , 7, , , , 3, , 163, 4, 12, 4, 144, 60, , 13 Reduce to lowest terms, , 8 Add the followings:, 3, , 2, , , , 11 Express as improper fractions, a, , f, , 7, , 9, , 14, 1, , x________, 24 12, , e, , 6, , 7, , b, , d, , 1, 7, 7,  1  3  _________, 4, 12, 9, , 9 Subtract, ,  4  __________, , 5, , d, ,  __________, , 7, , e, , 6, ,  ________, , 4, , 3, , 3 4 3,    _________, 5 5 8, ,  _________, , 4, , d, , c, , b, , c, , 12, 64, 12, 48, 144, 60, , Workshop Calculation & Science : (NSQF) Exercise 1.1.07, , Copyright free, under CC BY Licence, , 25

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14 Addition of decimals, , 0.168, , c, , a 4.56 + 32.075 + 256.6245 + 15.0358, b 462.492 + 725.526 + 309.345 + 626.602, , d, , 15 Subtract the following decimals, a 612.5200 – 9.6479, , 1.2, 1.54, 1.1, 27.2, , e, , b 573.9246 – 215.6000, , 1.6, , c 968.325 – 16.482, , f, , 31.5 ÷ 10.5, , d 5735.4273 – 364.2342, , g 1.54 ÷ 1.1, h 4.41 ÷ 2.1, , 16 Add and subtract the following, , 20 Change the fraction into a decimal, , a 56.725 + 48.258 – 32.564, b 16.45 + 124.56 + 62.7 - 3.243, , i, , 17 Multiplication of decimals, a By 10,100,1000, i, , ii, , 3.754, , ii 8.964 x 100, , 1, , 5, 8, , 12, 25, , 21 Find the value, , iii 2.3786 x 1000, , 20.5 x 40 ÷ 10.25 + 18.50, , iv 0.005 x 1000, , 22, , b By whole numbers, i, , 8.4 x 7, , ii 56.72 x 8, , A = 12.613 mm, X = __________mm., , c By another decimal figure (use calculator), i, , 15.64 x 7.68, , ii 2.642 x 1.562, 18 Divide the following, a, b, c, d, e, f, , 23, , 62.5, 25, , X = __________mm., , 14.4, 9, 64.56, 10, , 24, , 0.42, 100, , X = __________mm., , 48.356, 1000, , 25.5, , 25, , 15, , 19 Division, , 26, , a, , 16.8, 1.2, , b, , 1.68, 1.2, , B = __________mm., , Workshop Calculation & Science : (NSQF) Exercise 1.1.07, , Copyright free, under CC BY Licence

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Square root - Square and suare root, a basic number, 2 exponent, , Exercise 1.2.08, Extracting the square root procedure, , radial sign indicating the square root., , •, , Starting from the decimal point form groups of two, figures towards right and left. Indicate by a prime, symbol. 46,24.00, , a 2 square root of 'a' squared, a2 radicand, Square number, The square of a number is the number multiplied by itself., Basic number x basic number = Square number, , •, , Find the root of the first group, calculate the difference,, bring down the next group., , •, , Multiply the root by 2 and divide the partial radicand., , •, , Enter the number thus calculated in the divisor for the, multiplication., , a x a = a2, 4 x 4 = 42 = 16, , Splitting up, A square area can be split up into sub-areas. The largest, square of 36 is made up of a large square 16, a small, square 4 and two rectangles 8 each., Large square 4 x 4 = 16, , a2, , Two rectangles 2 x 4 x 2 = 16, , 2ab, , Small square 2 x 2 = 4, , b, , 2, , If there is a remainder, repeat the procedure., 68, 68, 6 46,24, 6 46,24, 36, 36, 128 1024, 128 1024, 1024, 1024, 0, 0, 46,24  68, , Basic number x basic number = Square, Square number  basic number, , Sum of sub-areas = 36 = a + 2ab + b2, 2, , Example, , 36  a 2  2ab  b2, , The cross-section of a rivet is 3.46 cm2. Calculate the, diameter of the hole., , Result: In order to find the square root, we split up the, square numbers., , Rivet cross-section is the hole cross-section., To find d1,, 3.46 cm2, d2 , Given that Area = 3.46 cm2, 0.785, Area = 0.785 x d2 (formula), 3.46, d, cm, 3.46 cm2 = d2 x 0.785, 0.785, d  2.1 cm (or) 21mm, , 27, , Copyright free, under CC BY Licence

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Square root - Simple problems using calculator, 1 a, , 2, , 3, , 2916 , , ., , b, , 45796 , , ., , c, , 8.2944 , , ., , d, , 63.845 , , ., , 6, , Exercise 1.2.09, A = 807.77 cm2, d = 140 mm, D=, , A = 2025 mm2, a=, , mm, , A = 176.715 mm2, d=, , 7, , a x a = 543169 mm2, a=, , 8, mm, , A = 65031 mm2, , A = 73.5 mm2, , 9, , d = 140 mm, D=, , mm, , A = 12.7%, A = 360 mm2, mm, , (d = diameter after the, increase in area), , I = 58 cm, b = 45 cm, A1 = A2, a=, , mm, , increase in area, , d=, , 5, , mm, , d : l = 1:1.5, d=, , 4, , mm, , cm, , 28, , Copyright free, under CC BY Licence

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Square root - Applications of pythagoras theorem and related problems, Exercise 1.2.10, 1 What is the side AC if AB = 15 cm, BC = 25 cm., AC = AB + BC, 2, , 2, , 2, , 5 What is the value side AC if AB = 6.45 cm, BC = 8.52, cm., , = 152 + 252, , AC2 = AB2 + BC2, , = 225 + 625 = 850, , AC2 = 6.452 + 8.522, AC2 = 41.60 + 72.59, , AC = 850 = 29.155 cm, , = 114.19, , 2 What is the side BC if AB = 10 cm, AC = 30 cm., AC2 = AB2 + BC2, , AC = 114.19 = 10.69 cm, 6 What is the value of side AB if BC = 3.26 cm, AC = 8.24, cm., , 302 = 102 + BC2, 900 = 100 + BC2, , AC2 = AB2 + BC2, , BC2 = 900 - 100 = 800, , 8.242 = AB2 + 3.262, , BC = 28.284 cm, , 67.9 = AB + 10.63, , 3 What is the side AB if BC = 20 cm, AC = 35 cm., , AB2 = 67.9 - 10.63, , AC2 = AB2 + BC2, 35, , 2, , = 57.27, , = AB + 20, 2, , 2, , AB =, , 1225 = AB + 400, , 57.27 = 7.57 cm, , 7 What is the value of side AB if AC = 12.5 cm, BC = 8.5, cm., , AB2 = 1225 - 400 = 825, AB = 28.72 cm, 4 What is the value of side BC if AB = 8 cm, AC = 24 cm., , AC2, , = AB2 + BC2, , 12.52, , = AB2 + 8.52, , AC2 = AB2 + BC2, , 156.25 = AB + 72.25, , 24 = 8 + BC, , AB2, , 2, , 2, , 2, , 576 = 64 + BC, , 2, , = 84, , BC = 576 - 64 = 512, 2, , BC =, , = 156.25 - 72.25, , AB, , =, , 84 = 9.17 cm, , 572 = 22.63 cm, , Assignment, 1 What is the value of side AB, in a right angled triangle, of side AC = 12.5 cm and BC = 7.5 cm., , 4 What is the value of side AC, in a right angled triangle, of side AB = 7 cm and BC = 5 cm., , 2 What is the value of side AC, in a right angled triangle, of side AB = 6.5 cm and BC = 4.5 cm., , 5 What is the value of side BC, in a right angled triangle, of side AC = 13.25 cm and AB = 8.75 cm., , 3 What is the value of side BC, in a right angled triangle, of side AC = 14.5 cm and AB = 10.5 cm., , 29, , Copyright free, under CC BY Licence

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Ratio and proportion, , Exercise 1.2.11, , Ratio, , •, , Introduction, It is the relation between two quantities of the same kind, and is expressed as a fraction., , •, , Expression, , •, , a, b two quantities of the same kind. a or a:b or a  b or, b, a in b is the ratio., , •, Ratio - relation of two quantities of the same kind., Proportion - equality between two ratios., , Ratio is always reduced to the lowest terms., Example, , Example, •, , A steel plate of 800 x 1400 mm is to be drawn to a scale, of 1:20. What will be the lengths in the Fig 1, , Proportion, It is the equality between the ratios, a : b is a ratio and c:, d is another ratio. Both ratios are equal. Then, a :b :: c : d or, Example, 250 : 2000 :: 1 : 8, Proportion fundamentals, If, •, , then, ad = bc, , The reduction ratio is, , ., , B is reduced from 800 to 800 x, , •, , L is reduced from 1400 x, , = 40 mm., , = 70 mm., , •, •, , Find the number of teeth of the larger gear in the gear, transmission shown in the Fig 2., , •, , •, •, , 3:4::6:8 or, •, , 3x8=6x4, , •, , 30, , Copyright free, under CC BY Licence

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Speed ratio = 400 : 300, , Find the ratio of A:B:C, , Teeth ratio = 24:T, , If A:B= 2:3 and B:C=4:5, A:B = 2:3, B:C = 4:5, A:B = 8 :12 (Ratio 2:3 multiply by 4), B:C = 12:15 (Ratio 4:5 multiply by 3),  A:B:C = 8:12:15, , Assignment, 1, , l1 : l2 = 2:3, , 5, , D : d = 1.75: 1, , L = 2.75 metres, , D = 35 mm, , l1 = ______ metres, , d = ______ mm, , l2 = ______ metres, , 2, , 3, , d: L of shaft = 2:7, , 6, , a:s = 5:1, , d = 40 mm, , s = 1.5mm, , L = ______ mm, , a =______ mm, , D : L=1:10, L=150mm, D=______ mm, , 7 A:B=9:12, B:C=8:10, Then A:B:C=___________, 8 A:B=5:6, B:C=3:4, Then A:B:C= ______, 9 A:55=9:11, , 4, , A = __________, I = 140 mm, h = _____ mm, , 10 15:9.3=40:x, x = __________, , Workshop Calculation & Science : (NSQF) Exercise 1.2.11, , Copyright free, under CC BY Licence, , 31

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Ratio and proportion - Direct and indirect proportions, , Exercise 1.2.12, , Proportion, Description, It is the equality between the ratios, a:b is a ratio and c:d, is another ratio. Both ratios are equal. Then, a : b::c : d or, , e.g. 250 : 2000::1 : 8, , Rule of three, A three step calculation, statement, single, multiple., Example, , Direct proportion, The more in one the more in the other - An increase in one, denomination produces an increase in the other. (Fig 1), , Four turners finish a job in 300 hours. How much time will, 6 turners take to do the same job?, Solution procedure in three steps:, Statement, 4 turners taken = 300 hours, The time will reduce if 6 turners to do the same job., Therefore this is inverse propotion., Multiple fraction, 6 Turners, , = 200 hours, , Result - The more the less., Problems involving both, , Example, 4 turners earn 300 Rupees. How much will 6 Turners earn?, , Example, Two turners need three days to produce 20 pieces. How, long does it take for six turners to produce 30 such pieces?, , Statement, 4 turners = 300 Rupees, , Statement, , Single, , 2 turners, 20 piece = 3 days, 6 turners, 30 pieces = how many days., , 1 Turner = 75 Rupees, Multiple, 6 Turners = 6 x 75 = 450 Rupees, , First step (Fig 3), , Result - The more the more., , Statement, , 2 turners for 20 pieces = 3 days, 1 turner for 20 pieces = 3 x 2 = 6 days, , Indirect or inverse proportion, The more in one the lesser other - Increase in one quantity, will produce a decrease in the other. (Fig 2), , Multiple 6 turners for 20 pieces =, , 32, , Copyright free, under CC BY Licence, , = 1 day

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, , Inverse proportion - The more the less., , Example, If 5 Fitter take 21 days to complete overhauling of 6 vehicles, how long 7 Fitters will take to over haul 8 vehicles (Assume, time of overhauling each vehicle is constant), , Second step (Fig 4), , In this both direct and indirect proportions are used., , Statement, , 6 turners for 20 pieces = 1 day, , Single, , 6 turners for 1 piece, , Multiple, , 6 turners for 30 pieces =, , =, , x 30 = 1.5 days, , Solve the probÎem by first writing the statement, and proceed to single and then to the multiple, according to the type of proportion that is, involved., , •, , Quantities (No. of days) are taken in last as that is, the answer required in this case., Ans: 7 Fitters will overhaul 8 vehicles in 20 days., , Inverse proportion, Some times proportions are taken inversely., Examples, •, , Proportional fundamentals, as applicable to motor vehicle, calculations are discussed below., Simple Proportion, , If one water pump fills the fuel tank in 12 minutes, two, pumps will take half the time taken., The time should not be doubled., , •, , Introduction, , If two pumps take 30 minutes to fill up a tank how long, will 6 similar pumps take this to fill this tank., Ans: Time taken by 6 pumps =, , Proportional parts in combustion equation, , Proportion, , This is an equality between two ratios, Examples, •, , 1 Fitter will over haul 1 vehicle in days (shorter time)., , days, , Direct proportion - The more the more., , •, , •, , If one vehicle fleet uses 30 litres of petrol per day how, much petrol is used by 6. Vehicles operating under, similar condition., , Introduction, Proportion of quantities form an important factor in the, combustion process of a fuel. The following happens during, the combustion process., , One vehicle uses petrol = 30 litres per day., , Fuel is a hydro carbon substance. The combustion air is, supplied from atmosphere and contains oxygen and nitrogen. Now the following chemical changes take place during, combustion of a fuel., , Then six vehicles will use = 6 Times as much, , •, , = 6 x 30 = 180 litres/day., •, , If 4 vehicles of a fleet use 120 gallons of petrol per day, how much petrol will be used by 12 vehicles operating, under the same condition., , 4 vehicles use 120 gallons per day, 1 Vehicle will use, , = 30 gallons/day, , 12 vehicles will use 12 x 30 = 360 gallons/day, Both examples are called simple proportion, because only two quantities were used and the, day is common for both ratios., Compound and Inverse proportions, •, , Carbon burns with oxygen and forms Co and Co2, (Carbon monoxide and carbon dioxide.), , •, , Hydrogen burns with oxygen and becomes water (H2O), , •, , Sulphur burns with oxygen and becomes sulphur, dioxide., , •, , Nitrogen is an inert gas and does not take part in, combustion., , Method of finding proportional parts in one lb of a, substance, To be found out now, •, , Proportion of oxygen and hydrogen in one lb/Kg of, water., , •, , Proportion of hydrogen and carbon in one lb/kg of fuel., , Compound proportions, , Workshop Calculation & Science : (NSQF) Exercise 1.2.12, , Copyright free, under CC BY Licence, , 33

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Examples, •, , The chemical formula for water is H2O. This means 2, atoms of hydrogen and one atom of oxygen combined, to make one molecule of water. If oxygen atom weighs, 16 times as much as hydrogen find out the proportions, in one kg of water., , Solution, Parts by weight of water are as below, Oxygen = 16/2 = 8kg, , •, , • For complete combustion of 1kg of carbon =, of oxygen, , • For complete combustion of 1 kg of hydrogen = 8kgs, of oxygen, , • For complete combustion of 1 kg of sulphur = 1 kg, of oxygen, , Hydrogen = 1/1 = 1kg, Total, , (by wt) are required for this purpose to supply sufficient, quantity of oxygen., , • Formula for calculation of mass of air for, , = 8+1= 9kg, , A hydrocarbon fuel has formula C6H14. This shows one, molecule of fuel contain 6 atoms of carbon and 14, atoms of hydrogen. If the carbon atom weighs 12 times, as the hydrogen atom, find the proportionate parts of, hydrogen and carbon in one kg of fuel., , complete combustion., Air contains 23% oxygen and 77% nitrogen, Mass of air = Mass of oxygen x for each constituent, 2 100, ,  11.6kg.of air, 3 23, , Solution, , For Carbon  2, , Parts of carbon by weight, = 6 x 12 = 72, Parts of hydrogen by weight = 14., , For hydrogen  8 , , Total No. of parts = 72 + 14 = 86., , For sulphur  1, , Weight of Carbon = 72/86 = 0.8372 kg, Weight of Hydrogen 14/86 = 0.1628 kg, , 23, , 100, 23, ,  34.8kg.of air, ,  4.35 kg.of air, , Total 50.75 Kg, , Ratio and Proportion, Proportion of air quantity required for combustion, process, Mass of air required for complete combustion of fuel, depends on the following factors and is called Air - fuel, Ratio, , • Carbon, Hydrogen, Sulphur are to burn with oxygen, in the combustion process., , • It has been found that the following quantities of air, , 34, , 100, , Hence 50.75 kg of air is to be supplied to the engine for, combustion of 1 kg of fuel., As the combustion process is not even more, quantity of air than 50.7 kg is to be supplied to, the engine., The calculations involved in the combustion, equations is beyond the scope of ITI students, as it involves chemistry and physics for computing the proportions of different elements., , Workshop Calculation & Science : (NSQF) Exercise 1.2.12, , Copyright free, under CC BY Licence

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Assignment, 1, , Length = 6.1 metre, Weight = 32 kgf, Weight of 1 metre of, the same channel =, _______ kgf, , 2, , d1 = 120 mm, d2 = 720 mm, n1 = 1200 rpm, n2 = ______ rpm, , 3, , Z1 = 42 T, n2 = 96 rpm, n1 = 224 rpm, , 9 In a gearing arrangement of a vehicle a gear having 26, teeth is meshing with a gear of 52 teeth. The dia of 52, teeth gear is 200mm. Find out the diameter of 26 teeth, gear wheel., 10 If two water pumps take 45 minutes to fill up a tank how, long will 4 similar pumps will take to fill this tank., 11 In a belt-pulley drive the driving pulley is of 12 cm, diameter and rotates at 360 rpm. Find the rpm of driven, pulley whose diameter is 20 cm., 12 To overhaul a gear box, 12 mechanics are needed to, complete the work in 5 days. If only 7 mechanics are, available, how many days they will be able to complete, the overhauling work., 13 Express in simple ratios the following, a, , 45  60, , b, , 40 paise  Rs4.00, , c, , 20mm, 4 metres, , d, , 4 C  100C, , Z2 = _______ T, , 4, D = 50 mm, H = 80 mm, h = 36 mm, d = ______ mm, , 14 Air contains 22% oxygen and 78% nitrogen by mass, (weight). Calculate the quantity of air (mass of air), required for complete combustion of unit mass fuel (The, main constitutents that take part in combustion process are carbon, hydrogen and sulphur), Note: Given the following data (Solve the problem), , 5 If a mechanic assembles 8 machines in 3 days, how, long he will take to assemble 60 machines., 6 In an autoshop the grinding wheel makes 1000 rpm and, the driven pulley is 200 mm dia. If the driving pulley is, 150 mm dia. Find out the rpm of the driving pulley., 7 In a gearing of a vehicle the following facts are found., A 180 mm dia of gear meshes with 60mm dia gear. If the, bigger gear makes 60 rpm. What will be the rpm of, smaller gear., 8 A vehicular job is completed by 5 mechanics in 4 days., If only 3 mechanics are available, in how many days the, work can be completed., , a 1 kg of carbon requires, , kg of oxygen., , b 1 kg of hydrogen requires 8 kg of oxygen., c 1 kg of sulphur requires 1 kg of oxygen., 15 A fuel is a hydro carbon substance of C7H14. This shows, each molecule of fuel contains 7 atoms of carbon and, 14 atoms of hydrogen. If carbon atomic weight is 12, times greater than hydrogen atom, find out the proportionate parts of hydrogen and carbon in one kg of fuel., 16 A vehicle worth Rs.20,000/- can be insured at a cost of, Rs.150/-. How much will it cost to insure a vehicle worth, Rs.24000/- for one year and 3 month at the same rate., (Compound proportion), , Workshop Calculation & Science : (NSQF) Exercise 1.2.12, , Copyright free, under CC BY Licence, , 35

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Percentage, , Exercise 1.2.13, , Percentage, Percentage is a kind of fraction whose denominator is, always 100. The symbol for percent is %, written after the, number. e.g. 16%., , Analyse the given data and proceed to arrive at, the answer through the unit., Example, A fitter receives a take-home salary of 984.50 rupees., , Ex., In decimal form, it is 0.16. Percentage calculation also, involves rule of three. The statement (the given data), for, unit, and then to multiple which is for calculating the, answer. (Fig 1), , If the deduction amounts to 24%, what is his total salary?, (Fig 3), Total pay 100%, , Deduction 24%, Take home salary 76%, , Example, The amount of total raw sheet metal to make a door was 3.6, metre2 and wastage was 0.18 metre2. Calculate the % of, wastage. (Fig 2), , If the take home pay is Rs.76, his salary is 100., For 1% it is, , 1, 76, , For Rs.984.50, it is, , For 100% it is, , 1, 76, , 984.50, 76, , x 984.50., , X100  1295.39, , 100% i.e. gross pay = Rs.1295.40., Example 1, 75 litres of oil is taken out from a oil barrel of 200 litres, capacity. Find out the percentage taken in this., Solution, % of oil taken = Oil taken out (litres) / Capacity of Barrel, (litres) x 100, , Solution procedure in three steps., Statement:, Area of door (A) =3.6 m2 = 100 %., Wastage = 0.18 m2, Single:, , 100, 1 m2, 3.6, , , , 100, x 0.18., 3.6, = 5%., , 75, x 100  37 21 %, 200, , Conclusion, , Example 2, A spare part is sold with 15%. Profit to a customer, to a, price of Rs.15000/-. Find out the following (a) What is the, purchase price (b) What is the profit., , The three steps involved are,, , Solution: CP =, , Multiple: for 0.18 m2=, Wastage, , step one, , :, , describe the situation (availability), , step two, , :, , decide for unit, , step three, , :, , proceed for the multiple., , x,, , CP = cost price, SP = sale price, SP=CP+15%of CP, 15000=, , +, , 36, , Copyright free, under CC BY Licence

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Original Price = 18000 x, , =, Profit = SP-CP = 15000-13043.47 = 1956.53, , = Rs.19565, , Purchase price = Rs.13,043/,Profit = Rs. 1957, , Example 5, , Example 3, Out of 80000 cars, which were tested on road, only 16000, cars had no fault. What is the percentage in this acceptance., , A Motor vehicle uses 100 litres of Petrol per day when, travelling at 30 kmph. After top overhauling the consumption falls to 90 litres per day. Calculate percentage of, saving., Solution, , =, , Percentage of saving = Decrease in consumption/Original, consumption x 100, , Example 4, The price of a motor cycle dropped to 92% of original price, and now sold at Rs.18000/- What was the original price., , = (100 – 90), , Solution, , =, , Present price of Motor cycle Rs.18000, , x100, , = 10% Saving in fuel., , This is the value of 92% of original price, , Assignment, 1, , a = 400mm (side of, square), , 5, , Weight of alloy = 140, Kgf, , d = 400 mm, , Weight of Sn 40%, , Wastage = ______ %., , Pb = ______ Kgf, Sn = ______ Kgf., , 2, , d = 26mm, , 6, , Shaded portion, = ______ %., , 7, , Compression length =, ______ %., , 8, , d = 360 mm, , 'a' depth of u/cut =, 2.4mm, reduction of area at, cross-section, = ______ %, 3, , Percentage of increase, = 36%, Value of increase, = 611.2 N/mm2, Original tensile strength, = ______ N/mm ., , a = 0.707 x d, , Copper in alloy = 27 kg, , Wastage = ______ %., , 2, , 4, , Zinc in alloy = 18 kg, % of Copper, = _______ %, % of Zinc = _______%., Workshop Calculation & Science : (NSQF) Exercise 1.2.13, , Copyright free, under CC BY Licence, , 37

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9, , Cu = 36 Kg, , 2 Average consumption per mile., , Zn = 24 Kg, , 3 Using the average consumption express maximum, consumption as a percentage of the average correct, to two decimal places. If the total petrol consumption of car on 4 different journeys each of 200 miles, are found to be as 6.65, 7.5, 6.85,7.05 gallons, respectively., , Cu = ______ %, Zn = ______ %, 10, , Cu = 42.3 Kg, Sn = 2.7 Kg, Cu ______ %, Sn = ______ %., , 11 What is the selling price, If a trader buys a spare part, for Rs.195/- This is 65% of selling price., , 17 In a Transport workshop, the following expenditure was, found to be occuring on the capital income., 1 40% income spent on tyres, 2 30% income spent on fuel and lubricants, 3 10% income spent on spare parts, If the month end saving comes to Rs.2000/- what is the, total income?, , 12 What is the purchase price if 25% profit is added to it,, If a Motor cycle tyre is sold for Rs.300/-., , 18 What is the final weight of the machined job if a casting, weighs 80kg. During preliminary machining weight is, reduced by 4% and final machining by 5%., , 13 How many m3 of elements of air present in 120 m3 of air,, If the composition of Air is 23% of Oxygen and 77% of, Nitrogen., , 19 What is the weight of zinc, copper and tin, casting, weight is 25kg. If a casting has 35% zinc, 40% copper, and 25% tin., , 14 How many kg of each of these elements are found. If an, Engine bearing made of alloy of 40 kg consists of the, following constitutents., a) Copper (Cu) - 86%, b) Tin (Sn) - 10%, c) Zinc (Zn) - 4%, 15 How much weight of these elements are found to exist., If a solder consists of 35%. Tin and 65% Lead. In a, solder of 40 kg., , 20 What is the total weight of a solder, if solder consists, of 35% tin 65% lead, and tin consists 14 grams., 21 What is the total annual income of the salesman, if a, salesman gets a monthly pay of Rs.1000 and a commission of 2.5% on his sale. In one year sale amount, is Rs.60,000., 22 What is the total income of a man, if he spends 15% of, his income on agriculture, 21% on family, 24% on, education of children and he saves Rs.360., 23 What is the percentage of his savings, if a person’s, monthly salary is Rs.450 and saves Rs.90 every month., , 16 Find out the following:, 1 Average consumption per journey., , 38, , Workshop Calculation & Science : (NSQF) Exercise 1.2.13, , Copyright free, under CC BY Licence

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Precentage - Changing percentage to decimal and fraction, Conversion of Fraction into Percentage, 1 Convert, , Solution: 0.35 x 100, = 35%, , 1, x 100, 2, , 2 Convert 0.375 into percentage., , = 50%, 2 Convert, , Solution: 0.375 x 100, = 37.5%, , 1, into percentage, 11, , Solution:, , Convert the following Decimal Fraction into Percentage, , 1, 100, x 100 =, 11, 11, , 1 0.2, 2 0.004, , = 9.01%, , 3 0.875, , Convert the following fraction into percentage., 1, , 1, 4, , 2, , 1, 5, , 3, , 2, 3, , 4, , 3, 8, , Conversion of Decimal Fraction into Percentage, 1 Convert 0.35 into percentage., , 1, into percentage., 2, , Solution:, , 4 0.052, Conversion of Percentage into Decimal fraction, 1 Convert 30% into decimal fraction., Solution:, , 30, = 0.3, 100, , 1, 2 Convert 33 % into decimal fraction., 3, , 1, 100, 3 =, 3  100  1, Solution:, 100, 100, 3 100, 33, , Conversion of Percentage into Fraction, 1 Convert 24% into fraction., Solution:, , =, , 24, 6, =, 100, 25, , 1 15%, 2 7%, , 1, 100, 3 =, 3  100  1, 100, 100, 3 100, , 33, , =, , 1, = 0.333, 3, , Convert the following percentage into decimal fraction, , 1, 2 Convert 33 % into fraction., 3, , Solution:, , Exercise 1.2.14, , 1, 3 12 %, 2, , 4 90%, , 1, 3, , Convert the following percentage into fraction, 1 15%, 1, 2 87 %, 2, , 3 80%, 4 12.5%, , 39, , Copyright free, under CC BY Licence

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Material science - Types metals, types of ferrous and non ferrous metals, Exercise 1.3.15, Types of metals, The metals is of two types:, 1 Ferrous metal, , 2 Non-ferrous metal, , 1 Ferrous metals : The metals that contains major part, of iron and contain carbon sre called ferrous metals such, as pig iron, mild steel, nickel etc., they have iron properties, such as rusting, magnetisations etc., 2 Non-ferrous metals : The metals that do not contains, iron or carbon and do not have the property of iron are called, non-ferrous metals such as copper, aluminum etc., Ferrous and Non ferrous alloys, Alloying metals and ferrous alloys, An alloy is formed by mixing two or more metals together, by melting., For ferrous metals and alloys, iron is the main constituent, metal. Depending on the type and percentage of the, alloying metal added, the property of the alloy steel will, vary., , Manganese steel can be used to harden the outer surface, for providing a wear resisting surface with a tough core., Manganese steel containing about 14% manganese is, used for making agricultural equipment like ploughs and, blades., Silicon (Si), Addition of silicon for alloying with steel improves resistance to high temperature oxidation., This also improves elasticity, and resistance against, corrosion. Silicon alloyed steels are used in manufacturing, springs and certain types of steel, due to its resistance to, corrosion. Cast iron contains silicon about 2.5%. It helps, in the formation of free graphite which promotes the, machineability of cast iron., Tungsten (W), The melting temperature of tungsten is 3380° C. This can, be drawn into thin wires., Due to this reason it is used to make filaments of electric, lamps., , Nickel (Ni), , Tungsten is used as an alloying metal for the production of, high speed cutting tools. High speed steel is an alloy of, 18% tungsten, 4% chromium and 1% vanadium., , This is a hard metal and is resistant to many types of, corrosion rust., , Stellite is an alloy of 30% chromium, 20% tungsten, 1 to, 4% carbon and the balance cobalt., , It is used in industrial applications like nickel, cadmium, batteries, boiler tubes, valves of internal combustion engines), engine spark plugs etc. The melting point of nickel, is 1450°C. Nickel can be magnetised. In the manufacture, of permanent magnets a special nickel steel alloy is used., Nickel is also used for electroplating. Invar steel contains, about 36% nickel. It is tough and corrosion resistant., Precision instruments are made of Invar steel because it, has the least coefficient of expansion., , Vanadium (Va), , Metals commonly used for making alloy steels, , Nickel-steel alloys are available containing nickel from 2%, to 50%., Chromium (Cr), Chromium, when added to steel, improves the corrosion, resistance, toughness and hardenability of steel. Chromium steels are available which may contain chromium, up to 30%., Chromium, nickel, tungsten and molybdenum are alloyed, for making automobile components and cutting tools., Chromium is also used for electroplating components., Cylinder liners are chrome-plated inside so as to have wear, resistance properties. Stainless steel contains about 13%, chromium. Chromium-nickel steel is used for bearings., Chrome-vanadium steel is used for making hand tools like, spanners and wrenches., Manganese (Mn), Addition of manganese to steel increases hardness and, strength but decreases the cooling rate., , This improves the toughness of steel. Vanadium steel is, used in the manufacture of gears, tools etc. Vanadium, helps in providing a fine grain structure in tool steels., Chrome-vanadium steel contains 0.5% to 1.5% chromium,, 0.15% to 0.3% vanadium, 0.13% to 1.10% carbon., This alloy has high tensile strength, elastic limit and, ductility. It is used in the manufacture of springs, gears,, shafts and drop forged components., Vanadium high speed steel contains 0.70% carbon and, about 10% vanadium. This is considered as a superior high, speed steel., Cobalt (Co), The melting point of cobalt is 1495°C. This can retain, magnetic properties and wear- resistance at very high, temperatures. Cobalt is used in the manufacture of, magnets, ball bearings, cutting tools etc. Cobalt high, speed steel (sometimes known as super H.S.S.) contains, about 5 to 8% cobalt. This has better hardness and wear, resistance properties than the 18% tungsten H.S.S., Molybdenum (Mo), The melting point of molybdenum is 2620°C. This gives, high resistance against softening when heated. Molybdenum high speed steel contains 6% of molybdenum, 6%, tungsten, 4% chromium and 2% vanadium. This high, speed steel is very tough and has good cutting ability., , 40, , Copyright free, under CC BY Licence

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Cadmium (cd), The melting point of cadmium is 320°C. This is used for, coating steel components., Alloying Metals and Non Ferrous Alloys, Non-ferrous Metals And Alloys, Copper and its alloys, Metals without iron are called non-ferrous metals. Eg., Copper, Aluminium, Zinc, Lead and Tin., Copper, , Bronze of different compositions are available for various, applications., Lead and its alloys, Lead is a very commonly used non-ferrous metal and has, a variety of industrial applications., Lead is produced from its ore ‘GALENA’. Lead is a heavy, metal that is silvery in colour when molten. It is soft and, malleable and has good resistance to corrosion. It is a, good insulator against nuclear radiation. Lead is resistant, to many acids like sulphuric acid and hydrochloric acid., , This is extracted from its ores ‘MALACHITE’ which contains about 55% copper and ‘PYRITES’ which contains, about 32% copper., , It is used in car batteries, in the preparation of solders etc., It is also used in the preparation of paints., , Properties, , Lead Alloys, , Reddish in colour. Copper is easily distinguishable because of its colour., , Babbit metal, , The structure when fractured is granular, but when forged, or rolled it is fibrous., It is very malleable and ductile and can be made into sheets, or wires., It is a good conductor of electricity. Copper is extensively, used as electrical cables and parts of electrical apparatus, which conduct electric current., Copper is a good conductor of heat and also highly, resistant to corrosion. For this reason it is used for boiler, fire boxes, water heating apparatus, water pipes and, vessels in brewery and chemical plants. Also used for, making soldering iron., The melting temperature of copper is 1083o C., The tensile strength of copper can be increased by hammering or rolling., , Babbit metal is an alloy of lead, tin, copper and antimony., It is a soft, anti-friction alloy, often used as bearings., An alloy of lead and tin is used as ‘soft solder’., Zinc and its alloys, Zinc is a commonly used metal for coating on steel to, prevent corrosion. Examples are steel buckets, galvanized, roofing sheets, etc., Zinc is obtained from the ore-calamine or blende., Its melting point is 420o C., It is brittle and softens on heating; it is also corrosionresistant. Due to this reason it is used for battery containers and is coated on roofing sheets etc., Galvanized iron sheets are coated with zinc., Tin and tin alloys, Tin, , Copper Alloys, Brass, It is an alloy of copper and zinc. For certain types of brass, small quantities of tin or lead are added. The colour of brass, depends on the percentage of the alloying elements. The, colour is yellow or light yellow, or nearly white. It can be, easily machined. Brass is also corrosion-resistant., Brass is widely used for making motor car radiator core and, water taps etc. It is also used in gas welding for hard, soldering/brazing. The melting point of brass ranges from, 880 to 930oC., Brasses of different composition are made for various, applications., Bronze, Bronze is basically an alloy of copper and tin. Sometimes, zinc is also added for achieving certain special properties., Its colour ranges from red to yellow. The melting point of, bronze is about 1005oC. It is harder than brass. It can be, easily machined with sharp tools. The chip produced is, granular. Special bronze alloys are used as brazing rods., , Tin is produced from cassiterite or tinstone. It is silvery, white in appearance, and the melting point is 231o C. It is, soft and highly corrosion-resistant., It is mainly used as a coating on steel sheets for the, production of food containers. It is also used with other, metals, to form alloys., Example: Tin with copper to form bronze. Tin with lead to, form solder. Tin with copper, lead and antimony to form, Babbit metal., Aluminium, Aluminium is a non-ferrous metal which is extracted from, ‘BAUXITE’. Aluminium is white or whitish grey in colour. It, has a melting point of 660o C. Aluminium has high electrical, and thermal conductivity. It is soft and ductile, and has low, tensile strength. Aluminium is very widely used in aircraft, industry and fabrication work because of its lightness. Its, application in the electrical industry is also on the increase., It is also very much in use in household heating appliances., , Workshop Calculation & Science : (NSQF) Exercise 1.3.15, , Copyright free, under CC BY Licence, , 41

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Material science - Physical and mechanical properties of metals, Exercise 1.3.16, Metal:, Metal is a mineral used in all types of engineering works, such as machineries, bridges, aero planes etc., so we, must have basic knowledge about the metals., Understanding the physical and mechanical properties of, metals has become increasingly important for a machinist, since he has to make various components to meet the, designed service requirements against factors, such as, the raise of temperature, tensile, compressive and impact, loads etc. A knowledge of different properties of materials, will help him to do his job successfully. If proper material/, metal is not used it may cause fracture or other forms of, failures, and endanger the life of the component when it is, put into function., Fig 1 shows the way in which the metals get deformed, when acted upon by the same load., Note the difference in the amount of deformation., , Physical properties of metals, , Conductivity (Figs 4 and 5), , – Colour, , Thermal conductivity and electrical conductivity are the, measures of ability of a material to conduct heat and, electricity. Conductivity will vary from metal to metal., Copper and aluminium are good conductors of heat and, electricity., , – Weight/specific gravity, – Structure, – Conductivity, – Magnetic property, , Magnetic property, , – Fusibility, Colour, , A metal is said to possess a magnetic property if it is, attracted by a magnet., , Different metals have different colours. For example,, copper is distinctive red colour. Mild steel is blue/black, sheen., , Almost all ferrous metals, except some types of stainless, steel, can be attracted by a magnet, and all non-ferrous, metals and their alloys are not attracted by a magnet., , Weight, Metals may be distinguished, based on their weights for, given volume. Metals like aluminium lighter weight (Specific, gravity 2.7) and metals like lead have a higher weight., (Specific gravity 11.34), Structure (Figs2 and 3), Generally metals can also be differentiated by their internal, structures while seeing the cross-section of the bar through, a microscope. Metals like wrought iron and aluminium, have a fibrous structure and metals like cast Iron and, bronze have a granular structure., 42, , Copyright free, under CC BY Licence

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Fusibility (Fig 6), It is the property possessed by a metal by virtue of which, it melts when heat is applied. Many materials are subject, to transformation in the shape (i.e) from solid to liquid at, different temperatures. Lead has a low melting temperature, while steel melts at a high temperature., Tin melts at 232°C., Tungsten melts at 3370°C., , Mechanical properties, –, –, –, –, –, –, –, , Ductility, Malleability, Hardness, Brittleness, Toughness, Tenacity, Elasticity, , Hardness (Fig 10), Hardness is a measure of a metal's ability to withstand, scratching, wear and abrasion, indentation by harder, bodies. The hardness of a metal is tested by marking by, a file etc., , Ductility (Fig 7), A metal is said to be ductile when it can be drawn out into, wires under tension without rupture. Wire drawing depends, upon the ductility of a metal. A ductile metal must be both, strong and plastic. Copper and aluminium are good, examples of ductile metals., Malleability (Figs 8 and 9), Malleability is the property of a metal by which it can be, extended in any direction by hammering, rolling etc., without causing rupture. Lead is an example of a malleable, metal., , Workshop Calculation & Science : (NSQF) Exercise 1.3.16, , Copyright free, under CC BY Licence, , 43

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Brittleness (Fig 11), , Toughness (Fig 12), , Brittleness is that property of a metal which permits no, permanent distortion before breaking. Cast iron is an, example of a brittle metal which will break rather than bend, under shock or impact., , Toughness is the property of a metal to withstand shock or, impact. Toughness is the property opposite to brittleness., Wrought iron is an example of a tough metal., , Tenacity, The tenacity of a metal is its ability to resist the effect of, tensile forces without rupturing. Mild steel, Wrought Iron, and copper are some examples of tenacious metals., Elasticity, Elasticity of a metal is its power of returning to its original, shape after the applied force is released. Properly heattreated spring is a good example for elasticity., , 44, , Workshop Calculation & Science : (NSQF) Exercise 1.3.16, , Copyright free, under CC BY Licence

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Material science - Introduction of iron and cast iron, , Exercise 1.3.17, , Ferrous Metals, , Cast Iron (Manufacturing process), , Metals which contain iron as a major content are called, ferrous metals. Ferrous metals of different properties are, used for various purposes., , The pig-iron which is tapped from the blast furnace is the, crude form of raw material for the cupola, and should be, further refined for making castings. This refining is carried, out in the cupola furnace which is a small form of a blast, furnace., , Introduction of Iron, Cast Iron, wrought Iron and steel, The ferrous metals and alloys used commonly are:, – Pig-iron, – Cast Iron, , Generally cupolas are not worked continuously like blast, furnaces but are run only as and when required., Cast Iron (Types), , – Wrought Iron, – Steels and Alloy steels, , Cast iron is an alloy of iron, carbon and silicon. The carbon, content ranges from 2 to 4%., , Different processes are used to produce iron and steel., , Types of cast iron, , Pig-iron (Manufacturing process), , The following are the types of cast iron., , Pig-iron is obtained by the chemical reduction of iron ore., This process of reduction of the iron ore to Pig-iron is known, as SMELTING., , – Grey cast iron, , The main raw materials required for producing Pig-iron are:, , – Malleable cast iron, , – Iron ore, , – Nodular cast iron, , – Coke, , Grey cast iron, , – Flux, , This is widely used for the casting of machinery parts and, can be machined easily., , Iron ore, The chief iron ores used are:, , – White cast iron, , Machine base, tables, slideways are made of cast iron, because it is dimensionally stable after a period of aging., , – magnetite, , Because of its graphite content, cast iron provides an, excellent bearing and sliding surface., , – hematite, – limonite, – carbonite., These ores contain iron in different proportions and are, naturally available., Coke, , The melting point is lower than that of steel and as grey cast, iron possesses good fluidity, intricate casting can be, made., Grey cast iron is widely used for machine tools because of, its ability to reduce vibration and minimize tool chatter., , Coke is the fuel used to give the necessary heat to carry, on the reducing action. The carbon from the coke in the, form of carbon monoxide combines with the iron ore to, reduce it to iron., , Grey cast iron, when not alloyed, is quite brittle and has, relatively low tensile strength. Due to this reason it is not, used for making components subjected to high stress or, impact loads., , Flux, , Grey cast iron is often alloyed with nickel, chromium,, vanadium or copper to make it tough., , This is the mineral substance charged into a blast furnace, to lower the melting point of the ore, and it combines with, the non-metallic portion of the ore to form a molten slag., , Grey cast iron is weldable but the base metal needs, preheating., , Limestone is the most commonly used flux in the blast, furnace., , White cast iron, , Properties and use of Pig-iron, Pig-iron is, therefore, refined and remelted and used to, produce other varieties of iron and steel., , This is very hard and is very difficult to machine, and for this, reason, it is used in components which should be abrasionresistant., White cast iron is produced by lowering the silicon content, and by rapid cooling. When cooled in this manner, it is, called chilled cast iron., White cast iron cannot be welded., 45, , Copyright free, under CC BY Licence

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Malleable cast iron, , 11.13 Types of steel, , Malleable cast iron has increased ductility, tensile strength, and toughness when compared with grey cast iron., , Main two types of steel are:, , Malleable cast iron is produced from white cast iron by a, prolonged heat-treatment process lasting for about 30, hours., Nodular cast iron, This is very similar to malleable cast iron. But this is, produced without any heat treatment. Nodular cast iron is, also known as: Nodular Iron - Ductile Iron - Spheroidal, Graphite Iron, This has good machinability, castability, resistance to, wear, low melting point and hardness., Mealleable and nodular castings are used for machine, parts where there is a higher tensile stress and moderate, impact loading. These castings are less expensive and are, an alternative to steel castings., Wrought Iron (Manufacturing process) (Fig 1), Wrought iron is the purest form of iron. The analysis of, Wrought iron shows as much as 99.9% of iron. (Fig 1), When heated, wrought iron does not melt, but only becomes pasty and in this form it can be forged to any shape., , 1 Plain steel, 2 Alloy steel, 1 Plain steel. In this carbon and iron are mixed. Accoring, to the percentage of carbon plain steels are classified, as:, A Low carbon steel, B Medium carbon steel, C High carbon steel, A Low carbon steel: It is also called mild steel. In, this. the percentage of carbon is from 0.15%to0.25%., Due to less quantity of carbon is sufficiently soft and, tolerates the strain. It can be put in different shapes, through forging and rolling. This is not very hard or, strong. This cannot be hardened or tempered by, ordinary methods. Nuts, bolts, rivets, sheets,, wires, T-iron and angle iron etc. are made out of it., B Medium carbon steel: The carbon content is from, 0.25% to 0.5%. Due to excess of carbon, it is harder, and tougher than mild steel. The tenacity is more., This can be hardened or tempered. Various things, are made by forging and rolling. This is used for, making high tensile tubes, wires, agricultural, implements, connecting rods, cam shafts, spanners, pulleys etc., C High carbon steel: It has carbon content from, 0.5% to 1.5%. It is very hard and wears least. This, can be hardened by heat treatment. This can, neither be cast nor rolled. This is very hard and, tough. It acquires permanent magnetic properties., This is used for making pointed tools, springs,, pumps, files, cutleries, cold chisels press die etc., 2 Alloy Steel, , Modern methods used to produce wrought iron in large, quantities are the, – puddling process, , Types of Alloy Steel, , – aston or Byers process, , Two types of alloy steel are:, , Steel, This is pure iron. Carbon content is more. Due to, excessive carbon it is harder and tougher. Carbon content, is from 0.15 to 1.5%. Besides there are other impurities, like sulphur, phosphorous etc. are there which cannot be, separated. This is hardened and tempered by heating it to, a definite temperature and cooling it in oil or water., The following methods are adopted for making different, types of steel:, 1 Cementation process, , 2 Crucible process, , 3 Bassemer process, , 4 Open hearth process, , 5 Electro thermo process, , 6 High frequency process., , 46, , When the steel is mixed with other metals like vinoleum,, manganese tungsten etc., it is called an alloy steel. Alloy, steel has properties of its ingredients., , A Low alloy steel, B High alloy steel, A Low Alloy steel: Besides carbon other metals are, in lesser quantity. Its tensile strength is more. The, welding can work on it. This can also be hardened, and tempered. It is used in manufacturing various, parts of an aeroplane and cam shaft etc., B High Alloy Steel: Besides carbon it has a high, percentage of the metals higher than low steel alloy., This is classified into following types:, , Workshop Calculation & Science : (NSQF) Exercise 1.3.17, , Copyright free, under CC BY Licence

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a High Speed Steel: It is also called high tungsten, alloy steel because it has more quantity of tungsten., According to the quantity of tungsten it is classified, into three types:, 1, , Tungsten 22%, Chromium4%, Vanadium 1%, , 2, , Tungsten 18%, Chromium 4%, Vanadium 1%, , 3, , Tungsten 14%, Chromium 4%, Vanadium 1%, , Cutting tools are made out of it because it is very, hard but becomes soft at low critical temperature., This temperature is raised out of cutting process of, tool, then the cutting tool becomes useless and is, unfit for work. But due to high percentage of, tungsten it keeps working upto high temperature. It, is used for cutting tools, drills, cutters, reamers,, hacksaw blades etc., b Nickel Steel: In this 0.3% carbon and 0.25 to, 0.35% nickel is pressent. Due to nickel its tensile, strength, elastic limit and hardness is increased. It, does not catch rust. Its cutting resistance increases 6 times more than plain carbon and steel, due to 0.35% nickel present in it. This is used for, making rivets, pipes, axle shafting, parts of buses, and aeroplanes. If 5% of cobalt is mixed with 3035% nickel, it becomes invar steel. It is mainly used, for making precious instruments., c Vanadium Steel: It contains 1.5% carbon 12.5%, tungsten, 4.5% chromium, 5% vanadium and 5%, cobalt. Its elastic limit, tensile strength and ductility, is more. It has strength to bear sharp jerks. It is, mainly used to manufacture of tools., d Manganese Steel: It is also called special high, alloy steel. It contains 1.6 to 1.9% of manganese, , and 0.4 to 0.5% carbon. It is hard and less wear. It, is not affected by magnet. It is used in grinders and, rail points etc., e Stainless Steel: Along with iron it contains 0.2 to, 90.6% carbon, 12 to 18% chromium, 8% nickel and, 2% molybdenum. It is used for making knives,, scissors, utensils, parts of aeroplane, wires, pipes, and gears etc., Properties of stainless steel:, 1 Higher corrosion resistance, 2 Higher cryogenis toughness, 3 Higher work hardening rate, 4 Higher hot strength, 5 Higher ductility, 6 Higher strength and hardness, 7 More attractive appearance, 8 Lower maintenance, f, , Silicon Steel: It contains 14% of silicon. Its uses, are multiferrous according to the percentage of, silicon. 0.5% to 1% silicon, 0.7 to 0.95% manganese, mixture is used for construction work. 2.5 to 4%, silicon content mixture is used for manufacturing, electric motors, generators, laminations of, transformers. In chemical industries 14% silicon, content mixture is used., , g Cobalt Steel: High carbon steel contains 5 to 35%, cobalt. Toughness and tenacity is high. It has, magnetic property therefore used to make permanent, magnets., , Workshop Calculation & Science : (NSQF) Exercise 1.3.17, , Copyright free, under CC BY Licence, , 47

