Phy MSBTE Study Material PDF

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Name HEAD Roll No. 20_______ SEMISTER I | DIPLOMA IN ENGINEERING AND TECHNOLOGY Secretary, Maharashtra State Board of TechnicalEducation 49, Kherwadi, Bandra (East), Mumbai - 400 051 A LEARNING MANUAL Maharashtra(INDIA) Tel: (022)26471255 (5 -lines) Fax: 022 - 26473980 FOR Web -www.msbte.org.in REGIONAL OFFICES: Basic Science MUMBAI Deputy Secretary (T), Mumbai Sub- PUNE Deputy Secretary (T), (22102) region, M.S. Board of Technical Education, 2n°Floor,Govt.PolytechnicBuilding, Regional Office, 49, Kherwadi, Bandra (East) 412-E, BahiratPatilChowk, Mumbai - 400 051 Shivaji Nagar, Pune Phone: 022-26473253/ 54 Phone: 020-25656994 / 25660319 Fax: 022-26478795 Fax: 020-25656994 Email: rbtemumbai @ msbte.com Email: rbtepn @ msbte.com NAGPUR Deputy Secretary (T), Deputy Secretary (T), M.S. Board of Technical Education M.S. Board of Technical Education, Regional Office, Regional Office, Mangalwari Bazar, Sadar, Nagpur - 440 001 Osmanpura, Aurangabad -431 001. Phone: 0712-2564836 / 2562223 Phone: 0240-2334025 / 2331273 Fax: 0712-2560350 Fax: 0240-2349669 MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, MUMBAI Email: rbteng @ msbte.com Email: rbteau @ msbte.com (Autonomous) (ISO 9001 : 2015) (ISO / IEC 27001 : 2013)  A Learning Manual For Basic Science (22102) Semester– (I) (First Semester Diploma in Engineering and Technology) Maharashtra State Board of Technical Education, Mumbai (Autonomous) (ISO-9001-2008) (ISO/IEC 27001:2013) Basic Science (22102) Maharashtra State Board of Technical Education, Mumbai (Autonomous) (ISO-9001-2008) (ISO/IEC 27001:2013) 4th Floor, Government Polytechnic Building, 49, Kherwadi, Bandra (East), Mumbai -400051. Page | 2 Basic Science (22102) Preface The primary focus of any engineering work in the technical education system is to develop the much needed industry relevant competencies and skills. With this in view, MSBTE embarked on this innovative ‘I’ Scheme curricula for engineering diploma programmes with outcome-based education. This Learning manual is designed to help all stakeholders, especially the students, teachers and instructors to develop in the student the pre-determined outcomes. The manual begins by identifying the competency, course outcomes,. The students will become aware about the concepts of Basic Science.. This manual also provides guidelines to teachers to effectively facilitate student-centered activities through each chapter by arranging and managing necessary resources ensuring the achievement of outcomes in the students. Diploma engineers (also called technologists) have to deal with various materials and machines. This course is designed with some fundamental information to help the technologists apply the basic concepts and principles of physics and chemistry to solve broad based engineering problems. The study of basic principles of sciences and the concepts related to various materials such as metals, alloys, inorganic salts, polymers, lubricants, paints, varnishes, adhesives, heat, electricity, magnetism, optics, semiconductors and others will help in understanding the technology courses where emphasis is on the applications of these in different technology The Learning manual development team wishes to thank MSBTE who took initiative in the development of curriculum re-design project and implementation and also acknowledge the contribution of individual course experts who have been involved in Learning manual as well as curriculum development (I scheme) directly or indirectly Although all care has been taken to check for mistakes in this learning manual, yet it is impossible to claim perfection especially as this is the first edition. Any such errors and suggestions for improvement can be brought to our notice and are highly welcome. Learning Manual Development Team Page | 3 Basic Science (22102) List of Content S. No Name of Unit Page No. SECTION – I [PHYSICS] 1. Units and Measurements 05 2. Electricity, Magnetism and Semiconductors 14 3. Heat and Optics 38 SECTION – II [CHEMISTRY] 4. Chemical Bonding and Catalysis 64 5. Metal Corrosion, its prevention and Electro 95 chemistry 6. Paint, Varnishes, Insulators, Polymer, Adhesives 134 and Lubricants Page | 4 Basic Science (22102) SECTION – I [PHYSICS] Topic1 Units and Measurements CO : Estimate errors in measurement of physical quantify UO: 1. Describe the given measurement device and its application 2. Describe with justification the device used to measure radius of curvature of the given object. 3. State with justification the error in the given measurement 4. Determine the procedure to determine the dimensions of the given physical quantity Introduction/ Rationale Measurement of any physical quantity in valves comparison with certain basic measurement are used by scientist for understanding nature phenomenon, it is used by society for transaction in business and practical purposes. engineers use measurements in building drawing, control systems, data processing etc. Unit is a standard in which a physical quantity is measured. The measurement of physical quantity is explored by a number accompanied by a unit. Ex 10 kg, 5 m where 10= magnitude kg= unit 5= magnitude m= unit The physical quantities are large in number, but we require limited number of units to express them, since they are related to one another. Significance All the devices used by engineers for measurement have magnitude and units and hence understanding of same is necessary for all branches of engineering Fundamental units/ base units. The units of fundamental quantities are called fundamental units fundamental quantifier are the physical quantities’, which do not depend on any other physical quantities for its measurements. There are seven base/ fundamental quantities with two more quantities (plane angle and solid angle) Table 1.1 Sr. No. Fundamental Physical Quantity Fundamental Unit Symbol 1 length metre m 2 mass kilogram kg Page | 5 Basic Science (22102) Sr. No. Fundamental Physical Quantity Fundamental Unit Symbol 3 time second s 4 electric current ampere A 5 absolute temperature kelvin K 6 luminous intensity candela cd 7 amount of substance mole mol Derived units The units of derived quantities are called as derived units derived quantities. These physical quantities are expressed as combination of one are more fundamental quantity. Table 1.2 Derived Quantity Formula Derived units Area Length * Length Square meter Volume Length * Length * Length Cubic meter Density Mass/ Volume Kg/m3 Velocity and Speed Length/ Time m/s Acceleration Velocity/ Time m/s2 Force Mass * Acceleration kg m/s2 Energy and Work Force * Length kg m2/s2 Power Energy/ Time kg m2/s3 Pressure and Thrust Force/ Area kg /m s2 Momentum Mass * Velocity Kg m/s Charge Current * time As Potential difference Potential energy /charge kg m2/A s3 Resistance Potential difference / current kg m2/A2 s3 Page | 6 Basic Science (22102) Systems of Unit The complete set of base/fundamental units