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Material science - Difference between iron & steel, alloy steel and carbon, steel, Exercise 1.3.18, Steel Plants in India, , Difference between iron and steel:, S.No, , Basic, distinction, , 1, , Formation, , 2, , 3, , Types, , Rusting, , Iron, , Steel, , Pure substance, , S.No Name of the Steel plant, , Made up of, iron and, carbon, , Cast iron,, Wrought, iron and steel, , Carbon steel, and alloy, steel, , Quickly gets, oxidised and, result in rust, , Have different, elements that, protect from, rusting, , 4, , Surface, , Its surface, is rusty, , 5, , Usage, , 6, , Existence, , Used in, Used in, buildings,tools, buildings,, and automobiles cars,, railways, and, automobiles, Available in, Has to be, nature, formed, , 1, , Tata Iron, , Bihar, , 2, , Indian Iron Steel, , West Bengal, , 3, , Visweshvaraiah Iron Steel, , Karnataka, , 4, , Bhilai Steel Plant, , Chattisgarh, , 5, , Durgapur Steel Plant, , West Bengal, , 6, , Alloy Steel Plant (Durgapur), , West Bengal, , 7, , Bokaro Steel Plant, , Bihar, , 8, , Rourkela Steel Plant, , Orissa, , 9, , Salem Steel Plant, , Tamilnadu, , 10 Vishakapatnam Steel Plant, , Its surface, stays shiny, , State, , Andhra Pradesh, , Comparison of the Properties of Cast Iron, Mild Steel and steel, Property, , Cast Iron, , Mild Steel, , Composition, , Carbon contents, from 2 to 4.5%, , Carbon contents, from 0.1 to 0.25%, , Carbon contents, from 0.5 to 1.7%, , Strength, , – High, compressive strength, – Poor tensile strength, – Poor shearing strength, , – Moderate, compressive strength, – Moderate tensile strength, – High shearing strength, , – High, compressive strength, – High tensile strength, – High shearing strength, , Malleability, , Poor, , High, , High, , Ductility, , Poor, , High, , High, , Hardness, , Moderately hard and can be, hardened by heating to, hardening temperature and, quenching, , Mild, , Hard, , Toughness, , Possesses poor toughness, , Very tough, , Toughness varies, with carbon content, , Brittleness, , Brittle, , Malleable, , Malleable, , Forgeability, , Cannot be forged, , Can be forged, , Can be forged, , Weldability, , Cannot be welded with, difficulty, , Can be welded very easily, , Can be welded, , Casting, , Can be easily cast, , Can be cast but not easily, , Can be cast, , Elasticity, , Poor, , High, , High, , 48, , Copyright free, under CC BY Licence, , Steel

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Ferrous metals, , Non Ferrous metals, , Difference between metals and non-metals, , 1 Iron content is more, , 1 Iron content is missing, , Metals, , Non Metals, , 2 The melting point is, high, , 2 The melting point is low., , Shiny, , dull, , 3 This is of brown and, black colour, , 3 This is of different colours, , 4 This catches rust, , 4 This doesn’t catch rust., , 5 This can be, magnetised, , 5 This cannot be, magnetised, , 6 This is brittle in cold, state., , 6 This becomes brittle in hot, state., , Usually good conductors Usually poor conductors of, , Difference between cast Iron and steel, Cast Iron, , Steel, , 1 Carbon content is high, , Carbon content is less, , 2 Carbon is in free state, , Carbon is mixed, , 3 Melting point is low, , Melting point is high, , 4 It cannot be magnetised, , It can be magnetised, , 5 Because it is brittle,, it cannot be forged, , In can be forged, , 6 It rusts with difficulty, , It rusts quickly, , 7 It cannot be welded, , It can be welded, , of heat and electricity, , heat and electricity, , Most are ductile, , not ductile, , Opaque (opposite of, ‘transparent’), , Transparent when as a thin, sheet, , Most are malleable, , Usually brittle when solid, , Form alkaline oxides, , Form acidic oxides, , Sonorous (make a bell not sonorous, -like sound when struck), Usually have 1-3 valency Usually have 4-8 valency, electrons, electrons, Most corrode easily, Usually high melting, points (usually solid at, room temperature, except for mercury), , Workshop Calculation & Science : (NSQF) Exercise 1.3.18, , Copyright free, under CC BY Licence, , 49

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Material science - Properties and uses of rubber, timber and insulating, materials, Exercise 1.3.19, Properties and uses of rubber, , •, , Cannot be used for high-voltage insulation., , Rubber, , •, , Low operating temperature (as it becomes brittle and, develops cracks at a temperature of 60°C), , •, , Sulphur in rubber reacts with copper. Hence, copper, wires are to be tinned., , Rubber is an elastic material. It can be classified into, •, , Natural rubber, , •, , Hard rubber, , Hard rubber, , •, , Synthetic rubber, , By increasing the sulphur content and prolonged, vulcanization, a rigid rubber product called hard rubber or, ebonite is obtained. It possesses good electrical and, mechanical properties., , Natural rubber, It is obtained from the secretion of plants. It softens on, heating, becomes sticky at 30°C and hardens at about, 5°C., Sulphur is added to rubber and the mixture is heated. This, process is called vulcanising. By this process, stronger,, harder and more rigid rubber is obtained. Further, it, becomes less sensitive to changes of temperature and, does not dissolve in organic solvents. Its oxidisation is also, minimised by increasing its weathering properties., By adding carbon black, oil wax, etc, the deformation, properties are minimised. Rubber is moisture-repellent and, possesses good electrical properties. The main disadvantages of the rubber are as given under., •, , Low resistance to petroleum oils., , •, , Cannot be exposed to sunlight., , Sl.No., , Name, , Uses, It is used for battery containers, panel boards, bushing,, ebonite tubes, etc., Synthetic rubber, This is similar to natural rubber and is obtained from, thermoplastic vinyl high polymers. Some of the important, synthetic rubbers are:, •, , Nitrite butadience rubber, , •, , Butyl rubber, , •, , Hypalon rubber, , •, , Neoprene rubber, , •, , Silicon rubber, , Properties, , Uses, , 1, , Nitrite, butadiene, rubber, , Good resilience, wear resistance, flexibility at low, temperature, resistance to ageing, oxidation, low, tensile strength, high thermal conductivity, low, hygroscopicity, , Automobile tyre inner tubes., , 2, , Butyl, , It is attacked by petroleum oils, gases and alcoholic, solvents. It has thermal and oxidation stability and, high resistance to ozone., , Used as insulation in hot and wet, conditions, used as tapes in repair, work., , 3, , Hypalon, rubber, , Resistance to deterioration when exposed to sunlight, and temperature (up to 150°C)., , Used in jacketing of electric wires, and cables, , 4, , Neoprene, rubber, , Better resistance to ageing, oxidation and gas, diffusion, better thermal conductivity and flame, resistance, poor mechanical properties., , Used for wire insulation and cable, sheating., , 5, , Silicon, , High operating temperature (200°C) flexibility, moisture, and corrosion resistance, resistance to oxidation,, ozone, arcing, good insulating properties and thermal, conductivity. It is a good insulator., , Insulation for power cables and, control wires of blast furnace coke, ovens, steel mills and nuclear, power stations high frequency, generators, boiler, airport lighting, cranes., , 50, , Copyright free, under CC BY Licence

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Properties and uses of timber, , Uses, , General properties, , Soft timber, , Timber should have the following properties, , •, , Because of its cheapness it is used for low grade, furniture, doors and windows for cheap type of houses., , •, , Used as fuel., , •, , Some timbers are used for baskets and mat making., , •, , The bark is used as garment is some places., , •, , Straight fibres., , •, , Silky lustre when planed., , •, , Uniform colour., , •, , Regular annual rings., , •, , Heaviness., , •, , Firm adhesion of fiber and compact modulary rays., , •, , Sweet smell., , •, , It should be free from loose or dead knots and shakes., , •, , The surface should not clog the teeth of the saw on, cutting but should remain bright., , Hard timber, , Classification, •, , •, , Used for high quality furniture such as chairs, tables,, sofas, dewans, beds, etc., , •, , Used for door, window frames for high quality houses as, they can take good polish and painting finish., , •, , Used for manufacturing katha., , Wood as an electrical insulator, , Timbers are classified as, , Wood is impregnated with oil or other substance, for use, as insulator., , a Softwood, , Example, , b hardwood, , It is used in electrical machine windings, as slot wedges., , Softwood timber, , Insulating materials, , •, , Description, , Usually all trees with needle leaves of softwood and, those with broad leaves are of hard-wood., , •, , The wood contains resins and turpentines., , •, , The wood has a fragrant smell., , •, , Fibres are straight., , •, , Texture is soft and regular., , •, , Tough for resisting tensile stresses., , These are the materials which offer very high resistance to, the flow of current and make current flow very negligible or, nil. These materials have very high resistance - usually of, may megohms (1 Megohm = 106 ohms) are centimetre, cubed. The insulators should also posseses high dielectric, strength. This means that the insulating material should, not break down or puncture even on application of a high, voltage (or high electrical pressure) to a given thickness., , •, , Weak across the fibres., , Properties of insulators, , •, , Annual rings are distinct, having one side soft, porous, and light coloured. The other side is dense and dark., , The main requirements of a good insulating material are:, , •, , The general colour of the wood is pale tinted or light, such as pine spruce, fir, ash, kail, deodar etc., , •, , High specific resistance (many megohms/cm cube) to, reduce the leakage currents to a negligible value., , •, , Good dielectric strength i.e. high value of breakdown, voltage (expressed in kilovolts per mm)., , •, , Good mechanical strength, in tension or compression, (It must resist the stresses set up during erection and, under working conditions.), , •, , Little deterioration with rise in temperature (The insulating, properties should not change much with the rise in, temperature i.e. when electrical machines are loaded.), , •, , Non-absorption of moisture, when exposed to damp, atmospheric condition. (The insulating properties,, specially specific resistance and dielectric strength, decrease considerably with the absorption of even a, slight amount of moisture.), , Properties of hardwood, •, , The wood generally contains a large percentage of acid., , •, , It is brightly coloured., , •, , Annual rings are not distinct., , •, , It is difficult and hard to work with., , •, , It resists shearing stress., , •, , Fibre are overlapped., , •, , The general colour is dark brown such as oak, walnut,, teak, mahagony, sishim, babul, sal etc., , Workshop Calculation & Science : (NSQF) Exercise 1.3.19, , Copyright free, under CC BY Licence, , 51

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Products and insulators, Insulators, , Uses in electric field, , 1 Mica, , In elements or winding (Slot insulation), , 2 Rubber, , Insulation in wires, , 3 Dry cotton, , Winding, , 4 Varnish, , Winding, , 5 Asbestos, , In the bottom of irons and kettles, etc., , 6 Gutta, parcha, , Submarine cables, , 7 Porcelain, , Overhead lines insulators, , 8 Glass, , -do-, , 9 Wood dry, , Cross arms in overhead lines, , 10 Plastic, , Wires insulation or switches body, , 11 Ebonite, , Bobbin of transformer, , 12 Fibre, , Bobbin making and winding insulation, , 13 Empire, cloth, , Winding insulation, , 14 Leathroid, paper, , -do-, , 15 Millimax, paper, , -do-, , 16 P.V.C., , Wire insulation, , 17 Bakelite, , Switch etc. making, for insulation, , 18 Shellac, , -do-, , 19 Slate, , Making panel board, , 20 Paraffin, Wax, , Sealing, , 52, , Workshop Calculation & Science : (NSQF) Exercise 1.3.19, , Copyright free, under CC BY Licence

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Mass, volume, density, weight and specific gravity, , Exercise 1.4.20, , Mass, , Weight (Fig 2), , Mass of a body is the quantity of matter contained in a, body. The unit of mass in F.P.S system is pound (lb), in, C.G.S. system gram (gr) and in M.K.S and S.I systems, kilogram (kg). 1ton which is 1000 kg is also used sometimes., The conversion factor is 1000. Three decimal places are, shifted during conversion.E.g.1 ton =1000 kg 1g = 1000mg., , Weight is the force with which a body is attracted by the, earth towards its centre. It is the product of the mass of the, body and the acceleration due to gravity. The weight of a, body depends upon its location., , m - mass of a body, g - acceleration due to gravity in metre/sec2 = 9.81 m/, sec 2, V - volume of the body,  - density (pronounced as `rho'), W or FG - weight or weight force, Mass (Fig 1), , weight = W or FG = mass x gravitational force, = mxg, System, , Density, Density is the mass of a body per unit volume. Hence its, unit will be gr/cm3 or kg/dm3 or ton/m3., Density =, , Absolute, unit, , Derived, unit, , Conversion, , F.P.S. system 1 poundal, , 1 Lbwt, , 32.2 poundals, (1 lb x 1 ft/sec2, = 1 pound), , C.G.S. system 1 dyne, 1 gr x 1, cm/sec2, , 1 Gr.wt, , 981 dynes, , M.K.S., , Newton, , 1 kg.wt, , 1 Newton =, , S.I.system, , Newton, , Newton, , 1 kg x 1 m/sec2, , 1 kg.wt = 9.81 Newton, ( approximately 10N), , 1 Newton = 105 dynes., , 53, , Copyright free, under CC BY Licence

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Difference between mass and weight, S. No, , Mass, , Weight, , 1, , Mass is the quantity, of matter in a body (ie), measurement of, matter in a body, , Weight is measure, of amount of force, acting on mass due, to acceleration, due to gravity, , 2, , It does not depend on, the position or space, , It depends on the, position, location, and space, , 3, , Mass of an object will, not be zero, , Weight of an object, will be zero if gravity, is absent, , 4, , It is measured using, by physical balance, , It is measured using, by spring balance, , 5, , It is a scalar quantity, , It is a vector quantity, , 6, , When immersed in, water mass does not, change, , When immersed in, water weight will, change, , The unit is in grams, and kilogram, , The unit is in, kilogram weight,, a unit of force, , 7, , Unit, The density is measured as below, MKS/SI= Kg/m3, CGS - 1 gm/cm3 FPS–lbs/c ft, Solids, 1, 2, 3, 4, , gm/cc3, , Aluminum, 2.7, Lead, 11.34, Cast iron, 6.8 to 7.8, Steel, 7.75 to 8.05, , Liquids, , gm/cc3, , Water, Petrol, Oxygen, Diesel Oil, , 1.00, 0.71, 1.43, 0.83, , The specific gravity of a substance is also called its relative, density., Formula, Specific gravity, (or) Relative density, , Mass and weight are different quantities., Mass of a body is equal to volume x density., Weight force is equal to mass x acceleration, due to gravity., , Comparison Between Density And Specific Gravity, (Relative Density), Density, , Weight , Density and Specific gravity, It is now seen that the mass of a substance is measued by, its weight only without any reference to volume. But if equal, weights of lead & aluminium, are compared the volume of, lead is much smaller than volume of aluminium. So we can, now say that lead is more dense than aluminium,. i.e In, other words the density of lead is greater than aluminium., (Fig 3 & 4), , Mass per unit volume of, a substance is called its, density, Its unit is gm per cu cm;, lbs per cu.ft and kg/cubic, meter, , The density of substance, to density of water at 4°C, is its relative density, It has no unit of measurement simply expressed in, a number, , Density =, , Relative density, , Solids, 1, 2, 3, 4, The relation of mass and volume is called density., The density expresses the mass of volume E.g. 1 dm3 of, water has the mass of 1kg - thus the density of 1kg/dm3, (Fig 2), , 54, , Relative density or, Specfiic gravity, , Sp.gy, , Aluminium 2.72, Lead, 11.34, Cast iron 6.8 to 7.8, Steel, 7.82, , Liquids, Petrol, Battery acid, Water, Diesel Oil, , Sp.gy, 0.71, 1.2 to 1.23, 1.00, 0.83, , From the above table, we can calculate the weight of any, given volume of a substance (say Diesel oil) in any units, provided we know the specific gravity of the substance., Also vice-versa for volume of density is known., , Workshop Calculation & Science : (NSQF) Exercise 1.4.20, , Copyright free, under CC BY Licence

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Related problems for mass, volume, density, weight and specific gravity, Exercise 1.4.21, 1 Gr. wt. = 981 dynes, , 1 Calculate the mass in kg of a rectangular steel plate of, dimensions 220 x 330 x 15 mm (Fig 1) (density of, steel = 7.82 gm/cm3), , 15 dynes =, , Mass = Volume x density, , Gr.wt, , Force = m x acceleration produced by the force, , = 22 x 33 x 1.5cm3 x 7.82 gm/cm3, , Gr.wt = mass x 2.5 cm/sec2, , = 1089 cm3 x 7.82 gm/cm3, , gr.cm/sec2 = mass x 2.5 cm/sec2, , mass = 8.516 kg, , mass =, , grams =, , mass = 0.00612 gram, 4 A force of 2 N acts on a mass of 10 kg. Find the, acceleration produced by the force on the mass., Force = 2 N (, , 1 N = 1kg.m/sec2), , Force = mass x acceleration, 2 kg.metre/sec2 = 10 kg x acceleration produced, , 2 A storage container holds 250 litres of water. What, weight in N will this amount of water exert on the surface, on which it is standing?(Fig 2), , 2 x 1 kg.metre/sec2 = 10 kg x acceleration, produced, , ( 1 litre of water = 1 kg of water ), , acceleration produced, , Density of water 1 gm/cm3 or 1 kg/dm3, , =, , metre/sec2, , = 0.2 metre/sec2, 5 Calculate the weight of a body having a mass of 1 kg if, the acceleration due to gravity is 9.81 metres/sec2, Weight force = mass x acceleration due to gravity, = 1 kg x 9.81 metres/sec2, (1 kg.metre/sec2 = 1 N), 9.81 kg metre/sec2 = 9.81 N, In the examples solved the value of `g' is taken, as 10 metre/sec2, unless specifically mentioned otherwise., •, , The outside and inside diametres of a hollow sphere are, 150 & 70mm respectively. Calculate its mass if the, density of material is 7.5 gm/cm3. (Fig 3), , Acceleration due to gravity is taken as 10, metre/sec2 (approximation)., Capacity = 250 litres = 250 dm3 in volume., Mass of water = volume x density of water, = 250 dm3 x 1 kg/dm3 = 250 kg, Weight extended = mass x acceleration due to gravity, = 250 kg x 10 metre/sec2, = 2500 kg.metre/sec2 = 2500 N(, , 1 kg.m/sec2=1N), , 3 A force of 15 dynes acting on a mass of `m' produces, an acceleration of 2.5 cm/sec2. Find the mass., , Mass = Volume x Density, = Volume x 7.5 gm/cm3, D= 150 mm = 15cm, , Copyright free, under CC BY Licence, , R= 7.5 cm, , 55

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d = 70mm = 7 cm, , r = 3.5 cm, , Now, density of battery acid, = Specific gravity x Density of water, = 1.3 x 1000 gm/cm3, =1587.5 cm3, , = 1300 gm/cm3, , Mass = 1587.5 cm X 7.5 gm/cm, 3, , = 11906.6 gm=11.9kg, , 3, , Determination of specific gravity of a substance, , say 12kg, , 6 A car has a mass of 800 kg. Find out its weight force, (Take 9.81 m/sec2), (, , 1n = 1kg.m/sec2), , The specific gravity of a substance may be determined by, 1 Archemedies principle, 2 Hydrometer, , The Wt. force of a car=Mass of car x gravitational, acceleration, = 800 x 9.81 N, = 7848 Newtons, 7 A cylindrical tank 2m dia x 3.5 m deep is filled with, petrol. Find the weight of petrol in Tonnes, Assume, density of petrol 720 Kg/m3.(Fig 4), , Archemedies Principle, Archemedies principle states that when a body is fully or, partially immersed in a liquid, the amount of liquid displaced by the body is equal to the loss of weight of the body, in the liquid., Weight of a body in a liquid = total weight of the body, - weight of the liquid displaced by the body, This quantity if it is zero then the body will float. It is, negative the body will rise up till the weight of liquid, displaced by the immersed portion of the body is equal and, equal to the weight of the body. If it is positive the body will, sink. Specific gravity of solids soluble in water, , Volume of Tank, , specific gravity of solids soluble in water, , 3.14 x 3.5 m3 = 10.99 m3, Since 1 m3, = 1000 litres, Volume of Tank = 10.99 x 1000 litres, Density of petrol = 720 Kg/m3 ., Weight of Petrol in Kg =10.99x1000 litresx720Kg, = 720 x 10990 Kg, , specific gravity of a liquid, , The solid chosen should be such that it is, insoluble in both water and the liquid whose, specific gravity is to be determined., , Weight of Petrol in Tonnes, , Example, , (Metric Units), Weight of Petrol, , 1, = 7912.8 Tonnes, , 8 If the battery acid specific gravity is 1.3, and this is, being filled up into a cylindrical tank. Find out its, density., (Density of water = 1000 gm/cm3 ), Specific gravity or Relative density, , 56, , An iron piece weighs 160 kgf in air and 133 kgf when, it is fully immersed in water. Determine the volume and, specific gravity of the iron piece., Weight of the solid in air, Weight of the solid in water, Loss of weight in water, , = 160 kgf, = 133 kgf, = 27 kgf, , By Archemedies principle the loss of weight of a solid, in water = volume of water displaced., Volume of water displaced = 27 cm3, , Workshop Calculation & Science : (NSQF) Exercise 1.4.21, , Copyright free, under CC BY Licence

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Volume of the solid= 27 cm3, , Weight of metal and wax in water, , = 17 kgf, , Weight of metal piece only in water, , = 19 kgf, , weight of wax in water, , = (17 -19) kgf, = - 2 kgf, , loss of weight of wax in, water, , = (21 - (-2)) kgf, = 23 kgf, , Specific gravity of iron piece = 5.93, 2 A metal piece weighs 6.5 kgf in air and 3.5 kgf in water., Find its weight when it is fully immersed in a liquid, whose specific gravity is 0.8 and also the S.G of the, metal., Weight of metal piece in air = 6.5 kgf, Weight of metal piece in water= 3.5 kgf, Loss of weight in water, = 3.00 kgf (6.5 - 3.5), Specific gravity of metal, , By applying the principle of Archemedes the above, results are derived., By using a hydrometer also, the specific gravity of a, liquid is determined. The most common type of, hydrometer is the Nicholson’s hydrometer which is a, variable weight but constant immersion type., Specific gravity of a liquid, wt. of hydrometer+ wt. required to sink the, hydrometer in the liquid to a fixed mark, ------------------------------------------------------------------------wt. of hydrometer+ wt. required to sink the, hydrometer in water up to the same mark, Let the weight of the metal piece in the liquid = W, loss of weight of the metal in the liquid = 6.5 kgf_W, , specific gravity of wax, , Sl.No, , Metal, , Density gm/cc, , 1, , Aluminium, , 2.7, , 2, , Cast Iron, , 3, , Copper, , 8.92, , 4, , Gold, , 19.32, , 5, , Iron, , 7.86, , 6, , Lead, , 11.34, , 7, , Nickel, , 8.912, , 8, , Silver, , 10.5, , 9, , Steel, , 7.75 - 8.05, , 10, , Tin, , 7.31, , 11, , Zinc, , 7.14, , 12, , Diamond, , 3.51, , 13, , Bismuth, , 9.78, , 14, , Brass, , 8.47, , 15, , Phosphrous, Bronze, , 16, , Ice, , 0.93, , 17, , Air, , 0.0013, , 18, , Mercury, , 13.56, , 6.8 - 7.8, , 8.7 - 8.9, , 19, , Petrol, , 0.71, , w = 6.5 kgf - 3 kgf x 0.8 = 4.1 kgf, , 20, , Diesel, , 0.83, , loss of weight of the metal in the liquid = 4.1 kgf., , 21, , Kerosene, , 22, , Water, , 3 A solid of wax weighs 21 kgf in air. A metal piece, weighing 19 kgf in water is tied with the wax solid and, both are immersed in water and the weight was found, to be 17 kgf. Find the specific gravity of wax., Weight of wax in air, , 0.78 - 0.81, 1.0, , = 21 kgf, , Workshop Calculation & Science : (NSQF) Exercise 1.4.21, , Copyright free, under CC BY Licence, , 57

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Assignment, 1, , l = 1800 mm, , 7, , b = 65 mm, , L = 200 mm, l 1 = 75 mm, , h = 12 mm, , l 2 = 50 mm, ,  = 7.85 g/cm3, , B = 80 mm, , m = ______ kg, , H = 110 mm, h = 45 mm, , 2, , Capacity = 36 litres, ,  = 2.7 gm/cm3, , d = 32 cm, , m = ______ kg, , H = ______ cm, 8, 3, , V = 320 cm3, ρ = 8.9 gm/cm3, , D = 74 mm, , g = 9.80665, metre/ sec2, , d = 68 mm, l = 115 mm, , m = ______ kg, ,  = 8.6 gm/cm, , 3, , FG = ______ N, , m = ______ gms, 9, , Capacity = 35 litres, g = 10 metres/sec2, , 4, , FG = ______ N, , D 1 = 80 mm, D 2 = 61 mm, d = 39 mm, L = 112 mm, l = 90 mm, , 10, ,  = 7.85 gm/cm3, , Total FG = 8 KN, , m = ______ kg, 5, , (m 1 ) mass of chain, = 150 kg, Load = ______ N, mass m2 = _____kg, , D = 44 mm, d = 20 mm, L = 120 mm, l 1 = 60 mm, , 11, , W (FG) = 22.5 N, V (volume) = ______, , l 2 = 40 mm,  = 7.85 gm/cm3, m = ______ kg, 6, , 58, , L = 120 mm, B = 90 mm, b 1 = 60 mm, b 2 = 30 mm, d = 55 mm, H = 42 mm, h = 18 mm,  = 7.85 gm/cm3, m = ______ kg, , 12, , Workshop Calculation & Science : (NSQF) Exercise 1.4.21, , Copyright free, under CC BY Licence, , F = 250 d N, side of cube, = ______ mm, (cubical counter, weight balances `F')

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13, , unbalanced load in, the set up = 16 cN, , 19, , V = 4 dm3, FG = 10.8 daN, , ø of balancing weight, = 20 mm, ,  = ______ gm/cm3, , l of balancing weight, = ______ mm, 14, , d 1 = 40 mm, , 20, , l = 500 mm, , m 1 = 9 x 10 –2 kg, , b = 300 mm, , r1=r2, , H = 250 mm, , d 2 = 60 mm, ,  of oil = 0.9 gm/cm3, , FG 2 = ______ N, , m = 2.5 kg, h = ______ mm, , 21, , Engine cooling, Data given, Water in Radiator = 10, litres, Find, , 15, , Mass of water =, __________ kg, , l x b = 1 m2, FG = 7.85 x 10–2 kN, , (Assume 1 litre = dm3, in volume), , s = ______ mm, , Density of water = 1 kg/, dm3, 16, F = 400 N, , 22, , m = ______ kg, , Cylinder Liner Dimension, Data given, OD = 111 mm, ID = 103 mm, , 17, , Length = 240 mm, , m 1 = 200 gms, , Material = C.I, , FG = 16 N, , Density of C.I = 7.259, gm/cm3, , F = ______ dN, , Find its mass___ in kg, 23, , Gudgeon Pin (Solid), Data given, , 18, , R = 14 kN, , Dia = 200 mm, , m = ______ kg, , Length = 70 mm, Material = M.S, Density = 7.85 gm/cm3, Find its mass = ___ gm, , Workshop Calculation & Science : (NSQF) Exercise 1.4.21, , Copyright free, under CC BY Licence, , 59

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24, , 28 Conversion of vehicle weights, , Data given, Dia = 80 mm, , Take g = 10 m/sec2, , Length = 100 mm, Density of Aluminum =, 2.7 g/cm3, Find its Mass _______, in kg, 25, , Hollow sphere (Cast, Brass), , Weight force, a 480 Newton, , _______, , b 14800 N, , _______, , c 2000 N, , _______, , d 7000 N, , _______, , 29 Coversion of mass of vehicle, Take g = 9.81 m/sec2, , Data given, , Mass of Vehicle Its weight, , O.D = 150 mm, I.D = 120 mm, , a 1200 kg, , ________ N, , Density of Brass = 6.89, gm/cc, , b 800 kg, , ________ N, , c 700 kg, , ________ N, , d 900 kg, , ________ N, , Use Vol = (, , (R3)3), , Find, , 30 Fill up the blanks, , Mass of Hollow sphere, = _______ kg, 26, , Mass, , Diesel Tank, Data given, Diameter = 400 mm, Depth of filling (h), = 600 mm, Spongy of oil = 0.8, , Comparison of Metals & Liquids, Material, , Sp.gy, , Density, , a, , Lead, , 11.34, , _______, , b, , Copper, , 8.92, , _______, , c, , Cast Iron, , 7.20, , _______, , d, , Petrol, , 0.71, , _______, , e, , Diesel, , 0.83, , _______, , f, , Sulphuric, Acid, , 1.84, , _______, , Density of water = 1000, kg/m3, , 31 Fill in the blanks with correct statement in a & b, , Find, Mass of oil in Tank =, ______ in kg, , a The density of water - 1000 kg/m3 specific gravity of, nitric acid = 1.2. The density of nitric acid = _______, b, , Material, , Density, , i, , Water, , 1000 kg/m, , Define the following term, , ii, , Aluminium, , 2.7 g/cm3, , _______, , a Mass, , iii, , Iron, , 8 g/cc, , _______, , b Weight, , iv, , Copper, , 8.7 g/cc, , _______, , 27 Definition, , Specific gravity, 3, , _______, , c Density, , c Mass of a body = Volume x _______, , d Specific gravity, , d Weight force = Mass x _______, e Give abbreviation for, i, , Mega newton _______, , ii Kilo newton per square metre _______, f, , 60, , 1 litre of water = _______kg., , Workshop Calculation & Science : (NSQF) Exercise 1.4.21, , Copyright free, under CC BY Licence

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Speed and velocity - Rest, motion, speed, velocity, difference between speed, and velocity, acceleration and retardation, Exercise 1.5.22, Body at rest, , unit = m/s2 (metre per square second), , When a body does not change its position, with respect to, its surroundings, it is said to be at rest., , u = Initial velocity in metre per second(m/sec), , Body at motion, When a body changes its position, with respect to its, surroundings, it is said to be in motion. The motion may, be linear if the body moves in a straight line or it may be, circular when it moves in a curved path., Terms relating to motion, , v = Final velocity in metre per second(m/sec), s = Distance in metre (m), t = Time in second (sec), a = Acceleration m/sec2 (positive value), R = Retardation m/sec2 (negative value of acceleration), Equations of motion, , Displacement, When a body is in motion from one place to another, the, displacement is the distance from the starting position to, the final position., , Then v = u + at, s = ut +, , at2 and v2 – u2 = 2as, , v2 = u2 + 2as, , Speed, It is the rate of change of displacement of a body in motion., It has got no direction and it is a scalar quantity., Speed = distance travelled per unit time, Unit = m/s, km/Hr.mile/Hr., Velocity, It is the rate of change of displacement of a body in motion, in a given direction. It is a vector quantity and can be, represented both in magnitude and direction by a straight, line. Velocity may be linear or angular. The unit of linear, velocity is metre/sec,, Velocity =, , Retardation, When the body has its initial velocity lesser than its final, velocity it is said to be in acceleration. When the final, velocity is lesser than the initial velocity the body is said to, be in retardation. Then the three equation of motion will be, v = u – at, s = ut – at2, u2–v2 = 2as, Average speed, Vm - Average speed in metre/min, (metre/sec), n - Revolutions per minute or number of strokes per, minute, s - Distance travelled, length of stroke., , Unit = m/s, km/Hr,mile/Hr., , Stroke speed (Fig 1), , Difference between speed & velocity, Speed, , Velocity, , The rate of change of, place of an object is, its speed., , The speed in a definite, direction is called velocity., , In the speed, direction is not indicated., Only the magnitude is, expressed., , Both the magnitude and direction are expressed., , Speed, , Velocity, , For one revolution of the point k, of the crank pin the, distance the power saw blade moves = 2 x s, Therefore ‘n’ revolutions in a minute the distance = 2 x s x, n. Since the stroke of the blade will be given in metre to, determine the average speed, Vm = 2 x s x n, , Acceleration, Rate of change of velocity is known as acceleration or it is, the change of velocity in unit time. Its unit is metre/sec2., It is a vector quantity., m/sec2, , Piston speed (Fig 2), As the piston moves backward and forward, its speed, constantly changes between the upper and lower dead, centres. Hence in this case also the average speed Vm =, 61, , Copyright free, under CC BY Licence

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2 x s x n. Since s is expressed in mm and n in number of, revolutions/per minute and since Vm is given in metre/sec,, we have, Vm = 2 x s x, , metre/min., , =, , Motion under gravity, A body falling from a height, from rest, has its velocity goes, on increasing and it will be maximum when it hits the, ground. Therefore a body falling freely under gravity has a, uniform acceleration. When the motion is upward, the, body is subjected to a gravitational retardation. The, acceleration due to gravity is denoted with ‘g’., Momentum, It is the quantity of motion possessed by a body and is, equal to the product of its mass, and the velocity with which, it is moving. Unit of momentum will be kg metre/sec., Momentum = mass x velocity, Newton’s laws, First law, Every body continues to be in a state of rest or of uniform, motion in a straight line unless it is compelled to change, that state of rest or of uniform motion by some external, force acting upon it., Second law, , If s is given in metres then, Vm = 2 x s x, , =sx, , The rate of change of momentum of a moving body is, directly proportional to the external force acting upon it and, takes place in the direction of the force., , metre/sec., , 2 x s denotes a double stroke., In case of the reciprocating motion the average, speed is taken into account for calculations., Vm = 2 x s x n metre/min if s is given in metres, Example (Fig 3), , Third law, To every action there is always an equal and opposite, reaction., In the rivet joint equal forces act on the strap and they, opposite force F2. (Fig 4), , An extrusion press has a crank radius of 20 cm and an rpm, of 30/min. Calculate the average speed in metre/min,, metre/sec., s = The diameter = 40 cm., One crank revolution makes the piston to travel in 2s=80cm, Vm = 2 x 400 x, , metre/min., Law of conservation of momentum, , = 24 metre/min = 0.4 metre/sec, , When two moving bodies have an intentional or unintentional impact, then sum of the momentum of the bodies, before impact = sum of the momentum after impact, or the, change in momentum after the impact is zero., m1 - mass of one body and, v1 - velocity with which it moves, m2 - mass of second body, , NEWTON’S LAWS OF MOTION, , v2 - velocity with which it moves, , Equations of motions under gravity, Downward, , V = u – gt, , v = u + gt, , s = ut –, , gt2, , u2–v2 = 2gs, 62, , Momentum = m x v= mass of the body x its velocity, , Upward, , s = ut +, , Rate of change of momentum = force acting on the body, , gt2, , v2–u2 = 2gs, , force = mass x acceleration, , Workshop Calculation & Science : (NSQF) Exercise 1.5.22, , Copyright free, under CC BY Licence

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Momentum of two bodies before impact = momentum after, impact, m1 x v1 + m2 x v2 = (m1 + m2)V, Terms - Some Examples in vehicles, Displacement, The piston displacement is the space between 2 dead, centres (TDC and BDC) where in the piston moves in the, cylinder. (Fig 5), Example, In circular motion bodies (like shafts, axles, gear-wheels,, pulleys, flywheels, grinding wheels) turn with constant, speed around its axis., The angular of circular motion is also called Angular, velocity or Peripheral speed., Expressed in Metre/sec or Radians per second., Bodies at rest and in motion, Speed, , Terms related to brake system, , This is reckoned in 2 ways in a vehicle, , Every vehicle has a brake system. When brakes are, applied on a moving vehicle (with certain velocity) its, velocity is reduced and vehicle is decelerated and it stops, at a certain distance. So the definition of the terms related, to Brake application are set forth below., , – Vehicle speed in kmh/mph, – Engine speed in rpm, Velocity, A motor vehicle, normally changes its speed and direction, on road. Hence used in velocity calculation., , Deceleration (a) (Fig 8), , Acceleration (Fig 6), When the speed of the vehicle is increased on road, it is, said to be accelerated., Deceleration (Fig 6), Deceleration or Retardation (this is further explained), During the application of brakes of a vehicle the speed of the, vehicle is decreased. Then it is said to be decelerated or, retarded., , This is the decrease in velocity within a certain time. e.g, A car travelling at 90 kmph stops after 10 Sec., The deceleration = 90 x, , x 1/10, , = 25 m/s/10 sec, = 2.5 m/sec2, Deceleration time, Circular or Angular motion (Fig 7), When a body rotates about an axis, it is said to have, angular motion or circular motion., , The time 10 seconds is called the above time to stop the, vehicle., , Workshop Calculation & Science : (NSQF) Exercise 1.5.22, , Copyright free, under CC BY Licence, , 63

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Stopping distance, , Bodies at rest or in Uniform motion, , During the deceleration time the car travels a distance, called i.e Stopping distance ‘d’., , The diesel engine piston remains at rest at TDC or BDC due, to its inertia. Expansion of gas pressure or flywheel, momentum moves the piston from TDC or BDC., , But the total stopping distance is reckoned as equal to, normal stopping distance and distance travelled by the car, during reaction time of the driver., The reaction time is explained as below, During the application of brakes, the driver takes sometime, to recognise the danger and then apply the brakes. The, time (thus elapsed) is called reaction time. During this time, the vehicle travels some more distance before coming to a, stop. So the total stopping distance actually varies due to, the reaction time of the driver and it is longer than the, normal stopping distance. The reaction time varies between driver to driver., Example, , Second law (with examples) (Fig 10), The rate of change of momentum of a moving body (say, Engine part or Vehicle)is directly proportinal to external, force acting take place in the direction of force., – A connecting rod in motion is brought to rest at BDC., – The direction of movement of a vehicle is altered by force, of wind., – When a vehicle travels in a down gradient its speed, increases., – The speed of vehicle is decreased when travelling up, gradient., , A car is travelling with a speed of 72 kmph and its, acceleration (a) = 5 m/sec2. The reaction time of driver to, apply brakes is 1.5 seconds. calculate the total stopping, distance., Solution, Third law (with examples) (Fig 11&12), , Velocity of car = 72 kmph, , = 20 m/sec, acceleration = 5 m/sec2, Normal stopping distance S =, , (m), , Total stopping distance, = 40 metre + Velocity x Reaction time, = 40 m + (20 x 1.5) m, = 70 metres., Newon's Law of Motion, Some Examples in vehicles, First law (with examples) (Fig 9), , To every action there is always an equal and opposite, reaction., All upward force = All downward forces, – Jack is lifting a differential, – Crane rope is lifting an engine., , 64, , Workshop Calculation & Science : (NSQF) Exercise 1.5.22, , Copyright free, under CC BY Licence

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Speed and velocity - Related problems on speed & velocity, Examples, •, , Time taken = 1 sec (t), , A body travels a distance of 168 metres in a straight line, in 21 secs. What velocity the body is travelling., Velocity = distance travelled/time, , = ut +, , = 168 metres/21 secs, , A train covers a distance of 150 kilometres, between, two stations, in 2 1/2 hours. Determine the average, velocity with which the train is moving., , •, , 150, 2,  150 , 5 Km/hr, = 150 Km/2 1/2 hrs = 5, 2, = 60 Km/hr, , s = 4.905 metres, , A force of 30 N acts on a body at rest. The mass, of the body is 50 kg. Determine the velocity of the body, after 4 secs, the distance it covers during that period, and the acceleration, F=mxa, 3 kg x metre/sec2 = 50 kg x a,  acceleration = 3/50 metre/sec2, , A vehicle accelerates uniformly from a velocity of 8 km/, hr to 24 km/hr in 4 secs. Determine the acceleration, and the distance travelled by it during that time., , = 0.06 metre/sec2, , a = 0.06 m/sec2, , v = u + at, = 0 + 0.06 metre/sec2 x 4 sec = 0.24 metre/sec, s = ut + 1/2 at2 = 0 + 1/2 x 0.06 metre/sec2 x 16 sec2, , Final velocity = 24 km/hr (v), time = 4 sec (t), , = 0.48 metre, , acceleration (a), •, , s = 0.48 metre, ,  4a sec = 16 km/hr = 16000 metre/3600 sec, , A stone is thrown vertically upwards with a velocity, of 120 metre/sec. Determine (a) the maximum height, to which it travels before starting to return to earth. (b), The total time taken by the stone to go up and come, down. (c) The velocity with which it will strike the, ground., ,  acceleration (a) = 16000 metre/3600 x 4 sec2, , Initial velocity of throw = 120 metre/sec (u), , v = u + at, 24 km/hr = 8 km/hr + a x 4 sec, (24km/hr - 8km/hr - 16km/hr), , 4a = 4.44, Acceleration (a) = 1.1 metre/sec, , Final velocity = 0 metre/sec (v) (taken g = 10 m/sec2), , 2, , Retardation due to gravity = 10 metre/sec2, , Distance travelled (4a) = 4 x 1.1m = 4.4 m, A car moving with a velocity of 50 km/hr is brought to, rest in 45 secs. Find out the retardation., Initial velocity = 50 km/hr, , (1km= 1000 metres), , Final velocity = 0 km/hr, , (1 Hour = 3600 seconds), , v = u – at, 0 = u – at, u = at, , u2–v2 = 2g.s,  1202 metre2/sec2 – 0 = 2 x 10 metre/sec2 x s,  s = 120 x120/2 x 10 metre =, = 720 metre, , Time = 45 secs, 50km/hr x, a, , v, t, , , , 5, 18, , m/sec  13.88 m/sec, , 13.88 m/sec, 45 sec, ,  0.3m/sec, , 2, , 50000/3600 metre/sec = a x 45 sec, , •, , x 9.81 metre/sec2 x 1 sec2, , 30 N = 50 kg x a, , Initial velocity = 8 km/hr (u), , •, , x 9.81 m/sec2 x 12 sec, , 1 Sec2 = 4.905 metres., , Average velocity = Distance travelled/time taken, , •, , gt2 0 x 1 sec=, , = 0 x 1 sec +, , = 8m/sec, •, , Exercise 1.5.23, , when it comes down its velocity at start = 0 metre/sec., The acceleration due to gravity = 10 metre/sec2 and the, distance travelled = 720 metre,  v2– u2 = 2as, , v2-0 = 2x 10 m/sec2 x 720 m, , Retardation = 50000/3600 x 45 metre/sec2, , v2– 0 = 2 x 10 x 720 metre2/sec2, , = 0.30 metre/sec2, ,  v = 120 metre/sec, , A body falling freely under the action of gravity reaches, the ground in one second. Determine the height from, which the body fell. Take g = 9.81 metre/sec2., , Time taken to go up and reach a velocity of 0 metre/sec, = u/g = 120 metre/sec/10 metre/sec2 = 12 sec., , Initial velocity = 0 metre/sec (U), , Time taken to start from rest and attain a velocity of 120, metre/sec = v/g = 12 sec., , Acceleration due to gravity = 9.81 metre/sec2 (g), ,  Total time taken = 24 sec., 65, , Copyright free, under CC BY Licence

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•, , Calculate the Angular velocity in radian/second of, an engine flywheel when it is rotating at 2800 rpm., (Fig 1 & 2), , Since wheel turns 120° angle, 120° = 120 x 2/360, = 2.094 radians, Distance moved by a point on tyre S – re, , Angular velocity (W) = This is the rate of change of, displacement or angle turned through per unit time., , [where r = 270 mm, , Solution, ,  = 2.094 radian], , Angular velocity of flywheel W= 2N/60 rad/sec., [N = 2800 rpm], , S = 270 x 2.094 mm, , = 2 x 2800/60 radian/sec., , Circumferential distance moved by the point = 565.38, mm, , = 565.38 mm, , = 293.3 radian/sec., •, , The rear wheels of a car have diameter of 600 mm. The, rear axle makes 250 rpm. Find out the peripheral speed, of rear wheels in m/sec., , Solution, Peripheral speed V =, , , •, , 3.14  600, , , , πdN, 1000, , , , 1, (m/s), 6, , 250, ,  7.85 m/sec, 60, 1000, Calculate the stopping distance of a car travelling with, a speed of 72 km/h and being accelerated with, a - 5 m/ sec2., , Solution, Va (initial speed of a car) = 72 kmph, (1 kmph , , •, , A motor car road wheel of dia 540 mm turns through an, angle of 120°. Calculate the distance moved by a point, on tyre thread of the wheel., , 3600, = 20 metres/sec, , 66, , m/sec)  72, , Stopping distance S =, , Solution, There are 2 radians in one turn of wheel. i.e 2 radians, = 360°, , 1000, , , , 18, , m/sec, , Va 2, (metre), 2a, , 20 2, 400, , 25, 10, ,  40metre, , Workshop Calculation & Science : (NSQF) Exercise 1.5.23, , Copyright free, under CC BY Licence, , 5

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Assignment, 1, , S = 180 mm, , 7, , Vm = 0.35 metre/sec, , n = 65 (double stroke), , s = 200 mm, , Vm = _____metre/min, , n = ______ rpm, , Vm is average cutting, speed), 8, 2, , V = 16 metre/min, , s = 650 mm, , s = 210 mm, , Vm = 90 metre/min, , n = ______, , n = ______ rpm, , 9 Vm1 = 5.2 metre/sec, Increased to, 3, , (V is the cutting speed), , Vm2= 6.3 metre/sec, , n = 22 strokes (Double, stroke)/min, , Increase in n (rpm) = ______ %, , V = 18 metre/min, , 10, , s = 250 mm, n = 45 (double, strokes), , s = ______ mm, , V = ______metre/min, 4, , s = 240 mm, , 11, , n = 30 (working stroke), , Is : Vm = 25 : 1, n =_____(double, strokes), , V = ______ metre/min, , Is = rack travel, , 5, , Vxm = stroke speed/, min, , n = 50 cutting strokes, V = 32 metre/min, d = ______ mm, , 12, , Vm = 10 metre/min., n = 12.5 / min., Rack travel = ______, , 6, , s = 64 mm, n = 3600 rpm, Vm = _____metre/sec, , 13, , Vm is the average, piston speed), , dia of crank = 100 mm, Rack, speed = 12 metre/min, Crank disc 'n'' = _____, rpm, , Workshop Calculation & Science : (NSQF) Exercise 1.5.23, , Copyright free, under CC BY Licence, , 67

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14, , spindle 'n' = 250 rpm, , 20, , Car wheel, , Average stroke speed, = 30 metre/min, , n = 720 rpm, Peripheral speed, = 18.84 m/sec, , stroke length =, ________mm, , d = ________, , 21, 15, , Angular speed, n = 2000 rpm, , Car Speed = 90 km/hr, , Angular velocity, = _______ radians/sec, , Time to stop = 10 sec, Deceleration, = ________ metre/, sec2, , Use, W = 2N/60 rad/sec, 22, , 16, , Car speed = 80 km/hr, Distance stopped = 60, metre, Deceleration of car, =________ metre/sec2, , 17, , Deceleration = 4.5 m/, sec2, Stopping distance = 50, metres, Velocity, of, car, =________ km/hr, , 18, , Distance travelled by, car = 600 km, Time = 8 hrs 20 min, Average velocity, = ________ km/hr, , 19, , S = 74 mm, n = 4500 rpm, Mean velocity = _______ m/sec, Maximum velocity = _______ m/sec, (Average Speed of Piston), 23 Total Stopping Distance , , Average, velocity, = 56.3 km/hr, , (Use = V2/2a), , Distance travelled, = 464.475 km, , Deceleration = 5m/sec2, , Travelling time, = ________ hrs, , 68, , Piston Velocity/Speed, , V2,  velocity  reaction time, 2a, , V = Vehicle speed = 80 km/hr, Reaction Time of driver = 2 seconds, Total Stopping Distance = _______ meter, , Workshop Calculation & Science : (NSQF) Exercise 1.5.23, , Copyright free, under CC BY Licence

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Work, power, energy, HP, IHP, BHP and efficiency, Work(Fig 1), , Exercise 1.5.24, , CGS = 1 gm x 1cm/second2, , Work is said to be done by a force, when it moves, its point, of application through a distance. Applied force ‘F’ moves, a body through a distance’s., , = Dyne, MKS = 1 kg x 1m/second2, = Newton., , Work done ‘W’ = F x s., The S.I. unit of work is 1 joule which is the work done by a, force of moving the body through a distance of 1 metre., , 1 Newton = 105 dynes, , Therefore joule = 1 N x 1 metre = 1 Nm, , 1 pound = 4.448N,, , Also 1 joule = 1 Nm = 105 dynes x 100 cm = 107 dynes cm, = 107 ergs., , Newton = 0.225 pound., , 1kg wt = 9.81N, , Absolute units, In C.G.S. system unit of work = 1 erg = 1 dyne x 1 cm, In F.P.S system unit of work = 1 Foot poundal = 1 poundal, x 1 foot, In M.K.S. system unit of work = 1 joule = 1 Newton x 1, metre, Derived units, C.G.S. system 1 Gm Wt x 1 cm = 981 ergs., F.P.S. system 1 ft LB = 981 foot poundal, M.K.S. system 1 kgf metre = 981 joule., , F - force or weight force in N, , Power(Fig 2), , s - distance the body on which force acts is moved in, metres, , It is the work done in unit time., , t - time in seconds, v - speed in metre/sec, w - work done by the force in joules, P - Power in Watts, Pout - Power output, Pin - Power input, Force, A Force is that which changes or tends to change the state, of rest or motion of a body., Force = Mass x Acceleration, F = Ma, Unit, F =Mxa, = kg x m/sec2, = 1 Newton (SI unit), (Newton: If 1 kg of mass accelerates at the rate, of 1m/sec2 then the force exerted on the mass, is 1 newton), FPS = 1 pound x 1 Feet/second, , 2, , which is equal to 1 Watt. Power in watts = w/t = F.s/t, =FxV, In M.K.S. system the unit is 1 kgf meter/sec. One horse, power is = 75 kg metre/sec or 4500 kgf metre/min., , = 1 pound, 69, , Copyright free, under CC BY Licence

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1HP (metric) = 735.5 Watts, efficiency , , 1HP (British) = 746 Watts = 0.746 KW, 1 KW, , = 1.34 HP, , power output, power input, ,  100%, , Energy, , Power input is the power given to a machine to do work., Power output is what we get out of the machine. Power, output is always less than power input due to friction in the, machine. The ratio between power output to power input, is efficiency of the machine and it is expressed in, percentage.(Fig 3), , The energy of a body is its capacity to do work. It is equal, to power x time. Hence the unit of energy is the same as, the unit of work in all systems., Forms of energy, Mechanical energy, Electrical energy, Atomic energy,, Heat energy, Light energy, Chemical energy, sound energy. Energy of one form can be transformed into energy, of another form., Law of conservation of energy, – The energy can neither be created nor destroyed., – Total energy possessed by a body remains the, same.(Fig 4), , efficiency , , Depending upon the position of the body or body in motion,, mechanical energy possessed by the body may be, potential energy or kinetic energy respectively., , power output,  100%, power input, , Indicated Horse Power and Brake Horse Power, The power actually generated by the engine or generator is, the indicated horse power which is indicated on the plate., The Brake horse power is the power available to do useful, work. B.H.P is always less than I.H.P. due to losses to, overcome frictional resistance.,  mechanical efficiency , , B.H.P,  100%, I.H.P, , Work done by a force = Magnitude of the force x distance, moved by the body, Power = Total work done / total time taken, , 70, , Workshop Calculation & Science : (NSQF) Exercise 1.5.24, , Copyright free, under CC BY Licence

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Assignment, 1, , m= 55 kg, , 7, , d = 3 metre, , a) s = 1.82 metres, W = ______ joules, , H = 2 metre, , b) s = 1.40 metres, W= ____ joules, , s = 6 metres, , t = 20 minutes, , c) s = 0.85 metres, , P = ______ kW, , W =Joules, , Water filled in the, tank. s is the pumping, , 2, , height, , t = 8 secs, a) P = _____ Watts, , 8, , d = 200 mm, , b) P = _____ Watts, , n = 750 rpm, , c) P = _____ Watts, , F = 700 N, P = _____ kW, , 3, , W = 1312.5 Joules, m = 350 kg, , 9, , P input = 4 kW, P output, , s = _____ metres, , = 3450 Joules/sec, , , 4, , m = 75 kg, , 10, , Volume of water, ‘V’ = 10 metre3, H = 18 metres, t = 20 sec, , P output, = ______ kW, , 11, , d = 225 mm, s = 450 mm, Piston pressure, ‘P’ = 4.5 bar, V = 2.5 metre/sec, (piston speed), , Power input, = _____ kW, , s = 100 metres, t = 12 secs, W = ______ Nm, P = ______ Watts, , 5, , V = 1 m3/min, H=2m, , = 0.75, Power input = ______, kW, , 6, , P = 12 kw, , 12, , s = 4 metres, t = 20 secs, m = ______ kg, , ‘V’ of water pumped, = 3 metre3/min, H = 6 metre,  = 0.8, Power input, = ______ kW, , Workshop Calculation & Science : (NSQF) Exercise 1.5.24, , Copyright free, under CC BY Licence, , 71

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Potential energy, kinetic energy and related problems with assignment, Exercise 1.5.25, Potential energy, , P.E = 10 kg x 9.81 metre/sec2 x 1.4 metre (Fig 1), , Potential energy is the energy possessed by a body by, virtue of its position (Fig 4). A body of mass ‘m’ kept at a, height ‘h’ from a datum possesses a potential energy of, mgh or Wh or Fh; where W or F are the Weight force. When, the body is allowed to fall it will be able to do a useful work, of Fh., , = 137.3 N metre, K.E =, , ( 1N = 1kg.m/sec2), , x 10 kg x 5.24, , 2, , metre, sec, , 2, , 2, , = 137.3 N metre., , Example, •, , Water stored in a Tank, , •, , Coil Spring., , Kinetic energy, It is the energy possessed by a body by virtue of its motion., If a body of mass ‘m’ starting from rest attains a velocity of, ‘v’ after covering a distance of ‘s’, by the action of an applied, force ‘F’, then work done on the body=F x s But F= m x a., Therefore work done on the body = m x a x s., But a x s =, , V2, because the body is starting from rest., 2, , Therefore Work done on the body =, , 1, , 2, , mv ., , 2, Since work done on the body = The energy possessed by, the body, , Kinetic Energy =, , 1, , Examples, •, , A pulley is used to lift a mass with a force of 900 N to, a height of 10 metres in 2 minutes. Find the work done, by the force and also the power.(Fig 2), , 2, , mv ., , 2, Energy possessed by a body = work done on the body, , Potential energy = mgh, 1, mv2, 2, If friction is neglected potential energy = Kinetic energy, , Kinetic energy =, , Example, •, , Rolling vehicle, , •, , Rotating fly wheel, , •, , Flowing water, , •, , Falling weight, , Work done = F x s = 900N x 10 metre, = 9000 Nm = 9000 joules., Power , , W, , Potential energy, Hammer head drops from height ‘h’ . m = 10 kg., , , h = 1.4 m., , u0, , V2 = 2 gs, V2 = 2 x 9.81 x 1.4, V2 = 27.468, V2 = 5.24 m/sec, , 9000 joules, 120sec, , 75 joules, sec, ,  75 watts, , metre, sec, , t, , , , •, , Determine the horse power required to drive a lift in, raising a load of 2000 kgf at a speed of 2 metre/sec, if, the efficiency is 70%., Useful workdone to raise the lift in 1 sec, Force = 2000 kgf, Work = F x d, , 72, , Copyright free, under CC BY Licence

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Power , , F  d 2000  2, , t, 1, = 4000 w, , Power output = 4000 w, Power input = Power output, , , , , 4000,  5714 w, 0.7, , Output, Input, , K.E.developed by a vehicle during acceleration, •, ,  100%, , 5714,  7.659  7.6 HP, 746, Power input = 7.6 HP, HP , , •, , Solution, Mass of motor vehicle= 1200 kg, , A mass of 100 gm is allowed to fall from a height of 10, metres. Determine the amount of Kinetic energy, gained by the body. (Take the value of g as 10 metre/, sec2, Since initial velocity is 0 and distance travelled is 10, metres. final velocity2, =V2 = 2 x g x s = 2 x 10x10 metre2/sec2, ,  K.E , , 1, 2, , 2, , mv , , 1, 2, , 2, , 2, , K.E. of the vehicle at 36 km/hr speed, =, , x 1200 x 362J, , v= 36km/hr, , KE=, , =36 x, , mv2 J, , = 10m/sec, , K.E of the vehicle at 48 km/hr speed, ,  100 gm  200 metres /sec, ,  10000 gm metre /sec, , A motor vehicle of 1200 Kg mass is being accelerated, from 36 km to 48 km/hr speed. Calculate the increase, in K.E during its acceleration., , 2, , =, , x 1200 x 482J, , 2, , (, (, , 1kg.m/sec2 = 1N), 1Nm = 1J), , 7, ,  10  10 ergs, v= 48 km/hr, ,  10 Joules., , =48 x=, , m/sec, , K.E. developed by the vehicle at a constant speed, •, , A motor vehicle of one tonne is travelling at 60 km/hr., Calculate K.E of the vehicle at this speed., K.E of the vehicle =, , 1, mv2, 2, , Where m = one tonne or 1000 kg, v = 60 km/hr, , KE =, , x1200 x 10 x10=60000J, , KE =, , x1200 x, , x, , =106666.67 J, , Increase in K.E of the vehicle = 106666.67 J - 60000J, = 46666.67 J, = 46.666 KJ., Workdone in vehicle operation, , Solution, Changing v into meter/sec we get,, (, , 1km = 1000m), , (, , 1hour = 3600sec), , The Mechanical Work performed by the motor vehicle for its, propulsion on road can generally be classified into two, major categories of work done., – Workdone by the IC engine in developing full power, under all condition of speed and load., – Workdone by the motor vehicle in performing different, operations on road like hill climbing/acceleration/braking/ towing and reversing operation, , Workshop Calculation & Science : (NSQF) Exercise 1.5.25, , Copyright free, under CC BY Licence, , 73

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Heat & Temperature - Concept of heat and temperature, effects of heat,, difference between heat and temperature, boiling point & melting point of, different metals and non-metals, Exercise 1.6.26, Heat, It is a form of energy. Heat energy can be transformed into, other forms of energies. Heat flows from a hotter body to a, colder body. (Fig 1), , Calorific value: The amount of heat released by the, complete combustion of unit quantity of the fuel (Mass or, volume) is known as calorific value of fuels., Water equivalent, It is the mass of water which will absorb the same amount, of heat as the given substance for the same temperature, rise. Water equivalent = Mass of the substance x specific, heat of the substance., Therefore water equivalent = ms, Types of heat, 1 Sensible heat and, 2 Latent heat, 1 Sensible heat, , Units of heat, Calorie: It is the quantity of heat required to raise the, temperature of 1 gram of water through 1°C., BTHU: It is the quantity of heat required to raise 1 lb of water, through 1°F. (British thermal unit)., C.H.U; It is the quantity of heat required to raise 1 lb of, water through 1°C., Joule : S.I. Unit (1 Calorie = 4.186 joule), , Sensible heat is the heat absorbed or given off by a, substance without changing its physical state. It is, sensible and can be obsorbed by the variation of temperature in the thermometers., 2 Latent heat, The heat gained or given by the substance during a change, of state (from solid to liquid to gas) is called latent heat or, hidden heat. The heat absorbed or given off does not cause, any temperature change in the substance., Types, 1. Latent heat of fusion of solid, , Effects of heat, , 2. Latent heat of vaporisation of solid., , •, , Change in temperature, , •, , Change in size, , •, , Change in state, , •, , Change in structure, , The amount of heat required per unit mass of a substance, at melting point to convert it from the solid to the liquid state, is called latent heat of fusion of solid. Its unit is cal/gram., , •, , Change in Physical properties, , Latent heat of fusion of ice, , 1 Latent heat of fusion of solid, , Specific heat, The quantity of heat required to raise the temperature of one, gm of a substance through 10C is called specific heat. It, is denoted by the letter ‘s’., Specific heat of water, Aluminium, Copper, Iron, , =1, = 0.22, = 0.1, = 0.12, , The amount of heat required to convert per unit mass of the, ice into water at 00C temperature is called latent heat of, fusion of ice., Latent heat of fusion of ice(L) = 80 cal/gram, 2 Latent heat of vapourisation of liquid, The amount of heat required to vaporise a unit mass of liquid, at its boiling point is called latent heat of vapourisation., , Thermal capacity:, , Latent heat of vaporisation of water or latent heat of, steam, , It is the amount of heat required to raise the temperature of, a substance through 10C is called the thermal capacity of, the substance., , The amount of heat required to convert into steam of a unit, mass of water at its boiling point (1000C) is called latent, heat of vaporisation of water or latent heat of steam., , Thermal capacity = ms calories., , Latent heat of steam(L) = 540 cal/gram, , 74, , Copyright free, under CC BY Licence

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Temperature, It is the degree of hotness or coldness of a body. The, temperature is measured by thermometers., Difference between heat and temperature, Heat, , Temperature, , 1 It is a form of energy., , This tells the state of heat., , 2 Its unit is calorie., , Its unit is degree., , 3 Heat is measured by calorimeter., , Temperature is measured by thermometer., , 4 By adding quantity of heat of two substances their, total heat can be calculated., , By adding two temperatures we cannot find the, temperature of the mixture., , 5 By heating a substance the quantity of heat is, increased regardless of increase in temperature., , Two substances may read the same temperature though, they might be having different amount of heat in them., , Boiling point, , Melting point, , Any substance starts turning into a gas shows the, temperature at which it boils this is known as the boiling, point. The boiling point of water is 1000C., , The temperature at which any solid melts into liquid or, liquid freezing to solid is called the melting point of, substance. ` The melting point of ice is 00C., , List of melting point and boiling point of metals and Non -metals, Metals and, Non-metals, , Melting, point °C, , Boiling, point °C, , Metals and, Non-metals, , Melting, point °C, , Boiling, point °C, , Aluminium, , 660.25, , 2519, , Manganese, , 1246, , 2061, , Argon, , -189.19, , -185.85, , Mercury, , -38.72, , 357, , Arsenic, , 817, , 614, , Molybdenum, , 2617, , 4639, , Barium, , 729, , 1897, , Nickel, , 1453, , 2913, , Beryllium, , 1287, , 2469, , Nitrogen, , -209.86, , -195.79, , Bromine, , -7.1, , 58.8, , Oxygen, , -226.65, , -182.95, , Cadmium, , 321.18, , 767, , Phosphorus (white), , 44.1, , 280, , Calcium, , 839, , 1484, , Plutonium, , 640, , 3228, , Carbon (diamond), , 3550, , 4827, , Potassium, , 63.35, , 759, , Carbon (graphite), , 3675, , 4027, , Radium, , 700, , 1737, , -100.84, , -34.04, , Silicon, , 1410, , 3265, , Cobalt, , 1495, , 2927, , Silver, , 961, , 2162, , Copper, , 1084.6, , 2562, , Sodium, , 98, , 883, , Gold, , 1064.58, , 2856, , Sulfur, , 115.36, , 444.6, , -, , -268.93, , Tin, , 232.06, , 2602, , -259.98, , -252.87, , Titanium, , 1660, , 3287, , Iodine, , 113.5, , 184.3, , Tungsten (wolfram), , 3422, , 5555, , Iridium, , 2443, , 4428, , Uranium, , 1132, , 4131, , Iron, , 1535, , 2861, , Zinc, , 419.73, , 907, , Lead, , 327.6, , 1749, , Lithium, , 180.7, , 1342, , 650, , 1090, , Chlorine, , Helium, Hydrogen, , Magnesium, , Workshop Calculation & Science : (NSQF) Exercise 1.6.26, , Copyright free, under CC BY Licence, , 75

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Heat & Temperature - Scales of temperature, celsius, fahrenheit, kelvin and, conversion between scales of temperature, Exercise 1.6.27, 2 Convert -400 C into 0F, , Temperature Scales, Temperatures are caliberated between two fixed reference, points namely the freezing point of water, and the boiling, point of water. These two fixed points on different temperature scales are:, Scale, , Freezing point, , Boiling point, , Centigrade (°C), , 0°C, , 100°C, , Fahrenheit (°F), , 32°F, , 212°F, , Kelvin (K), , 273°K, , 373°K, , Reaumur (°R), , 0°R, , 80°R, , Heat is a form of energy. Temperature is the, degree of hotness or coldness of a body. The, relationship for conversion from one temperature scale to the others is, , 0, , F  32, 180, , 0, , , , 100, 0, , 0, , 0, , F  32 , F  32 , , C, , C, , 100, ,  180, ,  40, 100, ,  180, , F - 32 = -72, °F = -72 + 32, = -400F, -400C = -400F, 3 Convert 370 C into K, , 0, , R, , 80, , 0, , , , C, , 100, , 0, , , , K  273, 100, , 0, , , , F  32, , 0, , 180, , C, , 100, , 0, , , , K  273, 100, , °K-273 = C, °K, , = C + 273, , °K, , = 37 + 273, = 310 K, , 37 C = 310K, 0, , 4 Convert 700 C into Reaumer, 0, , 100, , 1 Convert 00 C into 0F, 0, , F  32, 180, , 0, , , , 0, , F  32 , , F  32 , , 0, , C, , R, , 100, 0, , 0, , C, , 100, , 0, 100, , 0, , C, , 0, ,  180, , R, , , , R, , 80, , C, 100, 70, 100, ,  80, ,  80  56, , 700C = 560R,  180, , °F = 0 + 32, = 320F, 00C, , = 320F, , 76, , Copyright free, under CC BY Licence

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Heat &Temperature - Temperature measuring instruments, types of, thermometer, pyrometer and transmission of heat - Conduction, convection, and radiation, Exercise 1.6.28, Measuring heat energy, , •, , Ear thermometers (tympanic), , Energy can be released in chemical reactions as light,, sound or electrical energy. But it is most often released as, heat energy. This allows us to easily measure the amount, of heat energy transferred., , •, , Forehead thermometers (temporal), , •, , Digital thermometers, , •, , Mom's hand or lips, , They are based on the principle that liquids and solids, expand when they are subjected to heat. Mercury and, alcohol expand uniformly. When heat is applied the volume, of the liquid increases and the liquid rises in the capillary, tube integral with the container. Mostly mercury is used in, this type of thermometers because of its properties (Shiny, and will not adhere to the glass tubes and we can measure, up to 3000c., The bimetal thermometer consists of metals with different, coefficient of expansion. The bimetal is twisted into a spiral, which curls when the temperature rises., , The apparatus used to measure the amount of heat by, mixer method is called calorimeter. It is nothing but, cylindrical shaped vessel and a stirrer made out of mostly, copper., In a calorimeter when the hotter solid/liquid substance are, mixed with the cooler solid/liquid substances, heat transfer, takes place until both substances reach the same temperature. By the same time calorimeter also reaches the, same temperature. By mixing rule,, Loss of heat⎤, by solid/, liquid, , ⎡Heat absorb ed by⎤, ⎥ = ⎢ solid / liquid, ⎥ + ⎡Heat absorb ed⎤, ⎥ ⎢, ⎥ ⎢⎣by calorimeter ⎥⎦, ⎥⎦ ⎢⎣ substance, ⎥⎦, , Measurement, Temperature is generally measured in degrees Celsius. In, this system the freezing point of water is defined as 0°C and, the boiling point of water is defined as 100°C. The Kelvin, temperature scale begins from absolute 0. i.e.— 273°. The, temperature intervals are the same., ,  273K = 0°C, 20°C = 273K + 20°C = 293K., Instruments, , Pyrometer, Thermoelectric pyrometer is based on the principle that the, soldering point between the wires of different metals, when, heated a contact voltage is generated. The voltage depends, upon the temperature difference between the hot measuring, point and the cold end of the wire. Thermocouple elements, are constructed of copper and Constant (up to 600°C) or of, platinum and platinum-rhodium (up to 1600°C), Radiation pyrometers are used to measure temperatures, of red hot metals up to 3000°C. These concentrate thermal, rays through an optical lens and focus them on to a thermo, element. The scale of the ammeter is calibrated in degrees, Celsius or Kelvin., , The instruments used to measure and read temperature, takes into account changes in the properties of materials,, electrical phenomena incandescence, radiation and, melting., Thermometer, Types of thermometer, •, , Forehead strips, , •, , Wearable thermometers, , •, , Pacifier thermometers, , 78, , Copyright free, under CC BY Licence

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Radiation, , Transmission of Heat, Heat is a form of energy and is capable of doing work. Heat, flows from a hot body to a colder body or from a point of high, temperature to a point of low temperature. The greater is, the temperature difference the more rapidly will be the heat, flow. Heat is transmitted in three ways., •, , By Conduction, , •, , By Convection, , •, , By Radiation, , Heat is radiated or transmitted from one object to the other, in space without actually being in contact, by means of, electro-magnetic waves. These waves are similar to light, waves and radio waves. They can be refracted by lenses, and reflected by mirrors. This radiation is called infrared. It, requires no medium to carry the radiation. (e.g) The heat of, the sun travels through the space., , Conduction, Conduction is the name given to the transmission of heat, energy by contact. The heat source is in contact with the, Conductor. (metal rod). The rod is in contact with a, thermometer. Due to Conduction heat is transferred from, the heated end to the free end. In general good electrical, conductors are also good heat conductors and good, electrical insulators are also good heat insulators. A good, heat insulator does not necessarily withstand high, temperature., , Transmission of heat takes place in three ways, Conduction, Convection and Radiation., Expansion due to heat, When a solid, liquid or gaseous substance is heated, it, expands and volume is increased. Similarly when it is, cooled, it contracts (shrinks) and volume is decreased., E.g : small gaps are left in between the lines of railway track, to allow for expansion during summer. If this is not done,, the rails would expand and bend there by causing derailment of trains., Except a few substances, all solids, liquids and gases, expand. For the same amount of heat given, the expansion, of liquids is greater than solid and expansion of gas is more, than liquid., , Convection, Convection is the name given to the transmission of heat, energy by the up-ward flow. When heated, the fluid (liquid/, gas) becomes less dense and because of its mobility, is, displaced upwards, by a similar but colder and more dense, fluid. e.g., The domestic hot water system, The cooling, system in motor cars., , Volume of water is reducing while heating from 00C to 40C., After that volume is increasing. The data at 40C of water will, be taken as reference point for any calculations relating, with water., , Workshop Calculation & Science : (NSQF) Exercise 1.6.28, , Copyright free, under CC BY Licence, , 79

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Heat & Temperature - Co-efficient of linear expansion and related problems, with assignments, Exercise 1.6.29, Expansion of solids, A solid substance shows the following types of expansion, when heated., , Co-efficient of cubical expansion, = 3 x linear expansion, , , , 1 Linear expansion, , Examples, , 2 Superficial expansion and, , Find the co-efficient of linear expansion. If an 8 metre, long metal rod is heated from 300C to 800C. So that it, may produce an elongation of 0.84 mm., , 3 Cubical expansion, 1 Linear expansion, , Initial length (l), , = 8m, , When a solid is heated, its length increases. This is called, linear expansion. It depends upon the material, original, length and change in temperature., , Increased length, , = 0.84 mm, , Co-efficient of linear expansion, The co-efficient of linear expansion is the change in length, per unit original length per degree rise in temperature. It is, represented by (Alpha)., , Increased temperature(t) = 80 - 30 = 500C, Increased length, Co - efficient of linear , , expansion( ),  Initial length  Increased temp, , , Length of the solid at t10C = l1, Length of the solid at t20C = l2, Change in Temperature, , = t2 - t1 C, , Change in length, , = l2 - l1, , , , , , l2  l1, l1  t 2  t1 , , , , l2  l1, t 2  t1  t , l1t, , 0.84, 400000, , If iron bridge is 100 metre long at 00C. What will be, the length of bridge if the temperature is 400C and, the co-efficient of linear expansion is 12 x 10-6 per, degree., Initial length of iron bridge = 100 m, , Change in length, , linear expansion⎭, , Original length x change in temperature, , Increased temperature, , = 40 - 0 = 400C, , Increased length, Co - efficient of linear , , expansion( ),  Initial length  Increased temp, , Increased length l2-l1 = l1t, Final length, , 8000  50, , = 2.1 x 10-6 /0C, , Co − efficient of ⎫, , ⎬=, , , , 0, , 0.84, , l2 = l1(1 + t), , 2. Superficial expansion, When a solid is heated, its area increases is called superficial expansion., Co-efficient of superficial expansion, The increase in area per unit original area per degree rise, in temperature is called co-efficient of superficial expansion. It is represented by (Beta)., Co-efficient of superficial, Expansion = 2 x linear expansion, = 2, 3. Cubical expansion, , 12  10, , 6, , , , Increased length, 100  40, , Increased length , , 12, 1000000, ,  100  40, , = 0.048 m, Iron bridge at 400C, , = 100 + 0.048 = 100.048 m, , The length of a metal rod is 100 cm at 300C and 100.14, cm at 1000C. Calculate the co-efficient of linear expansion and the rod length in 00C., , When a solid is heated, its volume increases is called, cubical expansion., , Initial length at 300C, , = 100 cm, , Final length at 1000C, , = 100.14 cm, , Co-efficient of cubical expansion, , Increased length, , = 0.14 cm, , The increase in volume per unit original volume per degree, rise in temperature. It is represented by (Gama), , Increased temperature, , = 100 - 30 = 700C, , 80, , Copyright free, under CC BY Licence

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Increased length, Co - efficient of linear , , expansion( ),  Initial length  Increased temp, , , , , , , 100  70, , , , 14, 100  70  100, 2, 100000, , Length at 0°C, , = l0 (1 + 2 x 10 x 30), = l0 (1 + 0.0006), 100, 1  0.0006, , = 99.94 m, , Co-efficient of linear, expansion (), , = 10 x 10-6/0C, , Increased length, Co - efficient of linear , , expansion( ),  Initial length  Increased temp, , ,  0.015cm, , Find out the temperature that the rod will extend by, 0.54 mm in linear direction when a piece of metal, rod is 2.5 metre long in 200 C and the co-efficient of, linear expansion is 10.4 x 10-6 per degree centigrade., , = 200C, , expansion (), Increased length, Co - efficient of linear , , expansion( ),  Initial length  Increased temp, 10.4  10, , 6, , , , 0.54, 2500  Increased temp, , Increased temperature =, , = 100 cm, = 40 - 25 = 150C, , 10  10, , 1000, , Co-efficient of linear = 10.4 x 10-6, -5, , Increased temperature, , 6, , 15, , Initial temperature, , Find the change in length of metallic rod 100 cm, long, when its temperature is increased from 250C to, 400C and the co-efficient of linear expansion is 10 x, 10-6 / 0C., Initial length, , 1000000, , Increased length = 0.54 mm, , = l0 (1 + t), , l0 , , 10  100  15, , Initial length = 2.5 m = 2500 mm, , To find the length at 00C, , 100, 100, , , , 0.14, , = 2 x 10-5, , l1, , Increased length = 10 x 10-6 x 100 x 15, , , , , , 0.54, 2500  10.4  10, , 6, , 0.54  1000000, 2500  10.4, 5400, 260, ,  20.77 C, 0, , Final temperature = 20 + 20.77, = 40.770C, , Increased length, 100  15, , Assignment, Co-efficient of linear expansion, 1 Calculate the co-efficient of linear expansion of rod. If, rod is found to be 100m long at 200C and 100.14m long, at 1000C., , 4 Find the increase in length 100 cm iron rod if the temperature raise from 400C to 900C. The co-efficient of, linear expansion of the iron is 10x10-6/0C, , 2 Find the change in length if the co-efficient of linear, expansion of rod is 0.00024/oC and the temperature of, a rod of 3.6m length is raised by 1200C,, , 5 If micrometer reading is standardised at 150C. What, will be the true reading of the micrometer if the reading, taken at 350C is 20.20 mm?, , 3 Find the change in length if the co-efficient of linear, expansion of rod is 0.00024/0C. If the temperature of a, rod of 6m length is raised by 1200C,, , The co-efficient of linear expansion of material of micrometer is 11 x 10-6/0C., , Workshop Calculation & Science : (NSQF) Exercise 1.6.29, , Copyright free, under CC BY Licence, , 81

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Heat & Temperature - Problem of heat loss and heat gain with assignments, Exercise 1.6.30, Mixing of heat, , m1 - Mass of first substance, s1 - specific heat of first substance, m2 - mass of 2nd substance, s2 - specific heat of 2nd substance, tm - temperature of mixture, , m - mass, Q - Quantity of heat, , t/t - temperature difference, tm - temperature of the mixture., Unit of amount of heat, The derived unit for the amount of heat in S.I. unit is 1, joule (j)., Specific heat, It is also expressed as the amount of heat required to raise, the temperature of unit mass of a substance through 1°C., In S.I. unit in order to heat a mass of 1 kg of water through, 1°C,, the amount of heat needed or the, mechanical equivalent of heat, , = 4186 joules, = 4.2 kj/kg°C., , Quantity of heat needed for a substance to rise the, temperature, The amount of heat needed for heating 1 kg of the, substance through 1°C is equal to the specific heat of the, substance ‘s’. For heating a mass of ‘m’ kg of the, substance to attain a temperature difference of t,, the quantity of heat needed, Therefore Q, , = m x s x t, = m x s x t., , Mixing, When there is an exchange of temperatures, there is an, exchange in the amount of heat. When hotter bodies, involve with colder substances, heat transference takes, 82, , Copyright free, under CC BY Licence

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place from hotter substances to the colder substances, until the mixture or both the substances acquire the same, temperature., , 300 gram of water at 25°C is mixed with 200 gram of, water at 850C. Find out the final temperature of the, mixture assuming that no heat escapes., , Heat lost by the bodies at higher temperature, , i), , Initial temperature = 250C, , = Heat gained by the bodies at lower temperature and, hence the total amount of heat of the component, substances, = amount of heat in the mixture., , Weight of water = 300 gram, Final temperature = Assume ‘X’, Temperature gained = x - 250C, , ii), , Heat loss by hot substance =, , Mass of water = 200 gram, Initial temperature = 850C, , Heat gained by colder substance, S of the component amounts of heat =, amount of heat in the mixture, , Temperature lost = 850C - x, Heat gained by 300 gram water, , =mst, = 300 x 1 x (x - 25), , m1 x s1 x t1 + m2 x s2 x t2 = (m1s1 + m2s2)tm., , = 300 x -7500 cal., , Example, A bath tub contains 40 litres of water at 15°C and 80 litres, of water at 60°C is poured to it. What is the temperature, of the mixture., , Heat lost by 200 gram water = m s t, = 200 x 1 x (85 - x), = 17000 - 200 x cal., , m1 x s1 x t1 + m2 x s2 x t2 = (m1s1 + m2s2)tm.,  40 kg x, , 4.2 kj, kgC, , x 15C  80 kg x, , , , 4.2kj, , , , kg C, ,   40kg , , tm , , 0, ,  15  80kg , , 22680, , 0, , 120  4.2, , 4.2 kg, kgC, 4.2kj, 0, , kg C, , Heat gained = Heat lost, 300 x -7500, , x 60C, , 300 x + 200 x = 17000 + 7500, 500 x = 24500, , , ,  60  t m, , , , x=, , 0, , Examples, A container contains 25 kg of water. Initial temperature of container and water is 250C. Calculate the, heat required to heat the water to the boiling, temperature of water. Assume water equivalent of, container = 1 kg., = 25 Kg., , 20gm of common salt at 910C immersed in 250 gram, of turpentine oil at 130C. The final temperature is, found to be 160C. If the specific heat of turpentine oil, is 0.428. Calculate the specific heat of common salt., Mass of the salt(m), , = 20 gram, , Initial temperature(t), , = 910C, , Mass of the turpentine(m), Initial temperature(t), , = 250 gram, = 130C, , Initial temperature of water and container = 250C, , Specific heat of turpentine(s) = 0.428, , Final temperature of water and container = 1000C, , Final temperature of mixture = 160C, , Increased temperature (t), , = 100 - 25, , Water equivalent (m s), Required amount of heat to container, , Heat gained by turpentine(Q) = m s t, , = 750C, , = 250 x 0.428 x (16-13), , = 1 Kg., , = 250 x 0.428 x 3, , =mst, , = 321 calories., , = 25 x 1 x 75, Required amount of heat to container, , Heat lost by salt (Q) = m s t, , = 1875 K.cal., , = 20 x s x (91-16), , =mst, , = 20 x s x 75, , = 1 x 75, , = 1500 s calories, , = 75 K.cal., Total required amount of heat, , 24500,  49C, 500, , Final temperature = 490C, , C  45 C, , Mass of the water (m), , = 17000 - 200 x, , = 1875 + 75, , Heat lost = Heat gained, 1500 s = 321, , =1950 K.cal., Workshop Calculation & Science : (NSQF) Exercise 1.6.30, , Copyright free, under CC BY Licence, , 83

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s, , 80x (-1600) + 20x (-400) = 4000 - 100x, , 321, 1500, , Specific heat of salt, , 100x -2000, 100x+100x, , = 0.214, , x, , 6000, 200, , = 30, , Mass of the water in calorimeter = 80 gram, = 20 C, , = 4000 + 2000, , 200x = 6000, , If copper calorimeter contains 80 gram of water at, 200C. The water equivalent of calorimeter is 20 gm., What will be be resultant temperature of the mixture, when 100 gm of water at 400C is added to the, mixture?, Temperature, , = 4000 - 100x, , Final temperature = 300C, , 0, , Final temperature of the mixture = Assume ‘x’, Temperature raised in water = x - 20, Specific heat of calorimeter(ms) = 20 gram, Mass of water added, , = 100 gram, , Temperature, , = 400C, , Temperature lost, , = 40 - x, , Find the amount of heat required to boil 15 gram of, ice at -80C. Latent heat of ice = 336 joule/gm. Latent, heat of steam = 2268 J/gm. Relative specific heat of, ice = 0.5, Heat of ice cube, -80C to 00C Ice Q, , = mct kJ, = m x s x 4.2 x t kJ, = 0.015 x 0.5 x 4.2 x 8 kJ, , Heat gained, , = 0.252 kJ, , Heat gained by water in calorimeter = m s t, , 00C Ice to 00C water, , = 80 x 1 x (x - 20), , = 0.015 x 336 kJ, , = 80 x (-1600), Heat gained by calorimeter, , = 5.04 kJ, , =mst, , 00C water to 1000C water, , = 20 x (x - 20), , = 6.3 kJ, 100 C water to 100 C steam Q= m x hsfg kJ, 0, , Heat lost by added water, , =mst, , 0, , = 0.015 x 2268 kJ, , = 100 x 1 x (40 - x), = 4000 - 100x, Heat gained, , = m c t kJ, = 0.015 x 4.2 x 100 kJ, , = 20 x (-400), Heat lost, , = m x h s f kJ, , = 34.02 kJ, Total amount of heat Q = 0.252 + 5.04 + 6.3 + 34.02 kJ, , = Heat lost, , Answer = 45.612 kJ, , Assignment, Mixing of Heat, 1., , 84, , 2., m = 120 litres, t 1 = 20°C, t 2 = 85°C, s = 4.2, Q = ______ kj, , Workshop Calculation & Science : (NSQF) Exercise 1.6.30, , Copyright free, under CC BY Licence, , m 1 =80 litres of water, m 2 =40 litres of water, t 1 = 10°C, t 2 = 70°C, t m = ______ °C

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Heat & Temperature - Thermal conductivity and insulators, , Exercise 1.6.31, , Insulating materials: Heat will flow from high temperature to low temperature. Heat flow by radiation, conduction and convection method through the wall, door, ceiling, and glass door to the refrigerated space., , Expanded polystyrene(Thermocole): it is available as, a rigid board, beads, moulded into shape for pipe/curved, surface, can be cut easily with a saw, light weight allows, relatively low vapour transmission., , The material which restricts such heat flow is called insulating materials, , Polyurethane: available as a rigid board, flexible board,, liquid can be sprayed on surfaces and allowed to foam,, can be used for in site applications., , Properties of insulating materials, , Wood shaving/Saw dust: It needs good supporting compartment, can easily settle down. Fairly high conductivity, absorbs moisture/water., , -, , It has low conductivity, , -, , Resistance to fire, , -, , Less moisture absorption, , -, , Good rigidity, , Phenotherm: Available slabs with different facings, and, as performed pipe sections, can be easily cut with a saw., , -, , Odourless, , Insulting materials and properties/specifications:, , -, , Vapour permeability, , -, , Light in weight, , There are many insulating materials used in refrigeration, and air conditioning field. For our water tank only few of, them were in use., , Selection of insulating material: The following factors, are the prime importance in the selection of a proper insulating material., -, , Low thermal conductivity: Thermal conductance, value of a material is a measure of its effectiveness to, allow the flow of heat through it by conduction, obviously an insulating material should have a very low, thermal conductivity., , -, , Resistance to fire., , -, , Mechanical strength, , -, , Low moisture absorption capacity, , -, , Easy to lay, , -, , Cost, , -, , Easy of handling, , -, , Low cost, , Now-a-days the following insulating materials were in broad, use., -, , Thermocole, , -, , Glasswool/Tar felt, , -, , Puf, , -, , Fiberglass, , Thermocole: It is one of the insulting materials in normal, use. It is available in low and high density. This is available in various thicknesses ranging 0.25" to 5"., Thermocole is available in various shapes (moulded) of, necessity., Thermocole allows (Characteristically) low transmission, of vapour, thereby heat entry through is cut short. This, may vary with its low/high density., It can be cut very easily even with knife to a required shape., Thermocole withstands cool/heat for a longer time., , Types of insulating materials, Glass wool, PUF, Cork sheet, Thermocole, Insulating foil,, fiber glass., , The 'K' factor of an insulation material follows (thermocole)., , Types of insulating materials: Basic types of insulating, materials are inorganic fibrous or cellular materials., Example, glass wool, slag wool ceramic products,, asbestos, etc. Organic fibrous materials, cork, cotton,, rubber foam, saw dust, rice husk, polystyrene,, polyurethane, phenotherm, etc. The type and form available, as the applications of various insulations as follows., , Fibreglass: Also one of the insulating materials used for, is manufactured from inorganic materials (sand, dolomite,, limestone). Glass fibre insulation does not shrink due to, temperature variation., , Glass wool: Available as semi-rigid, resin bonded slabs/, sheets of different densities -higher density gives strength, and lower conductivity but allows vapour transmission., Available with foil or other coverings., , Fibreglass products does not absorb moisture from the, ambient air., , Cork: Compressed and moulded into a rigid block, light, but strong, can be cut easily with a saw, resists water but, allows relatively high rate of water vapour transmission., , Thermocole -0.20 btu/hr Ft2 deg.f°/inch, , This insulation materials used for higher temperatures also, upto 450°C (842°F)., , Glass wool: Normally glass wool material is heavily thin, weighted object in layers, soft (touching). It comes off in, various sizes (thickness from 0.5" to 2.5". it comes in, white, yellow colours mixed up with broken glass pieces., , 86, , Copyright free, under CC BY Licence

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Handling glass wool is hazardous and harmful (if it is, breathed). Always it is advisable to handle glass wool with, gloves and goggles (eye) while working on it. It also comes, off in various densities., Glass wools are of two types of uses. One type of glass, wool used for low temperature refrigeration/air conditioning purpose. The other type is used for boiler materials, (heat prevention) purposes., The 'K' factor of insulation material:, Glasswool: 0.230-.27 Btu/Hr ft2 deg. F°/inch., Puf: The other mode of insulating materials used in water, cooler at the evaporator tank's external body., For this kind of insulation two chemicals used namely, isocyanide-R11., Both available in liquid form in bottles, (for lesser capacities) and cans (for higher capacities)., Both the liquids (chemicals) should always kept cool. When, both of them added in a container and stirred in few minutes it becomes foamy (initially with thin and becomes, thicker and becomes hard (sticks with the unit)., We should be careful that there is no air gap in the tank, covered. It foams out with high density and uneven finish, at the outer level., Puf (materials) insulations are widely used by our, manufacturer's for their products as it keeps the temperature for a longer period., The main disadvantage of the insulation is as soon as the, chemicals are mixed and stirred it should be poured over, the evaporator coil (or) outside the evaporator tank within, the shortest period. If the time exceeds the solution starts, framing at the container itself and becomes useless., , The evaporator tank should be covered well with wooden/, steel boards with required gaps for insulation tightened all, the corners well giving small gaps to pour the solution., Method of laying duct insulation: when there is no, chance of moisture condensation on the duct, glass wool, can be used. Since it is economical and fire resistant., However if moisture condensation can occur greater care, should be exercised in case of glass wool. First a uniform, coat of bitumen is applied to the duct surface and the, wool is stuck to the bitumen. The insulation is then covered with a polythene sheet which acts as a vapour barrier. The surface can be plastered after spreading chicken, wire mesh as reinforcement., Expanded polystyrene can be laid easily as it is rigid., Bitumen is applied on the duct and the insulation is stuck, joints are also sealed with bitumen. No separate vapour, barrier is needed other than a coat of bitumen. The insulation can be finished with cement and plaster or metal cladding., Purpose of false ceiling: The conditioned air arrives, through the ducts at the supply air diffusers and enters, the conditioned space. Most diffusers are attached to the, false ceiling and a variety of diffusers are available for different air spreading needs. The return air grills will be fixed, to the false ceiling. The false ceiling prevents mixing of, conditioned air and return air., Return air usually flow into the plenum or return air box, through grill placed in the false ceiling. Since substantial, amount of energy goes into the air in the first place. It is, a practice to recycle the air. The air is therefore brought, back to the air conditioning. Plant room it is common to, route the return air through the gap between the false ceiling and the main ceiling. A space referred to as a plenum,, the false ceiling is also known as a return air duct., , Workshop Calculation & Science : (NSQF) Exercise 1.6.31, , Copyright free, under CC BY Licence, , 87

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Concept of pressure - Units of pressure, atmospheric pressure, absolute, pressure, gauge pressure and gauges used for measuring pressure, Exercise 1.6.32, Concept of pressure, , Measuring Instruments, , Continuous physics force exerted on or against an object, by something in contact with it., , I, , Manometers, a Simple manometer, , Definition, , i, , Pressure is an expression of force exerted on a surface, per unit area, i.e., the force applied is perpendicular to the, surface of object per unit area., , ii 'U' tube manometer, , Piezometer, , iii single column manometer, b Differential manometer, i, , As the amount of gas increases assuming the volume of, chamber and the temperature remain constant the pressure, increases., , 'U' tube differantial manometer, , ii Inverted 'U' tube manometer, II Mechanical Gauges, , Unit: Standard unit and also the S.I. unit of pressure is, Pascal (Pa) and Metric unit of pressure is Bar., , a Diaphragm pressure gauge, , 1 Pascal is defined as a force of one newton per square, metre, , c Dead weight pressure gauge, , i.e., 1 Pascal = 1 N/m2, 1 Bar, , b Bourdon's tube pressure gauge, d Bellows pressure gauge, Example, , = 105 N/m2, , A liquid gives force of 100 N over an area of 2m2. What is, the pressure?, , Pressure units in different systems, British unit, FPS, , Pounds per square inch, , Metric units, CGS, , Gram per square centimetre, , g/cm2, , MKS, , Kilogram per square metre, , kg/m2, , International Newtons per square metre, circuits SI, , lb/in2, , N/m2, , Force, , = 100 N, , Area, , = 2 m2, , Pressure = ?, F, = 100, A, 2, 2, = 50 N/m, , P =, , Types of Pressure, , Unit of pressure N/m2, 1 N/m2 = 1 pascal., This unit is too small (Pressure of a fly on a area of 1 cm2)., Hence ‘bar’ is introduced as the unit of pressure., 1 bar = 105 pascal., 105 Pa = 10, , 5, , N, , N, , 2 Atmosphere pressure, , = 1 bar, m, cm2, 1 bar = 1000 mbar. [SI unit of Pressure is Pascal (Pa) and, Metric unit of Pressure is bar], , 3 Gauge pressure, , Properties of Pressure, , 1 Absolute pressure, , 2, , = 10, , 1 The pressure in a liquid increases with increase in, depth., 88, , Copyright free, under CC BY Licence

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2 The pressure at a point increases with the density of, the liquid., , •, , 3 The pressure is same in all directions about a point in, liquid at rest, , Pressure Relationship, •, , Atmospheric pressure : The air surrounding the earth, exerts a pressure on the earth's surface. The pressure, prevailing directly on the earth's surface is known as, atmospheric pressure., , •, , The atmospheric pressure is also referred to as, reference pressure. Normally it considers the sea level, as its reference point., , •, , The atmospheric pressure may be calculated from the, fundamental principle of barometer which states that, the barometer reads the pressure due to the height of, mercury (Hg) in the tube and its weight., , 4 Upward pressure at a point in a liquid is equal to, downward pressure, Pascal's Law, A French scientist, Pascal stated that the pressure applied, at any point in liquid, at rest is transmitted equally in all, directions. This is known as Pascal' law., Applications of Pascal's law, Pascal's law is applied in many devices like the siphon,, hydraulic press, hydraulic lift, brahma press, air, compressor, rotary pump and hydraulic brake. These, hydraulic machines are based on the principle of, transmission of pressure in liquids., , The gaseous layer of air around the earth is known as, atmosphere, ,  Atmospheric pressure =  g h, Where (rho), ,  = Density of Hg = 13600 kg/m3, , g = Acceleration due to gravity = 9.81 m/s2,, , Principle of Hydraulic press, Two cylinders having different cross sectional area are, connected to each other by a horizontal connecting tube., The apparatus is filled with a liquid. The two cylinders are, fitted with air tight piston ., By giving a small input force on a plunger of a small cross, sectional area cylinder a large output force are produced, on the ram of large cross sectional area cylinder. According, to Pascal's law, small input pressure exerted on plunger, is transmitted by the liquid to the ram without any loss., Therefore a small force can be used to lift a much large, force or weight. (Fig 2), , and, h = height of Hg column = 760 mm of Hg at, normal sea level., Substituting the above values in equation, we get, Atmospheric pressure, , = 13600 x 9.81 x 0.76, , = 1,01,396 N/m2, = 1.013 bar, But for easy and simple calculation, we take the, atmospheric pressure as 1 bar., 1 Absolute pressure: absolute pressure is defined as, the pressure which ismeasured with reference in absolute, vacuum pressure., 2 Gauge pressure: It is defined as the pressure which, is measured with the help of a pressure measuring, instrument in which the atmospheric pressure is taken an, datum. The atmospheric pressure on the scale is marked, a zero., 3 Vacuum pressure: It is defined as the pressure below, the atmospheric pressure., Mathematically:, , Force on plunger(F) Weight on the ram(W), =, Plunger area(a), Ram area(A), F W, =, a, A, , Weight on the ram (W) =, , FxA, a, , Properties of Air, , i) Absolute pressure = Atmospheric pressure + Gauge, pressure, Pab = Patm + Pg, ii) Absolute pressure = Atmospheric pressure - Vacuum, pressure, Pab = Patm - Pvacc, iii) Vacuum pressure = Atmospheric pressure – Absolute, pressure, 1 Atmospheric pressure = 76 cm of mercury = 33.91 ft, of water, = 76 x 13.6 gm/cm2, = 76 x 13.6 x 10-3 kg/cm2, = 76 x 13.6 x 10-3 x 9.8 N/cm2, = 10.13 N/cm2, = 1.013 bar, , •, , Actually speaking, air is a mixture of gases. Air is, invisible, colourless, odourless, and tasteless., , •, , Composition: The main constituents of air by volume, are 78% nitrogen, 21% oxygen, and 1% other gases, such as argon and carbon dioxide., Workshop Calculation & Science : (NSQF) Exercise 1.6.32, , Copyright free, under CC BY Licence, , 89