and derived units is known as system of units. Different countries used different systems of units for measurement of physical quantity. The system of units is as following The following system of units are in used 1. C.G.S. system. They are centimeter, gram, second respectively 2. F.P.S. system. They are foot pound second respectively 3. M.K.S. system. They are meter kilogram system 4. S.I. system. System international The S.I. systems of units are now accepted worldwide for measurement. Multiples and Submultiples A measurement is expressed in terms of units. However, sometimes it finds more convenient to express a measurement in terms of multiples and submultiples of unit. For example, meter is convenient unit to express the length of table but it is too small to express the distance between two cities. It is more convenient to express the diameter of wire in millimeters and distance between two cities in kilometers. Thus Multiples and Submultiples of units are expressed by means of prefixes. Each prefix represents a multiplication factor in terms of a certain power of 10. Any measurement is expressed in a compact and convenient form by making use of an appropriate prefix for the unit. Thus while expressing very large as well as very small magnitude of physical quantity prefixes are used with unit of that physical quantity. They are called multiples and submultiples. Some of the most commonly used multiples and submultiples (prefixes) are listed below Table 1.3 Symbol Name Factor Symbol Name Factor Y yotta 1024 y yokto 10-24 Z zetta 1021 z zepto 10-21 E Exa 1018 a atto 10-18 P peta 1015 f femto 10-15 T tera 1012 p pico 10-12 G giga 109 n nano 10-9 M mega 106 µ micro 10-6 k kilo 103 m milli 10-3 h hecto 102 c centi 10-2 da deka 101 d deci 10-1 Page | 7 Basic Science (22102) Example: 1) 1 meter = x cm x= 100 or 102 2) 1 mm = x m x = 0.001 or 10-3 3 1 GHz =x MHz x= 1000 or 103 Dimension of physical quantity The physical quantities are described by its dimension. The physical quantities represented by derived units can be expressed in combination of fundamental quantities. The dimension is denoted with square brackets [ ]. Thus length has dimensions [L ] , Mass[M ], Time [T ] eclectic current[A ], Thermodynamic temperature [ K] luminous intensity [cd ] and amount of substance [mol]. The dimensions of physical quantities are the powers to which the fundamental quantities are raised to represent that quantity. Dimension formula The expression, which shows how which of fundamental quantities represent dimensions of physical quantity is called the dimensional formula. Table 1.4 Quantity Formula Dimensions Dimensional Formula Area Length x Length [L2] [L2 M0 T0] Volume Length x Length x Length [L3] [L3 M0 T0] Density Mass/ Volume [L-3 M1 ] [L-3 M1 T0] Velocity and Speed Length/ Time [L1 T-1] [L1 M0 T-1] Acceleration Velocity/ Time [L1 T-2] [L1 M0 T-2] Force Mass x Acceleration [L1 M1 T-2] [L1 M1 T-2] Energy and Work Force x Length [L2M1T-2] [L2 M1 T-2] Power Energy/ Time [L2 M1 T-3] [L2 M1 T-3] Pressure and Thrust Force/ Area [L-1 M1 T-2] [L-1 M1 T-2] Momentum Mass x Velocity [L1 M1 T-1] [L1 M1 T-1] Power Energy/ Time [L2 M1 T-3] [L2 M1 T-3] Errors The result of every measurement by a measuring instrument contains some uncertainty. This uncertainty is called as error. The accuracy and precision are two terms involved measurement. accuracy is a measure how close the measured value is to the true value while precision tells us to what resolution or limit the quantity is measured. Page | 8 Basic Science (22102) The errors are classified as instrumental systematic error, random errors. Instrumental error The errors which arise due to imperfect design or calibration of measuring instrument. ex. In Vernier caliper zero mark may not coincide with zero mask of main scale. This error can be minimize by selecting better Instruments Systematic errors These errors arise due to imperfection in experiment technique or procedure the pointer of voltmeter, which does not coincide with any mark. Random errors The errors which occur due to unpredictable. Change in experimental conditions. Which cannot be controlled. Change in temperature, voltage fluctuation, etc. Significant figures It is a trust worthy digit in measurement of physical quantify. the significant figure is related to accuracy of measuring instrument more the significant fingers is the accuracy in measurement Rules for significant figures All the non-zero digits are significant figures 1. All the zero between non zero digit are digit are significant figures 2. If the number is less than, the zeros on the right of decimal point but to the left of non zero digit are not significant figures.0.0028 Underlined zero are not significant figures 3. The terminal, number without a decimal point are not significant figure( 123m.12300cm=123000mm) has 3 significant Figures trailing zero are not significant 4. The trailing zero in number with a decimal point are significant 3.500 or 0.06900 have four significant figure. 5. The number should be generally reported in scientific notation (ie power of 10 ) 4.700m = 4.700x102cm=4.700x103mm = 4.700x10-3 km Power of 10 is irrelevant to determine significant figures. However, the base number in the significant notation is significant thus there are 4 significant figures. Table 1.5 Sr. No. Value Number of significant Figures 1 11.080 5 2 0.0196 3 Page | 9 Basic Science (22102) 3 200.0109 7 4 0.0005300 4 5 1.6×10-19 2 Estimation of error Absolute errors Suppose the values obtained in the measurement are a1, a2 , a3 , an the arithmetic mean of these values are taken as best possible value of quantity a1 + a2 + a3 + … an amean = 𝑛 The magnitude of difference between the individual measurement and true of the quantity is called the absolute error of measurement |∆a| ∆a1 = a1 − a mean ∆a2 = a2 − a mean ∆a3 = a3 − a mean : : : : : : ∆an = an − a mean ∆𝐚 calculated may be negative or positive but the absolute error is always taken positive. Mean absolute error the mean of all absolute errors is called as mean absolute error ∆a mean ∆a1 + ∆a2 + ∆a3 + … ∆an ∆a mean = 𝑛 Relative error The relative error is the ratio of mean absolute error ∆a mean to the mean value a mean of the measured quantify. ∆amean Relative error = amean Percentage error The relative error expressed in percent is called as percentage error Percentage error = relative error x 100% ∆amean = x 100% amean If the error ∆L in the measurement of length L then percentage error is given by, 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝐸𝑟𝑟𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑙𝑒𝑛𝑔𝑡ℎ 𝐿 =(∆𝐿/𝐿)×100% ……..