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= 1013 mbar [1 bar = 1000 mbar], , Effects of altitude on atmospheric pressure, , 1 Pascal = 1 N/m, , 2, , 1 bar = 10 Pascal = 10 N/m = 10 N/cm, 5, , 5, , 2, , 2, , 1 bar = 0.986923 atmosphere, , Atmospheric pressure changes according to altitude a, tabulation is shown here with variations., For every 11 meter above sea level drop in air, pressure is 1.3 m bar., , 1 millibar = 0.01 N/cm2 = 10-2 N/cm2, 1 atmospheric Pressure (FPS) = 14.7 Pound/inch2 (psi), 1 atmospheric Pressure (Metric) = 1.0336 Kg/cm2, , For every 1000 ft above sea level drop in air, pressure is 1” Hg (mercury), , 1 atmospheric Pressure (Metric) = 1.014 x 106 dyne/cm2, , S. No., , Place, , Unit of Pressure, , Mercury column, , Inch units, , 1, , Sea level, , 1013 m bar, , 750 mm, , 14.7 psi, , 2, , 520 metres, above sea level, , 951.5 m bar, , 700 mm, , 13.7 psi, , Pressure gauges, They are instruments or devices used to measure the, pressure of liquid steam or gas contained in a vessel., There are also known as mark meters., Types of manometers, •, , Open tube, , •, , Closed tube, , •, , Differential type, , •, , Inverted type, , For a barometer reading with reference to an atmospheric, pressure of 1 bar we have, Pa = 1 + Po (shown in Fig 3), Pa = 1 - Pu (shown in Fig 4), , Mechanical pressure gauges, •, , Bourden's pressure gauges, , •, , Diaphragm pressure gauges, , •, , Dead weight pressure gauges, , Open tube manometer (Fig 3), Example (Fig 5), A manometer is connected to an air pressure tank and it, indicates an over pressure of 615 mm pressure head. The, external air pressure is 1015 mbar. Calculate the absolute, pressure in bar and in Pascal (Fig 5)., , It is more suitable to measure pressure in vessels which, is having little variation to atmospheric pressure. It is a 'u', shaped tube containing mercury having its one end, connected to the vessel container in which the liquid is, there whose pressure is to be determined. The other end, is open. The manometer will show a difference in both the, limbs of the tube when the pressure inside the vessel is, more or less than the air pressure outside., 90, , Workshop Calculation & Science : (NSQF) Exercise 1.6.32, , Copyright free, under CC BY Licence

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Absolute Pressure = External Air Pressure + Over Pressure, , •, , Over-pressure: The gas pressure of the tyre is bigger, than the pressure of the atmosphere. In the tyre we, have over pressure., , •, , Under-pressure : The gas pressure in the cylinder, during the suction process is smaller. There is under, pressure in the cylinder., , •, , Absolute pressure: The absolute pressure = air, pressure + over-pressure. The pressure in vacuum is, 0 bar., , Pa = 1015 mbar + 100 mbar x 615 mm/750 mm, = 1835 mbar, Pa = 1.835 bar = 1.835 x 105 Pascal, If the barometer reading is in mm, it is always, necessary to convert into m bar., Pressure and Vacuum gauges, Bourdon tube pressure gauges (Fig 6), , The manometer indicates the over-pressure. The absolute, pressure on the earth is normally 1 bar. The measured, pressure plus 1 bar is the absolute pressure in normal, conditions., The unit for gas pressure is bar., Pa = absolute pressure, Po = over pressure, Pu = under pressure, Air pressure in technical calculations is assumed to be 1, bar., , In I.C. Engines, Bourdon tube pressure gauges are widely, employed for measurement of pressure, temperature and, vacuum. In these gauges, a Bourdon tube which is a, tempered, one is used and it attempts to straighten out, under pressure and temperature and contact under vacuum., The working is briefly described here refer to figure. A, phosphor bronze tube or elliptical cross section is used, and bent to an arc of a circle. The free end of the tube is, sealed under internal pressure (or temperature), it attempts, to straighten out. During this process, it operates sector,, pinion and needle which indicates pressure or temperature, over a calibrated dial. The entire unit is mounted on a, casing and covered with glass and frame and around it., Vacuum gauges, , A reading of 760 mm Hg is prefect vacuum (zero, absolute pressure), A zero of say 300 mm Hg means to say that 300, mm of vacuum is equivalent to (760-300) 460, milliHg absolute pressure., , •, , •, , Absolute pressure = over-pressure + air pressure, Pa = Po + 1 bar, Over-pressure = absolute pressure - air pressure, Po = Pa - 1 bar, Under-pressure = air pressure - absolute pressure, Pu = 1 bar - Pa, Absolute pressure = air pressure - under-pressure, Pa = 1 bar - Pu, Examples, •, , These are also of Bourdon tube type gauges where the, tube attempts to contract under vacuum thus recording, vacuum of the engine in mm Hg (millimeters of mercury), , •, , Rules and examples, , Vacuum gauges are often used by service mechanics, to find out the mechanical condition of the engine and, whether valves, ignition timing and carburettor setting, are correct and carry out fine adjustments to obtain, the best performance of the engine., Vacuum in Diesel Engine governors: This is, measured by water column methods in fuel injection, test bench, , What pressure is 2 bar over-pressure?, Pa = 2 bar + 1 bar = 3 bar, , •, , What over-pressure is 4 bar?, Po = 4 bar - 1 bar = 3 bar over-pressure, , •, , How many bar under pressure is 0.7 bar?, Pu = 1 bar - 0.7 bar = 0.3 bar under-pressure., , •, , How many bar is 0.3 bar under-pressure?, 0.7 bar., , Properties of gases, 1 Charle’s law, First law or law of volume, At constant pressure the volume (V) of a given mass of, gas is directly proportional to its absolute temperature(T), , Vacuum in manifolds of an engine: This can be, measured by using vacuum gauge, , V T V = K (K - Constant), T, , Workshop Calculation & Science : (NSQF) Exercise 1.6.32, , Copyright free, under CC BY Licence, , 91

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Second law or law of pressure, At constant volume the pressure (P) of a given mass of, gas is directly proportional to its absolute temperature, (T)., P T P = K (K - Constant), T, 2 Boyle’s law or Gas law, At constant temperature the volume (V) of a given mass of, gas is inversely proportional to its pressure. (P), V  1 ; PV = K (K - Constant), P, 3 Perfect gas equation, Since boyle’s law and charle’s law can not be applied, independently due to changes in pressure, volume and, temperature a combined law called “gas equation” has, been formulated. Gas equation is relating to pressure,, volume and temperature of perfect gas which obeys both, the boyle’s law and charle’s law. A gas which obeys boyle’s, and charle’s law is called ideal gas., As per boyle’s law, V , , 1, P, , PV = K (Constant) P1V1 = P2V2 = K, , In the perfect gas law, the P and T represents, absolute pressure and absolute temperature, (in °K) respectively., 1 1 kg of air at 5 kgf/cm2 and 30°C is expanded to, atmospheric pressure and 20°C. What will be the, volume occupied?, Solution: Assuming atmospheric pressure is 1.033 kgf/, cm2 and gas constant, R = 29.27 kg m/kg/°K;, P1 = 5 kgf/cm2 = 5 x 104 kg/m2, T1 = 30°C = 30 + 273 = 303° Kelvin, Mass of air = m = 1 kg, Applying formula:, P1V1 = m.R.T1, V1 =, , m.R.T1, P1, , = 1 x 29.27 x 303, 5 x 10 4, = 0.1774 cubic metre, The following information is ready:, , As per charle’s law, V T, , P1V1 P2 V2 P3 V3, PV, =, =, = .......... .... n n, T1, T2, T3, Tn, , P1 = 5 x 104 kg/metre2, , V, V, V, = K (Constant) 1 = 2 = K, T1, T2, T, , Combining the above two laws,, , P = Pressure (KN/m2), V = Volume (m3), m = Mass (Kg), R = gas constant (Kgf.m/kg/k), T = absolute temperature (K), , PV = RT, , T1 = 303°K, P2 = 1.033 kgf/cm2 = 1.033 x 104 kg/metre2, , P1V1 P2 V2, =, = R [R = gas constant], T1, T2, PV, =R, T, , V1 = 0.1774 cubic metre, , T2 = 20°C = 20 + 273 = 293°K, Let V2 = Volume occupied = To Find, Applying formula:, , P1V1 P2 V2, =, T1, T2, , If mass of the gas is m, then, V2 =, , PV = mRT, Gas constant R = 29.27 kgf.m / kg/ k, = 287 joule/Kg/k, , =, , True gas and its properties, 1 It has mass and volume. So, it has weight., 2 It can be compressed or expanded into a container., , (5 x 10 4 ) x 0.1774 x 293, (1.033 x 10 4 ) x 303, , = 0.8303 cubic metre, Volume occupied = 0.8303 cubic metre, 2 Find the volume of a gas, if its absolute, temperature doubled and the pressure is reduced, to one half., , 3 It is invisible., 4 General Gas Law, Boyle's, Charles', and Gay-Lussac's laws can be combined, to obtain the general gas law is given by,, , PV, = Constant (or), T, 92, , P1V1T2, P2 T1, , Solution: At initial stage:, Let initial pressure, , = P1, , Initial Volume, , = V1, , Initial temperature, , = T1, , Workshop Calculation & Science : (NSQF) Exercise 1.6.32, , Copyright free, under CC BY Licence

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At final stage:, P2 = Final pressure, Since pressure is reduced to one half of initial pressure,, we can say, 1, P2 =, P, 2 1, T2 = Final temperature, , 5 The column of mercury in a barometer is 76 cm. If, instead of mercury the kerosene oil is filled in the, barometer, what would be the state of column, when relative density of kerosene oil is 0.8?, Solution: Suppose the height of column of kerosene in, the mercury tube is h2, then the pressure of column of, kerosene = pressure of column of mercury. Assuming, relative density of mercury = 13.6, , Since temperature is doubled, we can say, , h2d2g = h1d1g, , T2 = 2T1, , h2 =, , V2 = Volume of gas required = To Find, Applying formula, , =, , P1V1 P2 V2, =, T1, T2, V2 =, , =, , P1V1T2, T1P2, , P1V12T1, 1, T1 P1, 2, , h1d1g h1d1, =, d2 g, d2, 76 x 13.6, 0.8, , = 1292 cm, = 12.92 metres, 6 The volume of a gas at 770 mm pressure is 403 cc., Find the pressure when the volume is reduced to, 341 cc., As per Boyles Law, P1V1 = P2 V2, , = 4V1, , 770 x 403 = P1 x 341, , V2 = 4V1, Final volume = 4 times of initial volume, 3 Find the pressure at the depth of 40 metres below, the surface of a lake in dynes per sq. cm., the, atmospheric pressure being neglected., Solution:, Depth of water level  = 40 m = 4000 cm, Density of water d = 1 gram/cc, Acceleration due to gravity g = 980 cm/sec2, Pressure below the surface of lake is given by the formula, = .d.g., , 770 x 403, 341, , Pressure when volume is reduced = 910 mm, 7 At 80 cm pressure the volume of a gas is 800 cu., cm. How much pressure be increased to bring the, volume of gas at 200 cu. cm?, Pressure P1 = 80 cm, Volume V1 = 800 cu. cm, Volume V2 = 200 cu. cm, As per Boyle's law, , Pressure = .d.g, = 4000 x I x 980, = 3920000 dynes per sq. cm., 4 At 5 atmospheric pressure 0.2 cu. metre air is filled, in a container. If the same air is filled at constant, temperature in a 1 cu. metre volume of container, then calculate the air pressure in the container., Solution: At constant temperature the gas follows Boyle's, Law, P1V1 = P2V2, Here, P1 = 5, V1 = 0.2 cu. metre, V2 = 1 cu. metre, P1V1 = P2V2, 5 x 0.2 = P2 x 1.0, 5 x 0.2, P2 =, =1, 1.0, , P1 =, , P1V1 = P2V2, 80 x 800 = P2 x 200, P2 = 80 x 800 = 640 = 320 cm, 200, 2, Pressure to be increased = 320 - 80 = 240 cm, = 2.4 metres, 8 A gas has a pressure of 2 kg/cm2 and volume of, 5m3. What will be the volume of gas if the pressure, is reduced to 1 kg/cm2 keeping the temperature, constant?, Gas pressure P1 = 2 kg/cm2, Volume V1 = 5 m3, Pressure P2 = 1 kg/cm2, , Air pressure in the container = 1 atmospheric., Workshop Calculation & Science : (NSQF) Exercise 1.6.32, , Copyright free, under CC BY Licence, , 93

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P1V1 = P2V2, , 10 An automobile tyre contains 0.14 kg of air at 2 kg/, cm2 gauge pressure at 27°C. What is the volume, of air in cubic cm?, , P1 x V1 2 x 5, =, P2, 1, Volume of gas = 10m3, , In this sum gauge pressure is given. To solve it absolute, pressure is necessary. Atmospheric pressure = 1.033 kg/, cm2, , As per Boyle's law, , V2 =, , 9 A gas at 1.5 kgf/cm2 occupying 0.2 m3 is at 20°C. It, is compressed to a pressure of 5 kgf/cm2 such that, its volume becomes 0.03m3. What will be final, temperature of the gas?, Temperature T1 = 20°C = 20 + 273 = 293° Kelvin, Volume of gas V1, , = 0.2 m, , Pressure of gas P1, , = 1.5 kgf/cm, , = 2 + 1.033, = 3.033 kg/cm2, Pressure P = 3.033 kg/cm2, , 3, , = 3.033 x 104 kg/m2, 2, , = 1.5 x 104 kgf/m2, = 0.03 m3, , Pressure P2, , = 5 kgf/cm2, , = 300° Kelvin, , = 5 x 104 kgf/m2, , AS per perfect gas equation, , P1V1 P2 V2, =, T1, T2, , =, , PV = mRT, , P2 V2, T, x 1, T2, P1V1, , = 5 x 104 x 0.03 x, , Gas constant R = 29.27 kgm kg/ Kelvin, (or) 287 Joule / kg/ Kelvin, , As per perfect gas equation, , T2 =, , Weight of air m = 0.14 kg, Temperature T = 27°C = 27+273, , Volume V2, , T2, , Absolute pressure = Gauge pressure + Atmospheric, pressure, , V =, , 293, 1.5 x10 4 x0.2, , 43.95, 0 .3, , 0.14 x 29.27 x 300, 3.033 x 10 4, , 1229.34, = 0.0405 cu.m, 30330, = 0.0405 x 106, , =, , Volume of air = 40532 cu. cm, , = 146.5 Kelvin, = 146.5 - 273, , Final temperature of gas = -126.5° C, , Assignment A, 1 At 5 atmospheric pressure 0.2 cu. metre air is filled in, a container. If the same air is filled at constant, temperature in a 1 cu. metre volume of container then, calculate the air pressure in the container, 2 The volume of a gas at 770 mm pressure is 403 cc., Find the pressure when the volume is reduced to 341cc., , 4 1 kg of air at 5 kgf /cm2 and 30°C is expanded to, atmospheric pressure and 20°C. What will be the, volume occupied?, 5 5 litre of air at 30°C and 1.1 atmospheric pressure is, compressed to one litre and 10 atmospheric pressure., Calculate the temperature after compression., , 3 A gas is transferred from one container of volume 100, cc. of a pressure of 1.5 kg/cm2 into another container, of capacity 200 cc. Find the pressure in the new, container., , 94, , Workshop Calculation & Science : (NSQF) Exercise 1.6.32, , Copyright free, under CC BY Licence

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Assignment B, 1, , P = 1 bar, , 8, , Po pressure head = 450, mm, Barometer reading =, 1040 m bar, a) Absolute pressure =, ______ m bar, Po = 500 mm, P1 = 1010 m bar, b) Pa = ______ m bar, , 9, , Pu = 540 mm, P1 = 1015 m bar, Pa = ______ m bar, Pu = 615 mm, P1 = 1.02 bar, Pa = ______ m bar, , 10, , Po = 1.5 bar, , (Rho) p = 103 kg/metre3, g = 9.81 metre/sec2, h of water, = ________ metre, 2, , P = 1 bar (Air pressure), (Rho_ ‘p’ = 13.6 x 103, kg/metre3, g = 9.81 metre/sec2, h of mercury, = _______ metre, , 3 Air Pressure = 1 bar, Force on 1 cm2 = ________ N, Force on 1 metre2 = _______ N, 4, , Pressure head ‘h’ =, a) 540 mm, = _______ mbar, b) 510 mm, = _______ mbar, c) 615 mm, = _______ mbar, , 5, , = ______ KN/m2, , 11, , Initial pressure, Po = 0.966 bar, , Absolute pressure ‘Pa’, = 1.75 bar, , Pressure drop = 0.08, bar, , Barometer reading =, , Final pressure = _____, bar, , 1040 m bar, Over pressure ‘Po’ =, , 12 Determine the missing values, bar, 1, -, , ________ pascal, 6, , External atmospheric, pressure P1 = 1010 m, bar, Over pressure ‘Po’ =, 16N/cm2, , 7, , N/Cm2, 5, -, , 13, , mbar, 1.3x103, -, , Pascal, 2x105, , Pa = 2.2 bar, , Absolute pressure, , Pu = _______ bar, , Pa = _______ m bar, , Po = _______ bar, , Barometer reading =, 995 m bar, under pressure Pu =, 320 m bar, , 14, , Absolute pressure ‘pa’, = _____ m bar, , Workshop Calculation & Science : (NSQF) Exercise 1.6.32, , Copyright free, under CC BY Licence, , Pressure head = 120, mm, Pressure = _____ m, bar, , 95

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15, , Pa = ______ m bar, , Bourdon Tube Pressure gauge, Pressure shown by, gauge = 500 kPa, Barometer pressure =, 985 mBar, Absolute Pressure = ________ kPa, , P1 = 1 bar, , 20, , P1 = 1 bar, Po = 220 mm head, , 16, , 19, , Gas equation, , Pu = 180 mm head, , Cylinder volume at BDC, = 480 cm3 of gas, , Pa = ________ m bar, , Gas pressure at BDC =, 96 kPa (Abs), Boyles Law, Cylinder Volume at, BDC - 400 CCS, Cylinder Volume at TDC, - 50 CCS, Compression Ratio 8:1, Absolute Pressure P1, = 101.3 kPa, Pressure at TDC =, __________, Note: Assume Temperature is constant, , Temperature of gas, 700°C, , 17, , 18, , Tyre pressure, Front wheel tyre pressure = 1.8 Bar, Rear wheel tyre pressure = 2.2 Bar, Pressure Increase during driving = 15%, Over pressure, (a) Front tyre ______ Bar, (b) Rear tyre _______ Bar, , Cylinder volume at TDC, = 80 cm3 of gas, Gas pressure at TDC =, 725 kPa (Abs), Temperature of gas at, TDC = ___________, , 21 Calculate the missing values in the table, a Data from Running Engine, Details, A, , B, , Force (Newtons), Dia (mm), Pressure(Bar), , 2380, 78, X, , X, 83, 42, , C, 5030, X, 10, , b Areas, Dia of Piston = 84 mm, Piston Head Area = ________, Total Area for 6 Piston heads = _________, c Piston force, Dia of Piston (D) = 84 mm, Combustion Pressure (P) = 50 Bar over pressure, Piston force FK = _______ Newtons, , C MCQ, 1 What is force per unit area., A Pressure, , B Force, , C Work, , D Power, , 2 How much pascal value for 1 bar., A 10, , B 10, , C 106, , D 107, , 4, , 96, , 5, , 3 What is the name of law if pressure applied at any, point in a liquid at rest is transmitted equally in all, directions., A Boyles, , B Charles, , C Pascal, , D Ohms, , 4 What is the height of mercury for one atmospheric, pressure., A 0.76 cm, , B 0.076 cm, , C 7.6 cm, , D 76 cm, , Workshop Calculation & Science : (NSQF) Exercise 1.6.32, , Copyright free, under CC BY Licence

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5 What is the name of law if at constant pressure the, volume of a given mass of gas is directly proportional, to its absolute temperature, A Charles law, , B Boyles law, , C Gay-lussac’s law, , D Newton’s law, , 8 How much milli bar is equal to 1 bar., A 10, , B 100, , C 1000, , D 105, , 9 What is equal to Gauge pressure + Atmospheric, pressure., , 6 Which degree scale used for absolute temperature is, calculated in perfect gas equation., A degree celsius, , B degree kelvin, , C degree fahrenheit, , D degree reumer, , A Absolute pressure, , B Gauge pressure, , C Vacuum pressure, , D Over pressure, , 10 How much bar equal to one atmospheric pressure., , 7 Which principle of law works in brahma press., A Boyles law, , B Charles lay, , C Pascal law, , D Gay-lussacs law, , A 0.1013, , B 0.01013, , C 10.13, , D 1.013, , Key Answers, 17 Use of boyles law (Temperature constant), , A, , Pressure at T.D.C of the engine = 810.4 kPa(Abs), , 1 1 atmospheric, , 18 Tyre pressure (Charles law), , 2 910 mm, 3 0.75 kg/cm2, , a Over pressure in front wheels = 2.07 bar, , 4 0.8303 m, , b Over pressure in rear wheels = 2.53, , 3, , 5 278ºC, , 19 Bourdon tube, Absolute pressure = 598.5 KPa, , B, , 20 Gas equation, , 1 10.19 metre, , Temperature at the end of compression = 196.5ºC, , 2 0.749 metre, , 21 a Force A = 22.7245 Newton, , 3 10 N, 105 N, 4 a 720, , b 680, , Diameter - C = 80 mm, , c 820, , Pressure - B = 5 bar over pressure, , 5 71000 Pascal, , b Piston head area (AK) = 55.42 cm2, , 6 2610 m bar, , Total piston head area (6 cylinder) = 332.51 cm2, , 7 675 mbar, 8 a 1040 m bar, , b 1676 m bar, , 9 a 295 m bar, , b 200 m bar, , c Piston force = 25/20 N, , 10 150 kN/m2, 11 1.88 bar, 12, , bar, , N/cm, , 1, , C, , MCQ, , 1, , A, , 2, , B, , 3, , C, , 4, , D, , 5, , A, , mbar, , pascal, , 10, , 1000, , 105, , 20, , 200, , 20000, , 20 x 10, , 0.5, , 5, , 500, , 0.5 x 105, , 6, , B, , 1.3, , 13, , 1.3 x 103, , 1.3 x 105, , 7, , C, , 2, , 20, , 2000, , 2 x 10, , 8, , C, , 13 1.2 bar, 0.4 bar, , 9, , A, , 14 160 m bar, , 10 D, , 2, , 5, , 5, , 15 1290 m bar, 16 760 m bar, Workshop Calculation & Science : (NSQF) Exercise 1.6.32, , Copyright free, under CC BY Licence, , 97

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Basic Electricity - Introduction and uses of electricity, molecule, atom, how, electricity is produced, electric current AC,DC their comparison, voltage,, resistance and their units, Exercise 1.7.33, Electricity is a kind of energy. It is the most useful sources, of energy which is not visible but its presense can be felt, by its effects. Electricity is obtained by conversion of, other forms of energy like heat energy, chemical energy,, nuclear energy, mechanical energy and energy stored in, water etc.,, To understand electricity, one must understand the structure of an atom., Basically an atom contains electrons, protons and neutrons. The protons and neutrons are located in the centre, of an atom and the electrons, a negative electric charge, particle revolving around the nucleus in an atom. The proton has a positive charge. Neutrons are neutral and have, no charge., Sources of electricity, Battery, Battery stores electrical energy in the form of chemical, energy and it gives power when required. Battery is used, in automobiles and electronics, etc.,, Generator, It is a machine which converts the mechanical enregy into, electrical energy., When a conductor rotates between a magnetic field using, prime mover an emf will be induced. By using this method, all types of AC and DC generator - generates power., E.g., , Thermal power station, Hydro power station, Nuclear power station, Wind power station, Solar power station, , 2. Magnetic effect, When an electric current passes through a coil, a magnetic field is produced around it., E.g. : Electromagnet Motor, Generator, Electric bell, 3. Chemical effect, When an electric current passes through an electrolyte,, chemical action takes place. Because of that, an electrical energy is stored in a battery as a chemical energy., E.g.: Electroplating, Cells and battery charging, refining, of metals etc.,, 4. Heating effect, When an electric current passes through any conductor,, heat is produced in the conductor due to its resistance., E.g. : Electric heater, Electric iron box, Electric lamp,, Geycer, Soldering iron, Electric kettles, Electric welding, etc.,, 5. X-ray and Laser rays effect, When a high frequency voltage is passed through a vacuum, tube, a special type of rays come out, which is not visible., These rays are called x-rays. Laser rays also can be, produced by electric current., 6. Gas effect, When electrons pass through a certain type of sealed, glass shell containing gas, then it emits light rays., E.g: Mercury vapour lamp, Sodium vapour lamp, Fluorescent lamp, Neon lamp etc.,, Uses of Electricity, 1. Lighting, 2. Heating, 3. Power, 4. Traction, 5. Communication, , -, , Effects of electric current, , 6. Entertainment, 7. Medical, 8. Chemical, 9. Magnetic, 10. Engineering, , -, , When an electric current flows through a medium, its presence can be felt by its effects, which are given below., , Classification, , 1. Physical effect, , •, , Static electricity, , Human body is a good conductor. when the body touches, the bare current carrying conductor, current flows through, the human body to earth and body gets severe shock or, cause even death in many cases., , •, , Dynamic electricity, , Thermo couple, If two dissimilar pieces of metals are twisted together and, its joined end is heated in a flame, then a potential difference or voltage will be induced across the ends of the, wires. Such a device is known as a Thermo couple., Thermo couple is used to measure very high temperature, of furnaces., , Lamps, Heaters, ovens, Motor, fan, Electromotives, lift, crane, Telephone, telegraph, radio,, wireless, Cinema, radio, T.V., x-rays, shock traeatment, Battery charging, electroplating, Temporary magnets, Magnetic chucks, welding,, x-rays of welding, , STATIC ELECTRICITY, If a dry glass rod is rubbed with silk cloth the glass rod gives, out negative electrons, and therefore, becomes positively, charged. The silk cloth receives negative electrons and, , 98, , Copyright free, under CC BY Licence

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therefore it becomes negatively charged. They acquire the, property of attracting small pieces of paper etc. because, like charges repel and unlike charges attract each other., The electric charge on the silk cloth is stationary and is, called static electricity. This type of electricity cannot be, transmitted from one place to another., DYNAMIC ELECTRICITY, The electrons in motion are called current electricity or, electric current. This type of electricity is carried through, wires and cables. Therefore, this electricity can be, transmitted from one place to another. This type of, electricity can be produced by cells, batteries, generators, alternators etc., , Types of electric current, •, , Direct current, , •, , Alternating current, , Direct current, In direct current (DC) the direction and magnitude of the, current does not change (Fig 1). The steady current flow, will be from the positive terminal to the negative terminal., (Fig 2), Examples, DC Sources : Cells, batteries and DC generators (Fig 2), , What is the difference between an atom and an element?, How are molecules different from atoms? I am often asked, these questions in my sessions over and over again and, so I finally decided to write a comprehensive post on them., Find answers to all your questions in this section that is, designed to help students explore and understand the, relationship between atoms, elements, molecules,, compounds and mixtures in a manner that is simple and, easy to understand. so, let’s begin!, What is an Atom?, atoms-imgAll the matter in the universe is made of tiny, particles called atoms. There are 92 different kinds of, atoms in nature. These 92 different atoms combine with, one another to form different kinds of matter that we see in, nature. (Fig 1), Gold, for example, is made of only gold atoms. When, matter is made of only one kind of atom, it is called an, element. In the same way, silver is another element which, is made of only silver atoms. Because there are 92 different, kinds of atoms in nature, there are 92 different kinds of, elements. Other examples of an atom are K (potassium), and Fe (iron)., , Altermating current (Fig 3), The current flow will be from the phase terminal to the, Neutral terminal. In the alternating current (AC) both the, direction and magnitude of the current will be changing at, definite intervals of time. The graph shows how an AC, current or voltage changes with time. The current increases to the maximum value in one direction, falls to zero, and increases to the maximum value in the other (opposite), direction before falling to zero again. Thus a cycle is one, complete series of changes. The normal supply frequency, is 50 cycles per second., , What is a Molecule?, molecule-imgA molecule is the smallest unit of a chemical, compound and it exhibits the same chemical properties, of that specific compound. As molecules are made up of, atoms jointly held by chemical bonds, they can vary greatly, in terms of complexity and size. The oxygen we breathe, has a molecular formula O2. Should we consider this as, an element or compound? When two or more atoms of, the same elements combine together, we call them, Molecules. So, we call O2 as an oxygen molecule. In the, same way, we find hydrogen molecules H2, chlorine, molecules Cl2 and others in nature., , Workshop Calculation & Science : (NSQF) Exercise 1.7.33, , Copyright free, under CC BY Licence, , 99

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Difference between AC and DC, AC, , DC, , 1., , It is generated in the ranges of 6,600 V, 11000 V and, 33,000 V., , It is generated up to 6,600 V only, , 2., , Voltage can be stepped up or stepped down by using, transformer, , It is not possible, , 3., , Transmission cost is less, , Cost High, , 4., , Less maintenance, , High maintenance, , 5., , Power up to 5,00,000 kw can be generated in a single, alternator., , Power up to 10,000 kw can be generated in a single, generator, , 6., , AC generator can run at high speeds. So, speed, control is not easy., , It can’t run at high speeds. Speed control is easy., , 7., , Slip rings and brushes are used to collect the current., , Commutator and brushes are used to collect the, current, , Advantages of A.C., , Electro motive force (EMF), , i, , It is the force which causes to flow the free electrons in any, closed circuit due to difference in electrical pressure or, potential. It is represented by ‘E.’ Its unit is Volt., , In transmission there is saving in copper wire., , ii Since there is no spark in A.C. machine there is no, interference in Radio sound., iii This can be produced to maximum voltage i.e. 33000, volts., , Potential difference (P.D), , iv Voltage can be dropped or raised with the help of, transformers., , This is the difference in electrical potential measured, across two points of the circuit. Potential difference is, always less than EMF. The supply voltage is called, potential difference. It is represented by V., , v Its mechanism is simple and cheap., , Voltage, , vi Output is more due to availability of more than one, phase., , It is the electric potential between two lines or phase and, neutral. Its unit is volt. Voltmeter is used to measure, voltage and it is connected parallel between the supply, terminals., , Disadvantages of A.C.:, i, , A single phase motor is not self-starter., , ii Due to thin wire in A.C., the voltage drop is more., , Volt, , iii It cannot be used for electroplating and in charging, secondary cells., , It is defined as when a current of 1 ampere flows through a, resistance of 1 ohm, it is said to have potential difference, of 1 volt., , iv The speed of motors run by it is difficult to change., , Current, , v There is danger to security due to high voltage., , It is the flow of electrons in any conductor is called current., It is represented by I and its unit is Ampere. Ammeter is, used to measure the current by connecting series with the, circuit., , Electrical terms and units, Quantity of electricity, The strength of the current in any conductor is equal to the, quantity of electrical charge that flows across any section, of it in one second. If ‘Q’ is the charge and ‘t’ is the time, taken, then, , I=, , Q, t, , Q=I x t, , The SI unit of current is coulomb. Coulomb is equivalent to, the charge contained in nearly 6.24 x 1018 electrons., Coulomb, In an electric circuit if one Ampere of current passes in one, second, then it is called one coulomb. It is also called, ampere second (As). Its larger unit is ampere hour (AH), , Ampere, When 6.24 x 1018 electrons flow in one second across any, cross section of any conductor, the current in it is one, ampere.(or) If the potential difference across the two ends, of a conductor is 1 volt and the resistance of conductor is, 1 ohm then the current through is 1 ampere., Resistance, It is the property of a substance to oppose to the flow of, electric current through it, is called resistance. Symbol: R,, Unit : Ohm (), Ohm meter is used to measure the, resistance., , 1 AH = 3600 As (or) 3600 coulomb, 100, , Workshop Calculation & Science : (NSQF) Exercise 1.7.33, , Copyright free, under CC BY Licence

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4 depends on the temperature of the conductor, , Ohm, If the potential difference across the two ends of conductor, is 1 volt and the current through it is 1 ampere, then the, resistance of the conductor is 1 Ohm., Laws of resistance, , R, , α L ; R α, , 1, ;, A, , R α, , L, ; R, A, , =, , ρ, , L, A, , Specific resistance, , The resistance of the conductor, , The specific resistance of a material is the resistance, offered to a current it passed between the opposite faces, of the unit cube of the material. Specific resistance is, measured in Ohm - m or micro ohm - cm., , 1 is directly proportional to the length of the conductor, (R  L), , Each mateiral has its own specific resistance or, resistivity., , 2 Varies inversely proportional to its cross sectional area, , E.g. : Copper - 1.72  cm, Silver - 1.64  cm,, Eureka - 38.5  cm, Iron - 9.8  cm,, , The resistance offered by conductor depends on the, following factors., , 1⎞, ⎛, ⎟, of the conductor ⎜ R α, A⎠, ⎝, , Aluminium - 2.8  cm, Nickel - 7.8 cm., , 3 Depends on the material with which it is made., , R =, , ρl, ohm cm, A, , R = Resistance in ohms, l, , = Length of the conductor in cm, , r, , = Specific Resistance in ohm cm, (symbol pronounced as rho), , A = Area of cross - section in cm2, , Workshop Calculation & Science : (NSQF) Exercise 1.7.33, , Copyright free, under CC BY Licence, , 101

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Basic electricity - Conductor, insulator, types of connections - series and, parallel, Exercise 1.7.34, Conductors, , Aluminium, , Some materials and metals readily allow passage for, electric current to flow. In such materials, called conductors, electrons are able to pass readily from atom to atom., , It is a metal light in weight. It is also ductile, malleable and, a good conductor of electricity. Nowadays, it is more, widely used (since it is cheaper than copper) for wires and, cables. All aluminium conductors (AAC) and aluminium, conductors (steel reinforced) (ACSR) are used in overhead, and transmission lines. (More details on copper and, aluminium are furnished under the topic ‘non-ferrous, metals and alloys as applicable to electrical trades’)., , Properties of conductors, A good conductor should have the following properties., Electrical properties, •, , The conductivity must be good., , •, , Electrical energy spent in the form of heat must be low., , •, , Resistivity must be low (to reduce voltage drop and, loss)., , These are conductors with very high resistance for specific, applications like filaments of incandescent lamps, heating, elements etc. The following are a few examples:, , •, , Increase in resistance with temperature must be low., , 1 Tungsten, , 2 Nichrome, , 3 Eureka, , Mechanical properties, , 4 German silver, , 5 Manganin, , 6 Platinum, , •, , Ductility (the property of being drawn into thin wires)., , 7 Mercury, , 8 Carbon, , 9 sBrass., , •, , Solderability: the joint should have minimum contact, resistance., , The resistance values of the metallic resistances will, increase with increase in temperature., , •, , Resistance to corrosion: should not get rusted when, used outdoors., , insulators, , •, , Should withstand stress and strain., , •, , It should be easy to fabricate., , RESISTANCE WIRES, , These are the materials which offer very high resistance to, the flow of current and make current flow very negligible or, nil. These materials have very high resistance - usually of, many megohms (1 megohm = 106 ohms) per centimetre, cube. The insulators should also posseses high dielectric, strength. This means that the insulating material should, not break down or puncture even on application of a high, voltage (or high electrical pressure) to a given thickness., , Economical factors, •, , Low cost., , •, , Easy availability., , •, , Easy to manufacture., , Classification of conductors, , Properties of insulators, The main requirements of a good insulating material are:, , Conductors, Metallic, , Electrolytic, , Description, , •, , high specific resistance (many megohms/cm cube) to, reduce the leakage currents to a negligible value, , •, , good dielectric strength i.e. high value of breakdown, voltage (expressed in kilovolts per mm), , •, , good mechanical strength, in tension or compression, (It must resist the stresses set up during erection and, under working conditions.), , •, , little deterioration with rise in temperature (The insulating properties should not change much with the rise in, temperature i.e. when electrical machines are loaded.), , •, , non-absorption of moisture, when exposed to damp, atmospheric condition. (The insulating properties,, specially specific resistance and dielectric strength, decrease considerably with the absorption of even a, slight amount of moisture.), , Gaseous, , The best conductors are metallic. The commonly used, conductors in electrical appliances and machines are, described hereunder., Silver, It is a soft and extremely malleable metal. Even though it, is the best conductor, its use is limited because of its high, cost., Copper, It is a very good conductor. It is mealleable and ductile, and, also has high resistance to corrosion by liquids. Therefore,, it is widely used for wires, cables, overhead conductors,, busbars and conducting parts of various electrical appliances., , 102, , Copyright free, under CC BY Licence

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Classification of insulators (Fig 1), , Air is an example of a gaseous insulator. Other examples, are hydrogen, nitrogen and inert gases., , Liquid insulators, Mineral oils, synthetic liquids, resins and varnishes are the, liquid insulators., Transformer oil, In transformers the oil is used as an insulator and also, forcooling of the transformer windings by convection., Therefore, the transformer oil should be dry and purified,, since the presence of moisture will reduce the dielectric, strength of the oil., Purpose of transformer oil, •, •, , Transfer of heat by convection, from winding and core to, the cooling surfaces., It maintains the insulation of winding and also, extinguishes fire that occurs due to faults occurring in, the windings., , Precaution, The insulating value of a transformer oil is reduced due to, the formation of sludge as a result of oxidation due to air and, temperature. To minimise oxidation, the oil should not be, exposed to air., Sludge is also formed due to the presence of acids and, alkalis., Sludge formation, •, , Reduces the rate of heat transfer., , •, , Blocks the ducts., , •, , Increases the operating temperature., , To prevent moisture from entering the oil, the whole, apparatus is made airtight, and calcium chloride, silicagel, fillets are used., , According to ISI specifications, the oil should be able to, withstand 40 kV for one minute with a gap (4 mm ± 0.02, mm) between the electrodes and with the diameter of the, electrodes as 13 mm., Moisture test : In this test, an oil sample is cooled in a, closed vessel down to 15-25°. A dry test tube, 12.5 mm in, diameter and 125 mm long, is taken and an adequate, quantity of oil is poured into it., The tube containing the oil is heated rapidly with the help, of an electric heater till the oil begins to boil. During the, process, oil should not produce cracking., The other tests are:, •, , acidity test, , •, , sludge resistance test., , Electrical insulating varnishes, They are of two types, Oil and resin varnishes., , Solid insulators/insulating materials, Sl. Classification, No., , Examples, , 1, , Mineral insulators, , Mica, marble, slate., , Testing of transformer oil as per ISI Standard (Fig 2), , 2, , Vitreous materials, , Glass, quartz, procelain., , Dielectrical strength test (Refer to Fig 2) : The oil should be, 40 mm above and 40 mm below the electrodes. The gap, between the two electrodes should be kept at 4 mm ± 0.02, mm)., , 3, , Rubber and rubber, products, , Rubber, vulcanised, (India) rubber (V.I.R), ebonite, , 4, , Waxes and compounds Paraffin wax, bitumen., , 5, , Fibrous materials, , Asbestos, paper, wood,, Press pahn, leatheroid,, cotton, silk, tapes etc., , 6, , Synthetic products, , Bakelite, shellac, oil, (for Transformer,, Switchgear etc)., , A high voltage is applied across the electrodes through a, step-up transformer, and increased till there is a spark in, between the electrodes. The voltage noted on the voltmeters,, when the spark occurs, is the breakdown voltage or, dielectric strength of the oil. This is the maximum voltage, the oil can withstand., , Workshop Calculation & Science : (NSQF) Exercise 1.7.34, , Copyright free, under CC BY Licence, , 103

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Paper, Various grades of insulating paper are available for use in, capacitors, cables, etc. Paper, if moist, loses its insulating property. Therefore, it is used in an impregnated, condition., Wood, It is impregnated with oil or other substance for use as an, insulator., For example, in machine windings, bamboo wood is used, as slot wedges., Press board, It is widely used in windings to insulate parts which support, windings. It is also used as spacers in electrical devices, and transformers., , tops of the batteries etc. It is waterproof, but it will crack, under certain conditions. It can be valcanised in the same, manner as rubber., Mica, It is a mineral and available as large slabs. It can be easily, separated into thin sheets. It is fireproof, waterproof, and, is a good insulator. It should be used carefully since it is, liable to crack. It is used in heating elements of electric iron, etc., Marble and slate, Marble and slate are mechanically strong insulators and, are non- hygroscopic. When polished they form good, mountings for switchboards, switches, resistance frames,, etc. Slate is used generally for low voltages., Micanite, , Asbestos, A fibrous, incombustible, fire- proof material - used for panel, boards and as frames for winding resistance wires of, regulators, rheostats etc., Cotton, It is soaked in paraffin to avoid moisture. It is a good, insulator for low voltages. It is used in conductors for, armatures and field coils., Silk, Like cotton, it is used for small jobs like telephone coils., Tapes, , It is made by sticking together pieces of mica with, insulating cement like shellac. It can be bent to any shape, by heating and pressing. Therefore, it is used as insulator, for slots of armatures and to insulate the commutator from, the shaft., Paraffin wax, It melts at 55°C and does not absorb water. It is used to, impregnate paper, wood, pressboard etc to reduce their, moisture absorption., Bakelite, , Tapes of various types are used, such as cotton, silk, jute, etc either pure or in impregnated form., , It can be moulded to any shape. It is heat-resistant and, highly insulating. It will not absorb oil and moisture. It is, used for bodies of switches, plugs, holders, regulators etc., , Empire cloth, , Rubber, , It is made by varnishing a cotton cloth, silk or paper. It is, not effected by moisture. It is available in yellow and black, colours in different sizes. It is used as slots insulation in, winding works and for coil insulation., , It has high insulating properties. It is used mainly on, lighting cables and for flexible cables. It deteriorates, gradually when exposed to atmosphere. Rubber is being, replaced now by elastic plastics such as PVC or polyethylene which can resist alkalis, acids and mineral oils., , Press pahn, Press pahn is a form of paper made from hemp, rags, and, wood pulp by special chemical treatment. It is widely used, for lining armature slots, insulating coil sides, etc., Leatheroid, It is a tough material made from cotton rags with chemical, treatment. It is unaffected by grease or oil and is used for, slot and coil insulation, transformer core coverings, etc., Adhesive tape, It is used widely for taping of ends of conductors, leads and, connections. Adhesive tape is made from cotton fabric, coated with a compound of rubber, bitumen, resin, gum,, etc. It dries when exposed to air. It is available in sizes, , ",, , 3/4", 1" etc. These are also available as P.V.C. adhesive, tape, cotton and bitumen tapes., Bitumen, It is used for filling cable jointing boxes and for sealing the, 104, , Valcanised India Rubber (VIR), This is manufactured by treating pure rubber with sulphur., It is stronger than pure rubber and is not affected much by, change in temperatures. It is used as coverings for low and, medium voltage wires and cables., Ebonite or vulcanite, Ebonite or Vulcanite is vulcanised rubber containing about, 30% to 50% of sulphur, and subjected to a prolonged, heating at 150°C. The material is hard and can be moulded, into different shapes. It is less affected by chemicals and, moisture. It is used for making containers of lead acid, batteries, cases for instruments and switchgears, terminal, plates and low voltage panel boards etc. It should not be, subjected to heat., Shellac, It is a good varnish which is used to improve the insulation, and moisture resisting properties of paper, cloth, wood,, slate etc., , Workshop Calculation & Science : (NSQF) Exercise 1.7.34, , Copyright free, under CC BY Licence

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Enamel, , Class C – maximum temperature above 180°C, , By this, an insulation coating is given on winding wires., , Mica, porcelain and other ceramics, glass, quartz, asbestos, treated glass fibre textile, treated asbestos, built up, mica treated with silicone resins possessing superior, thermal stability (limited stability up to 225°C)., , Polychloroprene (PCP), It is a plastic material used for insulation of cables. It is, resistant to oil and petrol. It can be used in conditions of, exposure to sulphur fumes, steam, ammonia, lactic acid, and direct sunlight., Glass, It is heat-resistant and suitable for high temperatures. It is, used as insulators, envelopes for lamps, radio tubes etc., , Series Connection, The total resistance is equal to the sum of all the resistances. In a series connection the end of the first load is, connected to the beginning of the second load and all loads, are connected end to end., , Quartz, Quartz (Silica) is a good insulator. As it has a very low, temperature coefficient of expansion, it does not crack with, sudden variations in temperature. It is used for pyrometer, sheaths, for heating elements, sparking plugs, etc., Porcelain, Porcelain is not so brittle as glass and is very widely used, for carrying bare conductors, for making fuse carriers and, other electrical fittings., Red fibre, Mainly used in motor and transformer winding work, for slot, insulation, separators etc., , Features of series connection:, •, , The same current flows through all the loads., , Insulators classified according to their temperature, limits, , •, , The voltage across each load is proportional to the, resistance of the load., , The permissible temperature limit at which the insulators, may be worked safely without deterioration, depends upon, the type and class of the insulation as detailed below., (IS:1271/1958), , •, , The sum of the voltages across each load is equal to the, applied voltage., , •, , The Total resistance is equal to the sum of all the, resistances., , l, , = l1 = l2 = ..., , Class Y – maximum temperature 90°C, Cotton, silk, paper products, press board, wood, valcanised, fibre - not impregnated or immersed in oil., , V = V1+ V2+ ..., , Class A – maximum temperature 105°C, , R = R1+ R2+ ..., , Cotton, silk, paper products, wood, valcanised fibre when, impregnated or immersed in liquid dielectric, varnished, paper and wire enamel (class A)., , Example, , Class E – maximum temperature 120°C, , Solution, , Three resistances of 3 ohms, 9 ohms and 5 ohms are, connected in series. Find their resultant resistance., , Wire enamel, cotton fabric and paper laminates treated, with oil, modified asphalt and synthetic resins, varnished, polyethylene, textile treated with suitable varnish., , R, , = R1 + R2 + R3, =3+9+5, , Total resistance = 17 , , Class B – Maximum temperature 130°C, Glass fibre, asbestos, varnished glass fibre, textile, varnished asbestos, built up mica treated with synthetic resin, varnishes., , Parallel connection, , Class F – maximum temperature 155°C, , Features of parallel connection:, , Similar to class B materials but treated with silicone, resins., , •, , The current flowing through each load depends upon, the resistance of the load., , Class H – maximum temperature 180°C, , •, , The voltage across each load is the same and is equal, to the voltage applied to the circuit., , Same as class F materials but treated with silicone resins, of higher thermal stability than class F., , In a parallel connection the beginning and the ends of the, loads are connected together., , Workshop Calculation & Science : (NSQF) Exercise 1.7.34, , Copyright free, under CC BY Licence, , 105

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Therefore R , , 24, ohms = 2.4 ohms, 10, , Example, Two resistors of 2 and 4 ohms are switched in parallel, to a 6V battery, – Calculate the total resistance, – Find the total current and partial current., , •, , The total resistance of a parallel connection is always, smaller than the smallest resistance in the circuit., , Solution, , •, , In parallel connection the reciprocal of the total, resistance is equal to the sum of the reciprocals of all, resistances in the circuit., , Total resistance, , l, , = l1 + l2 + ..., , 1 1 2 +1, + =, 2 4, 4, 3 Ω, =, 4, 4, 1, R tot = = 1 Ω, 3, 3, =, , V = V1 = V2 ..., 1, R, , , , 1, R, , , , 1, , 1, R, ,  .........., , 2, , Example, Two resistances of 4 ohms and 6 ohms are connected in, parallel. Determine the total resistance., 1, R, , , , 1, R, , , , 1, , Therefore, , 1, R, , 1, 1, 1, =, +, R tot R1 R 2, , I Total = I1 + I2 current, But I 1 =, , , , (since parallel connection), , 2, , 1, R, , I2 =, , , , 1, 4, , , , 1, 6, , , , 10, 24, , U, R1, , =, , 6V, = 3A, 2Ω, , U, 6V, =, = 1.5A, R2, 4Ω, , I total = 3A + 1.5A, = 4.5 Amp, , Assignment, 1, , R1 = 12 ohms, R2 = 22 ohms, R3 = 24 ohms in series, R = ______ ohms, , 2, , R1 = 15 ohms, R2 = 25 ohms, V = 220 v, V1 = ______ v, V2 = ______ v, , 106, , 3, , V = 220 v, R1 = 40 ohms, V1 = 100 v, (in series) R2, = ______ ohms, , 4, , V = 80 v, I =2A, R1 = 30 ohms, (in series), R2 = ______ ohms, , Workshop Calculation & Science : (NSQF) Exercise 1.7.34, , Copyright free, under CC BY Licence

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Basic electricity - Ohm’s law, relation between V.I.R & related problems, Exercise 1.7.35, Ohm’s law, V - Voltage in volts, l, , - Current in Ampere, , R - Resistance in ohms., In any closed circuit the basic parametres of electricity, (Voltage, Current and resistance) are in a fixed relationship, to each other., Basic values, To clarify the basic electrical values, they can be compared, to a water tap under pressure, Water pressure, , - electron pressure, , - Voltage, , Relationships, , Amount of water, , - electron flow, , - Current, , throttling of tap, , - obstruction to, , - Resistance, , If the resistance is kept constant and the voltage is, increased, the current is increased, IV, , electron flow, , If voltage is constant and the resistance is increased,, current is decreased, l , Ohm’s law, From the above two relationships we obtain Ohm’s law,, which is conveniently written as V = R.I., Ohm’s law states that at constant temperature, the current passing through a closed circuit is, directly proportional to the potential difference, and inversely proportional to the resistance., By Ohm’s law, EXAMPLE, A bulb takes a current of 0.2 amps at a voltage of 3.6 volts., Determine the resistance of the filament of the bulb to find, R. Given that V = 3.6 V and l = 0.2 A., To find ‘R’. Given that V = 3.6V and I = 0.2 A, Therefore V = l x R, 3.6 V, , = 0.2 A x R, , Therefore, , 108, , Copyright free, under CC BY Licence

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Example, , Resistance connections, , The voltage supply to a filament lamp is 10.8V. The voltage, should be 12V. Find out loss of voltage.(Fig 5), , V - Voltage (in volts), R - Resistance (in ohms), I, , - Current intensity (in Amperes), , Series Connection, The total resistance is equal to the sum of all the resistances. In a series connection the end of the first load is, connected to the beginning of the second load and all loads, are connected end to end., , Voltage drop = 12V – 10.8 = 1.2V, The supply voltage is called Potential difference., Example, The Internal resistance of a dynamo is 0.1 ohm. The, voltage of dynamo is 12V. What is the Voltage of dynamo, when a current of 20 amps being supplied to an outside, circuit., Solution, , Features of series connection:, •, , The same current flows through all the loads., , •, , The voltage across each load is proportional to the, resistance of the load., , •, , The sum of the voltages across each load is equal to the, applied voltage., , The Internal resistance of a Battery is 2 ohms. When a, resistance of 10 ohms is connected to a battery it draws 0.6, amps. What is the EMF of the battery., , •, , The Total resistance is equal to the sum of all the, resistances., , l, , = l1 = l2 = ..., , P.D = Current flowing x Resistance, , V = V1+ V2+ ..., , = 0.6 A x 10, , R = R1+ R2+ ..., , = 6 volts, , Example, , V.D = Current flowing x Internal resistance of battery, = 0.6 x 2 volts, , Three resistances of 3 ohms, 9 ohms and 5 ohms are, connected in series. Find their resultant resistance., , = 1.2 volts, , Solution, , Voltage drop = Current x Internal resistance, = 20 x 0.1 volts, = 2 volts, Example (Fig 6), , R, , EMF of the Battery = (6.00 + 1.2)V, , = R1 + R2 + R3, =3+9+5, , = 7.2 volts, , Total resistance = 17 , Parallel connection, In a parallel connection the beginning and the ends of the, loads are connected together., Features of parallel connection:, •, , The current flowing through each load depends upon, the resistance of the load., , •, , The voltage across each load is the same and is equal, to the voltage applied to the circuit., , Workshop Calculation & Science : (NSQF) Exercise 1.7.35, , Copyright free, under CC BY Licence, , 109

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Example, Two resistors of 2 and 4 ohms are switched in parallel, to a 6V battery, – Calculate the total resistance, – Find the total current and partial current., , Solution, •, , The total resistance of a parallel connection is always, smaller than the smallest resistance in the circuit., , •, , In parallel connection the reciprocal of the total, resistance is equal to the sum of the reciprocals of all, resistances in the circuit., = l1 + l2 + ..., , l, , V = V1 = V2 ..., 1, R, , , , 1, R, , , , 1, , 1, R, ,  .........., , Total resistance, 1, 1, 1, =, +, R tot R1 R 2, 1 1 2 +1, + =, 2 4, 4, 3 Ω, =, 4, 4, 1, R tot = = 1 Ω, 3, 3, =, , I Total = I1 + I2 current, , 2, , Example, , But I 1 =, , Two resistances of 4 ohms and 6 ohms are connected in, parallel. Determine the total resistance., 1, R, , , , 1, R, , , , 1, , Therefore, , 1, R, , , , (since parallel connection), , I2 =, , R, , =, , 6V, = 3A, 2Ω, , U, 6V, =, = 1.5A, R2, 4Ω, , I total = 3A + 1.5A, , 2, , 1, , U, R1, , = 4.5 Amp, , , , Therefore R , , 1, 4, , , , 1, 6, , , , Assume the given resistors in the assignment, as bulb with filaments and other current consuming devices like Horn, Wiper etc of the, vehicle., , 10, 24, , 24, ohms = 2.4 ohms, 10, , Assignment, 1, , 110, , V, , = 6 Volts, , = 6.5 Amps, , I, , = 0.5 Amps, , = ______ Volts, , R, , = ______ Ohms, , R, , = 40 Ohms, , I, V, , 2, , Workshop Calculation & Science : (NSQF) Exercise 1.7.35, , Copyright free, under CC BY Licence

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3, , 4, , 5, , R, , = 250 Ohms, , = 820 Ohms, , I, , = 0.44 Amps, , I, , = ______ Amps, , V, , = ______ Volts, , I, , = 4.5 Amps, , V, , V, , = 220 Volts, , R, , 7, , 8, , I, , = 11.5 Amps, , = 220 Volts, , V, , = 380 Volts, , R, , = ______ Ohms, , R, , =______ Ohms., , R, , = 50 Ohms, , V, , = 220 Volts, , I, , = _____ Amps, , 9, , R, , = 22 Ohms, = 7.8 Amps, , (Voltage drop), V, 6, , V, , = 110 Volts, , I, , = 4.55 Amps, , R, , =_____ Ohms, , Workshop Calculation & Science : (NSQF) Exercise 1.7.35, , Copyright free, under CC BY Licence, , = ______ Volt, , 111

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Basic electricity - Electrical power, energy and their units, calculation with, assignments, Exercise 1.7.36, Electric Power, In mechanical terms we defined power as the rate of doing, work. The unit of power is Watt. In an electrical circuit also, the unit of electrical power is 1 Watt. In mechanical terms, 1 Watt is the work done by a force of 1 N to move the body, through 1 metre in one second. In an electrical circuit, the, electromotive force overcomes the resistance and does, work. The rate of doing work depends upon the current, flowing in the circuit in amperes. When an e.m.f of one volt, causes a current of 1 ampere to flow the power is 1 Watt., Hence Power = Voltage x Current, P= V x l, Power in Watts = Voltage in Volts x Current in Amperes, , V - Voltage (Volts) V, i, , - Current Intensity (Ampheres) A, , P - Power (Watts, Kilowatts) W, kW, W - Work, Energy (Watt hour, Kilowatt hour) wh, Kwh, t, , - time (hours) h, , Electric work, energy, Electrical work or energy is the product of electrical power, and time, Work in Watt seconds = Power in Watts x time in sec, seconds, W=Pxt, Since 1 joule represents 1 Watt x 1 sec, which is very, small, larger units such as 1 Watt hour and 1 kilowatt hour, are used., 1 W.h, = 3600 Watt sec., 1 Kwh, = 1000 Wh = 3600000 Watt sec, Note: The charge for electric consumption is, the energy cost per Kwh and it varies according, to the country and states., Table of analogies between mechanical and electrical quantities, Mechanical quantity, , Unit, , Electrical quantity, , Unit, , Force 'F’, , N, , Voltage ‘V’, , V, , m/s, , Current I, , A, , Time t, , seconds, , Time t, , seconds, , Power P = F x v, , N, , Power P = V x i, , W=VxA, , Energy = F x v x t, , j = Nm, , Energy W = V x i x t, , j=Wxs, , Displaceme nt, , Velocity v =, , W, , Time, , =, =, , VI, I2 R, , m, sec, , V, , 2, , V, R, , =, , R, , =, , IR, , =, , W, I, , =, , WR, , =, , V, R, W, V, , =, , V, I, , =, , V2, W, , =, , W, , =, , =, , I, , 2, , I, , 112, , Copyright free, under CC BY Licence, , W, R

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Example, 1 Calculate the power rating of the lamp in the, circuit, if 0.25 amperes of current flows and the, voltage is 240 volts., , V, R, , Current (I), , =, , Power (w), , 220, 2 amperes, 110, = VxI, =, , P=VxI, V = 240 Volts, , = 220 x 2, , I, , = 440 watts, , = 0.25 Amperes, , Therefore Power = 240 Volts x 0.25 Amperes, = 60 Volts Ampere, But 1 Watt = 1 Volt x 1 Amphere, Therefore Power = 60 Watts, , 6 Find the total power if four 1000 W, 180 volt, heaters are connected in series across 240 V, supply and current carrying capacity is 15 amp., Find the total power., Connection, , =, , Series, , No. of heaters, , =, , 4, , Heater power (W), , =, , 1000 watts, , Given that R = 10 and I = 15A, , Heater voltage, , =, , 180 V, , Power = V x I = I x R x I = I2 x R, , Supply voltage =, , 240 V, , 2 A current of 15 amperes flow through a resistance, of 10 Ohms. Calculate the power in kilowatts, consumed., , Therefore Power = 152 x 10 = 2250 Watts = 2.25 kW, 3 At a line voltage of 200 Volts a bulb consumes a, current of 0.91 ampheres. If the bulb is on for 12, hour calculate the work in Wh to find the work, given that V = 200 Volts., I = 0.91 Amps., t, , = 12 hours, , Heater resistance (R) =, , V2, W, , =, , 180 x 180 324, =, 1000, 10, 32.4 ohms, , Total resistance, , =, , 32.4 x 4 = 129.6 ohms, , Total current (I), , =, , =, , V, R, 240, = 1.85 amperes, 129.6, VxI, , =, , 240 x 1.85 = 444 watts, , =, , =, Total Power (W), , 7 If a 40 watt fluorescent lamp draws a current of, 0.10 ampere. How much voltage will be required, to illuminate it?, Therefore Power=V x I = 200 Volts x 0.91 Amps, = 182 Watts, Therefore Work = P x t, , = 182 Watts x 12 hours, , Lamp power (W), , =, , 40 watt, , Current (I), , =, , 0.10 ampere, , Voltage (V), , =, , W, I, , =, , 40, = 400 volts, 0.1, , = 2184 Watt hour., 4 An adjustable resistor bears the following label:, 1.5 k Ohms/0.08 A. What is its rated power?, Given: R = 1.5 k Ohms; I = 0.08 A, , 8 Find the cost of running 15 HP motor for 15 days @, 6 hrs per day and the cost of energy is Rs. 3 per unit., , Find: P, , Motor power (w), , V = R x I = 1500 Ohms.0.08 A = 120 volts, P = V x I = 120 volts.0.08 A = 9.6 W alternatively:, P = 12.R = (0.08 A)2.1500 Ohms = 9.6 W., 5 Find the current and power consumed by an,  resistance when feed, electric iron having 110, from a 220 v supply, Resistance of electric iron (R) = 110 ohms, Voltage (V), , = 220 volts, , = 15 HP, = 15 x 746 = 11,190 watts, , Consumption per day, , = 11,190 x 6, = 67140 =, , 67.14 KWH, , Consumption for 15 days = 67.14 x 15, = 1007 KWH (or) unit, Cost per unit, Cost for total energy, , = Rs. 3, = 3 x 1007, , Workshop Calculation & Science : (NSQF) Exercise 1.7.36, , Copyright free, under CC BY Licence, , = Rs. 3021, 113

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Assignment, 1, , Current Consumed, , 9, , I = 0.136 A, , t = 1 hour, , Voltage ‘V’ = 220 V, , Energy consumption, = ______ kWh, , P = _____ Watts, 10, 2, , 3, , ‘W’, , I = 2.27 A, , Power ‘P’ = 100 W, , V = ______ v, , t = ______ hr, , P = 750 W, , 11, , I = ______ A, , P = 60 W, , = 1.5 kWh, , t, , = 45 min., , P, , = ______ kW., , 13, , Power consumed, ‘P’ = 6.2 kW, t = 8 hours, Charge per kwh, = 1.25 Rupees, Total cost, = ______ Rupees, , 14, , I = 5.45 A, , R = 484 ohms, P = ______ Watts, , 6, , W, , Energy metre reading, W1 = 6755.3 kWh, Increases to W2, = 6759.8 kWh, t = 45 min., P = ______ kW., , R = ______ W, , I = 0.455 A, , = 1 kWh, , 12, , V = 200 v, , 5, , Energy consumed, , P = 500 Watts, , V = 220 v, , 4, , P = 100 W, , P = 550 W, R = 22 ohms, I = ______ A, , V = 220 v, 7, , 8, , 114, , P consumed = 1.8 kW, R = 8 ohms, V = ______ v, I consumed = ____ A, P = 2 kW, V1 = 220V (Heating, element voltage), R = ______ W, , Workshop Calculation & Science : (NSQF) Exercise 1.7.36, , Copyright free, under CC BY Licence, , Energy consumed, = 1 kWh, t = ______ hr.

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Basic electricity - Magnetic induction, self and mutual inductance and EMF, generation, Exercise 1.7.37, Magnetic induction, When a magnet is brought near to an iron bar is brought, near to a magnet, a magnetism is produced in the iron bar., The phenomenon is known as magnetic induction. Actually, before attracting an iron bar towards it, a magnet, induces an opposite polarity in the iron bar and then due to, attraction between unlike poles, magnet attracts the iron, bar. The magnet need not to touch the iron bar for magnetic, induction., , Self-induction: The production of an electromotive force in, a circuit, when the magnetic flux linked with the circuit, changes as a result of the change in a current inducing in, the same circuit., At any instant, the direction of the magnetic field is, determined by the direction of the current flow., , In various electrical measuring instruments, soft iron pole, pieces are used along with bar magnets in order to given the, desired shape to the magnet used, such pole piece work, on the principle of magnetic induction., , With one complete cycle, the magnetic field around the, conductor builds up and then collapses. It then builds up, in the opposite direction, and collapses again. When the, magnetic field begins building up from zero, the lines of, force or flux lines expand from the centre of the conductor, outward. As they expand outward, they can be thought of, cutting through the conductor., , Intensity of magnetic field, , Self induction, , The force acting on a unit pole placed in a magnetic field, (attractive or repulsive force) is called the intensity of, magnetic field. It is denoted by letter H and its unit is, Wb/m., , According to Faraday’s Laws, an emf is induced in the, conductor. Similary, when the magnetic field collapses,, the flux lines cut through the conductor again, and an emf, is induced once again. This is called self-induction (Fig 1)., , Principles and laws of electromagnetic induction, Faraday’s laws of electromagnetic induction are also, applicable for conductors carrying alternating current., Faraday’s laws of elctromagnetic induction, Faraday’s first law states that whenever the magnetic, flux is linked with a circuit changes, an emf is always, induced in it., The second law states that the magnitude of the induced, emf is equal to the rate of change of flux linkage., Dynamically induced EMF, Accordingly induced emf can be produced either by moving, the conductor in a stationery magnetic field or by changing, magnetic flux over a stationery conductor. When conductor, moves and produces emf, the emf is called as dynamically, induced emf Example: Generators., Statically induced EMF, When changing flux produces emf the emf is called as, statically induced emf as explained below., Example:Transformer., Statically induced emf: When the induced emf is produced in a stationery conductor due to changing magnetic, field, obeying Faraday’s laws of electro magnetism, the, induced emf is called as statically induced emf., There are two types of statically induced emf as stated, below:, 1 Self induced emf produced with in the same coil., 2 Mutually induced emf produced in the neighbouring, coil., , Mutual induction, When two or more coils one magnetically linked together, by a common magnetic flux, they are said have the, property of mutual inductance. It is the basic operating, principal of the transformer, motor generators and any, other electrical component that interacts with another, magnetic field. It can define mutual induction on the current, flowing in one coil that induces as voltage in an adjacent, coil., In the Fig 2 current flowing in coil L1 sets up a magnetic field, around it self with some of its magnetic field line passing, through coil L2 giving in mutual inductance coil one L on, has a current of I, and N, turns while coil two L2, has N2, turns therfore mutual inductance M, of coil two that exists, with respect to coil one L, depend on their position with, inspect to each other., The mutual inductance M that exists between the two coils, can be greately measured by positioning them on a, common soft iron cone or by measuring the number of turns, of either coil on wound be found in a transformer., , 115, , Copyright free, under CC BY Licence

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permanently closed or opened, the flux produced by coil 1, becomes static or zero respectively and no emf will be, induced in coil 2. EMF will be induced only when there is, a change in flux which happens during the closing or, opening of the circuit of coil 1 by the switch in a DC circuit., Alternatively the battery and switch could be removed and, coil 1can be connected to an AC supply as shown in Fig4., Then an emf will be induced in coil 2 continuously as long, as coil 1 is connected to an AC source which produces, alternating magnetic flux in coil 1 and links with coil 2. THis, principle is used in transformers., The two coils are tightly wound one on top of the other over, a common soft iron core unit said to exist between them as, any losses due to the leakage of flux will be extremely, small. Then assuring a perfect flux leakage between the, two coils the mutual inductance M., Dynamically induced EMF, Generator: An electrical generator is a machine which, converts mechanical energy into electrical energy., Principle of the Generator: To facilitate this energy, conversion, the generator works on the principle of Faraday’s, Laws of Electromagnetic induction., Faraday’s laws of electromagnetic induction: There, are two laws, The first law states:, •, , whenever the flux linking to a conductor or circuit, changes, an emf will be induced., , The second law states:, •, , the magnitude of such induced emf (e) depends upon, the rate of change of the flux linkage., , Types of emf: According to Faraday’s Laws, an emf can, be induced, either by the relative movement of the conductor and the magnetic field or by the change of flux linking, on a stationary conductor., Dynamically induced emf: In case, the induced emf is, due to the movement of the conductor in a stationary, magnetic field as shown in Fig 3a or by the movement of the, magnetic field on a stationary conductor as shown in Fig, 3b, the induced emf is called dynamically induced emf., As shown in Fig 3a & 3b, the conductor cuts the lines of, force in both cases to induce an emf, and the presence of, the emf could be found by the deflection of the needle of the, galvanometer ‘G’. This principle is used in DC and AC, generators to produce electricity., Statically induced emf: In case, the induced emf is due, to change of flux linkage over a stationary conductor as, shown in Fig 2, the emf thus induced is termed as statically, induced emf. The coils 1 and 2 shown in Fig 2 are not, touching each other, and there is no electrical connection, between them., According to Fig 4, when the battery (DC) supply is used, in coil 1, an emf will be induced in coil 2 only at the time of, closing or opening of the switch S. If the switch is, 116, , Production of dynamically induced emf: Whenever a, conductor cuts the magnetic flux, a dynamically induced, emf is produced in it. This emf causes a current to flow if, the circuit of the conductor is closed., For producting dynamically induced emf, the requirements, are:, •, , magnetic field, , •, , conductor, , •, , relative motion between the conductor and the magnetic field., , If the conductor moves with a relative velocity ‘v’ with, respect to the field, then the induced emf ‘e’ will be, = BLV Sin Volts, , Workshop Calculation & Science : (NSQF) Exercise 1.7.37, , Copyright free, under CC BY Licence

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where, B = magnetic flux density, measured in tesla, L = effective length of the conductor in the field in metres, V = relative velocity between field and conductor in, metre/second.,  = the angle at which the conductor cuts the magnetic, field., , pattern of induced emf in a conductor when it rotates under, N and S poles of uniform magnetic field., The emf induced by this process is basically alternating in, nature, and this alternating current is converted into direct, current in a DC generator by the commutator., Fleming’s right hand rule: The direction of dynamically, induced emf can be identified by this rule. Hold the thumb,, forefinger and middle finger of the righ hand at right angles, to each other as shown in Fig 4 such that the forefinger is, in the direction of flux and the thumb is in the direction of, the motion of the conductor, then the middle finger indicates the direction of emf induced, i.e. towards the observer or away from the observer., , Likewise for every position of the remaining conductors in, the periphery, the emf induced could be calculated. If these, values are plotted on a graph, it will represent the sine wave, , Workshop Calculation & Science : (NSQF) Exercise 1.7.37, , Copyright free, under CC BY Licence, , 117

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Basic electricity - Electrical power, HP, energy and units of electrical energy, Exercise 1.7.38, Electric Power, , Therefore Power=V x I = 200 Volts x 0.91 Amps, , In mechanical terms we defined power as the rate of doing, work. The unit of power is Watt. In an electrical circuit also, the unit of electrical power is 1 Watt. In mechanical terms, 1 Watt is the work done by a force of 1 N to move the body, through 1 metre in one second. In an electrical circuit, the, electromotive force overcomes the resistance and does, work. The rate of doing work depends upon the current, flowing in the circuit in ampheres. When an e.m.f of one volt, causes a current of 1 amphere to flow the power is 1 Watt., Hence Power = Voltage x Current, P= V x l, Power in Watts = Voltage in Volts x Current in Ampheres, Electric work, energy, Electrical work or energy is the product of electrical power, and time, Work in Watt seconds, onds, , = Power in Watts x time in secW=Pxt, , Since 1 joule represents 1 Watt x 1 sec, which is very, small, larger units such as 1 Watt hour and 1 kilowatt hour, are used., 1 W.h, 1 Kwh, , = 182 Watts, Therefore Work = P x t, , = 182 Watts x 12 hours, , = 2184 Watt hour., 4 What is its rated power if an adjustable resistor, bears the following label: 1.5 k Ohms/0.08 A?, Given: R = 1.5 k Ohms; I = 0.08 A, Find: P, V = R.I = 1500 Ohms.0.08 A = 120 volts, P = V.I = 120 volts.0.08 A = 9.6 W alternatively:, P = I2.R = (0.08 A)2.1500 Ohms = 9.6 W., 5 Find the current and power consumed by an, electric iron having 110 resistance when feed, from a 220 v supply, Resistance of electric, iron (R), , = 110 ohm, , Voltage (V), , = 220 volt, , Current (I), , =, , Power (w), , 220, 2 ampere, 110, = VxI, =, , = 3600 Watt sec., = 1000 Wh = 3600000 Watt sec, , V, R, , EXAMPLE, , = 220 x 2, , 1 Calculate the power rating of the lamp in the, circuit, if 0.25 ampheres of current flows and the, voltage is 240 volts., P=VxI, V = 240 Volts, I, = 0.25 Amperes, Therefore Power = 240 Volts x 0.25 Amperes, = 60 Volts Ampere, But 1 Watt = 1 Volt x 1 Amphere, Therefore Power = 60 Watts, , = 440 watt, , 2 Calculate the power in kilowatts consumed. if a, current of 15 amperes flow through a resistance of, 10 Ohms., Given that R = 10 and I = 15A, , 6 Find the total power if four 1000 W, 180 volt heaters, are connected in series across 240 V supply and, current carrying capacity is 15 amp., Connection, No. of heaters, Heater power (W), Heater voltage, Supply voltage, , Heater resistance (R) =, , 3 calculate the work in Wh to find the work given, that V = 200 Volts if a line voltage of 200 Volts a bulb, consumes a current of 0.91 ampheres. If the bulb, is on for 12 hour, I = 0.91 Amps., t, , = 12 hours, , Series, 4, 1000 watt, 180 V, 240 V, , V2, W, , =, , 180 x 180 324, =, 1000, 10, , Total resistance, , =, =, , 32.4 ohm, 32.4 x 4 = 129.6 ohm, , Total current (I), , =, , Power = V x I = I x R x I = I2 x R, Therefore Power = 152 x 10 = 2250 Watts = 2.25 kW, , =, =, =, =, =, , Total Power (W), , V, R, , =, , 240, = 1.85 ampere, 129.6, , =, =, , VxI, 240 x 1.85 = 444 watt, , 118, , Copyright free, under CC BY Licence

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7 How much voltage will be required to illuminate, if a 40 watt florescent lamp draws a current of 0.10, ampere?, Lamp power (W), = 40 watt, Current (I), = 0.10 ampere, Voltage (V), , =, , W, I, , =, , 40, = 400 volt, 0.1, , 8 Find the cost if running 15 HP motor for 15 days @, 6 hrs per day. If the cost of energy is Rs. 3 per unit., Motor power (w), = 15 HP, = 15 x 746 = 11,190 watt, Consumsion per day, = 11,190 x 6, = 67140 =, 67.14 KWH, Consumsion for 15 days = 67.14 x 15, = 1007 KWH (or) unit, Cost per unit, = Rs. 3, Cost for total energy, = 3 x 1007 = Rs. 3021, 9 What is the percentage reduction in power, consumption and How much power is consumed, by series resistance if the rating of an electric iron, is 220 V and 500 watts. The equipment appears, abnormally hot. To reduce this a 10 W resistance, is connected in series?, , Electric iron power (W), Voltage (V), Resistance (R), , = 500 watt, = 220 volt, =, , V2, W, , =, , 220 x 220 484, =, 500, 5, , = 96.8 ohm, Circuit total resistance (R) = 96.8 + 10 = 106.8 ohm, Current (I), , Consumed power (W), , V, R, 220, = 106.8, , =, , = 2.06 ampere, , = I2 R, = 2.06 x 2.06 x 106.8, = 453 watt, , Reduction in power, consumsion = 500 - 453 = 47 watt, Percentage, Power consumed by, series resistance, , =, , 47, x 100 = 9.4 %, 500, , = I2 R, = 2.06 x 2.06 x 10, = 42.44 watt, , Assignment, 1, , Current Consumed, I = 0.136 A, Voltage ‘V’ = 220 v, P = _____ Watts, , 4, , P = 60 W, V = 200 v, R = ______ W, , 2, , P = 500 Watts, I = 2.27 A, V = ______ v, , 5, , I = 0.455 A, R = 484 ohms, P = ______ Watts, , 3, , P = 750 W, V = 220 v, I = ______ A, , 6, , P = 550 W, R = 22 ohms, I = ______ A, , Workshop Calculation & Science : (NSQF) Exercise 1.7.38, , Copyright free, under CC BY Licence, , 119

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7, , P consumed = 1.8 kW, R = 8 ohms, V = ______ v, , 9, , P = 100 W, t = 1 hour, Energy consumption, = ______ kWh, , 10, , Energy consumed, ‘W’, = 1 kWh, Power ‘P’ = 100 W, t = ______ hr, , 8, I consumed = ____ A, P = 2 kW, V1 = 220 v-Heating, element voltage., R = ______ W, , 120, , Workshop Calculation & Science : (NSQF) Exercise 1.7.38, , Copyright free, under CC BY Licence

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Mensuration - Area and perimeter of square, rectangle and parallelogram, Exercise 1.8.39, In Engineering field, an Engineer has to estimate the, material, manpower, machinery, etc. required to prepare, the geometrical objects. Hence we must be very, conversant with all relevant formulae connected with, geometrical objects., Length, , - l unit, - d unit, , Diameter, , - d unit, , - r unit, , Semiperimeter - S unit, Perimeter, , - P unit, , Circumference - C unit, Area, , Breadth or width - b unit, Diagonal, , Radius, , - A unit2, , Total surface area - T.S.A unit2, Lateral surface area - L.S.A unit2, Volume, , - V unit3, , Square, This is also a four sided figure, opposite sides are parallel., All the four sides are equal. All the sides are inclined at, 90º., , Area (A), Perimeter of square, Diagonal, , =, =, , a2, 18 x 18, , =, , 72 cm, , = 324 cm2, , = 25.45 cm ; Area = 324 cm2, , 2. If the diagonal of a square measure 10 cm. Find the, area of the square., Diagonal of the square (d), , 2a = 10 cm, , =, , d, , Side (a) =, A = a2 (or) unit2, Area (a2) =, , P = 4a unit, d=, a=, , a unit, , 2, , 100, 102, =, 2, 2, 2, = 50 cm, , d, , Perimeter(P) = 4a = 31.2 cm, , Area (A), , d2, 2, , =, , unit where 2 = 1.414, 2, Find the area of a brass sheet in the form of a square, whose perimeter is 31.2 cm., , a, , 2, , =, , 31.2, 4, , = 7.8 cm, , Area of the square, = 50 cm2, 3. The perimeter of one square is 748 cm and that of, another is 336 cm. Find the perimeter of a square, which is equal in area of the sum of the two., Side of the square (a), , = a2, , Perimeter, 4, , 1st square, , = 7.8 x 7.8 = 60.84 cm, , 2, , Side (a), , Examples, 1 Find out the circumference, diagonal and area of, a square, whose side is 18 cm., Side of the square (a)=, Perimeter (P), =, =, Diagonal (d), , =, , 18 cm, 4a, 4 x 18 = 72 cm, , =, , 2xa, , =, , 2 x 18 = 1.414 x 18, , =, , st, = Perimeter of I square, 4, , 748, = 187cm, 4, = a2, , =, Area (A), , = 187 x 187, = 34,969 cm2, 2nd square, Side (a), , =, , Perimeter of IInd square, 4, , =, , 336, = 84cm, 4, , 25.45 cm, , Copyright free, under CC BY Licence, , 121

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Area (A), , = a2, = 84 x 84, , Side (a), , = 42,025 = 205 cm, , Perimeter (P), , =4xa, , = 7,056 cm2, , = 4 x 205, , Total area of two squares = 34,969 + 7,056, = 42,025 cm, , 2, , = 820 cm, Perimeter of 3rd square = 820 cm, , Total area of two squares = 3 square area, rd, , 3rd square area, , = a2, , = 42,025 cm2, , Assignment, 1 Find the Area, Perimeter and diagonal of a square steel, plate whose side measures 28.1 cm., 2 Find the area of a square whose diagonal is equal to, 8.5 cm., 3 Find the area of the square if the side of the square is, 28 cm., , 4 Find its side if the area of the square field is 169 m2., 5 Find the area of the square if the diagonal of the square, is 20 cm., 6 Find the perimeter of a square whose diagonal is, 144 m., 7 Find the area if the perimeter of a square plot is 48 m., , Rectangle, This is a four sided figure. Opposite sides are parallel., Angles between adjecent sides are 90º ., A = Area = length x breadth = l.b.unit2, , Diagonal =, , l +b, , 2, , 2(l + 9) = 42, l + 9 = 42  2, , P = Perimeter = 2 ( l + b ) unit, 2, , P = 2(l + b) = 42, , l + 9 = 21, , unit, , Examples, 1 Find the Area, Perimeter and diagonal of a rectangle, whose length and breadth are 144 mm and 60 mm, respectively., Area = A, , l, , = 21 - 9, , l, , = 12 cm, , 3 The perimeter of a rectangle is 48 cm and its length is, 4 cm more than its width. Find the length and breadth, of the rectangle., , = l x b unit2, , P = 48 cm, , = 144 x 60 = 8640 mm2, Perimeter = P = 2 (l + b) unit, = 2(144 + 60), , 2, , 2, , =, , l +b, , =, , 144 + 60, , =, , 20736 + 3600, , =, , =x, , l, , =x+4, , 2(l + b) = Perimeter, , = 2 x 204 = 408 mm, Diagonal = d, , b, , unit, , 2(x + 4 + x), , = 48, , 2(2x + 4 ), , = 48, , 4x + 8, 2, , 2, , = 48, 4x = 48 - 8, , 24336 = 156 mm, , x, x, , 2 The perimeter of a rectangle is equal to 42 cm. If its, breadth is 9 cm. Find the length of the rectangle., , =, , 40, , = 10, 4, = breadth = 10 cm, , length = x + 4 = 10 + 4 = 14 cm, , P = 42 cm, , 122, , b, , = 9 cm, , l, , =?, , Workshop Calculation & Science : (NSQF) Exercise 1.8.39, , Copyright free, under CC BY Licence

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4 How many rectangular pieces of 50 cm x 20 cm can, be cut out from a sheet of 1000 cm x 500 cm., Sheet size = 1000 cm x 500 cm, , 1000, , No. of pieces to be cut in breadthwise =, Total no. of pieces to be cut out, , 50, , = 20, , 500, 20, , = 25, , = 20 x 25, = 500, , 5. The perimeter of a rectangle is 320 metre. Its sides, are in the ratio of 5:3. Find the area of the rectangle., Ratio, , = 5x, , breadth b = 3x, 2(l + b), , Size of the rectangular piece to be cut = 50 cm x 20 cm, No. of pieces to be cut in lengthwise =, , length l, , = Perimeter, , 2(5x + 3x) = 320, 2(8x), , = 320, , 16x, , = 320, =, , x, , 320, , = 20, , 16, , l = 5x = 5 x 20 = 100 m, b = 3x = 3 x 20 = 60 m, Area, , =lxb, = 100 x 60, , = 5:3 = l : b, , = 6000 m2, , Assignment, 1 Find the area of a rectangular plot whose sides are 24, metres and 20 metres respectively. Also find the, perimeter of the plot., 2 How many rectangular pieces of 5 cm x 4 cm will you, get out of 65 cm x 30 cm brass sheet?, 3 Find its breadth and area if the perimeter of a rectangle, is 400 metre and its length is 140 m. ., , 5 What is the width of the rectangle if a rectangle has an, area of 224 cm2 and length 16 cm., 6 What is the length of the diagonal of a rectangle with, sides 16 cm and 12 cm?, 7 Find the area of the rectangle if the perimeter of the, rectangle is 100 cm and the ratio of its length and, breadth is 3:2., , 4 Find its area, if the opposite sides of a rectangle are, 64 cm and 25 cm respectively., , Parallelogram, This is also a four sided figure, opposite side being parallel, to each other., , Examples, 1 The base and height of a paralleologram are 7.1 cm, and 2.85 cm. Calculate its area., A, , = base x height, = 7.1 x 2.85, = 20.235 cm2, , 2 Find the height of a parallelogram whose area is 20, cm2 and base is 10 cm., Area of parallelogram = base x height, or = 2x s (s − a) (s − b) (s − c), , A, , = base x height, , h, , =, , Where, s=, , =, , a+b+c, , area, base, 20, , 10, = 2 cm, , 2, a,b and c are adjacent sides., P = 2(a+b), , Workshop Calculation & Science : (NSQF) Exercise 1.8.39, , Copyright free, under CC BY Licence, , 123

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3. Two sides of a parallelogram are 12 cm and 8 cm. The, diagonal is 10 cm long. Find the area of the, parallelogram., A, , = 2x s (s − a) (s − b) (s − c), , s, , =, , =, , =, , a+b+c, , A, , = 2 x 15(15 - 12)(15 - 8)(15 - 10), = 2 x 15 x 3 x 7 x 5, = 2 x 1575, , 2, = 2 x 39.686, , 12 + 8 + 10, , = 79.37 cm2, , 2, 30, , 2, = 15, , Assignment, 1 Find the area of a parallelogram, if its base and height, are 8.1 cm and 30.8 cm respectively., , 7 Find the area of parallelogram if its base and height, are 25 cm and 12 cm., , 2 Find the area of a parallelogram, if the sides of a field, in the shape of parallelogram are 12 m and 17 m and, one of the diagonal is 25 m., , 8 Find the base of a parallelogram if height is 15 cm and, area is 150 cm2., , 3 Find the base of a parallelogram whose height is, 12 cm and area is 120 cm2., 4 Find the height of a parallelogram whose base is 40, cm and area is 320 cm2., 5 Find the area of the land if the sides of a land in the, shape of a parallelogram are 24 m and 28 m respectively, and one of the diagonal is 30 m., 6 What is the perimeter of parallelogram if base is 10 cm, and other side is 5 cm?, , 124, , 9 Find the area of parallelogram if side is 5 cm, diagonal, is 8 cm and diagonal bisects each other at right angles., 10 Find the height of a parallelogram if base is 80 cm and, area is 640 cm2., 11 Find the area of parallelogram if its base and height, are 15 cm and 8 cm., 12 Calculate the perimeter and area of parallelogram if, base, height are 12.7 cm, 5.5 cm and other side is 6.5 cm, 13 Find the height of parallelogram if the area is 20 cm2, and base is 10 cm, , Workshop Calculation & Science : (NSQF) Exercise 1.8.39, , Copyright free, under CC BY Licence

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Mensuration - Area and perimeter of Triangles, , Exercise 1.8.40, , Triangles, Tri means three. Hence tri- angle means three angled, figure. For construction of three angled figure, there should, be three sides. Hence triangle means three sided figure., Sum of the three angles of any triangle = 180º., i, , Any triangle., , Area of equilateral triangle =, , Area of any triangle =, , 1, 2, , =, x Base x Height unit2, , 3, 4, , x side2, , 3, , x a2 unit2, , 4, , Where 3 = 1.732, , ii Isosceles Triangle, , P = 3a unit, , In this triangle two of its sides are equal., , P =, , 3, , a unit, , 2, , iv Scalene triangle, In this triangle the sides are not equal. Angles between, the sides, are also not equal. we may also call this triangle as irregular triangle., , Area of isosceles triangle =, , 1, 2, , x Base x Height, , Where, base = 2.b, s = One of equal sides (or) Slant height, 2, , h = Height = s − b, Area of isosceles triangle =, , Area of triangle =x s (s − a) (s − b) (s − c) unit2, , 2, , 1, 2, , where, 2, , x 2b x s − b, 2, , = b . s −b, , 2, , 2, , a,b,c are sides of triangle, , unit2, , (Where b= half of base), , a+b+c, 2, , unit, , v Right angled triangle, , 1, 2, 2, (or) Area of Isosceles triangle = b 4a − b unit2, 4, a = Equal sides, b = Base, , In this triangle, angle between one of two adjacent sides, is 900. Right angle means ninety degrees. That’s why right, angled triangle means, one of the angles of this triangle is, definitely ninety degrees., Area of right angled triangle, , iii Equilateral triangle, In this triangle all the three sides are equal. Hence angle, between adjacent sides is 60º (because no. of angles), total = 180 ; angle between sides =, º, , s = Semi perimeter =, , 180, 3, , = 60, , º, , =, =, , 1, 2, 1, 2, , x Base x Height, bh unit2, 125, , Copyright free, under CC BY Licence

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h, , =, , 60 x 2, 10, , height h = 12 cm, 4 Find the area of an isosceles triangle whose base is, 6 cm long and each of the other two sides 5 cm long., , Hypotenuse, , 2, , = Base + Height, , 2, , Where hypotenuse means, the diagonal or largest length, of the side of right angled triangle., Examples, 1 Calculate its area if the base and height of a triangle, are 10 cm and 3.5 cm respectively., Base (b), , = 10 cm, , Height (h), , = 3.5 cm, , Area (A), , =?, A, , =, , 1, , xbxh, , Base (b), , = 6 cm =, , x 10 x 3.5, , Equal sides or, slant height ‘s’, , = 5 cm, , Area (A), , =?, , 2, =, , 1, 2, , 6, 2, , = 3 cm, , = 17.5 cm, , 2, , 2 Calculate the base of a triangle having an area of, 15 cm2 and height is 3.5 cm., Area (A), , = 15 cm2, , Height (h), , = 3.5 cm, , Base (b), , =?, , 1, 2, 1, 2, , xbxh, , A, , = b x s −b, , 2, , 2, , =3x, , 2, , 2, , = 3 x 25 − 9, =3x, , = 12 cm2, x b x 3.5, , or, , = 15, =, , A, , 15 x 2, , =, , 3.5, = 8.57 cm, , =, , 3 Calculate the height of a triangle whose area is, 60 cm2 and base is 10 cm., Area, , (A), , = 60 cm2, , Base, , (B), , = 10 cm, , Height (h), , 2, 1, 2, 126, , 16, , =3x4, , = A, , b, , 1, , 5 −3, , =?, , xbxh, , = A, , x 10 x h, , = 60, , =, , 1, 4, , 1, 4, 1, 4, , b 4a 2 − b 2, 2, , x 6 4x5 − 6, , 2, , x6x8, , = 12 cm2, 5 Find its height if an isosceles triangle has base of 200, mm and its area is 2000 mm2., Base, , = 200 mm, , Area, , = 2000 mm2, , h, , =?, , Workshop Calculation & Science : (NSQF) Exercise 1.8.40, , Copyright free, under CC BY Licence

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=, , 1939.4375, , = 44.03 cm2, 9 Find the cost of polishing on both sides of a triangular, metal plate has sides 60 cm, 50 cm and 20 cm at the, rate of Rs.1.35 per 100 cm2, Semi Perimeter =, , =, , 1, 2, 1, 2, , xbxh, , = A, , =, , 200, , P, , =, , = 468.4 cm2, = 936.8 cm2, , a2 unit2, , 4, , Cost of polish per 100 cm2, , 1.732, 4, , x5x5, , = Rs. 1.35, ,  Cost of polish is 936.8 cm2, , =, , 2, , =?, = 3a unit, , Equal sides or, slant height, , = 8 cm, , = 3 x 55, , Area A, , =?, , Area A, , =, , =, unit, , 9 + 10 + 12, , 2, = 15.5 cm, , =, , 1, , x base x height unit2, , 1, 2, , x 20 x 8, , = 80 cm2, , 31, 2, , =x s (s − a) (s − b) (s − c) unit2, , 11 Find the area of the right angled triangle if the sides, containing the right angle being 10.5 cm and 8.2 cm., Area A, , = 15.5(15.5 - 9)(15.5 - 10)(15.5 - 12), =, , x 1.35, , 2, , 8 Find the area of the triangle having its sides 9cm, 10cm, and 12 cm., , 2, , 100, , 10 Find the area of the right angled triangle with base 20, cm and height 8 cm., = 20 cm, , a+b+c, , 936.8, , = Rs. 12.65, , = 165 mm, , Area A, , 2, , 65 x 5 x 15 x 45, , Base b, , =, , 130, , 65(65 - 60)(65 - 50)(65 - 20), , = 55 mm, , Semi Perimeter =, , =, , Area of polish on both sides = 2 x 468.4, , 3, , 7 Calculate its perimeter if one side of an equilateral, triangle is 55 mm long., Perimeter, , 2, , =, , = 10.825 cm, , Side, , 60 + 50 + 20, , 2000 x 2, , 6 Find the area of an equilateral triangle whose side is, 5 cm., , =, , unit, , =x s (s − a) (s − b) (s − c) unit2, , Area A, , = 20 mm, , Area, , 2, , = 65 cm, , x 200 x h= 2000, h =, , a+b+c, , 15.5x 6.5 x 5.5 x 3.5, , =, , =, , 1, 2, 1, 2, , x 10.5 x 8.2, , = 43.05 cm2, , Workshop Calculation & Science : (NSQF) Exercise 1.8.40, , Copyright free, under CC BY Licence, , x base x height unit2, , 127

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12 Calculate the perpendicular height of the triangle if the, area of the right angled triangle is 19.44 m2 and its one, of the adjacent side containing the right angle being, 5.4 m., , 1, 2, , x base x height unit2, , 1, 2, , 1 Calculate the hypotenuse of a right angled triangle, whose base is 5 cm and height 12 cm., , = Area A, , x 5.4 x h, , = 19.44, , h, , =, , 19.44 x 2, 5.4, , = 7.2 m, , As per pythagoras theorem,, AC2, , 13.Calculate the base of a right angled triangle having an, area of 15 cm2. If its height is 3.5 cm., , 1, 2, , x base x height unit2, , = 122 + 52, = 144 + 25, = 169, , = Area A, AC, , 1, 2, , x b x 3.5, , = 15, , b, , =, , = AB2 + BC2, , =, , 169, , = 13 cm, , 15 x 2, 3.5, , 2 What is the length of the hypotenuse of a right angled, triangle, when the sides containing the right angles, are 10 cm and 12 cm., , = 8.57 cm, Pythagoras theorem, , As per pythagoras theorem,, AC2, , = AB2 + BC2, = 102 + 122, = 100 + 144, , In a right angled triangle the area of the square drawn with, the hypotenuse as the side is equal to the sum of the, areas of the squares drawn with the other two sides., , ∠B, , = 90º, , AC, , = Hypotenuse, , As per pythagoras theorem,, , 128, , AC, , =, , 244, , = 15.62 cm, , AB & BC = Adjacent sides, , , , = 244, , AC2, , = AB2 + BC2, , AC, , =, , 2, , AB + BC, , 3 Find the height of a right angled triangle whose base is, 15 cm and hypotenuse is 21 cm., As per pythagoras theorem,, AB2 + BC2, , = AC2, , 2, , Workshop Calculation & Science : (NSQF) Exercise 1.8.40, , Copyright free, under CC BY Licence

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AB2 + 152 = 212, AB2, , = 441 - 225, = 216, , AB, , =, , 216, , = 14.7 cm, , Assignment, I, 1 Find the area of a triangle whose base is 85.4 mm and, its height 29 mm respectively., , 4 Find the area and perimeter of the triangle if the three, sides of a triangle are 5 mm, 12 mm and, 13 mm respectively., , 2 The area of a triangle is 30 sq. cm. Its base is 10 cm., Find its height., , 5 Find the area and perimeter of the triangle if the sides, of a triangle are 15 mm, 17 mm and 8 mm respectively., , 3 Calculate the base of a triangle having an area of, 80 cm2 and height 8 cm., , V, , 4 Calculate the height of triangle whose area is 160 cm, and base is 20 cm., , 2, , II, 1 Find the area of an isosceles triangle whose base is, 16 cm long and each of the other two sides are 10 cm, long., 2 Find the area of an isosceles triangle whose side is 7, cm and base is 5 cm., 3 Find the area of an isosceles triangle whose side is 10, cm and base is 8 cm., III, 1 Find out the area of an equilateral triangle whose base, is 2.8 cm., 2 Find the area of an equilateral triangle whose sides are, 8 cm each., 3 Find the area of an equilateral triangle whose one side, is 64 mm., 4 Find the area of a triangle whose all sides are equal, and sum of the three sides is equal to 12 cm., , 1 Find the area of a right angled triangle whose base is, 15 cm and perpendicular height is 21 cm., 2 Find the area of a right angled triangle has its base, side 60 mm and height 75 mm., 3 Find the area of a right angled triangle the adjacent, sides to the right angle being 13.7 cm and 9.2 cm., 4 Calculate the height of triangle whose area is 60 cm2, and base is 10 cm., 5 Calculate the height of triangle whose area is 160 cm2, and base is 20 cm., 6 Calculate the base of a triangle having an area of, 80 cm2 and height is 8 cm., VI, 1 What is the length of the third side in a right angled, triangle the two small sides are 30 cm and 40 cm., 2 Find the length of the hypotenuse of a right angled, triangular frame having 60 cm base and 18 cm height., 3 Find the height of an equilateral triangle whose side is, 60 cm., , IV, , 4 ABC is a right angled triangle. If AB = 15 cm and, BC = 17 cm. Find the length of AC., , 1 Find the area of a triangle whose sides are 6 cm,, 7 cm and 9 cm., , 5 Find out the length of AC in a right angled triangle ABC,, AB=30 cm, BC = 40 cm., , 2 Calculate the area of the triangle if sides of a triangle, are 3 cm, 4 cm and 6 cm., , 6 Find the height of a right angled triangle whose base is, 20 cm and hypotenuse is 30 cm., , 3 Find the area of a triangle whose sides are 20 cm,, 16 cm and 10 cm, , Workshop Calculation & Science : (NSQF) Exercise 1.8.40, , Copyright free, under CC BY Licence, , 129