(1) If the error ∆L in the measurement of L then percentage error Ln is given by, 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝐸𝑟𝑟𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝐿𝑛 =𝑛×(∆𝐿/𝐿)×100% ………. (2) Page | 10 Basic Science (22102) If the error ∆L in the measurement of length L, ∆B in the measurement of breadth B then surface area given by A = L×B and percentage error in the measurement of surface area is given by, 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝐸𝑟𝑟𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝐴 =% error in L + % error in B 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝐸𝑟𝑟𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 A =(∆𝐿/𝐿)×100+ (∆𝐵/𝐵)×100% ………(3) If acceleration due to gravity using simple pendulum is given by g = 4π2 L/T2, 4π2 being constant are not considered in calculation hence percentage error in ‘g’ is given by, 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝐸𝑟𝑟𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑔 =% error in L + 2(% error in T) 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝐸𝑟𝑟𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑔 = (∆𝐿/𝐿)×100% +2 ×(∆𝑇/𝑇)×100% Example : In an experiment a venier caliper of least count 0.01cm is used for measurement. The radius of the sphere measured was 2.0 cm. Find the percentage error in measurement of radius, area, volume of the sphere? Solution : The percentage in measuring the radius of the sphere is given by equation 1 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝐸𝑟𝑟𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 radius =(∆r/r)×100% = (0.01/2) ×100% =0.5 % The percentage in measuring the area of the sphere is given by equation 2 Area of sphere is 4π r2 4π being constant are not considered in calculation 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝐸𝑟𝑟𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 r2 =𝑛×(∆r/r)×100% = 2(0.01/2) ×100% =1 % The percentage in measuring the volume of the sphere is given by equation 2 Area of sphere is (4/3)π r3 (4/3)π being constant are not considered in calculation 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝐸𝑟𝑟𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 r3 =𝑛×(∆r/r)×100% = 3(0.01/2) ×100% =1.5 % Example : Find percentage error in measuring the density of wood cube. When the mass of the mass of the block is 50 ± 0.1g , and the length of the cube is 2 ± 0.01cm. Solution : Density of the wooden block is given as density = (mass /volume) Volume of the block is V = L3 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝐸𝑟𝑟𝑜𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 density = (∆m/m)×100% +3 ×(∆l/l)×100% = (0.1 /50)×100% +3 ×(0.01 /2)×100% Page | 11 Basic Science (22102) =0.2 +1.5 =1.7% Application 1. Measurement of length breadth height of given object 2. Measuring the radius of circular object 3. Dimensional analysis to deduce relation between different physical quantities 4. Estimate error for quality control E learning website http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html https://en.wikipedia.org/wiki/Unit_of_measurement https://www.youtube.com/watch?v=HQdy2Z-GmdY https://www.youtube.com/watch?v=k_CoXqKDyuQ Sample MCQ 1. ______of the following is the fundamental quantity? a) Meter B) length C) force d) speed Ans (b) 2. There are_____ significant figure in 0.26x102 m. 2 B) 3 C) 4 d) 1 Ans (a) 3. _____is correct dimension of density. a) [L-1 M1 T0] b) [L-3 M1 T0] c) [L3 M1 T0] d) [L-3 M1 T1] Ans (b) 4. Relative error is given as ___. ∆amean ∆amean amean amean a) b) c) d) amean 𝑎 ∆amean 𝑎 Ans (a). 5. ____ is not a system of units. C,G,S b) C.I.S c) M.K.S. d) S.I Ans (a) Reference: 1. Physics Textbook part 1, J.V.Naralikar, A.W.Joshi et al 11th NCERT New Delhi ISB N81-7450-508-3 Self-learning 1 Read the definitions of fundamental units. Page | 12 Basic Science (22102) Micro projects 1 Measure the object using Vernier caliper of different least counts. 2 Measure the object using micrometer screw gauge of different least counts. 3 Measure the circular object using spherometer of different least counts. 4 List down various measuring instrument in your core programs Page | 13 Basic Science (22102) Topic 2 Electricity, Magnetism and semiconductors. Course outcomes -Apply the principals of electricity and magnetism to solve engineering problems. Unit outcomes a) Calculate electric field, potential, and potential difference of the given static charge. b) Describe the concept of given magnetic intensity and flux with relevant units. c) Explain the heating effect of the given electric current. d) Apply Laws of series and parallel combination in the given electric circuits. e) Distinguish the given conductor, semi-conductor and insulators on basis of energy bands. f) Explain IV characteristics and applications of given P-N junction diode. Introduction / Rationale The concepts of electricity and magnetism is the foundation of engineering and technology The devices for communication such as radio, television, electric motors, medical instruments, transportation are based on electricity and magnetism. Semiconductor devices are small in size, consume less power and have long life and reliability. The following topic we will introduce the basic concepts of semiconductor physics. Semiconductor basic of electronic gadgets Significance All devices like Electric motors, electrical equipment, transportation devices need knowledge and understanding of electricity and magnetism. Moreover, electronic gadgets such as mobiles, computers, power supplies, electronic circuits, sensors require understanding of semiconductors. 2.1 Electric field and electric potential: We have experience that, when we remove out sweater in winter, we hear crackling sound, and the sweater appears to stick to our body. Similarly when dry hair is combed with a comb crackling sound is produced. This is because of electric charges produced due to friction between two bodies. Due to friction, electrons get transferred from one substance to other substance, making them charged. There are two types of charges. The substance receiving electrons becomes negatively charged and the other which losses electrons become positively charged. Charge is measured in coulomb. Page | 14 Basic Science (22102) 2.1.1: Coulomb’s Inverse Square Law:- There are two types of charges, namely, positive and negative Charges. It was observed that- a) Like charges repel each other and unlike charges attract each other. b) The force of attraction or repulsion between two charge q1 and q2 separated by distance ‘r’ is given by – F 𝛼 q1 q2 1 F 𝛼 𝑟2 𝑞1 𝑞2 F𝛼 𝑟2 𝑞1 𝑞2 F=K … (1) 𝑟2 K is a constant of proportionality depends on the nature of medium. 1 K = 4 𝜋∈ 0 ∈𝑟 Where, ∈0 Permittivity of free space = 8.854X10 -12 F/m ∈𝑟 Relative permittivity of the medium between two charges = 1 for air Above equation (1) becomes 𝑞1 𝑞2 F= … (2) 4 𝜋 ∈𝑜𝑟 2 In equation (2), if q1 = q2 = 1 C And if r = 1 m , we have 1 F = 4𝜋∈𝑜 = 8.99 x 109 N = 9 x 109 N From this we can define one coulomb or unit charge. It is that charge which when placed in air at a distance of 1m from and equal and similar charge, experiences a force of 8.99x109 N Example: The distance between the electron and proton in the atom is 5.8x 10-11 m. Calculate the electrostatic force between them. Given ∈𝑟 = 1 𝑎𝑛𝑑 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 = 1.6 𝑥 10−19 𝑐 Formula: 𝑞 𝑞 F = 9 x 10q x 𝑟1 2 2 1.6𝑥10−19 𝑥1.6𝑥10−19 F = 9 x 10q x (5.8𝑥10−11 )2 F = 6.84x10-8 N Page | 15 Basic Science (22102) 2.1.2: Electric field A charged body placed near a charge will experience a force. The region in which a charge experiences a force is called electric field. 2.1.3 : Electric field Intensity Consider a charge of ‘q’ coulomb. Electric field intensity at point A in its electric field is the force experienced by unit positive charge placed at that point. Fig:2.1.3 𝐹 E= 𝑞 In above Fig:2.1.3 the electric intensity at point A is given by 𝐹𝑜𝑟𝑐𝑒 𝐼𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 = 𝑐ℎ𝑎𝑟𝑔𝑒 1 𝑞 ∈= 𝑓𝑜𝑟 𝑎𝑖𝑟 ∈𝑟 = 1 4𝜋 ∈0 ∈𝑟 𝑟 2 1 𝑞 ∈= 4𝜋 ∈0 𝑟 2 𝑞 ∈ 9𝑥109 𝑟2 Electric Intensity is measured in N/C Example: A charged sphere of 60 micro coulomb is placed in air. Find the electric field intensity at a point 30cm from the center of sphere. Given :- q = 60µ𝑐 = 60𝑥 10-6 c , r = 30 cm = 0.3m , ∈𝑟 = 1 𝑞 ∈ = 9 × 109 × 𝑟 2 60×10−6 ∈ = 9 × 109 × (0.3)2 ∈ = 6 X 106 N/C 2.1.4 : Electric potential Consider a point charge + q Let A is the point in its electric field. Imagine a unit positive charge at infinity as shown in Fig : 2.1.4.1 Page | 16 Basic Science (22102) Fig : 2.1.4.1 The electric potential at point A is the amount of work done in bringing unit positive charge from infinity to that point A against the electric field 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 = 𝑐ℎ𝑎𝑟𝑔𝑒 𝑤 V= 𝑞 The S.I. unit of potential is joule/ coulomb The potential at point A is, 1 𝑞 VA = 4 𝜋 ∈ ×𝑟= 0 ∈𝑟 𝑞 ------ For air V= 9 x 10 9 x 𝑟 One volt: 1 𝑗𝑜𝑢𝑙𝑒 1 volt = 1 𝑐𝑜𝑢𝑙𝑜𝑚𝑏 The potential at point is said to be one volt if one joule of work is done in bringing unit positive charge from infinity to that point. Concept of potential difference: Consider two water containers A and B, Connected by a tube. The water will flow through the tube only when there is a water level difference between A and B. As water level is the factor deciding the flow of water, similarly charge level i.e. potential is the factor which decides the flow of charges. Consider a point charge +q as shown in Fig: 2.1.4.2. A and B are two points in the electric field of charge +q. Fig : 2.1.4.2 Potential difference between the two points A and B is the work done in bringing unit positive charge from point B to point A Page | 17 Basic Science (22102) The potential at point A is given by, 1 𝑞 VA = 4𝜋∈ 0 ∈𝑟 𝑟1 The potential at point B is, 1 𝑞 VB = 4𝜋∈ 0 ∈𝑟 𝑟2 The potential difference between A and B is, V = VB – VA 1 𝑞 𝑞 V = 4𝜋∈ x(𝑟 -𝑟) 0 ∈𝑟 2 1 𝑞 𝑞 V = 9 × 109 x ( - ) 𝑟2 𝑟1 Example : Calculate the potential at point 10 cm away from a charge of 300 𝜇𝑐, in air Given : Q = 300 𝜇𝑐 = 30 X 10-6 C r = 10 cm = 0.1m, ∈𝑟 = 1 Formula :- 𝑄 V = 9 × 109 × 𝑟 300𝑥10−6 = 9 × 109 × 0.1 = 27 x 104 v 2.2 Magnetism In physics, magnetism is defined as the property by virtue of which a magnetic material is able to create an attraction or repulsion force. It is the behavior of matter in magnetic field. Magnetism is a force that can attract or repel objects that have a magnetic material like iron inside them magnetism is caused by the motion of electric charges. Every substance is made up of tiny units called atoms. Each atom has electrons, particles that carry electric charges. The first magnet was discovered from a naturally occurring mineral called magnetite. A material with its molecular alignment has one effective North Pole and an effective South Pole. The most basic law of magnetism is that like poles repel each other and unlike poles attract each other. If a bar magnet is broken, into two or more pieces then each piece behaves like an independent magnet with somewhat weaker magnetic field. When a bar magnet/ magnetic needle is suspended freely or is pivoted, it aligns itself in geographically North-South direction. 2.2.1 Magnetic field: - A magnetic field can be produced by either movement of charge or a magnetized material.The region around a magnetic material or a moving electric charge within which the force of magnetism acts. Magnetic field has magnitude and direction at each point. So it is a vector field. Magnetic field is denoted as ‘B’.Its SI unit is tesla (T) and CGS unit is gauss. 2.2.2 Magnetic field intensity (H) Intensity of a magnetic field is the force which a unit North Pole experiences when placed within the magnetic field. Number of magnetic lines passing through unit area is called Page | 18 Basic Science (22102) magnetic field intensity. It is also called magnetic field strength or magnetizing force. Its unit is ampere/ meter (A/M) or ampere turn. 2.2.3 Magnetic lines of force These are imaginary lines representing direction of magnetic field such that the tangent at any point is the direction of the field vector at that point. The path along which unit North Pole moves when placed in magnetic field is called magnetic lines of force. 2.3.1 Properties of magnetic lines of force Unlike poles attract each other Like poles repel each other Courtesy: physics stack Exchange 1) The magnetic lines of force of a magnet or a solenoid form closed loops. 2) Magnetic lines of force originate from North Pole and end at the South Pole outside the magnet. 3) The direction of magnetic field within a magnet is from South Pole to North Pole. 4) They never intersect each other. 5) The magnetic field strength is proportional to magnetic line density. 6) The magnetic field is tangent to the magnetic lines of force. 7) The magnetic lines of force are denser near the north or south poles indicating large value of magnetic field. 8) All the lines of force have same strength. 2.2.4 Magnetic Flux It is the number of magnetic field lines passing through a closed loop surface. Or The amount of magnetic lines of forces set up in magnetic circuit is called magnetic flux. It is analogous to electric current in electric circuit. SI unit of magnetic flux is weber (wb) The description of magnetic flux allows engineers to easily calculate voltage generated by an electric generator even when the magnetic field is complicated. If we choose a simple flat surface with area ‘A’ as our test area and there is an angle ‘ϴ’ between normal to the surface and magnetic field vector (B) then magnetic flux is given by Page | 19 Basic Science (22102) Fig 2.2.4 (Courtesy: IB physics: Magnetic flux, Saburchill.com) 2 Example: A circular antenna of area 3m is installed at a point. The plane of the area of antenna is inclined at 470 with the direction of earth’s magnetic field. If magnitude of earth’s magnetic field at a place is 40773.9 nT. Find magnetic flux linked with antenna. B = 40773.9 nT = 40773.9 x 10-9 T 𝜃 = 90 – 47 = 430 A = 3 m2 ∅ = B A cos∅ ∅ = 40.773.9x10-9 x 3 x cos430 ∅ = 89.47x 10-6 wb ∅ = 89.47 µ Wb 2.3 : Current Electricity 2.3.1: Electric Current: Consider a metal rod AB as shown in Fig 2.3.1.1. There are number of free electrons moving randomly. The number of electrons moving in one particular direction is same as the number of electrons moving in exactly opposite direction. As a result the net number of electrons passing in direction is zero. Fig 2.3.1.1 (Courtesy: Circuit Globe) Page | 20 Basic Science (22102) Now if potential difference is applied across the conductor AB, by connecting battery or cell as shown in Fig 2.3.1.2 below, all the electrons will move towards the positive terminal of the battery. We say the current is flowing through the conductor. The conventional direction of current I opposite to the direction of flow of electrons Fig 2.3.1.2 (Courtesy: Circuit Globe) The rate of flow of charge through a conductor is called as current If, ‘Q’ is charge flowing through the conductor in time ‘T’ second Then, 𝐶ℎ𝑎𝑟𝑔𝑒 Current = 𝑇𝑖𝑚𝑒 𝑄 I=𝑇 the unit of current is ampere. 