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Mensuration - Area and perimeter of circle, semi-circle, circular ring, sector, of circle, hexagon and ellipse, Exercise 1.8.41, Circle, It is the path of a point which is always equal from its, centre is called a circle., , r, , r = radius of the circle, , C, , =, , 196, , = 14 m, = 2r unit, , d = diameter of the circle, , =, , 22, 7, , =2x, , = 3.14, , 22, 7, , x 14, , = 88 m, , Area of the circle = r2, , 3. Find the side of square into which it can be bent if a, wire is in the form of a circle of radius 49 cm., radius of circle r, , = 49 cm, , side of square, , =?, , Perimeter of the square = Perimeter of the circle, , (or), , =, , π, 4, , d2 unit2, , 4a, , = 2r, , 4a, , =2x, , 4a, , = 308, , a, , =, , Circumference of the circle 2r (or) d unit, Examples, 1. Find the area of a circle whose radius is 1.54 m. Also, find its circumference., radius r, , = 1.54 cm, , Area A, , =?, , Circumference C, , 22, 7, , 308, 4, , = 77 cm, 4. Find its radius if the difference between the, circumference and diameter of a circle is 28 cm., Circumference - Diameter = 28 cm, , =?, , 2r - d = 28, , = r2 unit2, , A, , =, , 2r - 2r = 28, , 22, , x 1.54 x 1.54, , 7, , = 7.4536 m2, C, , =2x, , 22, 7, , 2r ( - 1) = 28, 2r (, , = 2r unit, x 1.54, , 2r (, , = 9.68 m, 2. Find out the circumference if the area of a circular shape, of land is 616 m2., A, = r2 unit2, r2, , =, , =, , x 49, , 2r x, , 616, , 22, 7, , 22 - 7, 7, 15, 7, r, , π, , - 1) = 28, , ) = 28, , = 28, , =, , 28x7, 15x2, , = 6.53 cm, , 616x7, 22, , = 196, 130, , Copyright free, under CC BY Licence

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5. What is the side of the largest square cut out from a, circle of 50 cm dia.?, , 1, x 50 x 16 cm2, 2, = 400 cm2, , =, , Diagonal of a square = Diameter of the circle, , 2a, , = 50, , a, , =, , =, , Area of Semi circle, , =, , πr 2, unit2, 2, , 50, , 2, , 1, cm2, 2, , 2, , = π x 15 x, , 50, , = 353.57 cm2, , 1.414, , Area of the figure, , = 1253.57 cm2, , = 35.36 cm, 6. Calculate the area of the figure given below., , = 500 + 400 + 353.57, , 7. Find the area of remaining steel plate if in a rectangular, steel plate 16 cm x 12 cm, there are 6 holes each 4, cm in diameter., Area of a rectangular plate = length x breadth unit2, = 16 x 12, = 192 cm2, No. of holes, , =6, , Radius of hole, , = 2 cm, , Area of 6 holes, Area of rectangle, , = 6 x r2 unit2, , = lb unit2, =6x, , = 25 x 20 cm2, = 500 cm2, Area of Trapezium =, , 1, x (a + b) h, 2, , 22, 7, , x 2 x 2 unit2, , = 75.43 cm2, Area of remaining plate = 192 - 75.43, = 116.57 cm2, , 1, = x (30 + 20) 16 cm2, 2, , Semi circle, A semi circle is a sector whose central angle is 180º., Length of arc of semi circle., Length of arc, , = 2r x, , + 2r, 2, =r + 2r, , 180, 360, , =r ( + 2) unit, , 1, 2, = r unit, , Examples, , = 2r x, , Area of semi circle =, , 2πr, , Perimeter of a semi circle =, , 1. Calculate the circumference and area of a semi circle, whose radius is 6 cm., , πr 2, Sq. units, 2, , radius r, , = 6 cm, , Area A, , =?, , Circumference c = ?, A, , =, , =, , πr 2, unit2, 2, , 22, 7, , Workshop Calculation & Science : (NSQF) Exercise 1.8.41, , Copyright free, under CC BY Licence, , x, , 1, x 62, 2, , 131

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22, , Area (A) =, , 7, , 396, , =, , 1, x 36, 2, , x, , 7, , = 56.57 cm2, , 22, Perimeter of a semicircle = 6( 7 x 2), , ⎛ 22 + 14 ⎞, ⎟, = 6⎜, ⎝ 7 ⎠, 36, , =6x, , Plate length AB, , = 100 mm, , Breadth BC, , = 50 mm, , Radius, , = 50 mm, , Waste area = Plate area - Area of semi circle, , 7, , = lb -, , 216, , =, , 7, = 30.86 cm, , πr 2, 2, , = 100 x 50 -, , 22 x 50 x 50, 7x2, , = 5000 - 3928.57, , 2. From the figure given below ABCD is a steel plate, a, semi circular plate of radius 50 mm has been prepared, by gas cutting. Find the waste area., , = 1071.43 mm2, , Circular ring, Solution:, Area of cross section of pipe =  (R + r) (R - r) unit2, =  (8.5 + 7) (8.5 - 7), =, R = Outer radius of circular ring, , = 73 cm2, , r = Innner radius of circular ring, Area of circular ring =, A=, , , ,  (R, , 2, , 2, ,  r ) unit, , 2 Find the distance between the boundaries and the area, of the circular ring, if the circumference of two concentric, circle are 134 cm and 90 cm., , 2, , or, (R + r) (R - r) unit2, , Given:, , 1 Calculate the area of cross section of pipe having, outside dia of 17 cm and inside dia of 14 cm., Given:, , Circumference of outer circle = 134 cm, Circumference of inner circle = 90 cm, To find:, , Outer dia of pipe = 17 cm, Outer radius of pipe (R) =, , Distance between the circles = ?, = 8.5 cm, , Area of circular ring = ?, Solution:, , Inner dia of pipe = 14 cm, Inner radius of pipe (r) =, , x 15.5 x 1.5 cm2, , = 7 cm, , To find:, Area of cross section of pipe = ?, , Circumference of outer circle = 134 cm, 2 R, = 134 cm, R, , =, , Circumference of inner circle = 90 cm, 2 r, = 90 cm, 132, , Workshop Calculation & Science : (NSQF) Exercise 1.8.41, , Copyright free, under CC BY Licence

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r, , 4 Calculate the length of material required , if a piece of, 12mm dia bar is to be bent round to form a ring 150 mm, inside dia., , =, , Given:, , Distance between the circle = R - r, = 21.32 - 14.32 cm, , dia of bar, , = 12 mm, , = 7 cm, , inner diameter, , = 150 mm, , =  (R + r) (R - r) unit2, , Area of circular ring, , To find:, , =  (21.32 + 14.32) (21.32 - 14.32) cm, , length of bar, , 2, , =, , x 35.64 x 7 cm2, , Solution:, , = 784.08 cm2, , Inner diameter, , = 150 mm, , Outer diameter, , = dia of bar + inner diameter, + dia of bar, , 3 Calculate the inner and outer diameter of the circular, ring. If the area of a circular ring is 176 cm2 and width of, the circular ring is 4 cm., , = 12 + 150 + 12 = 174 mm, Average diameter =, , Given:, , =?, , inner diameter  outer diameter, 2, , Area of circular ring = 176 cm, , 2, , width, , =, , = 4 cm, , To find:, Average radius, , Outer diameter = ?, Inner diameter = ?, Solution:, , =, , 150  174, 2, , , , 324,  162 mm, 2, , 162, = 81 mm, 2, , Length of the material required = average circumference, , Inner radius (r), , = x cm, , Outer radius (R), , = inner radius + width, , = 2r unit, = 2 x  x 81 mm, , = x + 4 cm,  (R + r) (R - r),  (x + 4 + x) (x + 4 - x),  (2x + 4) (4), x 4 x (2x + 4), , = 509 mm, , = area, = 176 cm2, = 176 cm2, , 5 A wire can be bend in the form of a circle of radius 56, cm. If it is bend in a form of a square, find the side., Given:, , = 176 cm2, , Radius of circle, , = 56 cm, , Side of square, , =?, , Radius of circle, , = 56 cm, , To find:, x (2x + 4), , = 176 cm2, Solution:, , 2x + 4, 2x, , =, , 176  7, , = 14, 88, = 14 - 4 = 10, , inner radius (r), , = 5 cm, 2, = x = 5 cm, , outer radius (R), , = x + 4 = 5 + 4 = 9 cm, , inner diameter, , = 2 x 5 = 10 cm, , outer diameter, , = 2 x 9 = 18 cm, , x, , =, , 10, , Circumference of circle = 2r unit, = 2 x 56 cm, Side of square, , = x cm, , Wire can be bend from the form of round to square, Perimeter of square, , = circumference of circle, , 4xa, a, , Workshop Calculation & Science : (NSQF) Exercise 1.8.41, , Copyright free, under CC BY Licence, , = 352 cm, =, , 133

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Sector of Circle, , 2 Find the radius of the circle if the angle is 600 and the, area of a sector of a circle is 144 cm2,, Given:, Area of sector of circle (A) = 144 cm2, Angle of sector of circle  = 600, To find:, Radius of circle = ?, Solution:, ,  = Angle of sector of circle, , =, , x r2 unit2, , 144, , =, , x, , r2, , = 274.91 cm2, , r, , =, , Area (A), , l = Arc length, r = radius, =, , Length of Arc, , Perimeter P = 2r +, Area =, , x, , x2, unit, , = 16.58 cm, , r2 unit2, 3 Find the area of the sector whose angle is 1050, and the, perimeter of sector of circle is 18.6 cm., , or, , Given:, , unit2, , A =, , x r2 cm2, , r unit, , 1 Find the perimeter and area of a sector of circle of radius, 7 cm and its angle is 1200., , Perimeter of a sector of a circle = 18.6 cm, Angle of sector of circle = 1050, To find:, , Given:, Angle of sector of circle, , Area = ?, , = 1200, , Radius, , Solution:, , = 7 cm, , To find:, , Length of Arc ( ) =, , Perimeter = ? , Area = ?, , x 2r unit, , Solution:, =, Length of arc ( ) =, , x2x, , xr, , x 2  r unit, = 1.83r, , =, , x2x, , x 7 cm, , = 14.67 cm, , Perimeter (P), , =, , + 2r unit, , 18.6, , = 1.83r + 2r, , 3.83r, , = 18.6 cm, , r, , =, , = 4.86 cm, , Area A, , =, , x, , Perimeter = 2r + unit, = 2 x 7 + 14.67 cm, = 28.67 cm, Area, , =, , x r2 unit2, , =, , x, , =, x 7 cm, , 2, , = 21.65 cm2, , = 51.33 cm2, , 134, , Workshop Calculation & Science : (NSQF) Exercise 1.8.41, , Copyright free, under CC BY Licence, , r2 unit2, , x (4.86) cm2

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4 Find the area, if the radius is 12.4 cm and the perimeter, of a sector of a circle is 64.8 cm., Given:, Perimeter P, Radius r, , = 64.8 cm, = 12.4 cm, , To find:, Area A = ?, Solution:, Perimeter P, , , Length  A =, , =  + 2r unit, = P - 2r unit, , =, , = 64.8 - 24.8 = 40 cm, =, , =, , r, 2, , 40  12.4, 2, , x2x, , 360 0, , Length  B =, , unit2, , r unit, , 210 0, , = 64.8 - 2 (12.4) cm, , Area A, , x2, , x2, , x 30 = 110 cm, , r unit, , 105 0, =, , = 248 cm2, , 360 0, , x2x, , x 5 = 91.7 cm, , =  A +  B + 2 x 214 cm, , 5 Find out the length of the belt , if the arrangement of a, belt is shown in the figure below., , = 110 + 9.17 + 428 cm, , Solution:, , = 547.17 cm, , Hexagon, To Find: P = ?, A = ?, DAF = ?, DAC = ?, Solution:, Perimeter of hexagon (P), , = 6a unit, , = 6a unit = 6 x 2 cm = 12 cm, Side = a unit, , Area of hexagon A, , = 6, , Perimeter P = 6a unit, = 6, , 3, 2, Area A = 6 ,  a units2 (Area of 6 equilateral triangle), 4, , 3, 2,  a unit2, 4, , 1.732, 2, 2, 4, , = 10.392 cm2, DAF (Distance Across Flats) =, , 3  a unit, , DAF (Distance Across, , DAC (Distance Across Cormers) = 2 x a unit, , Flats), , Example 1, Find out the perimeter, area, DAF and DAC of a regular, hexagon whose side is 2cm., (DAF - Distance Across Flats), , =, , 3  a unit, , =, , 3  2 = 1.732 x 2, , = 3.464 cm, DAC (Distance Across, Corners) = 2 x a unit, , (DAC - Distance Across Corners), , = 2 x 2 = 4 cm, , Given: Side of hexagon (a) = 2cm, , Workshop Calculation & Science : (NSQF) Exercise 1.8.41, , Copyright free, under CC BY Licence, , 135

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Ellipse, , Minor axis 2b, b, , = 8 cm, =, , Area A =, =, , 8, 2, , , , = 4 cm, x a x b unit2, x 6 x 4 cm2, , = 75.43 cm2, Major axis AB = 2a, Perimeter (P), , Half of Major axis OB = a,, , =, , unit, , Minor axis CD = 2b, Half of Minor axis OC = b, Area of ellipse A =, , =, , x a x b unit2, , Perimeter of ellipse P =, , unit, , =, , unit, , unit, , Example 1, =, , Find its area and perimeter, if the major and minor axis of, an ellipse are 12 cm and 8 cm respectively., Solution:, , =2x, Major axis 2a, a, , x 5.1 = 32.06 cm, , = 12 cm, =, , = 6 cm, , Assignment, Circle, 1 Find the circumference and area of a circle whose radius, is 10.00 metre., 2 Find its diameter if the area of a circle is 330 cm2., 3 Find its radius if area of a circle is 498 m2., 4 Find its area if the circumference of a circle is 50 cm., 5 Find its area if the circumference of a circle is 44 cm., 6 Find out the area and circumference of a circle of, diameter is 50 cm., 7 A wire is in the form of a circle whose radius is 42 cm., Find the side of that square which can be made by, bending the same wire., 8 A square of side 22 cm is made from a wire. Calculate, the diameter of circle which will be made from the same, length of wire as that of square., 9 Find its radius if the difference between the, circumference and diameter of a circle is 30 cm., , 136, , 10 Calculate the side of the largest square that can be, obtained if a 150 mm dia. round bar is milled to a square, bar., 11 What is the maximum size of square which can be cut, from a circular sheet of diameter 100 mm?, 12 From a brass sheet 270 cm x 100 cm. Calculate how, many pieces of size 15 cm x 10 cm may be cut., 13 Find the area of remaining plate if in a 48 cm x 18 cm, rectangular plate there are 5 holes of 4 cm diameter., 14 Find the area of remaining steel plate if in a rectangular, steel plate 36 cm x 24 cm. There are 54 holes of 4 cm, in diameter., 15 Find the radius of circle if a rectangle with sides 14, metre length and 11 metre breadth has area equal to, that of circle., Semi circle, 1. Calculate the circumference and area of semi circle, whose radius is 14 cm., , Workshop Calculation & Science : (NSQF) Exercise 1.8.41, , Copyright free, under CC BY Licence

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2. Find the area of the figure given below., , 5 What length of rod did you require to make a ring of, 300cm inside dia out of 12 mm round bar?, Sector of circle, 1 Find the perimeter and area of a sector of a circle of, radius 5cm and its angle is 960., 2 Find the radius of the circle if the angle is 900 and the, area of sector of a circle is 196 cm2., 3 Find the area of a sector of a circle whose perimeter is, 82.4 mm and radius is 16.2 mm., , 3. Calculate the area of the given figure., , 4 Find out the length of the saw blade, if the arrangement, of a band saw is shown in the figure below., , 4. Calculate the area of the figure given below., Hexagon, 1 Find out the Area, perimeter, DAF,and DAC of hexagon, of side 4cm., 2 Find the area of cross section of a regular hexagon rod, whose side is 7.5 cm., 3 Find out the size across the flats of a hexagonal piece, having 15 mm each sides., 4 Find out the distance across the flats, if the distance, across the c orners of a hexagonal bar 40 mm., Circular ring, 1 Find out area of a ring washer, whose inner radius and, outer radius are 13 cm and 15 cm respectively., 2 Find the area of a ring portion of a washer whose outer, dis is 30 m and inner dis is 20 m. Also calculate the, difference between the circumference of circles., 3 Find the thickness of the pipe and area of cross section, of the pipe, if the outside and inside circumference of the, steel pipe is 70 cm and 45 cm., , 5 What will be the area of a largest hexagon which is, inscribed in a circle of radius 10 cm?, Ellipse, 1 Find the area of the biggest ellipse that can be inscribed, in a rectangle of length 18 cm and breadth 12 cm. Also, calculate its perimeter., 2 How much fenching will be required to enclose an, elliptical plot of ground the axes of the ellipse being 200, and 170 meter respectively., , 4 Calculate the inner and outer diameter of the washer.if, the cross sectional area of a washer is 264 cm2 and the, width of the washer is 2cm., , Workshop Calculation & Science : (NSQF) Exercise 1.8.41, , Copyright free, under CC BY Licence, , 137

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Mensuration - Surface area and volume of solids - cube, cuboid, cylinder,, sphere and hollow cylinder, Exercise 1.8.42, Cube, , 2, , All sides of cube are same i.e length,breadth and height, have same value. It is bounded by six equal square faces., , Volume of cylinder = π r h, , or, , π, 4, , 2, , d h, , Volume of cube = side x side x side, = a3 unit3, Lateral surface area = 4a2 unit2, Total surface area = 6 x side x side, = 6a2 unit2, Rectangular solid (or) cuboid, Rectangular soild is bounded by six rectangular surfaces, and opposite surfaces are equal and parallel to each other., , Curved area of cylinder = 2rh unit2, Total surface area of cylinder = 2r(h+r) unit2, r = Radius of base, d = Diameter of base, h = Height of cylinder, Hollow cylinder, Hollow means empty space. In hollow cylinder there is an, empty place. Water pipe is an example of hollow cylinder., , Volume of rectangular solid, = Length x breadth x height, = l . b . h unit3, Lateral surface area = 2h(l+b) unit2, Total surface area = 2lb + 2bh + 2hl, = 2(lb+bh+hl) unit2, , Volume of hollow cylinder = (R2 - r2) h (or), = (R + r) (R - r) h (or), =, , l = length, b = breadth, , π, 4, , (D2 - d2) h unit3, , π, , (D + d)(D − d) h, 4, Total surface area of hollow cylinder =, =, , h = height, Cylinder, This is a prism whose top and bottom surfaces are equal, and circular., , Inner + outer curved area + area of top, and bottom circular part,  TSA : 2Rh + 2rh + 2(R2 - r2), , 138, , Copyright free, under CC BY Licence

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L.S.A, , R = outer radius, , = 4a2 unit2, , r = inner radius, , = 4 x 4.5 x 4.5, , D = outer diameter, , = 81 cm2, , d = inner diameter, , T.S.A, , = 6a2 unit, , h = height of cylinder, , = 6 x 4.5 x 4.5, , t = thickness, , = 121.5 cm2, , Mean dia =, , D−d, , V, , = a3 unit3, = 4.5 x 4.5 x 4.5, , 2, , If thickness given then:, , = 91.125 cc., , Volume of hollow cylinder = x mean dia x thickness x, height, , 2 Calculate volume of a cube where side is 9 cm, , Finding out volumes of solids, The space occupied by a body is known its volume. The, volume of a body indicates the capacity to hold substance, in it., , a, , = 9 cm, , V, , =?, , V, , = a3, =9x9x9, , The general form of Lateral surface area Total surface area, and Volume is :, Lateral surface area = perimeter of the base x height, Total surface area, Volume, , = LSA + 2 (base area ), , = 729 cm3, 3 Find out side of the cube if a cube has volume of, 3375cm3., , = Area of base x height, , Important and commonly used solids are described below, one after another:, , V, , = 3375 cm3, , a, , =?, , a, , = 3375, , a, , =, , 3, , Cube, All sides of cube are same i.e length,breadth and height, have same value. It is bounded by six equal square faces., , 3, , 3375, 3x3x3x5x5x5, , =, , Volume of cube = side x side x side, , =3x5, , = a3 unit3, , = 15 cm, , Lateral surface area = 4a2, Total surface area = 6 x side x side, , Surface area = T.S.A = 216 cm2, , = 6a2 unit2, Diagonal d, , 4 Find the side of a cube, if its surface area is 216 cm2, , = 3 a unit where 3 = 1.732, , 6a2, a2, , = 216, =, , 216, 6, , = 36, a, , =, , 36, , = 6 cm, , 1 Find the diagonal, lateral surface area,, total surface, area and volume of a cube of side 4.5 cm., side a, , = 4.5 cm, , diagonal d, , = 3 a unit, , 5 Find the side of the square tank, if its height is 2 metre, and has the capacity to hold 50,000 litre of water., Height of square shape tank (h) =, , = 1.732 x 4.5, , Capacity, 1000 litre, , 50,000 Litre =, , = 7.794 cm, Workshop Calculation & Science : (NSQF) Exercise 1.8.42, , Copyright free, under CC BY Licence, , =, =, , 2m, 50,000 litre, 1 m3, 50000, 1000, , 139

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=, Capacity of tank =, a2 x h, =, 2, a x2, =, a2 =, a, , 50 m3, 50 m3, 50, 50, 50, = 25 m2, 2, , =, , v, , = 10500 cm3, , h, , =?, , l.b.h = volume, 26 x 18 x h = 10500, h =, , 25 = 5 m, , Side of the square tank = 5 m, , Rectangular solid (or) cuboid, Rectangular soild is bounded by six rectangular surfaces, and opposite surfaces are equal and parallel to each other., , 10500, 26 x 18, , = 22.44 cm, 3 How many litres of water it can store if a water tank, has the following dimensions length = 1 metre, width =, 0.8 metre and height = 1.2 metre?, Volume, , = l x b x h unit3, = 1 x 0.8 x 1.2, , Volume of rectangular solid, , = 0.96 m3, , = Length x breadth x height, , [1 m3 = 1000 litres], , = 0.96 x 1000, , = l . b . h unit3, , = 960 litres of water can store in the tank., 4 Find its volume if the base of a prism is a rectangle, having 5m length, 4m breadth and the height of the, prism is 15m., The base of prism is rectangle, Area of base = length x breadth, =5x4, = 20 square m, , Lateral surface area = 2h(l+b), , Volume of prism = Area of base x Height, , Total surface area = 2lb + 2bh + 2hl, , = 20 x 15, , = 2(lb+bh+hl) unit3, , = 300 cubic metres, , Examples, 1 Find its volume and T.S.A if a tank is 20 m long, 15 m, broad and 12 m high., l, , = 20m, , b, , = 15 m, , h, , = 12 m, , v, , =?, , T.S.A, , =?, , Volume v, , Cylinder, This is a prism whose top and bottom surfaces are equal, circular., π 2, 2, d h, Volume of cylinder = π r h or, 4, , = lbh unit3, = 20 x 15 x 12, = 3600 m3, , T.S.A, , = 2(lb + bh + hl) unit2, = 2((20 x 15) + (15 x 12) + (20 x 12)), , Curved area of cylinder = 2rh, , = 2 (300 + 180 + 240), , Total surface area of cylinder = 2r(h+r), r = Radius of base, , = 1440 m2, 2 Find out its height if the cross section is 260 mm length, and 180 mm wide rectangular and the capacity of a, fuel tank is 10500 cm3., , 140, , l, , = 260mm = 26 cm, , b, , = 180 mm =18 cm, , d = Diameter of base, h = Height of cylinder, , Workshop Calculation & Science : (NSQF) Exercise 1.8.42, , Copyright free, under CC BY Licence

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Examples, 1 Find the volume and total surface are of a cylinder having, 9cm diameter and 15 cm height., diameter, = 9 cm, radius r, = 4.5 cm, height h, = 15 cm, Volume V, =?, T.S.A, =?, V, = r2 h unit 3, =, , 22, , r, , =2x, , 5 A cylindrical vessel is to be made of 3 metre long and, 1.9994 metre diameter. Calculate its surface area, if it, is in a closed form on one end., , x 4.5 x 4.5 x 15, , 22, 7, , =2x, , r, , = 48 cm2, = 10 cm, =?, = 48, = 48, , = (2 x, , r2, r2, , r, , =v, = 5544, =, , r, , = 0.9997 m, , 22, 7, , x 0.9997 x 3) + (, , 5544, 3.14 x 16, , 5544, 50.24, = 110.35, , V, , = 68.46, , r2, , =, , r2, , = 10.9, , 68.46, 3.14x2, , x 0.999972), , = r2 h unit3, = 3.142 x 75 x 75 x 100, = 1767375 cm3, =, , 1767375, 1000, , [1000 cc = 1 litre], , = 1767.375 litres., 7 Calculate the height of cylindrical tin if a closed, rectangular box 40 cm long, 30 cm wide and 25 cm, deep has the same volume as that of cylinder tin of, radius 17.5 cm., r2 h, , 22, 7, , = Volume of rectangular box, =lxbxh, , x 17.5 x 17.5 x h= 40 x 30 x 25, h, , = 10.5 cm, , r2 h, , 7, , 6 How many litres of water a cylinder of radius 75 cm, and height 100 cm can hold., , =, , 40 x 30 x 25 x 7, 22 x 17.5 x 17.5, , =, , 210000, 6737.5, , 110.35, , 4 Find the diameter of the tank if the volume of a circular, tank is 68.46 m3, its height is 2 m., , 22, , = 21.99 m2, , Volume of cylinder, , =, , =, , = 1.9994 m, , = 18.85 + 3.14, , 48 x π, =, 2 x π x 10, , 3 Find its radius if the volume of a cylinder is 5544 cm3, and its height is 16 cm., r2 h, 3.14 x r x 16, , d, , = 2rh + r2, , 22, , = 2.4 cm, , 2, , = 3m, , T.S.A = C.S.A + Base area, , x 4.5 x 19.5, 7, = 551.4 cm2, , C.S.A, length, radius, 2rh, 2 x x r x 10, , h, , x 4.5 (15 + 4.5), , 2 Calculate the radius if the curved surface area of a, cylindrical roller is 48 cm2 and the roller is 10 cm long, , 10.9, , = 3.3 m, = 2r, = 2 x 3.3, = 6.6 m, , diameter, , 7, = 954.4 cm3, = 2r(h+r) unit2, , T.S.A, , =, , = 31.17 cm, 8 An oxygen cylinder is 15 cm in diameter and 100 cm, in length. It is filled with gas under pressure so that, every cm3 of the cylinder contains 120 cm3 of gas. How, much cc of oxygen does this hold?, Volume of cylinder, , = r2 h unit3, , Workshop Calculation & Science : (NSQF) Exercise 1.8.42, , Copyright free, under CC BY Licence, , 141

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=, , 22, , 4, , x 7.5 x 7.5 x 100, , 7, = 17678.57 cm3, , 3, , 22, , x, , r3, , Gas contain in 1 cm3 = 120 cm3 of gas, , x r3 = 15625, , 7, =, , 15625 x 3 x 7, 4 x 22, , =, , 328125, 88, , Gas contain in 17678.57 cm3 = 17678.57 x 120, = 2121428 cm, , 3, , Volume of oxygen = 2121428 cc., , = 3728.69, , Sphere, , r, , =, , Sphere is a solid circular body., , 3, , 3728.69, , = 15.51 cm, diameter, , = 2 x radius, = 2 x 15.51, = 31.02 cm, , 3 How many spherical balls of 1 cm radius can be made, from a sphere of 32 cm diameter., No. of balls x volume of small sphere = Volume of, bigger sphere, Nx, , 4, Volume of sphere = r3, 3, , or, Nx, , π 3, d unit3, =, 6, , 4, 3, , x r3 =, , 4, , 4 Three brass balls of diameters 3 cm, 4 cm and 5 cm, are melted and make into one solid ball, if there is no, wastage. Find the diameter of the solid ball., , 1, of diameter, 2, Examples, , Radius =, , 1st ball d1, , 1 Find the volume and surface area of a sphere of 3 cm, radius., , 3, , = 3 cm,, , r1, 2nd ball d2, r2, , r3 unit3, , 3 ball d3, , Volume of new ball = Volume of 3 spherical balls, , 4, , = 4r2 unit2, =4x, , 22, , x3x3, 7, = 113.1 cm2, , 2 Find the diameter of sphere having volume of, 15625 cc., , 3, 142, , = 2.5 cm, , Diameter of new ball = ?, , = 113.1 cm3, , 4, , = 2 cm, = 5 cm,, , rd, , =, , = 1.5 cm, = 4 cm,, , r1, , Total Surface Area, , 4, , = 4096 balls, , d = Diametre of sphere, , 4, , r3, , x π x r3, 3, = 16 x 16 x 16, , 3, , N, , Where r = Radius of sphere, , =, , 3, , x π x 13 =, , Total surface area of sphere = 4r2 unit2, , V, , 4, , 3, , r3 =, , 4, 3, , 4, 3, , r13 +, , π r3, , =, , 4, 3, , 3, , r23 +, , 4, 3, , r =, , 3, , 27, , Workshop Calculation & Science : (NSQF) Exercise 1.8.42, , Copyright free, under CC BY Licence, , r33, , π (1.53 + 23 + 2.5)3, , r3 = 3.375 + 8 + 15.625, r3 = 27, , r3 = Volume, , 4

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r =, , 3, , 4, πR3 cubic cm, 3, Assuming no wastage in material, we can say, , 3x3x3, , Volume of new sphere =, , = 3 cm, Diameter of ther ball, =2xr, =2x3, , Volume of new sphere = Total volume of two spheres, , = 6 cm, , 4, 4, πR3 = π x 54, 3, 3, , 5 Calculate the number of spheres that can be made if a, solid metal cylinder of radius 14 cm and height, 21 cm is melted and recast into spheres, each of radius, 3.5 cm., Cylinder radius, , = 21 cm, , Sphere radius, , = 3.5 cm, , No. of sphere, , =?, , R3 = 54, R = Cube root of 54, = 3 54, , No. of balls x Volume of sphere = Volume of cylinder, Nx, , Nx, , 4, 3, , 3, , r3, , = r2 h, , π x 3.53 = π x 142 x 21, , N, , 14 x 14x21x3, =, 4 x 3.5x3.5x3.5, =, , 4, πRfrom both sides (because of similarity),, 3, , we can write:, , = 3 cm,, , height, , 4, , Deleting, , 12348, 171.5, , = 3.780 cm, Radius of new sphere formed = 3.780 cm, 7 Find the ratio of the total surface of all the smaller balls, to that of the original one if a spherical lead ball is, melted and made into smaller balls of one third the, radius of the original one. (i) How many such balls can, be made?, Let the radius of the original ball = R, The radius of the small ball = R/3, Volume of the original ball =, , 4, 3, , And, Volume of the small ball =, , = 72 balls., , 3 4, 3, πr = π x (3), 3, 3, , No. of small balls =, , 4, , Volume of 2 spheres = 2 x, , 4, 3, , 4, 3, , =, , 6 Two spheres of 3 cm radius melted together and formed, into single sphere. Find the radius of the new sphere., Volume of each sphere =, , π ⋅R, , 3, π x (3), , 4, 3, , 3, , π ⋅R ÷, , 3, , 3, π ⋅ (R/3), , 4, 3, 4, 3, , x, , π ⋅ R3, 27, , π⋅, , R3, 27, , = 27, Now, surface of the original ball = 4. R2, And, total surface of 27 small balls = 27 x 4  (R/3)2, Surface ratio = 4 . R2 : 27.4 (R/3)2, , 4, π x 54 cubic cm, 3, Let R = Radius of new sphere made, =, , = 1:3, , Assignment, 2, , 1, , V = 5832 cm3, l = 60 mm, l = ______ mm, V = ______ cm3, , Workshop Calculation & Science : (NSQF) Exercise 1.8.42, , Copyright free, under CC BY Licence, , 143

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3, , V = 1800 cm3, l : b = 3:1, H = 1500 mm, l = ______ mm, b = ______ mm, , 4, , l = 1.5 metre, h = 0.8 metre, b = 0.45 metre, Capacity, = ______ litres, 1 m3 = 1000 litres, , 8 How many litres of water it can hold if a water tank has, the following dimensions of Length = 4m, breadth =, 3.2 m and height = 1400 cm?, Cylinder, 1 Find the curved surface area of cylinder where, a diameter 18 cm and height 34 cm, b diameter 28 cm and height 42 cm, 2 Find the total surface area of cylinder whose, a diameter 24 cm and height 40 cm, b diameter 42 cm and height 60cm, c diameter 14 cm and height 35 cm, 3 Find out the volume of cylinder whose, a base is 10 cm radius and height is 40 cm, , 5, , V = 140000 cm3, l, , = 60 cm, , b = 40 cm, H = ______ cm, , b base is 7 cm radius and height is 12 cm, c base is 35 cm diameter and height is 100 cm, 4 Find the volume, C.S.A and T.S.A of a cylinder having, diameter 10 cm and 20 cm height., 5 A cylindrical tank has 22000 cc water. If the depth of, water is 70 cm. Calculate the diameter of the tank., 6 Find out its radius. if the volume of cylinder is 5544, cm3 and height is 16 cm., , Cube, 1 Find the diagonal, lateral surface area, total surface, area and volume of cube, whose side is 15 cm., , Sphere, 1 Find the volume of sphere using the following dimension., , 2 Find the volume of 10 cubes where each side is 5 cm., , a 3.5 cm diameter, , 3 Find its volume if a solid cube has each of its sides 60, mm long., , b 4 cm diameter, , 4 What is its side if the total surface area of a cube is, 384 m2., , d 20 cm diameter, , 5 Find out side of the cube if a cube has volume 422 cc., Cuboid, 1 Find the volume of the tank in m3, if the length is, 60 m, breadth 40 m and height 20 m., 2 Find the volume of a C.I. casting of a rectangular block, having 25 cm x 20 cm x 8 cm size., 3 Calculate the total surface area of a box whose length,, width and height are 120 cm, 50 cm and 60 cm, respectively., 4 Find the volume of the sheet if a brass sheet is of 25, cm square and 0.4 cm thick., 5 Express its capacity in litres if a vessel measures 3m, x 4m x 5m., 6 A milk tank with square base has a volume of holding, 10 m3 of milk. What will be the height of the tank, if its, side is 2.583 m., 7 Find out its height if the cross section is 420 mm length, and 230 mm wide rectangular and the capacity of the, tank is 48 litres., 144, , c 7 cm diameter, e 5 cm diameter, 2 Find the diameter of a sphere having volume of 512cc., 3 Find the total surface area of a sphere having, a radius 1.75 cm., b radius 12 cm, c radius 56 cm, d diameter 20cm, e radius 3 cm, 4 How many spherical balls of 1 cm radius can be made, from a sphere of 16 cm diameter., 5 Three balls of diameter 2m, 4cm and 6 cm are melted, and made into one solid ball. If there is no wastage,, find the diameter of solid ball., 6 How many solid spheres each radius 3 cm can be, moulded from a solid metal cylinder whose length is, 45 cm and base radius is 2 cm., 7 Calculate the number of balls made if a ball of 10 cm, radius is to be converted into small ball of 2 cm radius., , Workshop Calculation & Science : (NSQF) Exercise 1.8.42, , Copyright free, under CC BY Licence

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Mensuration - Finding the lateral surface area, total surface area and capacity in, litres of hexagonal, conical and cylindrical shaped vessels, Exercise 1.8.43, Hexagonal bar, Volume of Hexagonal bar = Area of hexagonal x heigth, Lateral surface area of hexagonal bar, = 6 x length of the bar x side of hexagon, or, , = 3.464 x length of the bar x flat of hexagon, , Total surface area of hexagonal bar, = lateral surface area + (2 x area of hexagon), , TSA = l (R + r) + A1 + A2 unit2, V=, , π, 3, , h (R2 + Rr + r2) unit3, , [A1 = Top aera ; A2 = Bottom area ], Cylinder, , Cone is a pyramid with a circular base., , This is a prism whose top and bottom surfaces are equal, and circular., π 2, 2, d h, Volume of cylinder = π r h or, 4, , 1, Volume of cone = r2h, 3, , Curved area of cylinder = 2rh, , Cone, , or, , =, , Curved area =rs, , π, 12, , d, , 2, , Total surface area of cylinder = 2r(h+r), h, , r = Radius of base, d = Diameter of base, , Total surface area = r(s+r), , h = Height of cylinder, , Where r = radius of base, , Hollow cylinder, , d = diametre of base, h = vertical height of cone, , Hollow means empty space. In hollow cylinder there is an, empty place. Water pipe is an example of hollow cylinder., Volume of hollow cylinder = (R2 - r2) h (or), = (R + r) (R - r) h (or), , s = slant height r 2 + h2, Frustum of a cone, When a cone is cut by a plane parallel to the base, and, upper part is removed, the formation appears, is termed, as frustum of a cone. Buckets, oil cans etc.are such, frustums in shape., L.S.A = l (R + r) unit2, , =, , π, 4, , (D2 - d2) h, , π, , (D + d)(D − d) h, 4, Total surface area of hollow cylinder =, =, , Inner + outer curved area + area of top, and bottom circular part, 145, , Copyright free, under CC BY Licence

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2 Calculate the height. Also find the lateral surface, area if a cone has a base diameter of 210 mm and, its volume is 3056 cm3., , Volume of a cone =, 3056 cm3 =, , 1, 3, , 1, 3, , x 0.785 x 2102mm2 x H, , 3056 x 3 x 1000 mm3, H=, = 264.82 mm, 0.785 x 2102 mm2, ,  TSA : 2Rh + 2rh + 2(R2 - r2), , R = outer radius, , L = Slant height =, , r = inner radius, Lateral surface area =, , D = outer diameter, d = inner diameter, , 1,  X 210 X 284.9mm2, 2, , = 94017 mm2 = 940.17 cm2, , h = height of cylinder, , 4 Determine its diameter in mm if the height of a rod, of 1.6 metres and its volume is 1.017 metre3., , t = thickness, Mean dia =, , x Area of base x height, , D−d, , V=AxH, , 2, , 1.017 metre3 = 0.785d2 x 1.6 metres, , If thickness given then:, Volume of hollow cylinder = x mean dia x thickness x, height, , 0.785d2, , =, , 1.017, metre2, 1.6, , Example, , d2, , =, , 1.017, metre2, 1.6 x 0.785, , =, , 1.017, metre2, 1.6 x 785, , =, , 10170, metre, 16 x 785, , 3 2, xa xh, 4, , =, , 10170, 12560, , 3, , = 0.8097, , 1 Find the volume of an hexagonal prism having, its side 20 cm and height 200 cm., Side of hexagonal prism (a) = 20 cm, Height (h), = 200 cm, Volume (V) = Base side area x Height, =, , 6x, , =, , 6x, , x 20 x 20 x 200, , 4, = 1,20,000 x 3, = 1,20,000 x 1.732, =, , d, , = 0.8998, = 899.8mm, , 2,07,840 cm3, , Volume of the hexagonal prism = 2,07,840 cm3, , 146, , Workshop Calculation & Science : (NSQF) Exercise 1.8.43, , Copyright free, under CC BY Licence

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Simple machines - Effort and load, mechanical advantage, velocity ratio,, efficiency of machine, relationship between efficiency, velocity ratio and, mechanical advantage, Exercise 1.9.44, Introduction:, A machine is a tool containing one or more parts that, uses energy to perform an intended action. Machines are, usually powered by mechanical, chemical, thermal, or, electrical means, and are often motorized. Historically, a, power tool also required moving parts to classify as a, machine. However, the advent of electronics has led to, the development of power tools without moving parts that, are considered machines., A simple machine is a device that simply transforms the, direction or magnitude of a force, but a large number of, more complex machines exist. Examples include, 1 Levers, , Efficiency =, , Output, Input, , % Efficiency =, , Relation between M.A., V.R. and , Efficiency =, , Load x Distance moved by the load, Output, = Effort x Distance moved by, Input, the effort, , Load Distance moved by the load, = Effort x Distance moved by, the effort, , = Mechanical advantage x, , 2 Screw Jack, Efficiency () =, , 3 Wheel and axle, 4 Pulleys, Load (or) Weight, The force overcome by the effort is called load or weight, (W)., Effort (or) power :, The force applied to lift the load is called effort or power, (P)., Fulcurm :, It is a fixed point in the machine around which the machine, rotates (F)., Mechanical advantage, In a simple machine when the effort (P) balances a load, (W) the ratio of the load to the effort is called the mechanical, advantage of the machine. It is simply expressed in a, number., Mechanical advantage (M.A), , 1, Velocity ratio, , Mechanical advantage M.A., =, %, Velocity ratio, V.R., , Ideal Machine, In an ideal machine the mechanical advantage is equal to, the velocity ratio. so, efficiency is 100% or unity., , 5 Inclined plane, etc.,, , Load, , W, , = Effort = P, , Velocity ratio, It is the ratio between the distances moved by the effort to, the distance moved by the load. It is also experssed in a, number., Velocity ratio =, , Output, x 100 %, Input, , Distance moved by the effort (dp), Distance moved by the load (dw), , Efficiency of Machine, The ratio of output to the input of machine is known as, efficiency. In simple machines, the ratio of mechanical, advantage to the velocity ratio is also known as efficiency, of machine. Efficiency is generally expressed in, percentage., , Examples, 1 Calculate, mechanical advantage, velocity ratio, and efficiency if a machine mass of 120 kg is lifted, to a height of 5 metre by a force of 60 kg. moving, 15 m., Load (w) = 120 kg, Distance moved by load = (dw) = 5 m, Power (P) = 60 kg, Distance moved by power (dp) = 15 m, MA =, , W 120 kg, =, =2, P, 60 kg, , dp 15, =, =3, dw 5, MA, Efficiency (η) =, x 100%, VR, 2, = x 100%, 3, VR =, , = 66.66 %, 2 Calculate the mechanical advantage and, efficiency of machine, if effort applied 300 kg a, load of 900 kg is lifted by a simple machine having, a velocity ratio of 4., Load (W), Effort (P), Velocity ratio (V.R), , = 900 kg, = 300 kg, = 4, , Mechanical advantage (M.A.) =, , 148, , Copyright free, under CC BY Licence, , Load (W), Effort (P)

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900, 300, = 3, =, , Efficiency (), , M.A., x 100%, V.R., 3, x 100%, =, 4, = 75 %, , Power, , W, P, M.A., Work done, HP, Height, t, , =, =, =, =, =, =, =, , 180 kg, 15 kg, ?, ?, ?, 6m, 27 second, , M.A., , =, , M.A. =, , Work done, , =, , F x d (Force x Distance), , =, , 15 kg x 6 m, , =, , 90 m - kg, , W 180 kg, =, = 12, P, 15 kg, , Workdone / time, , =, , 90 m - kg / 27 s, , = 90/27m-kg/s [75m-kg/sec=1HP], , =, , 3 Find Mechanical advantage. Calculate the work, done and horse power required if it is required to, be raised to a height of 6 m in 27 sec.and using a, pulley block, a weight of 180 kg is raised with a, force 15 kg., , =, , =, , 90/27 x 1/75 H.P., , =, , 0.04444 H.P., , 4 Calculate the applied force if a Load of 400 kg is, lifted by a machine having an  of 72%. If velocity, ratio = 6?, W, , V.R., , =, =, =, , 400 kg, 72 %, 6, , , , =, , M.A., x 100%, V.R., , 72, , =, , M.A., x 100 %, 6, , M.A., , =, , 72 x 6, 100, , =, , 4.32, , =, , 4.32, , =, , 400, = 92.59 kg., 4.32, , W, P, 400 kg, P, , Applied Force P, , Assignment, 1 Calculate the efficiency of a machine having velociy, ratio 5 if a force of 275 kg applied to lift a weight of, 1100kg with the help of simple machine., 2 Calculate (i) Mechanical advantage (ii) Efficiency of, machine if the effort applied is 250 kg and a load of 1000, kg is lifted by a simple machine having a velocity ratio, of 5., 3 What effort would be required and what would be the, mechanical advantage if a lifting machine having a, velocity ratio of 25, lifts a load of 40 Kg with an efficiency, of 54.4% ?, 4 Find out the effort required if the velocity ratio of a weight, lifting machine is 20. If the efficiency of the machine is, 40%., , 5 Find the mechanical advantage if using a pulley block, a load of 350 N is raised with a force of 25N., 6 Calculate M.A.  of the machine, if the effort applied is, 250 kg and a load of 1000 kg is lifted by a simple, machine having a velocity ratio 5., 7 Find out the  of the machine of effort applied is 300 kg, if load of 1200 kg is lifted using simple machine having, a velocity ratio of 5., 8 Calculate M.A. and efficiency if in a simple machine the, velocity ratio is found to be 20. An effort of 20 kg is, required to lift a load of 400 kg., 9 What is the velocity ratio, if its efficiency is 0.75 and in, a lifting machine an effort of 31 kg just raises a load of, 1000 kg?, , Workshop Calculation & Science : (NSQF) Exercise 1.9.44, , Copyright free, under CC BY Licence, , 149

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Lever & Simple machines - Lever and its types, , Exercise 1.9.45, , Lever, A lever is a rigid rod which rotates about a fixed point, called the fulcrum., E.g. : Cutting plier, A pair of scissors, Crow bar, Beam, balance, Hand pump., The distance of the load from the fulcrum is called the, load arm. The distance of the effort from the fulcrum is, called the effort arm., Principle of Lever, •, , All levers are functioning in the following principle, Load x Load arm = Effort x Effort arm, , •, , Classification of lever, , In this type of lever, the mechanical advantage will be, greater than 1 (M.A. > 1). Less effort is used to lift more, load., Third order lever, In this type, the effort lies between the fulcrum and the, load., , 1. Straight lever, 2. Curved lever, , E.g. The human force arm, forceps, broom, fire tongs,, fishing rod., , 1. Straight lever, There are three types :, 1. First order lever, 2. Second order lever, 3. Third order lever, First order lever, In this type the fulcrum lies between the load and the, effort., E.g : A pair of scissors, See-saw, Crow bar, Beam balance, Hand pump, etc.,, , In this type of lever, the mechanical advantage will be less, than 1 (M.A < 1) more effort is used to lift less load., Bell cranked levers (Curved levers) (Fig 5), In addition to the above types of levers, two rods may be, joined together at an angle to increase leverage without, utilising much space. Such levers are cranked levers and, the special form inwhich included angle is 90°, is called the, bell cranked lever., E.g : Motor cycle breaks system clutch pedal., , In this type of lever the mechanical advantage will be equal, or less than or greater than 1 (M.A < = > 1), Second order lever, In this type, the load lies between the fulcrum and the, effort., E.g : Nut crakers, Wheel barrow, Paper sheet cutter, Bottle, openers, Lime squeezer, etc.,, , 150, , Copyright free, under CC BY Licence

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Examples, 1 Calculate the load at B, if the load is in the balance, condition if a rod AB is 8 metre long and has got a, weight of 10 kg at A. The fulcrum is 3 metre fromB., Load x Load arm, 10 x 5, 50, P, , =, =, =, =, =, , Effort x Effort arm, Px3, 3P, 50 / 3, 16.67 kg, , When load and effort are not given separately, in the sum consider which one having more, weight is as a load., 2 Find the effort required and mechanical advantage of the system if a weight of 3000 kg is to be, lifted by a bar of length 3 metre. The load arm is, 1 metre and the effort arm is 2 metre., , As per lever principle, Load x Load arm, 3000 x 1, 3000, P, , =, =, =, =, =, , Effort x Effort arm, Px2, Px2, 3000/2, 1500 kg, , Workshop Calculation & Science : (NSQF) Exercise 1.9.45, , Copyright free, under CC BY Licence, , 151