1 𝐶𝑜𝑢𝑙𝑜𝑚𝑏 1 ampere = 1 𝑠𝑒𝑐𝑜𝑛𝑑 One ampere is the current flowing through the conductor, when one coulomb charge is flowing through the conductor in one second. When the electrons are flowing through the conductor, the motion of electrons is opposed by the conductor. This opposition offered by the conductor to the flow of current is called as resistance. A good conductor have low resistance and poor conductor has high resistance 2.3.2: Ohm’s law According to ohm’ law, the current flowing through the conductor is directly proportional to potential difference applied across its ends, provided physical state of the conductor ( material, length, cross-sectional area, temperature) remains same. If, ‘I’ is current flowing through the conductor and ‘V’ is the potential difference across the conductor then, Then V𝛼I V = IR Where, R is the constant called as resistance of the conductor. The value of R depends on the nature, dimension, and temperature of the conductor Page | 21 Basic Science (22102) 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 Resistance = 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 The unit of resistance is ohm (Ω) 𝑉 R= 𝐼 1 𝑣𝑜𝑙𝑡 One ohm = 1 𝑎𝑚𝑝𝑒𝑟𝑒 One ohm is the resistance, when one ampere current is flowing through a conductor and the potential difference of one volt is applied across the conductor. 2.3.3: Specific resistance or resistivity (ρ) When temperature & nature of material of conductor remain same, the resistance of a conductor depends on : (1) Length :- It is observed that resistance of the conductor is directly proportional to its length. R𝛼 L (2) Area of cross-section:- the resistance of the conductor is inversely proportional to its cross- sectional area. 1 R𝛼 𝐴 𝐿 Therefore, R 𝛼 𝐴 𝐿 R = constant x 𝐴 𝐿 R=𝜌× ρ 𝐴 Where ‘ρ’ is constant of proportionality, called as resistivity or specific resistance of the conductor, It depend on nature of 𝑅×𝐴 conductor and temp, ρ= 𝐿 If A = 1 m2 and L = 1 m then ρ = R Specific resistance of the material is the resistance of that material having unit length and unit cross- section area. Unit of ρ is Ω m Conductance: - Conductance is the reciprocal of resistance. 1 Conductance = 𝑅𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 The unit of conductance is mho ( Ʊ) Effect of temperature on resistance:- Resistance of conductor increases with increase in temperature. Page | 22 Basic Science (22102) i. e. R 𝛼 temperature In case of bad conductors of electricity, resistance decreases with increase in temperature. 2.3.4 : Law of series combination of resistances Consider three resistances R1 R2 and R3 connected end to end as shown in Fig. 2.3.4. This is called as series combination. Fig. 2.3.4 I Current flowing through each resistance V Total voltage applied across series combination V1 Voltage drop across R1 V2 Voltage drop across R2 V3 Voltage drop across R3 We have by ohm’s law V = IR ---------------------- Similarly , V1 = IR1 V2 = I R2 V3 = I R3 But, V= V1+V2+V3 Substituting these values in above equation, IR = IR1 + IR2 + IR3 Rs = R1 + R 2 + R3 This is the effective resistance of series combination. Law of resistance in series combination: It states that the effective resistance offered by two or more resistors connected in series combination is the algebraic sum of resistances of the individual resistors. 2.3.5 : Law of parallel combination of resistances Page | 23 Basic Science (22102) Consider three resistances R1 R2 and R3 connected as shown in Fig. 2.3.5. This is called as parallel combination. Fig. 2.3.5 I Total current flowing through each resistance V Total voltage applied across parallel combination I1 Current flowing through resistor R1 I2 Current flowing through resistor R2 I3 Current flowing through resistor R3 We have by ohm’s law 𝑉 𝐼 = 𝑅- Similarly 𝑉 𝐼1 = 𝑅1 𝑉 𝐼2 = 𝑅2 𝑉 𝐼3 = 𝑅3 But, in parallel circuit current gets divided. Therefore total current, I= I1+I2+I3 𝑽 𝑽 𝑽 𝑽 = + + 𝑹 𝑹𝟏 𝑹𝟐 𝑹𝟑 𝟏 𝟏 𝟏 𝟏 = + + 𝑹𝒑 𝑹𝟏 𝑹 𝟐 𝑹𝟑 This is the effective resistance of parallel combination. Law of resistance in parallel combination: It states that the reciprocal of effective resistance offered by two or more resistors connected in parallel combination is equal to the algebraic sum of the reciprocals of the resistances of individual resistors. 2.3.6: Heating effect of electric current Page | 24 Basic Science (22102) Introduction: When an electric current is passed through a conductor, the conductor becomes hot after some time and produce heat. This happens due to the conversion of some electric energy passing through the conductor into heat energy. This effect of electric current is called heating effect of current. Fig 2.3.6 The heating effect of current was studied experimentally by Joule in 1941. After doing his experiments Joule came to the conclusion that heating effect of an electric current depends on three factors: 1. The resistance, R of the conductor. A higher resistance produces more heat. 2. The time, t for which current flows. The longer the time the larger the amount of heat produced 3. The amount of current, I. the higher the current the larger the amount of heat generated. Statement of Joule’s law: When current flows through a conductor, the heat produced in a conductor is directly proportional to the product of square of current (I2), resistance of the conductor (R) and the time (t) for which current is passed. Thus, H ∝ I2Rt Derivation of Formula To calculate the heat produced in a conductor, consider current I is flowing through a conductor AB of resistance R for time t. also consider that the potential difference applied across its two ends is V. Now, total amount of work done in moving a charge q from point A to B is given by: W = q X V ………… (1) Now, we know that charge = current × time or q= IX t and V = I X R ………..