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Mechanical advantage, , =, , Load, 3000, =, Effort 1500, , = 2, 3 According to Fig. the lever has to support a 100 kg, load with a 17 kg equivalent force supplied to it., Find the distance between the load and point of, force., , W, , = 2.1 kg, , Tension, , = T kg, , P x dp, , = 2.1 x dv, 80, , T kg x (80 - 15) cm = 2.1 kg x 2 cm, Solution., , T x 65, , = 2.1 x 40, , T, , =, , Tension, , = 1.292 kg, , Load = 100 kg; Effort = 17 kg., Load arm = 50 cm, Let effort arm = x cm, As per principle of levers:, , 2.1 x 40, kg., 65, , 5 In the figure given below in bell cranked lever AFB, on perpendicular AF the force P is 40 kg. Weight, W is on perpendicular FB. Find the measure of W., , Effort x Effort arm = Load x Load arm, 17x, , =, , 100 x 50, , x, , =, , 100 x 50, = 294.1 cm, 17, , x, , =, , 294.1 cm, , Distance between the load and point of force = 294.1 - 50, =, , 244.1 cm, , =, , 2.4410 m, , 4 Find the tension of the string if an uniform bar of, length 80 cm and weighing 2.1 kg is supported on, a smooth peg at one end and by a vertical string, at a distance of 15 cm from the other end., , Solution. By principle of momentum, P x AF, 40 x 20, , =, =, , W x BF, W x 15, , W, , =, , 160 = 53.3 kg., 40 x 20, =, 3, 15, , Assignment, 1, , 1st Order Lever (Pliers), Data given, Effort at F1 = 90 N, Arm r1 = 380 mm, Arm r2 = 36 mm, , Find, , 2, , Cutting Force F2 = _______ N, , 152, , Workshop Calculation & Science : (NSQF) Exercise 1.9.45, , Copyright free, under CC BY Licence, , Brake Lever, Foot force = 500 N, Arm r1 = 210 mm, Arm r2 = 70 mm, Find, Force on Master Cylinder = ______ N

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3, , 2nd Order Lever (Brake lever), Data given, , b Which order belongs to a pair of sugar tongs., , Load Arm = 60 mm, , c Which order belongs to carburettor Throttle Valve., , Effort Arm = 270 mm, , d Which order belongs to a common balance., , Foot force F1 = 600 N, , e Which order belongs to a pair of scissors., f, , Find, Force on MC Piston F2 = _______ N, 4, , Brake Lever, Data given, Lever Arm Ratio = 250:50, Force on MC Piston = 1800N, Find, Foot force = _______N, , 5, , 7 a Which order belongs to forearm of a human body., , Brake Lever, Data given, , Which order belongs to a safety valve., , g Which order belongs to a Crow bar., h Which order belongs to a Brake lever., 8 Find out the values aganist the question mark., Types of lever, , Load Effort Load Effort M.A., arm arm, , Ist order, , 30 kg 20 kg, , 3m, , ?, , ?, , IInd order, , 25 kg 15 kg, , ?, , 2m, , ?, , 1m, , 2m, , ?, , Bell cranked, lever, , ?, , 25 kg, , 9 a What is the principle of levers?, , Effort Arm = 30 cm, , b Write two examples of first order lever., , Load Arm = 6 cm, , c Write two examples of second order lever., , Pedal force = 500 N, , d Write two examples of third order lever., , Dia of MC Piston =, 3.2 cm, , e Which order belongs to bell cranked lever., , Find, , f, , a, , Force on MC piston = _______ N, , g What is the Velocity ratio?, , b, , Pressure in the line = ______ N/cm2, , h What is the Efficiency?, , 6 a Which order both arms are equal in length., b Which order Effort arm is longer., c Which order Effort arm is shorter than load arm., , What is the Mechanical advantage?, , 10 A forceps of 8 cm length is used to apply a foce of 100, gm. Find out the force required if the forceps are held, at 5 cm from the fulcurm., , Workshop Calculation & Science : (NSQF) Exercise 1.9.45, , Copyright free, under CC BY Licence, , 153

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Trigonometry - Measurement of angles, , Exercise 1.10.46, , Introduction:, , Circumference makes an angle (2r) = 360°, , Trigonometry is the branch of mathematics which deals, with the study of measurement and relationship of the, three sides and three angles of a triangle., , Radius of the circle makes an angle (r) = 1 Radian, C, , ie :, , Units:, , r, , Measurement of Angles, , 2π r, , There are three systems of measuring the angle:, , =, , r, , (i) Sexagesimal System, This is called British System. In this system, one right, angle is divided into 90 equal parts which are called, degrees.Each part is divided into 60 parts which are called, minutes.Each minute is divided into 60 parts which are, called seconds.The parts so divided respectively are called:, , 360, , =, , 2, , 1Radian, , 360, , 1Radian, , 360, , =, , 180, , 1 Radian =, , 1 degree (1°) = 60' (60 minutes), , (ii) Centesimal System, This is called French System. In this system, the right, angle is divided into 100 equal parts which are called, grades. Each grade is divided into 100 minutes and each, minute is divided into 100 seconds., Parts so divided are respectively called:, One grade (1 g), one minute (1' ), one second (1")., It means 1 right angle = 100 grades (100g), 1 grade (1 g) = 100 minutes (100’), 1 minute (1') = 100 seconds (100"), 90° = 100g (because each is a right, angle), This system is easier than Sexagesimal System. But to, use this system many other systems will have to be devised that is why this system is not used., , , , 1Radian, , 1° =, , , , π, , It means 1 right angle = 90° (90 degrees), , In Trigonometry, mostly this system is used., , , , 2 Radian = 360°,  Radian = 180°, , One degree (1°), one minute (1') and one second (1"), , 1 minute (1') = 60" (60 seconds), , , , π, 180, ,  Radian, , Examples, 1 Convert 45°36’20” into degree and decimal of degree., 60 second, , = 1 minute, 20, , 20 second, , =, , 60 minute, , = 1 degree, , 36.333 minute =, 45036’20”, , 60, , = 0.333’, , 36.333, 60, , = 0.606°, , = 45.606°, , 2 Convert 24.59° into degree, minute and second, 1 degree, , = 60 minute, , 0.59 degree, , = 0.59 x 60 = 35.4’, , 1 minute, , = 60 second, , 0.4 minute, , = 60 sec x 0.4, , (iii) Circular System, , = 24”, , In this system, the unit of measuring angles is radian. It, is that angle which is formed at the centre and is formed, of an arc of length equal to radius in a circle., There is one constant ratio between the circumference, and dia of a circle. This is represented by  ., , Circumference, = constant point = , Diameter, Circumference, , =  x dia, = 2r (where r is radius, of the circle), ,  =, , Therefore 24.59° = 24035’24”, 3 Change 50037’30” into degrees, By changing angle degrees into decimals, 30”, , =, , 30, = 0.50’, 60, , 37’30” = 37.5’, 37.5’, , =, , 37.5, = 0.6250, 60, , 50037’30” = 50.6250, , 154, , Copyright free, under CC BY Licence

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4 Convert 230 25’ 32” into radians, , 4, 180 4, π radian =, x π degree, 7, π 7, , We know 10 = 60’ = 3600”, Therefore 23025’32”, , 25, 32 ⎞, ⎛, = ⎜ 23 +, +, ⎟ degrees, 60 3600 ⎠, ⎝, 82800 + 1500 + 32, =, 3600, 84332, =, 3600, = radians, , But 1800, , = 102.9 degree, = 1020 0.9 x 60’, = 1020 54’, 8 Convert 0.8357 radian into degrees, 1 radian =, , 180, 0.8357 radian = π x 0.8357 degree, , Therefore 23.4255 dgrees, , = 47.880, = 470 0.88 x 60’, = 470 52.80’, = 470 52’0.8 x 60”, = 470 52’48”, , 23.4255, π radians, 180, 23.4255 22, x, radians, =, 180, 7, =, , = 0.4089 radians, 5 Convert 87 19’ 57” into Radian., , 180, degree, π, , 9 Convert 2.752 radian into degrees, , 0, , 19’57”, , =, , 1 Radian, , = 19’ + 0.95’, , 2.7520 radian =, , = 19.95’, 87°19.95’ = 87° +, , 87.33°, , =, =, , π, radian, 180, , 10 Convent, , 3, π radian into degrees, 5, , 1 Radian =, , π, x 87.33 radian, 180, , = 1080, , 6 Convert 67°11’43” into Radian, = 11’ +, , 43 ′, 60, , = 11’ + 0.716’, = 11.72’, 67°11.72’, , 1°, 67.2°, , 7 Convert, , 4, 7, , 11.72o, = 67° +, 60, = 67° + 0.195°, = 67.2°, , =, , π, radian, 180, , π, =, x 67.2 radian, 180, = 1.173 radian, , π radian into degrees, , 1 radian =, , 180, degree, π, , 3, 3, 180, π radian =, x π degree, 5, 5, π, , = 1.524 radian, , 11’43”, , 180, x 2.752 degree, π, , = 157.70, = 157.70 x 60’, = 1570 42’, , 19.95o, 60, , = 87° + 0.332° = 87.33°, 1°, , 180, degree, π, , 57 ′, = 19’ +, 60, , Assignment, Convert into Degree, 1. 12 Radian, Convert into Radians, 2. 78o, 3. 47020', 4. 520 36' 45", 5. 250 38", Convert into degree, minute and seconds, 6. 46.723o, 7. 68.625o, 8. 0.1269 Radians, 9. 2.625 Radians, , 180, 10.3/5 Radians, degree, π, Workshop Calculation & Science : (NSQF) Exercise 1.10.46, , Copyright free, under CC BY Licence, , 155

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Trigonometry - Trigonometrical ratios, Dependency, The sides of a triangle bear constant ratios for a given, definite value of the angle. That is, increase or decrease in, the length of the sides will not affect the ratio between them, unless the angle is changed. These ratios are trigonometrical ratios. For the given values of the angle a value of the, BC AC BC AB AB, AC, ,, ,, ,, ,, and, do not change even, AB AB AC BC AC, BC, when the sides AB, BC, AC are increased to AB', BC' and, AC' or decreased to AB", BC" and AC"., , ratios, , For the angle, AC is the hypotenuse, , Exercise 1.10.47, AC, 1, 1, , , AB AB cos θ, AC, , sec θ , , 1, AB, 1, , , BC BC tanθ, AB, , cot θ , , sideBC a, , sideAC b, , sin θ , , side AB c, , sideAC b, , cos θ , , a, a b a, b,   x , cos θ c b c c, b, , AB is the adjacent side, , sin θ, , BC is the opposite side., The ratios, , side BC,  tan θ, side AB, , =, sin θ , , 1, 1, or cosec θ , or sin θ.cosec θ  1, cosec θ, sin θ, , cos θ , , 1, 1, or sec θ , or cos θ . sec θ  1, sec θ, cos θ, , tan θ , , 1, cot θ, , or cot θ , , 1, tan θ, , or cot θ . tan θ  1, , By pythogoras theorem we have, AC2 = AB2 + BC2, The six ratios between the sides have precise definitions., Sine θ , , BC, AC, , Cosine θ , , AB, AC, , Tangent θ , , Cosecant θ , , Secant θ , , , , Opposite side, Hypotenuse, , AB, AC, BC, , AC, AB, , Cotangent θ , , Adjecent side, , , , BC, , , ,  Sin θ, , Hypotenuse, , Opposite side, , , , Adjecent side, , , , Hypotenuse, Opposite side, , Hypotenuse, Adjecent side, , AB, BC, , , ,  Tan θ, ,  Cosec θ, ,  Sec θ, , Adjecent side, Opposite, , Relationship between the ratios, Cosec θ , ,  Cos θ, , 1, AC, 1, , , BC BC sin θ, AC, , side, ,  Cot θ, , Dividing both sides of the euation by AC2, we have, AC 2, AC, , 2, , =, , AB 2, AC, , 2, , +, , BC 2, AC, , 2, , ⎡ AB ⎤ ⎡ BC ⎤, =, ⎢⎣ AC ⎥⎦ + ⎢⎣ AC ⎥⎦, , 2, , 2, , 1 = (cos  ) 2 + (sin ) 2, sin 2  + cos 2 = 1, Sine, Cosine, Tangent, Cosec, Sec and, Cotangent are the six trigonometrical ratios, tan , , Sin , Cos , , and sin 2  + cos 2 , = 1, , 156, , Copyright free, under CC BY Licence

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sin2  + cos2  = 1, It can be transformed as, sin2  = 1 – cos2 , sin  =, , tan C , , AB, BC, , A +, , B+, , C = 1800, , 1 – cos 2 , , or cos2  = 1 – sin2 , cos  =, , tan  =, , C = 900 -, , A, , B = 900, 1 – sin , , sin , 1 - cos 2 , =, cos , cos, , sinA =, , 1 – sin 2 , , BC, AC, , = CosC, , sinA = cosC = cos (90 - A), Sin  = cos (90 - ), , sin , , We know sin2  + cos2  = 1, , cos A =, , AB, = Sin C, BC, , Cos A = Sin C = Sin ( 90- A), , Dividing both sides by cos2 ., =, , C, , 2, , tan  =, , 2, 2, sin θ cos θ, +, 2, 2, cos θ cos θ, , A = 900 -, , Cos  = Sin 90 - , , 1, 2, , cos , , or 1 + tan2  = sec2 , , The values of the trigonometrical ratios, When  = 0° (Fig 4), , Using the same euation, sin2  + cos2  = 1., Dividing both sides by sin2 ,, , , , 1, , cos 2 , 2, , sin , , =, , 1, sin 2 , , 1 + cot2  = cosec2 , 1 + tan  = sec , 2, , 2, , When  comes closer and closer to 0°, point C approaches, B closer and closer and when  = 0°, point C coincides on, B so that BC = 0 and AB = AC., , Relation between the trignometrical ratio (Fig 3), =0, , sin0° =, cos0° =, tan0° =, , AB, AC, BC, AB, , =1, =, , 0, AB, , =0, , When  = 30° (Fig 5), , CB is extended to D to make BD = CB. AD is joined. The, two right angled triangles ACB and ADB are congruent., AC = AD. The triangle ACD is an euilateral triangle., If AC = x, CB = x, , 2, , Workshop Calculation & Science : (NSQF) Exercise 1.10.47, , Copyright free, under CC BY Licence, , 157

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3, , Then AB =, , 2, , Extend AB to D such that BD = AB. Join CD . The two, triangles ACB and DCB are congruent., , x, , side AC = side DC. The triangle ACD becomes an, euilateral triangle., , 1x, sin  = CB = 2 = 1, AC, x, 2, , Let side AC = x ,, , 3x, 3, cos  = AB, 2, =, =, AC, x, 2, , Then AB =, , 3x, 3, 2, =, =, AC, x, 2, , sin  = BC, , 1, , x, 1, 2, 1, = 2 = x, =, 2, AB, 3, 3, 3, x, 2, AC = x, 1, x, CB =, 2, CB, , tan  =, , 3, , AB =, , 2, , 1, 3, x, and BC =, x, 2, 2, , 1x, , cos  = AB = 2, , AC, , x, , 3, , tan  =, , x, , BC, , = 2, 1, AB, , =, , 3, , 2, , x, , 2, , 1, , 1, 2, , x, , x, , 2, , When  = 45° (Fig 6), , =, , 3, , =, , 1, , = 3., , When  = 90° (Fig 8), , CAB = 45°, , ACB = 45°, , Triangle ACB is a right angled isosceles triangle., side AB = BC . Let it be x ., Then AC the hypotenuse =, 1, x, = ., sin =, 2, 2x, cos =, tan =, AB = BC = x, AC =, , x, 2x, , =, , x, = 1., x, 2x, , 1, 2, , 2x, , When  becomes closer and closer to 90°, point A goes, closer and closer to B and when  = 90° point A coincides, with B , making AC = BC and AB = 0., sin  =, cos  =, tan  =, , When  = 60° (Fig 7), , 158, , =1, , Ratio, , 0o, , sin , , 0, , cos , , 1, , tan , , 0, , =0, BC, , =, , AB, , BC, , 30o, , 0, , =, 45o, , 60o, , 90o, , 1, , 3, 2, , 1, , 2, , 1, 2, , 0, , 1, , 3, , ¥, , 2, , 3, 2, 1, 3, , Workshop Calculation & Science : (NSQF) Exercise 1.10.47, , Copyright free, under CC BY Licence, , 1

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When  increases,, Sine value increases;, Cosine value decreases;, Tangent value increases to more than 1 when the, angle is more than 45o (tan60o = 1.732), Sine of an angle = Cosine of its complementary, angle, Cosine of an angle = Sine of its complementary, angle, , Angles and ratios at four quadrants, Examples, , 1st quadrant (0° to 90°) (Figs 11 & 12), , 1, If sin 300 =, find the value of sin 600, 2, , By applying pythagores theorem, BC2 = AC2 - AB2, , BC2 = 22 - 12, =4-1, =3, BC =, sin 600 =, Cosθ , , 3, Find the other trigonometrical ratios, 5, , By applying pythagores theorem, AB2 = AC2 - BC2, = 5 2 - 32, , = 25 - 9, , = 16, , + PQ, = + sin, + OP, , AB, Now, , = 16 = 4, , + OQ, = + cos, + OP, + PQ, = + tan, + OQ, , Workshop Calculation & Science : (NSQF) Exercise 1.10.47, , Copyright free, under CC BY Licence, , 159

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sec (3600 - ) = sec , , Simplify:, , sin ( 1800 + ) = - sin , , Cot  + tan ( 1800 + ) + tan (900 + ) + tan (3600 - ), , cot (900 + ) = - tan , , tan (1800 - ) = tan , , cos 90  θ  sec  θ  tan 180  θ , , sec 360  θ  sin180  θ  cot 90  θ , , tan (900 + ) = - cot , tan (3600 - ) = - tan , cot + tan (1800 + ) + tan (900 + ) + tan ( 3600 - ), , =, , ( −sinθ)( secθ)( tanθ), (sec θ)( −sinθ)( − tanθ), , cot + tan  - cot  - tan = 0, , =1, , Assignment, 1 Given sin 300 = 1/2, find the value of tan 600, , 9 What is the value of, , 2 If cos  = 4/5, find the other radios, 3 If sin A = 3/5, find cos ,tan & sec , 4 If tan  = 24/7, find sin andcos , 5 Find the value of cos  and tan if sin = 1/2, 6 If cos  = 5/13, find the value of tan , , Simplify :, 1, , tan (90 + A) + (tan 180 + A) tan (90 + A), , 2, , cos (90 + θ ) ⋅ sec (-θ ) ⋅ tan (180 - θ ), sec (360 + θ ) ⋅ sin (180 + θ ) ⋅ cot (90 + θ ), , 7 If sin  = 1/2, find the value of sin2cos2, 8 i, , What is the value of cos  and tan , sin, , 4, 5, , ii What is the value of sin  and cos , tan , , What is the value of, 3, , Sin 1600, , 4, , Sin 4500, , 5, , Cos 1350, , 6, , tan 2600, , 12, 5, , Workshop Calculation & Science : (NSQF) Exercise 1.10.47, , Copyright free, under CC BY Licence, , 161

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Trigonometry - Trigonometrical tables, cos 43°– 41', , Use of trigonometrical tables, Minutes from 0 to 4, , Mean difference, , Deg. 0' 6' 12' 18' 24' 54' 1', , 2' 3' 4' 5', , 0, , ., , ., , 1, , ., , ., , 2, , ., , ., , 3, , ., , ., , ., , ., , ., , ., , ., , ., , ., , ., , ., , 26, , .. .. .., , x .., , .., , Exercise 1.10.48, , .., , = value for cos 43°.36 ', – the value given for, mean difference of 5 ' = 0.7242 – 0.0010, = 0.7232, When reading sine value add the mean difference value. When reading cosine value subtract the mean difference value., , Arrangement, Values of trigonometrical ratios can be taken from mathematical tables., The left hand vertical column consists of degrees., The top horizontal column is arranged in minutes in steps, of 6' from 0' to 54'. In the extreme right horizontal columns, the mean differences are written in minutes from 1' to 5' in, steps of 1' to account for angles with minutes between the, interval of 6'., , 5, , ., ., , •, , ., 89, , •, Sine value for 26°– 20', •, , Refer to Natural sine table., Degrees column go up to 26° down, , •, , Minutes column 18' horizontal and under this note the value, which is given as 0.4431., , •, , Under mean difference for 2' in the same horizontal line 5, is given. Add this to the extreme right number noted for, 26°– 18'., Sine 26° – 20' = 0.4431 + .0005 = 0.4436, Cosine value for 43° – 41', Referring to the Natural cosines table for 43° –36' it is, given as 0.7242 and the mean difference for 5' minutes is, given as 10., , The values of cosine, cosecant and cotan, gent decrease when the value of the angle, increases., For sine, secant and tangent, the value, increases when the angle increases., The value of sine and cosine will never be, more than 1., The value of secant and cosecant will never, be less than 1., The value of Tan and Cot ranges from 0 to , , EXAMPLE, From the tables obtain the cosine of 45°–20 '., cos 45°–18 ' = 0.7108, mean difference for 2 ' = 0.0004, cos 45°– 20 ' = 0.7104, , SINE TABLE, 1, , Sin 250 = 0.4226, , 2, , Sin 170 5', , 0, , sin 17 45' 13", sin 17 46' = 0.3051, 0, , sin17 45' = 0.3048, Difference 1’ = 0.0003, 0, , sin 17 = 0.2924, Difference 5’ =, 14, 0, , sin 17 5' = 0.2938, 0, , Ans, , 162, , Copyright free, under CC BY Licence

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1' (or) 60", 13", , 0.6800 = sin 420 51’, 0.00015 =, 30”, ----------------------------------0.68015 = sin 420 51’30”, ---------------------------------- = 420 51’30”, , = 0.0003, 0.0003, =, × 13, 60, 0.0039, =, 60, 0.00039, =, 6, = 0.000065, , 4. sin , = 0.84756, 0.8471 = sin 570 54’, 0.0003 =, 2’, -----------------------------0.8474 = sin 570 56’, 0.8476 = sin 570 57’, -----------------------------, , sin 170 45' = 0.3048, 13" = 0.000065, sin170 45'13" = 0.304865, , Ans., , Difference 0.0002 = 1’ (or) 60”, 60, X 0.00016, 0.0002, = 60 × 16, 20, = 48", , 0.00016 =, , 0, , 4 sin82 14' 18", sin 82 15', 0, , = 0.9908, , sin 82 14' = 0.9908, Difference 1’ = 0, 0, , 0.8474 = sin 570 56’, 0.00016 =, 48”, ----------------------------------0.84756 = sin 570 56’48”, -----------------------------------, , 1' (or ) 60" = 0, = 0, sin18", sin820 14' = 0.9908, 18" = 0.0000, sin820 41'18" = 0.9908, ,  = 570 56’48”, 5. sin , = 0.6, , Ans., , Finding the corresponding angles when sine, values are given:, 1. Sin  = 0.9925,  = 830, 2. Sin  = 0.8791, 0.8788 =, , Sin 610 30’, , 0.0003 =, 2’, ------------------------------------------0.8791 =, Sin 610 32’, 3. sin  = 0.68015, 0.6794 = sin 420 48’, 0.0006 =, 3’, -------------------------------0.6800 = sin 420 51’, 0.6803 = sin 420 52’, --------------------------------Difference 0.0003 = 1’ (or) 60”, , 0.00015 =, =, =, , 0.5990, = sin 36o 48’, 0.0009, =, 4’, -------------------------------------0.5999, = sin 36o 52’, 0.6002, = sin 36o 53’, -------------------------------------difference 0.0003, 0.0001, , =, =, , 1’ (or) 60’, 60, 0.0003, , x 0.0001=, , 60, x1, 3, , 0.5999, = sin 36o 52’, 0.0001, =, 20”, ---------------------------------------------0.6000, = sin 36o 52’20”, ---------------------------------------------, , =, , 36o 52’ 20”, , Calculations involving tapers, D - Big diameter of the taper, , 60, X 0.00015, 0.0003, 60 × 15, 30, 30", , d - small diameter of the taper, C - Taper Ratio - 1:x, , Workshop Calculation & Science : (NSQF) Exercise 1.10.48, , Copyright free, under CC BY Licence, , 163

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or, 1 Ratio of inclination =, , of the taper ratio., , Setting angle, One of the methods of turning taper is by swivelling the, compound slide to an angle known as setting angle and, feeding the tool at an angle to the axis of work., , =, α, is the setting angle which is eual to half of the, 2, included angle of the taper., , NOTE:, , Taper Ratio C = 1 : x or D : l or (D –d) : l, Ratio of inclination, , C, =1:2x =, 2, , : l or, , :l, , Setting angle determination is by the formula, , C, Ratio of inclination - 1:2 x, 2, l - length of taper, , a - included angle of taper, , , , - setting angle, 2, Taper ratio = Ratio of inclination (for wedges)., EXAMPLE, Taper ratio, The ratio between the difference in diameter to the length, of the taper is known as taper ratio. D is the difference in, larger diameter shown in the sketch as the small diameter, of taper is 0. Taper ratio is D : l . In the sectioned portion, the difference in diameter is 1 and the length of taper is, shown as x ., C = D: d = 1 : x as per Fig 1 (a), C =, , C = 1:8 =, , as per Fig 1 (b), , 1 80, =, = 10 mm., 8, 8, D – 30 mm = 10 mm, ∴D − d =, , Ratio of inclination, Taking half of the taper,, , A pivot in the form of a frustum of a cone has a taper ratio, 1:8. If the small diameter is 30 mm and length of taper is, 80 mm, find its large diameter., , is the difference in diameter, , for a taper length of l, if d = 0., , D = 10 mm + 30 mm = 40 mm, Large diameter D = 40 mm, , if the small diameter is 0, 164, , Workshop Calculation & Science : (NSQF) Exercise 1.10.48, , Copyright free, under CC BY Licence

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Cos Table, , tan Table, , 1. Cos 38, Cos 38o, , 1. tan 35O 37', , o, , = 0.7880, , tan 35O 36', , 2. Cos 83o 12', Cos 83o 12' = 0.1184, , 2. tan 50O 5', Tan 50O 0', , Cos 31o 18' = 0.8545, 2' =, 3 (-), -----------------------------------------Cos 31o 20' = 0.8542, ------------------------------------------, , 1. tan , , , , 1.1918, , = 0.3972, , 0.3959, , Find the corresponding angles when cos values are, given:, , 2. Cos , , =, , 5', =, 0.0036, --------------------------------------------Tan 50O 5' =, 1.1954, --------------------------------------------Find the corresponding angles when tan values are, given, , 4. Cos 31o 20', , , , 0.7159, , 1' =, 0.0004, --------------------------------------------tan 35O 37' =, 0.7163, ---------------------------------------------, , 3. Cos 26o 40', Cos 26o 36' = 0.8942, 4' =, 5 (-), -------------------------------------------Cos 26o 40' = 0.8937, --------------------------------------------, , 1. Cos , , =, , = 0.5150, = 59o, = 0.0192, , = tan 21O 36', , 0.0013 =, 4', --------------------------------------------0.3972 = tan 21O 40', --------------------------------------------2. tan , = 1.0065, 1.0035, , = tan 45O 6', , 0.0030 =, 5', -------------------------------------------1.0065 = tan 45O 11', --------------------------------------------, , = 88o54', , 3. Cos , = 0.9682, 0.9686 = cos 14o24', (-), 4 =, 5', -------------------------------------------0.9682 = cos 14o 29', ------------------------------------------- = 14o 29', , Problems Related with Trigonometrical tables, 1. A 250 mm Sine bar is used to measure an angle., If the difference in height is 5 cm, find the angle., , 4. Cos , = 0.8476, 0.8480 = cos 32o 0', (-) 0.0003 =, 2', -------------------------------------------0.8477 = cos 32o 2', 0.8475 = cos 32o 3', -------------------------------------------0.0002 = 1' (or) 60", 60, 0.0001 = ----------- x 0.0001, 0.0002, 60, =, ------ x 1, 2, =, 30", 0.8477 =, Cos 32o 2', (-) 0.0001 =, 30" (+), ---------------------------------------------------0.8476 =, Cos 32o 2' 30", ----------------------------------------------------, , Sin θ =, , Opp.side h, =, Hyp., l, 5 cm, 250 mm, , =, , =, , 50 mm, 250 mm, , = 0.2000,  = 11o 32', 2 Find the height of the slip gauge if a Sine bar with, plugs of 10" centre is set up to inspect a taper, having an included angle of 9o 56"., , Sin θ =, , Opp.side h, =, Hyp., l, , Sin 9o56" =, , h, 10, , h = 10 x sin 9o56", , Workshop Calculation & Science : (NSQF) Exercise 1.10.48, , Copyright free, under CC BY Licence, , 165

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3 Find the angle which the ladder makes with the, ground if the foot of a 4.5 m long ladder is placed, at 1 m away from the wall., In Right angled D, , Sin 9o 0" = 0.1564, Sin 9o 1" = 0.1567, -------------------------------------1' (or) 60" = 0.0003, , 56"=, , Cos C, , =, , BC, AC, , Cos , , =, , 1m, = 0.2222, 4.5 m, , 0.0003, x 56, 60, , 0.0168, 60, 0.00168, =, 6, =, , Cos = 0.2222, 0.2233 = cos 77o 6', (-), 0.0011 =, 4' (+), ----------------------------------0.2222 = cos 77o 10', --------------------------------- = 77o 10', , = 0.00028, sin9 0' 00" = 0.1564, o, , 56" = 0.00028, -----------------------------------sin9o 0' 56" = 0.15668, -----------------------------------h = 10 x sin9o 56", = 10 x 0.15668, = 1.5668 cm, Height of slip gauge = 1.5668", , Assignment, I, , Find the values of the given angles, , 6, , Cos  = 0.8926, , 1, , Sin 65, , 7, , Cos  = 0.11773, , 2, , Sin 42 23', , 8, , Cos  = 0.21646, , 3, , Sin 66 35' 32", , 9, , Tan  = 0.3411, , 4, , Sin 7o 15' 41", , 10 Tan  = 2.3868, , 5, , Sin 27 27", , III, , 6, , Cos 47 39', , 7, , Cos 47 39', , 1 Calculate its base. if the slant height of a cone is 12.25, cm and the vertex angle is 1100., , 8, , Cos 79o 31' 53", , 9, , Tan 28o 45', , o, , o, , o, , o, , o, o, , 2 A ladder 2.5 m long makes an angle of 600 with the, ground. Find the height of the wall where the ladder, touches the wall., , 10 Tan 67o 27' 36", II Find corrosponding angles for given values, 1, , Sin , , = 0.3062, , 2, , Sin , , = 0.6002, , 3, , Sin , , = 0.22453, , 4, , Sin , , = 0.04802, , 5, , Cos  = 0.6446, , 166, , 3 A sine bar of 200 mm is to be set at an angle of 15015'3"., Select the slip gauge block to built up the reuired, height., 4 In a right angled triangle ABC, C = 90o, If AB =, 50 mm and B = 75o, Find the remaining sides., 5 Calculate the reuired length of the bar for this point if, a centre point having an included angle of 60o is to be, turned at the end of a 50 mm dia bar., , Workshop Calculation & Science : (NSQF) Exercise 1.10.48, , Copyright free, under CC BY Licence

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Natural Sines, O, , 0', 0.0O, , 6', 0.1O, , 12', 0.2O, , 18', 0.3O, , 24', 0.4O, , 30', 0.5O, , 36', 0.6O, , 42', 0.7O, , 48', 0.8O, , 54', 0.9O, , 1', , 2', , 3', , 4', , 5', , 0, 1, 2, 3, 4, , 0.0000, 0.0175, 0.0349, 0.0523, 0.0698, , 0.0017, 0.0192, 0.0366, 0.0541, 0.0715, , 0.0035, 0.0209, 0.0384, 0.0558, 0.0732, , 0.0052, 0.0227, 0.0401, 0.0576, 0.0750, , 0.0070, 0.0244, 0.0419, 0.0593, 0.0767, , 0.0087, 0.0262, 0.0436, 0.0610, 0.0785, , 0.0105, 0.0279, 0.0454, 0.0628, 0.0802, , 0.0122, 0.0297, 0.0471, 0.0645, 0.0819, , 0.0140, 0.0314, 0.0488, 0.0663, 0.0837, , 0.0157, 0.0332, 0.0506, 0.0680, 0.0854, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 6, , 9, 9, 9, 9, 9, , 12, 12, 12, 12, 12, , 15, 15, 15, 15, 14, , 5, 6, 7, 8, 9, , 0.0872, 0.1045, 0.1219, 0.1392, 0.1564, , 0.0899, 0.1063, 0.1236, 0.1409, 0.1582, , 0.0906, 0.1080, 0.1253, 0.1426, 0.1599, , 0.0924, 0.1097, 0.1271, 0.1444, 0.1616, , 0.0941, 0.1115, 0.1288, 0.1461, 0.1633, , 0.0958, 0.1132, 0.1305, 0.1478, 0.1650, , 0.0976, 0.1149, 0.1323, 0.1495, 0.1668, , 0.0993, 0.1167, 0.1340, 0.1513, 0.1685, , 0.1011, 0.1184, 0.1357, 0.1530, 0.1702, , 0.1028, 0.1201, 0.1374, 0.1547, 0.1719, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 6, , 9, 9, 9, 9, 9, , 12, 12, 12, 11, 11, , 14, 14, 14, 14, 14, , 10, 11, 12, 13, 14, , 0.1736, 0.1908, 0.2079, 0.2250, 0.2419, , 0.1754, 0.1925, 0.2096, 0.2267, 0.2436, , 0.1771, 0.1942, 0.2113, 0.2284, 0.2453, , 0.1788, 0.1959, 0.2130, 0.2300, 0.2470, , 0.1805, 0.1977, 0.2147, 0.2317, 0.2487, , 0.1822, 0.1994, 0.2164, 0.2334, 0.2504, , 0.1840, 0.2011, 0.2181, 0.2351, 0.2521, , 0.1857, 0.2028, 0.2198, 0.2368, 0.2538, , 0.1874, 0.2045, 0.2215, 0.2385, 0.2554, , 0.1891, 0.2062, 0.2232, 0.2402, 0.2571, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 6, , 9, 9, 9, 8, 8, , 11, 12, 11, 11, 11, , 14, 14, 14, 14, 14, , 15, 16, 17, 18, 19, , 0.2558, 0.2756, 0.2924, 0.3090, 0.3256, , 0.2605, 0.2773, 0.2940, 0.3107, 0.3272, , 0.2622, 0.2790, 0.2957, 0.3123, 0.3289, , 0.2639, 0.2807, 0.2974, 0.3140, 0.3305, , 0.2656, 0.2823, 0.2990, 0.3156, 0.3322, , 0.2672, 0.2840, 0.3007, 0.3173, 0.3338, , 0.2689, 0.2857, 0.3024, 0.3190, 0.3355, , 0.2706, 0.2874, 0.3040, 0.3206, 0.3371, , 0.2723, 0.2890, 0.3057, 0.3223, 0.3387, , 0.2740, 0.2907, 0.3074, 0.3239, 0.3404, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 5, , 8, 8, 8, 8, 8, , 11, 11, 11, 11, 11, , 14, 14, 14, 14, 14, , 20, 21, 22, 23, 24, , 0.3420, 0.3584, 0.3746, 0.3907, 0.4067, , 0.3437, 0.3600, 0.3762, 0.3923, 0.4083, , 0.3453, 0.3616, 0.3778, 0.3939, 0.4099, , 0.3469, 0.3633, 0.3795, 0.3955, 0.4115, , 0.3486, 0.3649, 0.3811, 0.3971, 0.4131, , 0.3502, 0.3665, 0.3827, 0.3987, 0.4147, , 0.3518, 0.3681, 0.3843, 0.4003, 0.4163, , 0.3535, 0.3697, 0.3859, 0.4019, 0.4179, , 0.3551, 0.3714, 0.3875, 0.4035, 0.4195, , 0.3567, 0.3730, 0.3891, 0.4051, 0.4210, , 3, 3, 3, 3, 3, , 5, 5, 5, 5, 5, , 8, 8, 8, 8, 8, , 11, 11, 11, 11, 11, , 14, 14, 13, 13, 13, , 25, 26, 27, 28, 29, , 0.4226, 0.4384, 0.4540, 0.4695, 0.4848, , 0.4242, 0.4399, 0.4555, 0.4710, 0.4863, , 0.4258, 0.4415, 0.4571, 0.4726, 0.4879, , 0.4274, 0.4431, 0.4586, 0.4741, 0.4894, , 0.4289, 0.4446, 0.4602, 0.4756, 0.4909, , 0.4305, 0.4462, 0.4617, 0.4772, 0.4924, , 0.4321, 0.4478, 0.4633, 0.4787, 0.4939, , 0.4337, 0.4493, 0.4648, 0.4802, 0.4955, , 0.4352, 0.4509, 0.4664, 0.4818, 0.4970, , 0.4368, 0.4524, 0.4679, 0.4833, 0.4985, , 3, 3, 3, 3, 3, , 5, 5, 5, 5, 5, , 8, 8, 8, 8, 8, , 11, 10, 10, 10, 10, , 13, 13, 13, 13, 13, , 30, 31, 32, 33, 34, , 0.500, 0.5150, 0.5299, 0.5446, 0.5592, , 0.5015, 0.5165, 0.5314, 0.5461, 0.5606, , 0.5030, 0.5180, 0.5329, 0.5476, 0.5621, , 0.5045, 0.5195, 0.5344, 0.5490, 0.5635, , 0.5060, 0.5210, 0.5358, 0.5505, 0.5650, , 0.5075, 0.5225, 0.5373, 0.5519, 0.5664, , 0.5090, 0.5240, 0.5388, 0.5534, 0.5678, , 0.5105, 0.5255, 0.5402, 0.5548, 0.5693, , 0.5120, 0.5270, 0.5417, 0.5563, 0.5707, , 0.5135, 0.5284, 0.5432, 0.5577, 0.5721, , 3, 2, 2, 2, 2, , 5, 5, 5, 5, 5, , 8, 7, 7, 7, 7, , 10, 10, 10, 10, 10, , 13, 12, 12, 12, 12, , 35, 36, 37, 38, 39, , 0.5736, 0.5878, 0.6018, 0.6157, 0.6293, , 0.5750, 0.5892, 0.6032, 0.6170, 0.6307, , 0.5764, 0.5906, 0.6046, 0.6184, 0.6320, , 0.5779, 0.5920, 0.6060, 0.6198, 0.6334, , 0.5793, 0.5934, 0.6074, 0.6211, 0.6347, , 0.5807, 0.5948, 0.6088, 0.6225, 0.6361, , 0.5821, 0.5962, 0.6101, 0.6239, 0.6374, , 0.5835, 0.5976, 0.6115, 0.6252, 0.6388, , 0.5850, 0.5990, 0.6129, 0.6266, 0.6401, , 0.5864, 0.6004, 0.6143, 0.6280, 0.6414, , 2, 2, 2, 2, 2, , 5, 5, 5, 5, 4, , 7, 7, 7, 7, 7, , 9, 9, 9, 9, 9, , 12, 12, 12, 11, 11, , 40, 41, 42, 43, 44, , 0.6428, 0.6561, 0.6691, 0.6820, 0.6947, , 0.6441, 0.6574, 0.6704, 0.6833, 0.6959, , 0.6455, 0.6587, 0.6717, 0.6845, 0.6972, , 0.6468, 0.6600, 0.6730, 0.6858, 0.6984, , 0.6481, 0.6613, 0.6743, 0.6871, 0.6997, , 0.6494, 0.6626, 0.6756, 0.6884, 0.7009, , 0.6508, 0.6639, 0.6769, 0.6896, 0.7022, , 0.6521, 0.6652, 0.6782, 0.6909, 0.7034, , 0.6534, 0.6665, 0.6794, 0,6921, 0.7046, , 0.6547, 0.6678, 0.6807, 0.6934, 0.7059, , 2, 2, 2, 2, 2, , 4, 4, 4, 4, 4, , 7, 7, 6, 6, 6, , 9, 9, 9, 8, 8, , 11, 11, 11, 11, 10, , 45, 46, 47, 48, 49, , 0.7071, 0.7193, 0.7314, 0.7431, 0.7547, , 0.7083, 0.7206, 0.7325, 0.7443, 0.7558, , 0.7096, 0.7218, 0.7337, 0.7455, 0.7570, , 0.7108, 0.7230, 0.7349, 0.7466, 0.7581, , 0.7120, 0.7242, 0.7361, 0.7478, 0.7593, , 0.7133, 0.7254, 0.7373, 0.7490, 0.7604, , 0.7145, 0.7266, 0.7385, 0.7501, 0.7615, , 0.7157, 0.7278, 0.7396, 0.7513, 0.7627, , 0.7169, 0.7290, 0.7408, 0.7524, 0.7638, , 0.7181, 0.7302, 0.7420, 0.7536, 0.7649, , 2, 2, 2, 2, 2, , 4, 4, 4, 4, 4, , 6, 6, 6, 6, 6, , 8, 8, 8, 8, 8, , 10, 10, 10, 10, 9, , 50, 51, 52, 53, 54, , 0.7660, 0.7771, 0.7880, 0.7986, 0.8090, , 0.7672, 0.7782, 0.7891, 0.7997, 0.8100, , 0.7683, 0.7793, 0.7902, 0.8007, 0.8111, , 0.7694, 0.7804, 0.7912, 0.8018, 0.8121, , 0.7705, 0.7815, 0.7923, 0.8028, 0.8131, , 0.7716, 0.7826, 0.7934, 0.8039, 0.8141, , 0.7727, 0.7837, 0.7944, 0.8049, 0.8151, , 0.7738, 0.7848, 0.7955, 0.8059, 0.8161, , 0.7749, 0.7859, 0.7965, 0.8070, 0.8171, , 0.7760, 0.7869, 0.7976, 0.8080, 0.8181, , 2, 2, 2, 2, 2, , 4, 4, 4, 3, 3, , 6, 5, 5, 5, 5, , 7, 7, 7, 7, 7, , 9, 9, 9, 9, 8, , 55, 56, 57, 58, 59, , 0.8192, 0.8290, 0.8387, 0.8480, 0.8572, , 0.8202, 0.8300, 0.8396, 0.8490, 0.8581, , 0.8211, 0.8310, 0.8406, 0.8499, 0.8590, , 0.8221, 0.8320, 0.8415, 0.8508, 0.8599, , 0.8231, 0.8329, 0.8425, 0.8517, 0.8607, , 0.8241, 0.8339, 0.8434, 0.8526, 0.8616, , 0.8251, 0.8348, 0.8443, 0.8536, 0.8625, , 0.8261, 0.8358, 0.8453, 0.8545, 0.8634, , 0.8271, 0.8368, 0.8462, 0.8554, 0.8643, , 0.8281, 0.8377, 0.8471, 0.8563, 0.8652, , 2, 2, 2, 2, 1, , 3, 3, 3, 3, 3, , 5, 5, 5, 5, 4, , 7, 6, 6, 6, 6, , 8, 8, 8, 8, 7, , Workshop Calculation & Science : (NSQF) Exercise 1.10.48, , Copyright free, under CC BY Licence, , 167

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Natural Sines, O, , 0', 0.0O, , 6', 0.1O, , 12', 0.2O, , 18', 0.3O, , 24', 0.4O, , 30', 0.5O, , 36', 0.6O, , 42', 0.7O, , 48', 0.8O, , 54', 0.9O, , 1', , 2', , 3', , 4', , 5', , 60, 61, 62, 63, 64, , 0.8660, 0.8746, 0.8829, 0.8910, 0.8988, , 0.8669, 0.8755, 0.8838, 0.8918, 0.8996, , 0.8678, 0.8763, 0.8846, 0.8926, 0.9003, , 0.8686, 0.8771, 0.8854, 0.8934, 0.9011, , 0.8695, 0.8780, 0.8862, 0.8942, 0.9018, , 0.8704, 0.8788, 0.8870, 0.8949, 0.9026, , 0.8712, 0.8796, 0.8878, 0.8957, 0.9033, , 0.8721, 0.8805, 0.8886, 0.8965, 0.9041, , 0.8729, 0.8813, 0.8894, 0.8973, 0.9048, , 0.8738, 0.8821, 0.8902, 0.8980, 0.9056, , 1, 1, 1, 1, 1, , 3, 3, 3, 3, 3, , 4, 4, 4, 4, 4, , 6, 6, 5, 5, 5, , 7, 7, 7, 6, 6, , 65, 66, 67, 68, 69, , 0.9063, 0.9135, 0.9205, 0.9272, 0.9336, , 0.9070, 0.9143, 0.9212, 0.9278, 0.9342, , 0.9078, 0.9150, 0.9219, 0.9285, 0.9348, , 0.9085, 0.9157, 0.9225, 0.9291, 0.9354, , 0.9092, 0.9164, 0.9232, 0.9298, 0.9361, , 0.9100, 0.9171, 0.9239, 0.9304, 0.9367, , 0.9107, 0.9178, 0.9245, 0.9311, 0.9373, , 0.9114, 0.9184, 0.9252, 0.9317, 0.9379, , 0.9121, 0.9191, 0.9259, 0.9323, 0.9385, , 0.9128, 0.9198, 0.9265, 0.9330, 0.9391, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 4, 3, 3, 3, 3, , 5, 5, 4, 4, 4, , 6, 6, 6, 5, 5, , 70, 71, 72, 73, 74, , 0.9397, 0.9455, 0.9511, 0.9563, 0.9613, , 0.9403, 0.9461, 0.9516, 0.9568, 0.9617, , 0.9409, 0.9466, 0.9521, 0.9573, 0.9622, , 0.9415, 0.9472, 0.9527, 0.9578, 0.9627, , 0.9421, 0.9478, 0.9532, 0.9583, 0.9632, , 0.9426, 0.9483, 0.9537, 0.9588, 0.9636, , 0.9432, 0.9489, 0.9542, 0.9593, 0.9641, , 0.9438, 0.9494, 0.9548, 0.9598, 0.9646, , 0.9444, 0.9500, 0.9553, 0.9603, 0.9650, , 0.9449, 0.9505, 0.9558, 0.9608, 0.9655, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 3, 3, 3, 2, 2, , 4, 4, 3, 3, 3, , 5, 5, 4, 4, 4, , 75, 76, 77, 78, 79, , 0.9659, 0.9703, 0.9744, 0.9781, 0.9816, , 0.9664, 0.9707, 0.9748, 0.9785, 0.9820, , 0.9668, 0.9711, 0.9751, 0.9789, 0.9823, , 0.9673, 0.9715, 0.9755, 0.9792, 0.9826, , 0.9677, 0.9720, 0.9759, 0.9796, 0.9829, , 0.9681, 0.9724, 0.9763, 0.9799, 0.9833, , 0.9686, 0.9728, 0.9767, 0.9803, 0.9836, , 0.9690, 0.9732, 0.9770, 0.9806, 0.9839, , 0.9694, 0.9736, 0.9774, 0.9810, 0.9842, , 0.9699, 0.9740, 0.9778, 0.9813, 0.9845, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 3, 3, 2, 2, 2, , 4, 3, 3, 3, 3, , 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, , 0.9848, 0.9877, 0.9903, 0.9925, 0.9945, 0.9962, 0.9976, 0.9986, 0.9994, 0.9998, 1.0000, , 0.9851, 0.9880, 0.9905, 0.9928, 0.9947, 0.9963, 0.9977, 0.9987, 0.9995, 0.9999, , 0.9854, 0.9882, 0.9907, 0.9930, 0.9949, 0.9965, 0.9978, 0.9988, 0.9995, 0.9999, , 0.9857, 0.9885, 0.9910, 0.9932, 0.9951, 0.9966, 0.9979, 0.9989, 0.9996, 0.9999, , 0.9860, 0.9888, 0.9912, 0.9934, 0.9952, 0.9968, 0.9980, 0.9990, 0.9996, 0.9999, , 0.9863, 0.9890, 0.9914, 0.9936, 0.9954, 0.9969, 0.9981, 0.9990, 0.9997, 1.0000, , 0.9866, 0.9893, 0.9917, 0.9938, 0.9956, 0.9971, 0.9982, 0.9991, 0.9997, 1.0000, , 0.9869, 0.9895, 0.9919, 0.9940, 0.9957, 0.9972, 0.9983, 0.9992, 0.9997, 1.0000, , 0.9871, 0.9898, 0.9921, 0.9942, 0.9959, 0.9973, 0.9984, 0.9993, 0.9998, 1.0000, , 0.9874, 0.9900, 0.9923, 0.9943, 0.9960, 0.9974, 0.9985, 0.9993, 0.9998, 1.0000, , 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, , 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, , 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, , 2, 2, 1, 1, 1, 1, 1, 1, 0, 0, , 2, 2, 2, 2, 1, 1, 1, 1, 0, 0, , Quadrant, , Angle, , sinA =, , Examples, , First, , 0 to 90, , sin A, , sin 34 38' = 0.5683, , Second, , 90o to 180o, , sin(180o – A), , sin 145o22', , o, , o, , = sin(180o – 145o 22'), , = sin 34o38' = 0.5683, Third, , 180o to 270o, , –sin(A – 180o), , sin 214o38', , = –sin(214o38' – 180o), , = –sin34o38' = –0.5683, Fourth, , 270o to 360o, , –sin(360o – A), , sin 325o22', , = –sin(360o – 325o22'), , = – sin 34o38' = –0.5683, , 168, , Workshop Calculation & Science : (NSQF) Exercise 1.10.48, , Copyright free, under CC BY Licence