(Ohm’s law) Putting the values of q and V in equation (1), we get Page | 25 Basic Science (22102) W = (I X t) X (I X R) Or W = I2Rt Now, assuming that all the work done is converted into heat energy we can replace symbol of ‘work done’ with that of ‘heat produced’. So, H = I2RT This equation is called the Joule’s equation of electrical heating. Applications of Heating Effect of Current The heating effect of current is used in various electrical heating appliances such as electric bulb, electric iron, room heaters, geysers, electric fuse etc. Electrical energy and power The work done in pushing a charge round an electrical circuit is given by 𝑊 =𝑉×𝐼×𝑡 𝑊 So that power, 𝑃 = =𝑉×𝐼 𝑡 The electrical power consumed by an electrical appliance is given by 𝑉2 𝑃 = 𝑉 × 𝐼 = 𝐼2𝑅 = 𝑅 Example: An electrical bulb is labeled 100W, 240V. Calculate: 1. The current through the filament when the bulb works normally 2. The resistance of the filament used in the bulb. Solution: 𝑃 100 𝐼= = = 0.4167 𝐴 𝑉 240 𝑃 100 𝑅= 2= = 576.04 Ω 𝐼 (0.4167)2 2.4 :Semiconductors 2.4.1: Conductors The material which allow electric current to flow through them easily are called conductor. All metals are almost good conductor like gold, silver, copper, aluminum etc. Conductor possess larger numbers of free electrons. They possess very low resistivity (ρ) or they have high conductivity (σ) 2.4.2 : Insulators The material which does not allow electric current to flow through them are called insulators. The electrons are tightly bounded to their nucleus as such there are no free electrons to move and conduct electricity. All nonmetals are insulators like quartz, mica, glass, rubber etc. They have high resistivity or they have low conductivity. Page | 26 Basic Science (22102) 2.4.3: Semiconductor The material which has conductivity in between metal & insulator is called semiconductor. The semiconducting materials are neither conductor no insulator like germanium, silicon, carbon etc. Semiconductor are insulators at low temperature and act as conductors at high temperature. They have resistivity or conductivity intermediate to metals & insulators. 2.4.4: Electron volt: The kinetic energy gained by an electron when accelerated trough potential difference of one volt is called as one electron volt. Energy gained by charged particle = magnitude of charge * potential difference 1eV = charge on electron * 1 volt =1.6*10-19 *1V = 1.6*10-19 J 2.4.5: Valance band (Ev) The electron in the outermost shell are called as valance electron. The band formed by the series of energy levels containing the valance electrons is known as valance band. The valance band may also be defined as a band which is occupied by the valance electrons. A valance band has highest occupied band energy. The valance band may be partially or completely filled depending upon nature of crystal. 2.4.6: Conduction band (Ec) After the valance band the next higher band is conduction band. The electrons occupying this band are known as conduction electrons or free electrons. The conduction band may also be defined as the lowest unfilled energy band. The band may be empty or partially filled in the conduction band the electron can move felly thus there electrons are known as free elections. 2.4.7: Forbidden gap/Energy gap Eg The gap between the top of valance band and bottom of conduction band is called the energy band gap/Forbidden gap.no election can exit in the forbidden gap. The energy gap may be large small or zero depending on the material. Fig. 2.4.7.1 Case I Conductors. The conduction band is partially filled and the valance band is partially empty or the conduction and valance band overlap. The energy gap is zero (0 eV). In overlap electrons from valance band Page | 27 Basic Science (22102) can move easily into the conduction band Thus we have electrons available for electrical conduction. The resistance of such materials is low or the conductivity is high. Fig. 2.4.7.2 Case II Insulators: In this case a large band gap exists (Eg >3eV) There are no electrons in the conduction band and therefore no electrical conduction is possible the energy gap is so large that the electrons cannot be excited from valance band to conduction by thermal energy. Fig. 2.4.7.3 Case III Semiconductor: The energy band gap is finite & small (Eg Ө2 The amount of heath conducted through the in steady state is directly proportional to 1) cross sectional area a 2) temperature difference between layers C and D i.e. Ө1- Ө2 3) time for which heat flows i.e. T and inversely proportional to 4) distance d between two planes C & D QαA Q α (Ө1- Ө2) Qαt Q α 1/d Combining above we can write A (Ө1− Ө2) t Qα 𝑑 𝐴× (θ1 − θ2 ) ×𝑡 Q = constant x 𝑑 𝐴× (Ө1− Ө2) × 𝑡 Q = K× 𝑑 Where K is coefficient of thermal conducting. Form above equations we get 𝑄.𝑑 K = 𝐴(Ɵ 1 −Ɵ2 )𝑡 If, d= 1, A= 1 , (Ɵ1-Ɵ2)=1, t=1 then K =Q 3.2.7: Definitions of coefficient of thermal conductivity- It is defined as the amount of heat conducted in steady of temp of the rod per unit time, per unit cross-sectional area, for unit temperature gradient Ɵ1 − Ɵ2 Here is called as the temperature gradient. 𝑑 Temp gradient is the change in temperature per unit length of the rod. Unit of temperature gradient is 0c/m or 0k/m Page | 45  Basic Science (22102) Example: A windows pane with glass material has a dimension 100cm x 50cm x 5mm. Amount of heat conducted in one hour is Q. calculate Q if the temperature difference is 50C between outside & inside. given k for glass =1 W/m/0K Given A = 100 cm x 50cm = 1m x 0.5 m = 0.5m2 d= 5mm = 5x10-3m t = 1hr = 60 x 60 = 3600sec Ɵ1 – Ɵ2 = 50c k= 1 w/m / 0k 𝐴 (Ө1− Ө2) 𝑡 Formula: Q=k 𝑑 1𝑥0.5𝑥5𝑥3600 = 5𝑥10−3 = 18 x 105 J 3.3 : Gas laws 3.3.1: Boyle’s law This law gives the relation between pressure and volume of constant temperature. Fig:3.3.1 Statement of Boyle’s law “For fixed mass of a gas temperature of a gas remaining Constant its pressure is inversely proportional to volume” 1 i.e P∝ (at const. temp) 𝑉 1 P = const , x 𝑉 PV = const. P1 V1 = P2 V2 = const Where P1, P2 are pressure & V1, V2 are volumes at constant temperature of fixed mass of gas. Page | 46 Basic Science (22102) 3.3.2: Charle’s law This law gives relation between volume & temperature at constant pressure. Fig: 3.3.2 Statement of Charle’s law: “For fixed mass of a gas, if pressure of a gas is kept constant then its volume is directly proportional to absolute temperature” i.e V ∝ T ----------( at constant pressure) 𝑉 = constant 𝑇 𝑉1 𝑉 = 𝑇2 =constant 𝑇1 2 Where V1, V2 are volumes and T1, T2 are temperatures of given mass of gas. 3.3.3: Gay lussac’s law This law gives the relation between pressure and temperature at constant volume of gas. Fig:3.3.3 Page | 47  Basic Science (22102) Statement of Gay Lussacç law: “For fixed mass of a gas if volume is kept constant, then its pressure is directly proportional to absolute temperature.” i.e P ∝ T (at constant volume ) 𝑃 = constant, 𝑇 𝑃1 𝑃 = 𝑇2 = const 𝑇1 2 Where P1, P2 are pressure & T1, T2 are temperature of given mass of gas. 3.3.4: General Gas Equation/Perfect Gas Equation The relation between pressure, volume and temperature is given by gas laws. Each law gives the relations of two variables by keeping third variable constant. The relation between all the three variables is given by Perfect Gas Equation. According to Gay- Lussac’s Law PαT And according to Charles law VαT Combining above two equations we get, PxVαT P x V= Const. x T PV=KT ----------------------------(1) Where, K = Specific gas constant. (constant K is different for different gases) This constant depends on mass and nature of gas. Equation- 1 is known as ideal gas equation. If mass of gas is 1 kg-mole OR 1 gm-mole then proportionality constant K will have same value for all gases. Therefore, k is replaced by R 3.3.5: Perfect Gas Equation From above two equations we can write PV α T PV = constant x T PV=KT Where K is constant called as specific gas constant, value of K is different for different gases, since it depends on mass and nature of gas. If we consider molecular weight of gas (gram-mole or kg. mole) then constant K will have same value for all gases. Replacing K by another constant R, we get PV=RT This is called General gas equation OR Perfect gas equation OR Universal gas equation Page | 48 Basic Science (22102) Where R- universal gas constant, Value of R is same for all gases, because one-gram mole of any gas occupies same volume under NTP. ( P = 76 cm of mercury, T= 273°K) R= 8314.91J/0k.kg mole or R=8.314 J/mole0k Examples:- 1) The volume of a gas at 20 0C is 500 cm3.What will be its volume at 80 0C. Solution :- Given :- T1 = 20+273= 293 0K V1 = 500 cm3 T2 = 80+273 = 353 0K V2 = ? Formula :- 𝑉1 𝑉 = 𝑇2 𝑇1 2 𝑉1 𝑇2 V2 = 𝑇1 500 𝑥 353 = 293 = 602. 39cm3 2) A gas at 20 0C and pressure of 70 cm. of mercury has volume 2.5 litres. Find its volume at 30 0 C and pressure of 85cm of mercury. Solution:- Given:- T1 = 20 + 273 = 293 0K P1 = 70 cm V1 = 2.5 Litre T2 = 30 + 273 = 303 0k P2 = 85 cm V2 = ? 𝑃1 𝑉1 𝑃2 𝑉2 Formula:- = 𝑇1 𝑇2 70𝑥2.5 85 𝑥 𝑣2 ⸫ = 293 303 V2 = 2.12 Lit. 3.3.6: Application of gas laws Boyle’s law 1) Spray paint 2) Soda can 3) The syringe Page | 49 Basic Science (22102) Charles law 1) Helium ballon 2) Hot air bollon 3) Deodorant spray bottle 4) Type pressure- low in winter (cold) Gay Lussacs 1) Firing a bullet 2) Heating a closed a aerosol can-may cause the container to explode. 3.4 Heat capacity (s) The change in temperature of substance, when a given quantity of heat is absorbed or rejected by it, is characterized by a quantity called ‘heat capacity’ of a substance Heat capacity (s) of substance is given by 𝑄 S= 𝑡 ………………………. (i) ∵ Q= amount of heat supplied to the substance to change its temperature by t. If equal amount of heat is added to equal masses of different substances, the resulting temperature changes will not be same i.e each substance has a unique value for the amount of heat absorbed or rejected to change the unit mass of it by a unit. This refers to the specific heat capacity of the substance it is given by 𝑠 1 ∆𝑄 S= = x ……………………(ii) 𝑚 𝑚 ∆𝑡 3.4.1 Specific heat of gases The specific heat of a substance is defined as the amount of heat required to increase the temperature of unit mass of the substance through 1°c. The specific heat capacity is the property of substance which determines the change in the temperature of substance when a given quantity of heat is absorbed or rejected by it. Specific heat capacity “s” is defined as the amount of heat per unit mass absorbed or rejected by the substance to change its temperature by one unit. It depends on nature of substance and its temperature. SI unit of specific heat capacity is J/kg°k If amount of substance is specified in terms of mole ‘µ’ instead of mass ‘m’ kg then the molar specific heat capacity ‘c’ is given by 𝑠 1 𝑄 C= = x …………………… (iii) µ µ 𝑡 SI unit of molar heat capacity is J/mol°k When we heat a substance it increases in volume. Thus surrounding atmosphere increases through a small distance. As a result of this external mechanical work is done against the atmospheric pressure. In case of solids and liquids the change in volume is very small hence the external mechanical work done while changing the temperature is very less. But, in case of gases, there is Page | 50 Basic Science (22102) a large change in volume and pressure and part of this work is utilized to do the external mechanical work. In case of solids and liquids heating takes place at constant pressure while in case of gases heating can be done either at constant pressure or at constant volume. Therefore gas has two specific heats, namely i) Specific heat at constant volume (Cv ) ii) Specific hat at constant pressure (Cp) These two specific heats are called principle specific heats of gases. 3.4.2 Specific heat of gas at constant volume (Cv) The amount of heat required to raise the temperature of unit mass of gas through 1°c when its volume is kept constant is called specific heat of the gas at constant volume and denoted by cv. It we heat ‘m’ gram of gas enclosed in a cylinder fitted with a piston, the volume of the gas will not increase but only pressure and temperature will increase. As there is no increase in volume there will be no external work done and total heat energy supplied will be used to increase the temperature of the gas. Let the initial temperature of gas be Ɵ1 and final temperature at constant volume be Ɵ2, then the heat supplied to the ‘m’ gm of the gas. Q1 = mass x specific heat at constant volume X rise in temperature. ∴ Q1 = m x cv (Ɵ2-Ɵ1) ∴ Q1 = m cv dT ∴ dT = change in temperature. 3.4.3 Specific heat of a gas at constant pressure (CP) The amount of heat required to increase the temperature of unit mass of a gas through 10c when its pressure is kept constant is called specific heat at constant pressure denoted as CP. Fig. 3.4.3 Heating of a gas at constant pressure. Page | 51 Basic Science (22102) Consider a cylinder fitted with piston of negligible mass and friction as shown in fig 3.4.3. Fill the cylinder with ‘m’ gram of gas and start heating. The temperature of the gas increases with increase in volume of gas. Piston moves upward keeping pressure constant, equal to atmospheric pressure. As a volume is increased some external work is done and the heat energy supplied gets utilized in two ways, i) Increase the temperature of gas. This is equal to mcv (Ɵ2-Ɵ1) ii) Fording work by increasing the volume of gas. As during heating the gas at constant pressure, extra work is done therefore cp is greater than cv 3.4.4 Work done at constant pressure Work done by heating the gas at constant pressure is given by, W = R(T2-T1) ∵ W= work done R = universal gas constant T2 – T1 = Ɵ2-Ɵ1= change in temperature of the gas at expansion. S-I unit of work done is joule Gas unit of work done is erg. 1J = 107 erg. 3.4.5: Relation between Cp and Cv (Mayor’s relation) 𝑅 Cp – Cv = 𝐽 ∵ Cp = specific heat of gas at constant pressure. Cv = specific heat of gas at constant volume R = universal gas constant. J = mechanical equivalent of heat. 3.4.6 Ratio of two specific heats ( 𝜸 ) The ratio of Cp and Cv is denoted by ‘𝛾’. It is also known as adiabatic index. 