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Natural Cosines, Numbers in different columns to be subtracted, not added, O, , 0', 0.0O, , 6', 0.1O, , 12', 0.2O, , 18', 0.3O, , 24', 0.4O, , 30', 0.5O, , 36', 0.6O, , 42', 0.7O, , 48', 0.8O, , 54', 0.9O, , 1', , 2', , 3', , 4', , 5', , 0, 1, 2, 3, 4, , 1.0000, 0.9998, 0.9994, 0.9986, 0.9976, , 1.0000, 0.9998, 0.9993, 0.9985, 0.9974, , 1.0000, 0.9998, 0.9993, 0.9984, 0.9973, , 1.0000, 0.9997, 0.9992, 0.9983, 0.9972, , 1.0000, 0.9997, 0.9991, 0.9982, 0.9971, , 1.0000, 0.9997, 0.9990, 0.9981, 0.9969, , 0.9999, 0.9996, 0.9990, 0.9980, 0.9968, , 0.9999, 0.9996, 0.9989, 0.9979, 0.9966, , 0.9999, 0.9995, 0.9988, 0.9978, 0.9965, , 0.9999, 0.9995, 0.9987, 0.9977, 0.9963, , 0, 0, 0, 0, 0, , 0, 0, 0, 0, 0, , 0, 0, 0, 1, 1, , 0, 0, 1, 1, 1, , 0, 0, 1, 1, 1, , 5, 6, 7, 8, 9, , 0.9962, 0.9945, 0.9925, 0.9903, 0.9877, , 0.9960, 0.9943, 0.9923, 0.9900, 0.9874, , 0.9959, 0.9942, 0.9921, 0.9898, 0.9871, , 0.9957, 0.9940, 0.9919, 0.9895, 0.9869, , 0.9956, 0.9938, 0.9917, 0.9893, 0.9866, , 0.9954, 0.9936, 0.9914, 0.9890, 0.9863, , 0.9952, 0.9934, 0.9912, 0.9888, 0.9860, , 0.9951, 0.9932, 0.9910, 0.9885, 0.9857, , 0.9949, 0.9930, 0.9907, 0.9882, 0.9854, , 0.9947, 0.9928, 0.9905, 0.9880, 0.9851, , 0, 0, 0, 0, 0, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 1, 1, 1, 2, 2, , 1, 2, 2, 2, 2, , 10, 11, 12, 13, 14, , 0.9848, 0.9816, 0.9781, 0.9744, 0.9703, , 0.9845, 0.9813, 0.9778, 0.9740, 0.9699, , 0.9842, 0.9810, 0.9774, 0.9736, 0.9694, , 0.9839, 0.9806, 0.9770, 0.9732, 0.9690, , 0.9836, 0.9803, 0.9767, 0.9728, 0.9686, , 0.9833, 0.9799, 0.9763, 0.9724, 0.9681, , 0.9829, 0.9796, 0.9759, 0.9720, 0.9677, , 0.9826, 0.9792, 0.9755, 0.9715, 0.9673, , 0.9823, 0.9789, 0.9751, 0.9711, 0.9668, , 0.9820, 0.9785, 0.9748, 0.9707, 0.9664, , 1, 1, 1, 1, 1, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 2, 2, 2, 3, 3, , 3, 3, 3, 3, 4, , 15, 16, 17, 18, 19, , 0.9659, 0.9613, 0.9563, 0.9511, 0.9455, , 0.9655, 0.9608, 0.9558, 0.9505, 0.9449, , 0.9650, 0.9603, 0.9553, 0.9500, 0.9444, , 0.9646, 0.9598, 0.9548, 0.9494, 0.9438, , 0.9641, 0.9593, 0.9542, 0.9489, 0.9432, , 0.9636, 0.9588, 0.9537, 0.9483, 0.9426, , 0.9632, 0.9583, 0.9532, 0.9478, 0.9421, , 0.9627, 0.9578, 0.9527, 0.9472, 0.9415, , 0.9622, 0.9573, 0.9521, 0.9466, 0.9409, , 0.9617, 0.9568, 0.9516, 0.9461, 0.9403, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 2, 2, 3, 3, 3, , 3, 3, 3, 4, 4, , 4, 4, 4, 5, 5, , 20, 21, 22, 23, 24, , 0.9397, 0.9336, 0.9272, 0.9205, 0.9135, , 0.9391, 0.9330, 0.9625, 0.9198, 0.9128, , 0.9385, 0.9323, 0.9259, 0.9191, 0.9121, , 0.9379, 0.9317, 0.9252, 0.9184, 0.9114, , 0.9373, 0.9311, 0.9245, 0.9178, 0.9107, , 0.9367, 0.9304, 0.9239, 0.9171, 0.9100, , 0.9361, 0.9298, 0.9232, 0.9164, 0.9092, , 0.9354, 0.9291, 0.9225, 0.9157, 0.9085, , 0.9348, 0.9285, 0.9219, 0.9150, 0.9078, , 0.9342, 0.9278, 0.9212, 0.9143, 0.9070, , 1, 1, 1, 1, 1, , 2, 2, 2, 2, 2, , 3, 3, 3, 3, 4, , 4, 4, 4, 5, 5, , 5, 5, 6, 6, 6, , 25, 26, 27, 28, 29, , 0.9063, 0.8988, 0.8910, 0.8829, 0.8746, , 0.9056, 0.8980, 0.8902, 0.8821, 0.8738, , 0.9048, 0.8973, 0.8894, 0.8813, 0.8729, , 0.9041, 0.8965, 0.8886, 0.8805, 0.8721, , 0.9033, 0.8957, 0.8878, 0.8796, 0.8712, , 0.9026, 0.8949, 0.8870, 0.8788, 0.8704, , 0.9018, 0.8942, 0.8862, 0.8780, 0.8695, , 0.9011, 0.8934, 0.8854, 0.8771, 0.8686, , 0.9003, 0.8926, 0.8846, 0.8763, 0.8678, , 0.8996, 0.8918, 0.8838, 0.8755, 0.8669, , 1, 1, 1, 1, 1, , 3, 3, 3, 3, 3, , 4, 4, 4, 4, 4, , 5, 5, 5, 6, 6, , 6, 6, 7, 7, 7, , 30, 31, 32, 33, 34, , 0.8660, 0.8572, 0.8480, 0.8387, 0.8290, , 0.8652, 0.8563, 0.8471, 0.8377, 0.8281, , 0.8643, 0.8554, 0.8462, 0.8368, 0.8271, , 0.8634, 0.8545, 0.8453, 0.8358, 0.8261, , 0.8625, 0.8536, 0.8443, 0.8348, 0.8251, , 0.8616, 0.8526, 0.8434, 0.8339, 0.8241, , 0.8607, 0.8517, 0.8425, 0.8329, 0.8231, , 0.8599, 0.8508, 0.8415, 0.8320, 0.8221, , 0.8590, 0.8499, 0.8406, 0.8310, 0.8211, , 0.8581, 0.8490, 0.8396, 0.8300, 0.8202, , 1, 2, 2, 2, 2, , 3, 3, 3, 3, 3, , 4, 5, 5, 5, 5, , 6, 6, 6, 6, 7, , 7, 8, 8, 8, 8, , 35, 36, 37, 38, 39, , 0.8192, 0.8090, 0.7986, 0.7880, 0.7771, , 0.8181, 0.8080, 0.7976, 0.7869, 0.7760, , 0.8171, 0.8070, 0.7965, 0.7859, 0.7749, , 0.8161, 0.8059, 0.7955, 0.7848, 0.7738, , 0.8151, 0.8049, 0.7944, 0.7837, 0.7727, , 0.8141, 0.8039, 0.7934, 0.7826, 0.7716, , 0.8131, 0.8028, 0.7923, 0.7815, 0.7705, , 0.8121, 0.8018, 0.7912, 0.7804, 0.7694, , 0.8111, 0.8007, 0.7902, 0.7793, 0.7683, , 0.8100, 0.7997, 0.7891, 0.7782, 0.7672, , 2, 2, 2, 2, 2, , 3, 3, 4, 4, 4, , 5, 5, 5, 5, 6, , 7, 7, 7, 7, 7, , 8, 9, 9, 9, 9, , 40, 41, 42, 43, 44, , 0.7660, 0.7547, 0.7431, 0.7314, 0.7193, , 0.7649, 0.7536, 0.7420, 0.7302, 0.7181, , 0.7638, 0.7524, 0.7408, 0.7290, 0.7169, , 0.7627, 0.7513, 0.7396, 0.7278, 0.7157, , 0.7615, 0.7501, 0.7385, 0.7266, 0.7145, , 0.7604, 0.7490, 0.7373, 0.7254, 0.7133, , 0.7593, 0.7478, 0.7361, 0.7242, 0.7120, , 0.7581, 0.7466, 0.7349, 0.7230, 0.7108, , 0.7570, 0.7455, 0.7337, 0.7218, 0.7096, , 0.7559, 0.7443, 0.7325, 0.7206, 0.7083, , 2, 2, 2, 2, 2, , 4, 4, 4, 4, 4, , 6, 6, 6, 6, 6, , 8, 8, 8, 8, 8, , 9, 10, 10, 10, 10, , 45, 46, 47, 48, 49, , 0.7071, 0.6947, 0.6820, 0.6691, 0.6561, , 0.7059, 0.6934, 0.6807, 0.6678, 0.6547, , 0.7046, 0.6921, 0.6794, 0.6665, 0.6534, , 0,7034, 0.6909, 0.6782, 0.6652, 0.6521, , 0.7022, 0.6896, 0.6769, 0.6639, 0.6508, , 0.7009, 0.6884, 0.6756, 0.6626, 0.6494, , 0.6997, 0.6871, 0.6743, 0.6613, 0.6481, , 0.6984, 0.6858, 0.6730, 0.6600, 0.6468, , 0.6972, 0.6845, 0.6717, 0.6587, 0.6455, , 0.6959, 0.6833, 0.6704, 0.6574, 0.6441, , 2, 2, 2, 2, 2, , 4, 4, 4, 4, 4, , 6, 6, 6, 7, 7, , 8, 8, 9, 9, 9, , 10, 11, 11, 11, 11, , 50, 51, 52, 53, 54, , 0.6428, 0.6293, 0.6157, 0.6018, 0.5878, , 0.6414, 0.6280, 0.6143, 0.6004, 0.5864, , 0.6401, 0.6266, 0.6129, 0.5990, 0.5850, , 0.6388, 0.6252, 0.6115, 0.5976, 0.5835, , 0.6374, 0.6239, 0.6101, 0.5962, 0.5821, , 0.6361, 0.6255, 0.6088, 0.5948, 0.5807, , 0.6347, 0.6211, 0.6404, 0.5934, 0.5793, , 0.6334, 0.6198, 0.6060, 0.5920, 0.5779, , 0.6320, 0.6184, 0.6046, 0.5906, 0.5764, , 0.6307, 0.6170, 0.6032, 0.5892, 0.5750, , 2, 2, 2, 2, 2, , 4, 5, 5, 5, 5, , 7, 7, 7, 7, 7, , 9, 9, 9, 9, 9, , 11, 11, 12, 12, 12, , 55, 56, 57, 58, 59, , 0.5736, 0.5592, 0.5446, 0.5299, 0.5150, , 0.5721, 0.5577, 0.5432, 0.5284, 0.5135, , 0.5707, 0.5563, 0.5417, 0.5270, 0.5120, , 0.5693, 0.5548, 0.5402, 0.5255, 0.5105, , 0.5678, 0.5534, 0.5388, 0.5240, 0.5090, , 0.5664, 0.5519, 0.5373, 0.5225, 0.5075, , 0.5650, 0.5505, 0.5358, 0.5210, 0.5060, , 0.5635, 0.5490, 0.5344, 0.5195, 0.5045, , 0.5621, 0.5476, 0.5329, 0.5180, 0.5030, , 0.5606, 0.5461, 0.5314, 0.5165, 0.5015, , 2, 2, 2, 2, 3, , 5, 5, 5, 5, 5, , 7, 7, 7, 7, 8, , 10, 10, 10, 10, 10, , 12, 12, 12, 12, 13, , Workshop Calculation & Science : (NSQF) Exercise 1.10.48, , Copyright free, under CC BY Licence, , 169

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Natural Cosines, Numbers in different columns to be subtracted, not added, O, , 0', 0.0O, , 6', 0.1O, , 12', 0.2O, , 18', 0.3O, , 24', 0.4O, , 30', 0.5O, , 36', 0.6O, , 42', 0.7O, , 48', 0.8O, , 54', 0.9O, , 1', , 2', , 3', , 4', , 5', , 60, 61, 62, 63, 64, , 0.5000, 0.4848, 0.4695, 0.4540, 0.4384, , 0.4985, 0.4833, 0.4679, 0.4524, 0.4368, , 0.4970, 0.4818, 0.4664, 0.4509, 0.4352, , 0.4955, 0.4802, 0.4648, 0.4493, 0.4337, , 0.4939, 0.4787, 0.4633, 0.4478, 0.4321, , 0.4924, 0.4772, 0.4617, 0.4462, 0.4305, , 0.4909, 0.4756, 0.4602, 0.4446, 0.4289, , 0.4894, 0.4741, 0.4586, 0.4431, 0.4274, , 0.4879, 0.4726, 0.4571, 0.4415, 0.4258, , 0.4863, 0.4710, 0.4555, 0.4399, 0.4242, , 3, 3, 3, 3, 3, , 5, 5, 5, 5, 5, , 8, 8, 8, 8, 8, , 10, 10, 10, 10, 11, , 13, 13, 13, 13, 13, , 65, 66, 67, 68, 69, , 0.4226, 0.4067, 0.3907, 0.3746, 0.3584, , 0.4210, 0.4051, 0.3891, 0.3730, 0.3567, , 0.4195, 0.4035, 0.3875, 0.3714, 0.3551, , 0.4179, 0.4019, 0.3859, 0.3697, 0.3535, , 0.4163, 0.4003, 0.3843, 0.3681, 0.3518, , 0.4147, 0.3987, 0.3827, 0.3665, 0.3502, , 0.4131, 0.3971, 0.3811, 0.3649, 0.3486, , 0.4115, 0.3955, 0.3795, 0.3633, 0.3469, , 0.4099, 0.3939, 0.3778, 0.3616, 0.3453, , 0.4083, 0.3923, 0.3762, 0.3600, 0.3437, , 3, 3, 3, 3, 3, , 5, 5, 5, 5, 5, , 8, 8, 8, 8, 8, , 11, 11, 11, 11, 11, , 13, 13, 13, 14, 14, , 70, 71, 72, 73, 74, , 0.3420, 0.3256, 0.3090, 0.2924, 0.2756, , 0.3404, 0.3239, 0.3074, 0.2907, 0.2740, , 0.3387, 0.3223, 0.3057, 0.2890, 0.2723, , 0.3371, 0.3206, 0.3040, 0.2874, 0.2706, , 0.3355, 0.3190, 0.3024, 0.2857, 0.2689, , 0.3338, 0.3173, 0.3007, 0.2840, 0.2672, , 0.3322, 0.3156, 0.2990, 0.2823, 0.2656, , 0.3305, 0.3140, 0.2974, 0.2807, 0.2639, , 0.3289, 0.3123, 0.2957, 0.2790, 0.2622, , 0.3272, 0.3107, 0.2940, 0.2773, 0.2605, , 3, 3, 3, 3, 3, , 5, 6, 6, 6, 6, , 8, 8, 8, 8, 8, , 11, 11, 11, 11, 11, , 14, 14, 14, 14, 14, , 75, 76, 77, 78, 79, , 0.2588, 0.2419, 0.2250, 0.2079, 0.1908, , 0.2571, 0.2402, 0.2233, 0.2062, 0.1891, , 0.2554, 0.2385, 0.2215, 0.2045, 0.1874, , 0.2538, 0.2368, 0.2198, 0.2028, 0.1857, , 0.2521, 0.2351, 0.2181, 0.2011, 0.1840, , 0.2504, 0.2334, 0.2164, 0.1994, 0.1822, , 0.2487, 0.2317, 0.2147, 0.1977, 0.1805, , 0.2470, 0.2300, 0.2130, 0.1959, 0.1788, , 0.2453, 0.2284, 0.2113, 0.1942, 0.1771, , 0.2436, 0.2267, 0.2096, 0.1925, 0.1754, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 6, , 8, 8, 9, 9, 9, , 11, 11, 11, 11, 11, , 14, 14, 14, 14, 14, , 80, 81, 82, 83, 84, , 0.1736, 0.1564, 0.1392, 0.1219, 0.1045, , 0.1719, 0.1547, 0.1374, 0.1201, 0.1028, , 0.1702, 0.1530, 0.1357, 0.1184, 0.1011, , 0.1685, 0.1513, 0.1340, 0.1167, 0.0993, , 0.1668, 0.1495, 0.1323, 0.1149, 0.0976, , 0.1650, 0.1478, 0.1305, 0.1132, 0.0958, , 0.1633, 0.1461, 0.1288, 0.1115, 0.0941, , 0.1616, 0.1444, 0.1271, 0.1097, 0.0924, , 0.1599, 0.1426, 0.1253, 0.1080, 0.0906, , 0.1582, 0.1409, 0.1236, 0.1063, 0.0889, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 6, , 9, 9, 9, 9, 9, , 11, 11, 12, 12, 12, , 14, 14, 14, 14, 14, , 85, 86, 87, 88, 89, 90, , 0.0872, 0.0698, 0.0523, 0.0349, 0.0175, 0.0000, , 0.0854, 0.0680, 0.0506, 0.0332, 0.0157, , 0.0837, 0.0663, 0.0488, 0.0314, 0.0140, , 0.0819, 0.0645, 0.0471, 0.0297, 0.0122, , 0.0802, 0.0628, 0.0454, 0.0279, 0.0105, , 0.0785, 0.0610, 0.0436, 0.0262, 0.0087, , 0.0767, 0.0593, 0.0419, 0.0244, 0.0070, , 0.0750, 0.0576, 0.0401, 0.0227, 0.0052, , 0.0732, 0.0558, 0.0384, 0.0209, 0.0035, , 0.0715, 0.0541, 0.0366, 0.0192, 0.0017, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 6, , 9, 9, 9, 9, 9, , 12, 12, 12, 12, 12, , 14, 15, 15, 15, 15, , Quadrant, , Angle, , cos A =, , Examples, , First, , 0 to 90, , cos A, , cos 33 26' = 0.8345, , Second, , 90o to 180o, , –cos(180o – A), , sin 146o34', , o, , o, , = –cos(180o – 146o 34'), , = –cos 33o26' = –0.8345, Third, , 180o to 270o, , cos(A – 180o), , cos 213o26', , = –cos(213o26' – 180o), , = –cos 33o26' = –0.8345, Fourth, , 270o to 360o, , cos(360o – A), , cos 326o34', , = cos(360o – 326o34'), , = cos 33o26' = 0.8345, , 170, , Workshop Calculation & Science : (NSQF) Exercise 1.10.48, , Copyright free, under CC BY Licence

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Natural Tangents, O, , 0', 0.0O, , 6', 0.1O, , 12', 0.2O, , 18', 0.3O, , 24', 0.4O, , 30', 0.5O, , 36', 0.6O, , 42', 0.7O, , 48', 0.8O, , 54', 0.9O, , 1', , 2', , 3', , 4', , 5', , 0, 1, 2, 3, 4, , 0.0000, 0.0175, 0.0349, 0.0524, 0.0699, , 0.0017, 0.0192, 0.0367, 0.0542, 0.0717, , 0.0035, 0.0209, 0.0384, 0.0559, 0.0734, , 0.0052, 0.0227, 0.0402, 0.0577, 0.0752, , 0.0070, 0.0244, 0.0419, 0.0594, 0.0769, , 0.0087, 0.0262, 0.0437, 0.0612, 0.0787, , 0.0105, 0.0279, 0.0454, 0.0629, 0.0805, , 0.0122, 0.0297, 0.0472, 0.0647, 0.0822, , 0.0140, 0.0314, 0.0489, 0.0664, 0.0840, , 0.0157, 0.0332, 0.0507, 0.0682, 0.0857, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 6, , 9, 9, 9, 9, 9, , 12, 12, 12, 12, 12, , 15, 15, 15, 15, 15, , 5, 6, 7, 8, 9, , 0.0875, 0.1051, 0.1228, 0.1405, 0.1584, , 0.0892, 0.1069, 0.1246, 0.1423, 0.1602, , 0.0910, 0.1086, 0.1263, 0.1441, 0.1620, , 0.0928, 0.1104, 0.1281, 0.1459, 0.1638, , 0.0945, 0.1122, 0.1299, 0.1477, 0.1655, , 0.0963, 0.1139, 0.1317, 0.1495, 0.1673, , 0.0981, 0.1157, 0.1334, 0.1512, 0.1691, , 0.0998, 0.1175, 0.1352, 0.1530, 0.1709, , 0.1016, 0.1192, 0.1370, 0.1548, 0.1727, , 0.1033, 0.1210, 0.1388, 0.1566, 0.1745, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 6, , 9, 9, 9, 9, 9, , 12, 12, 12, 12, 12, , 15, 15, 15, 15, 15, , 10, 11, 12, 13, 14, , 0.1763, 0.1944, 0.2126, 0.2309, 0.2493, , 0.1781, 0.1962, 0.2144, 0.2327, 0.2512, , 0.1799, 0.1980, 0.2162, 0.2345, 0.2530, , 0.1817, 0.1998, 0.2180, 0.2364, 0.2549, , 0.1835, 0.2016, 0.2199, 0.2382, 0.2568, , 0.1853, 0.2035, 0.2217, 0.2401, 0.2586, , 0.1871, 0.2053, 0.2235, 0.2419, 0.2605, , 0.1890, 0.2071, 0.2254, 0.2438, 0.2623, , 0.1908, 0.2089, 0.2272, 0.2456, 0.2642, , 0.1926, 0.2107, 0.2290, 0.2475, 0.2661, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 6, , 9, 9, 9, 9, 9, , 12, 12, 12, 12, 12, , 15, 15, 15, 15, 16, , 15, 16, 17, 18, 19, , 0.2679, 0.2867, 0.3057, 0.3249, 0.3443, , 0.2698, 0.2886, 0.3076, 0.3269, 0.3463, , 0.2717, 0.2905, 0.3096, 0.3288, 0.3482, , 0.2736, 0.2924, 0.3115, 0.3307, 0.3502, , 0.2754, 0.2943, 0.3134, 0.3327, 0.3522, , 0.2773, 0.2962, 0.3153, 0.3346, 0.3541, , 0.2792, 0.2981, 0.3172, 0.3365, 0.3561, , 0.2811, 0.3000, 0.3191, 0.3385, 0.3581, , 0.2830, 0.3019, 0.3211, 0.3404, 0.3600, , 0.2849, 0.3038, 0.3230, 0.3424, 0.3620, , 3, 3, 3, 3, 3, , 6, 6, 6, 6, 7, , 9, 9, 10, 10, 10, , 13, 13, 13, 13, 13, , 16, 16, 16, 16, 16, , 20, 21, 22, 23, 24, , 0.3640, 0.3839, 0.4040, 0.4245, 0.4452, , 0.3659, 0.3859, 0.4061, 0.4265, 0.4473, , 0.3679, 0.3879, 0.4081, 0.4286, 0.4494, , 0.3699, 0.3899, 0.4101, 0.4307, 0.4515, , 0.3719, 0.3919, 0.4122, 0.4327, 0.4536, , 0.3739, 0.3939, 0.4142, 0.4348, 0.4557, , 0.3759, 0.3959, 0.4163, 0.4369, 0.4578, , 0.3779, 0.3979, 0.4183, 0.4390, 0.4599, , 0.3799, 0.4000, 0.4204, 0.4411, 0.4621, , 0.3819, 0.4020, 0.4224, 0.4431, 0.4642, , 3, 3, 3, 3, 4, , 7, 7, 7, 7, 7, , 10, 10, 10, 10, 11, , 13, 13, 14, 14, 14, , 17, 17, 17, 17, 18, , 25, 26, 27, 28, 29, , 0.4663, 0.4877, 0.5095, 0.5317, 0.5543, , 0.4684, 0.4899, 0.5117, 0.5340, 0.5566, , 0.4706, 0.4921, 0.5139, 0.5362, 0.5589, , 0.4727, 0.4942, 0.5161, 0.5384, 0.5612, , 0.4748, 0.4964, 0.5184, 0.5407, 0.5635, , 0.4770, 0.4986, 0.5206, 0.5430, 0.5658, , 0.4791, 0.5008, 0.5228, 0.5452, 0.5681, , 0.4813, 0.5029, 0.5250, 0.5475, 0.5704, , 0.4834, 0.5051, 0.5272, 0.5498, 0.5727, , 0.4856, 0.5073, 0.5295, 0.5520, 0.5750, , 4, 4, 4, 4, 4, , 7, 7, 7, 8, 8, , 11, 11, 11, 11, 12, , 14, 15, 15, 15, 15, , 18, 18, 18, 19, 19, , 30, 31, 32, 33, 34, , 0.5774, 0.6009, 0.6249, 0.6494, 0.6745, , 0.5797, 0.6032, 0.6273, 0.6519, 0.6771, , 0.5820, 0.6056, 0.6297, 0.6544, 0.6796, , 0.5844, 0.6080, 0.6322, 0.6569, 0.6822, , 0.5867, 0.6104, 0.6346, 0.6594, 0.6847, , 0.5890, 0.6128, 0.6371, 0.6619, 0.6873, , 0.5914, 0.6152, 0.6395, 0.6644, 0.6899, , 0.5938, 0.6176, 0.6420, 0.6669, 0.6924, , 0.5961, 0.6200, 0.6445, 0.6694, 0.6950, , 0.5985, 0.6224, 0.6469, 0.6720, 0.6976, , 4, 4, 4, 4, 4, , 8, 8, 8, 8, 9, , 12, 12, 12, 13, 13, , 16, 16, 16, 17, 17, , 20, 20, 20, 21, 21, , 35, 36, 37, 38, 39, , 0.7002, 0.7265, 0.7536, 0.7813, 0.8098, , 0.7028, 0.7292, 0.7563, 0.7841, 0.8127, , 0.7054, 0.7319, 0.7590, 0.7869, 0.8156, , 0.7080, 0.7346, 0.7618, 0.7898, 0.8185, , 0.7107, 0.7373, 0.7646, 0.7926, 0.8214, , 0.7133, 0.7400, 0.7673, 0.7954, 0.8243, , 0.7159, 0.7427, 0.7701, 0.7983, 0.8273, , 0.7186, 0.7454, 0.7729, 0.8012, 0.8302, , 0.7212, 0.7481, 0.7757, 0.8040, 0.8332, , 0.7239, 0.7508, 0.7785, 0.8069, 0.8361, , 4, 5, 5, 5, 5, , 9, 9, 9, 9, 10, , 13, 14, 14, 14, 15, , 17, 18, 18, 19, 20, , 22, 23, 23, 24, 24, , 40, 41, 42, 43, 44, , 0.8391, 0.8693, 0.9004, 0.9325, 0.9657, , 0.8421, 0.8724, 0.9036, 0.9358, 0.9691, , 0.8451, 0.8754, 0.9067, 0.9391, 0.9725, , 0.8481, 0.8785, 0.9099, 0.9424, 0.9759, , 0.8511, 0.8816, 0.9131, 0.9457, 0.9793, , 0.8541, 0.8847, 0.9163, 0.9490, 0.9827, , 0.8571, 0.8878, 0.9195, 0.9523, 0.9861, , 0.8601, 0.8910, 0.9228, 0.9556, 0.9896, , 0.8632, 0.8941, 0.9260, 0.9590, 0.9930, , 0.8662, 0.8972, 0.9293, 0.9623, 0.9965, , 5, 5, 5, 6, 6, , 10, 10, 11, 11, 11, , 15, 16, 16, 17, 17, , 20, 21, 21, 22, 23, , 25, 26, 27, 28, 28, , 45, 46, 47, 48, 49, , 1.0000, 1.0355, 1.0724, 1.1106, 1.1504, , 1.0035, 1.0392, 1.0761, 1.1145, 1.1544, , 1.0070, 1.0428, 1.0799, 1.1184, 1.1585, , 1.0105, 1.0464, 1.0837, 1.1224, 1.1626, , 1.0141, 1.0501, 1.0875, 1.1263, 1.1667, , 1.0176, 1.0538, 1.0913, 1.1303, 1.1708, , 1.0212, 1.0575, 1.0951, 1.1343, 1.1750, , 1.0247, 1.0612, 1.0990, 1.1383, 1.1792, , 1.0283, 1.0649, 1.1028, 1.1423, 1.1833, , 1.0319, 1.0686, 1.1067, 1.1463, 1.1875, , 6, 6, 6, 7, 7, , 12, 12, 13, 13, 14, , 18, 18, 19, 20, 21, , 24, 25, 25, 27, 28, , 30, 31, 32, 33, 34, , 50, 51, 52, 53, 54, , 1.1918, 1.2349, 1.2799, 1.3270, 1.3764, , 1.1960, 1.2393, 1.2846, 1.3319, 1.3814, , 1.2002, 1.2437, 1.2892, 1.3367, 1.3865, , 1.2045, 1.2482, 1.2938, 1.3416, 1.3916, , 1.2088, 1.2527, 1.2985, 1.3465, 1.3968, , 1.2131, 1.2572, 1.3032, 1.3514, 1.4019, , 1.2174, 1.2617, 1.3079, 1.3564, 1.4071, , 1.2218, 1.2662, 1.3127, 1.3613, 1.4124, , 1.2261, 1.2708, 1.3175, 1.3663, 1.4176, , 1.2305, 1.2753, 1.3222, 1.3713, 1.4229, , 7, 8, 8, 8, 9, , 14, 15, 16, 16, 17, , 22, 23, 24, 25, 26, , 29, 30, 31, 33, 34, , 36, 38, 39, 41, 43, , 55, 56, 57, 58, 59, , 1.4281, 1.4826, 1.5399, 1.6003, 1.6643, , 1.4335, 1.4882, 1.5458, 1.6066, 1.6709, , 1.4388, 1.4938, 1.5517, 1.6128, 1.6775, , 1.4442, 1.4994, 1.5577, 1.6191, 1.6842, , 1.4496, 1.5051, 1.5637, 1.6255, 1.6909, , 1.4550, 1.5108, 1.5697, 1.6319, 1.6977, , 1.4605, 1.5166, 1.5757, 1.6383, 1.7045, , 1.4659, 1.5224, 1.5818, 1.6447, 1.7113, , 1.4715, 1.5282, 1.5880, 1.6512, 1.7182, , 1.4770, 1.5340, 1.5941, 1.6577, 1.7251, , 9, 10, 10, 11, 11, , 18, 19, 20, 21, 23, , 27, 29, 30, 32, 34, , 36, 38, 40, 43, 45, , 45, 48, 50, 53, 56, , Workshop Calculation & Science : (NSQF) Exercise 1.10.48, , Copyright free, under CC BY Licence, , 171

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Natural Tangents, O, , 0', 0.0O, , 6', 0.1O, , 12', 0.2O, , 18', 0.3O, , 24', 0.4O, , 30', 0.5O, , 36', 0.6O, , 42', 0.7O, , 48', 0.8O, , 54', 0.9O, , 1', , 2', , 3', , 4', , 5', , 60, 61, 62, 63, 64, , 1.7321, 1.8040, 1.8807, 1.9626, 2.0503, , 1.7391, 1.8115, 1.8887, 1.9711, 2.0594, , 1.7461, 1.8190, 1.8967, 1.9797, 2.0686, , 1.7532, 1.8265, 1.9047, 1.9883, 2.0778, , 1.7603, 1.8341, 1.9128, 1.9970, 2.0872, , 1.7675, 1.8418, 1.9210, 2.0057, 2.0965, , 1.7747, 1.8495, 1.9292, 2.0145, 2.1060, , 1.7820, 1.8572, 1.9375, 2.0233, 2.1155, , 1.7893, 1.8650, 1.9458, 2.0323, 2.1251, , 1.7966, 1.8728, 1.9542, 2.0413, 2.1348, , 12, 13, 14, 15, 16, , 24, 26, 27, 29, 31, , 36, 38, 41, 44, 47, , 48, 51, 55, 58, 63, , 60, 64, 68, 73, 78, , 65, 66, 67, 68, 69, , 2.1445, 2.2460, 2.3559, 2.4751, 2.6051, , 2.1543, 2.2566, 2.3673, 2.4876, 2.6187, , 2.1642, 2.2673, 2.3789, 2.5002, 2.6325, , 2.1742, 2.2781, 2.3906, 2.5129, 2.6464, , 2.1842, 2.2889, 2.4023, 2.5257, 2.6605, , 2.1943, 2.2998, 2.4142, 2.5386, 2.6746, , 2.2045, 2.3109, 2.4262, 2.5517, 2.6889, , 2.2148, 2.3220, 2.4383, 2.5649, 2.7034, , 2.2251, 2.3332, 2.4504, 2.5782, 2.7179, , 2.2355, 2.3445, 2.4627, 2.5916, 2.7326, , 17, 18, 20, 22, 24, , 34, 37, 40, 43, 47, , 51, 55, 60, 65, 71, , 68, 73, 79, 87, 95, , 85, 92, 99, 108, 119, , 70, 71, 72, 73, 74, , 2.7475, 2.9042, 3.0777, 3.2709, 3.4874, , 2.7625, 2.9208, 3.0961, 3.2914, 3.5105, , 2.7776, 2.9375, 3.1146, 3.3122, 3.5339, , 2.7929, 2.9544, 3.1334, 3.3332, 3.5576, , 2.8083, 2.9714, 3.1524, 3.3544, 3.5816, , 2.8239, 2.9887, 3.1716, 3.3759, 3.6059, , 2.8397, 3.0061, 3.1910, 3.3977, 3.6305, , 2.8556, 3.0237, 3.2106, 3.4197, 3.6554, , 2.8716, 3.0415, 3.2305, 3.4420, 3.6806, , 2.8878, 3.0595, 3.2506, 3.4646, 3.7062, , 26, 29, 32, 36, 41, , 52, 58, 64, 72, 81, , 78, 87, 96, 108, 122, , 104, 116, 129, 144, 163, , 131, 145, 161, 180, 204, , 75, 76, 77, 78, 79, , 3.7321, 4.0108, 4.3315, 4.7046, 5.1446, , 3.7583, 4.0408, 4.3662, 4.7453, 5.1929, , 3.7848, 4.0713, 4.4015, 4.7867, 5.2422, , 3.8118, 4.1022, 4.4374, 4.8288, 5.2924, , 3.8391, 4.1335, 4.4737, 4.8716, 5.3435, , 3.8667, 4.1653, 4.5107, 4.9152, 5.3955, , 3.8947, 4.1976, 4.5483, 4.9594, 5.4486, , 3.9232, 4.2303, 4.5864, 5.0045, 5.5026, , 3.9520, 4.2635, 4.6252, 5.0504, 5.5578, , 3.9812, 4.2972, 4.6646, 5.0970, 5.6140, , 46, 53, , 93, 107, , 139, 160, , 186, 213, , 232, 267, , 80, 81, 82, 83, 84, 85, , 5.6713, 6.3138, 7.1154, 8.1443, 9.5144, 11.43, , 5.7297, 6.3859, 7.2066, 8.2636, 9.677, 11.66, , 5.7894, 6.4596, 7.3002, 8.3863, 9.845, 11.91, , 5.8502, 6.5350, 6.3962, 8.5126, 10.02, 12.16, , 5.9124, 6.6122, 7.4947, 8.6427, 10.20, 12.43, , 5.9758, 6.6912, 7.5958, 8.7769, 10.39, 12.71, , 6.0405, 6.7720, 7.6996, 8.9152, 10.58, 13.00, , 6.1066, 6.8548, 7.8062, 9.0579, 10.78, 13.30, , 6.1742, 6.9395, 7.9158, 9.2052, 10.99, 13.62, , 6.2432, 7.0264, 8.0285, 9.3572, 11.20, 13.95, , 86, 87, 88, 89, 90, , 14.30, 19.08, 28.64, 57.29, ×, , 14.67, 19.74, 30.14, 63.66, , 15.06, 20.45, 31.82, 71.62, , 15.46, 21.20, 33.69, 81.85, , 15.89, 22.02, 35.80, 95.49, , 16.35, 22.90, 38.19, 114.6, , 16.83, 23.86, 40.92, 143.2, , 17.34, 24.90, 44.07, 191.0, , 17.89, 26.03, 47.74, 286.5, , 18.46, 27.27, 52.08, 573.0, , Quadrant, , Angle, , DIFFERENCES, UNTRUSTWORTHY, HERE, , tan A =, , Examples, , First, , 0 to 90o, , tan A, , tan 56o17' = 1.4986, , Second, , 90o to 180o, , –tan(180o – A), , tan 123o43', , = –tan(180o – 123o 43'), , = –tan 56o17' = –1.4986, Third, , 180o to 270o, , tan(A – 180o), , tan 236o17', , = tan(236o17' – 180o), , = tan 56o17' = 1.4986, Fourth, , 270o to 360o, , –tan(360o – A), , tan 303o43', , = –tan(360o – 303o43'), , = – tan 56o17' = –1.4986, , 172, , Workshop Calculation & Science : (NSQF) Exercise 1.10.48, , Copyright free, under CC BY Licence

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Trigonometry - Application in calculating height and distance, (Simple applications), Exercise 1.10.49, Heights and distances, , y = x tan60o, , One of the practical applications of Trigonometry is to find, distances and heights of distant and inaccessible objects., Two angles are often used in the practical applications of, Trigonometry and they are defined as follows:, The angle between a horizontal plane through an observer’s, eye and line joining the eye to an object is called, i, , The angle of elevation when the object is higher than, the eye. (Fig 1), , Hence, (x + 720) tan30o = x tan60o, , ( x  720), , i.e., , 1, 3, , x 3, , =, , x + 720 = 3x, 720 = 3x - x, , ii The angle of depression when the object is lower, than the eye. (Fig 2), , 2x = 720, x =, , 720, , = 360, 2, Hence y = 360 tan60 o, =, tan 600 =, , tan 600 =, , y, x, y, 360, , y = tan 600 x 360, If OX be a horizontal line through ‘O’ the observer’s eye and, ‘P’ any point above OX, then XOP is the angle of elevation, of P at O. If ‘Q’ below OX, then XOQ is called the angle of, depression of Q at O., Example, 1 At a certain point on the ground, it is found that the, angle of elevation of the top of a tower is 30o. On, walking 720 m. towards the foot of tower, the, angle of elevation is found to be 60o. Find the, height of the tower., , Let CD be the tower and A and B be the points from which, the tower is observed and let BC be x and CD be y., In triangle ADC,, tan 30 =, o, , Hence the height of the tower is, = 360 x 1.732, Height of the tower = 623.5 metres, 2 A flag pole stands on the top of a building when, viewed from a distance of 50 m. (measured horizontally) the angle of elevation of the top and, bottom of the flag staff are 24 and 32o respectively., Find the height of the flag pole., , Let CD be the flag pole, A the point of observation and B a, point on the same level as A and directly underneath the, flag pole., In the triangle DAB, , CD, y, , AC x  720, , y = (x + 720) tan 300, Angle BDC, tan 60o =, , CD y, , BC x, , Copyright free, under CC BY Licence, , tan24o =, , BD, AB, , , , BD, 50, , BD = 50. tan24o, = 50 x 0.4452, = 22.26, 173

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In the triangle CAB,, tan 32o =,  BC =, =, =, Hence DC =, =, =, , CB, AB, , , , BC, 50, , 50 . tan32o, 50 x 0.6249, 31.25, 31.25 – 22.26, 8.99, 9 m nearly., , Find the distance between ship and cliff, From right angled triangle CAD, , 3 A divider having legs of equal length of 10 cm is, opened so that its points are 4.5 cm apart. Using, trigonometrical tables. Find the angle between, the legs., , tan θ =, , Opposite side, CD, =, Adjacent Side, AC, , 10, tan 20o =, AC, 0.3640 =, AC =, , 10, AC, , 10, = 27.473 m, 0.3640, , Height of cliff BD = DC + CB, Find CB in right angled triangle BAC, Distance between the two legs of divider = BC = 4.5 cm, , tan θ =, , AC = AB ; Length of leg = 10 cm, The perpendicular line drawn from the centre of BC, (point, E) to point A makes two right angled triangle ABE and, AEC. If the angle between two legs of divider is , BAE =, = Sin θ =, , Sin, , θ, 2, , tan 42o =, , BC, 27.473, , 0.9004 =, , BC, 27.473, , BC=, =, Height of cliff BD =, =, =, , Opposite side, Hypotenuse, , θ BE, =, 2 AB, , 0.9004 x 27.473, 24.737 m, DC + CB, 10 + 24.737, 34.737 m, , Distance between ship and cliff = 27.473 m, , 2.25, = 0.225, =, 10, Find the  value of 0.225 from sin table, 0.225 =, 130, θ, =, 2, , 130, ,  =, , 130 x 2 = 26o, , Height of cliff = 34.737 m, 5 The angle of depression from the top of a hill to the, bottom of a tower is 60o and the angle of depression from the top of the tower to the bottom of hill, is 30o. If the height of tower is 50m, then find the, height of the hill., Note :, , Angle between two leg of divider is 26, , o, , 4 A man on the deck of a ship is 10 m above water, level. He observes that angle of elevation of a cliff, is 42o and angle of depression of its base is 20o., Calculate (i) the distance of the cliff from the ship., (ii) the height of the cliff., , 174, , Opposite side, BC, =, Adjacent Side, AC, , 1 If the angle of depression at A is 60o, then the angle of, elevation at C is 60o., 2 If the angle of depression at D is 30o, then the angle of, elevation at B is 30o., Let, the height of hill is 'h' and distance between bae of hill, and base of tower is x., , Workshop Calculation & Science : (NSQF) Exercise 1.10.49, , Copyright free, under CC BY Licence

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cos ,  0.2222, 0.2233 = cos 77o 6', (-), 0.0011 =, 4' (+), ----------------------------------0.2222 = cos 77o 10', ----------------------------------, , From right angled triangle ACB, tan θ =, , Opposite side, Adjacent Side, , tan 60o =, ,  = 77o 10', Ladder makes the angle = 77o 10’, , h, x, , h, 3=, x, , .................. equation 1, h = 3x, From right angled triangle DBC, , tan 30° =, 1, 3, , 50, x, , AC and CD are Ladder, , 50, x, , =, , AB and DE are lamp post, In  ABC, , 3, 1, , x = 50x, , 7 A Line man who is working on a road places his, ladder which is 12 m in length at a point on the, road such that it makes an angle of 600 with the, ground, when it is placed against a lamp placed, against another lamp post directly on the opposite, side of the road it makes an angle of 300. Find the, distance between the 2 lamp posts., , x = 50 3, , Cos 60o =, ....................equation 2, , x, , BC, x, =, AC 12 m, , = cos 60o x 12 m, = 0.5000 x 12 m, , Substitute the value of = x = 50 3 in equation 1., , =6m, , h= 3 x x, = 3 x 50 3, = 3 x 50 = 150 m, Height of hill = 150 m, 6 The foot of a 4.5 m long ladder is placed at 1 m, away from the wall. Find the angle which the, ladder makes with the ground, In Right angled , Cos C, Cos , , BC, =, AC, =, , 1m, 4.5 m, , In  CDE, Cos 30o =, y, , CE, y, =, CD 12 m, , = cos 30o x 12 m, = 0.8660 x 12 m, = 10.392 m, , Workshop Calculation & Science : (NSQF) Exercise 1.10.49, , Copyright free, under CC BY Licence, , 175

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Distance between two lamp posts, =x+y, = 6 + 10.392 m, Distance between two lamp posts = 16.392 m, 8 A pole stands 10 metre above the ground and stay, wire is fixed to the pole at 0.5 metre from the top., If the stay wire is to make an angle of 600 with the, horizontal, find the distance of stay rod from the, base of the pole. Also find the length of the stay, wire., , 9 An electrician standing 30metres away from the, base of a transmission line tower looks at the top, of the tower. His line of sight makes an angle of, 33.690 with the horizontal. If the height of his eyes, above the ground level is 1.6metres. Find the, height of the tower?, , Given:, Distance between the Electrician = EB = DC = 30m, and the tower, stay wire fixed 0.5m below from the top of the pole, Sin θ =, , OPP, HYP, , Height of his eyes above the = CB = DE = 1.6m, ground level, Height of the tower AB (h) = ?, 33.690, , 9.5, Sin 60, = x, x x sin 60o = 9.5 m, o, , x, , =, , = 330 + 41’, 33.690, , Sin 60°, , =, Stay wire (x) =, , 10.97 m, , tan 60o, , =, , y tan 60o, , =, , y, , = 10.9699, , 9.5, 1.7321, , AC, , = tan x DC, , DC, , = 19.995 m, Height of the tower AB (h), Height of the tower (h), , = 5.48 m, , Length of the stay wire, = 10.97 metre, distance of the stay wire from, the base of the pole, = 5.48 metre, , 176, , = AC + CB, = 19.995 + 1.6, , 9.5, , =, , =, , = 0.6665 x 30 m, , 9.5, y, , =, , AC, , Tan , , = tan 330 41’ x 30, , Opp, Adj, , 9.5, tan 60°, , = 330 41’, , In Right angled triangle ADC, , 9.5, , 9.5, 0.8660, , tan θ =, , = 330 + (0.69 x 60’), , Workshop Calculation & Science : (NSQF) Exercise 1.10.49, , Copyright free, under CC BY Licence, , = 21.595 m

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1 A person walking along straight observed that at, consecutive mile stones the angle of elevation of a hill, in front of him 30o and 75o. Find the height of the hill., , 8, , 2 A shadow of an electric pole is reduced by 4 metres, when the sun changes its angle of elevation from 30o to, 45o. If the pole is burried in the ground by 2 metres, find, the total length of the pole?, 3 The angle of elevation of the top on an unfinished tower, at a point distance 120 metre from its base is 45o. How, much higher must the tower be raised so that its angle, of elevation at the same may be 60o., 4 Two objects on horizontal plane in the same line of the, foot of the cliff form with the top of cliff angles of elevation, of 30o and 45o. If the height of the cliff is 100 m, calculate, the distance between the two objects., 5 From a point 'A' on the ground at unknown distance from, the bae of the radio tower, the angle of elevation of the, top of mast is 65o. Proceeding in the same straight line, to the point 'B' 50 m from 'A', the angle of elevation is, reduced to 50o. Find the height of the mast and distance, of 'A' from the mast., ,  = 75°, b = 3 metres, H = ______ metres, L = ______ metres, 9 The height of the pole,, AB = __________ m., , 6, ,  = 40°, , 10 Find the values of the given angles., , L = 120 metres, , 1 Sin 65o, , h = 1.5 metres, , 2 Sin 42o 23', , H = ______ metre, , 3 Sin 66o 35' 32", 4 Cos 47o 39', , 7, , 5 Tan 28o 45', 11 Find corresponding angles for given values., 1 Sin , , = 0.3062, , 2 Sin , , = 0.04802, , 3 Cos  = 0.6446, ,  = 25°,  = 50°, l = 100 metres, H = ______ metre, , 4 Tan , , = 0.3411, , 6 Tan , , = 2.3868, , 12 The slant height of a cone is 12.25 cm and the vertex, angle is 1100. Calculate its base., 13 A ladder 2.5 m long makes an angle of 600 with the, ground. Find the height of the wall where the ladder, touches the wall., , Workshop Calculation & Science : (NSQF) Exercise 1.10.49, , Copyright free, under CC BY Licence, , 177