𝑐𝑝 𝛾 = 𝐶𝑣 As Cp > Cv , 𝛾 > 1 𝛾 = 1.66 for monoatomic gas Γ = 1.4 for diatomic gas (air) Example The ratio of two specific heats of a gas is 1.4 and the difference is 0.0808. Find values of cp and cv 𝐶𝑝 Given:- = 1.4 𝐶𝑣 Cp –Cv = 0.0808 ∴ Cp = 1.4x Cv ………………. ( I ) ∴ Cp –Cv = 0.0808 ∴1.4Cv – Cv = 0.0808 Page | 52  Basic Science (22102) ∴ 0.4 cv = 0.0808 0.0808 ∴Cv = = 0.202 unit 0.4 ∴Cp = 1.4 x 0.202 ∴ Cp = 0.2828 unit 3.5: Optics 3.5.1: Reflection of light- When light reaches reflecting media then it bounces (rebounds) back in the same media. Let, PQ=incident ray QR=reflected ray NN’=normal to reflecting media  i = angel of incidence  r = angle of reflection Fig:3.5.1 Laws of reflection 1) Angle of incidence is equal to angle of refection  i =  r 2) Incident ray, reflected ray & normal lie in same plane. 3.5.2: Refraction of light When light is allowed to fall on transparent medium such as glass then some light reflects back & some light travels in other medium. When light travels from one medium into other medium then it bends (deviates) this bending of light is called refraction Refraction is defined as property of light on account of which light changes its path (direction) when it enters form one medium into other medium. If incident light is perpendicular to the surface, then light passes without bending and for all other angle of incidence bending of light takes place. When light travels form one medium to other then there is change in velocity direction as well as wavelength of light but the frequency remains unchanged. Page | 53 Basic Science (22102) Snell’s law or laws of refraction: Snell’s law state that for any two media the ratio of sine angle of incidence to sine angle of refraction is constant for a given light beam. The incident ray Refraction ray & normal lie in same plane Thus sin i  cons tan t sin r This constant is called refractive index of second medium with respect to the first medium & is denoted by 1µ2. sin i  µ 1 2 sin r Fig:3.5.2.1 Case 1: when light enters form air (rare) to (denser) medium ray bends toward the normal (Fig:3.5.2.1) sin i  Costrant >1 sin r The constant is called refractive index of glass with respect to air and is denoted by aµg sin i aµg = >1 sin r Case 2 when light enters form denser (glass) to air (rarer) medium then ray bends away from normal (Fig:3.5.2.2) thus i < r sin i =Constant e i.e. angle of incidence should be greater than critical angle 3.5.4.1: Structure (Construction) of Optical fiber Communication optical fiber has cylindrical core surrounded with cylindrical coat of cladding coated with protective skin (insulation, jacket). When light enters into core it propagates by means of total internal reflection at core cladding interface and emerges out form the other end Core- The light is transmitted within core. The core is innermost layer. The refraction index of core is slightly greater than cladding. The typical value of refraction index of core is 1.48 Cladding- The cladding keeps light waves with core because refractive index of cladding is less than that of core. Typical value of refractive index of cladding is 1.46 it also provides some strength to core. Protective skin- The protective skin protects fiber form moisture. This protective skin provides mechanical strength to optical cable. Optical fiber is made from either glass (silk) or plastic Fig:3.5.4 3.5.4.2: Dimensions The length of optical fiber is normally 1km they can be joined using suitable connectors. Outer diameter of fiber ranges from 0.1 mm to 0.15mm. Core diameter ranges from 5µm to 600 µm (typical value is approximately 50µm) cladding diameter varies from 125µm to 750µm. Thickness of protective skin varies from 30 to 50µm 3.5.4.3: Conditions for T.I.R. or Propagation of light through optical fiber 1) Refractive index of core should be greater than refractive index of cladding (i.e. µcore>µcladding) 2) Entering light must have angle of incidence greater than critical angle ( C) Critical angle (𝜽C) - It is the angle of incidence at which angle of refraction is 900 OR The critical angle is defined as the particular value of angle of incidence at which ray emerges along the interface Acceptance angle (𝜽a max)- The maximum value of external incident angle for which light will propagate in the optical fiber It is also called acceptance cone half angle Page | 56  Basic Science (22102) Acceptance cone- If 𝜽a is rotated around the core axis then cone is formed which is called acceptance cone. If light is allowed to fall within this cone (tunnel) then it will propagate till far end Numerical aperture (NA)- It is the sine of maximum acceptance angle NA = sin 𝜽a max It measures the light gathering capacity of optical fiber. Greater the numerical aperture greater the numerical aperture greater is the amount of light the optical fiber will accept. Path (propagation or transmission of light through optical fiber) A beam of light is focused at one end of cable if the angle of incidence of light greater than critical angle then total internal reflection takes place & light beam are reflected through the fiber. The beam bounces back and passes the cable & it exit at the other end the light which travels from one end to other end is called “light is guided” through fiber. Because of total internal reflection light beam will continue to propagate through the fiber even though it is bent number of times. More bending may cause angle of incidence to change & hence loss of light may take place. With long little bends the light with stay within the cable 3.5.5: Path (propagation) of light through different types of optical fibers Depending on variation of refractive index of core & mode of propagation, the optical fibers are of different types. Path of light through different types of optical fibers is different 3.5.5.1: Type Depending on variation of refractive index of core 1) Step index optical fiber 2) Graded index optical fiber Type depending on mode of propagation 1) Single mode optical fiber 2) Multimode optical fiber Thus the different types of optical fiber are a) Step index single mode optical b) Step index multimode optical fiber c) Graded index multimode optical fiber Step index optical fiber – In this types refractive index of core is uniform through the fiber thus if we move radially outward form core axis then there is step (sudden) change in refractive index at core cladding interface. Propagation of light through step index optical fiber is zig- zag Page | 57 Basic Science (22102) Single mode step index optical fiber Fig:3.5.5.a Core diameter of this fiber is very small it is about 10µm. There is only one path of ray of propagation hence it is called single mode & the path is zig-zag i.e. in this type there is only one zig-zag path Multimode step index optical fiber- Core diameter of this is larger than single mode & it is 50-200µm. There are many paths of propagation hence it is called multimode & paths are zig-zag. Thus there are many paths that are zig-zag. Some light paths are longer & some are shorter hence some light rays exit latter some rays exit earlier Fig 3.5.5.b Graded index fiber- In this type refractive index of core is not uniform throughout the fiber. Refractive index of core at core axis is maximum & if we see radially outward from core axis there is gradual decrease in the refractive index. Refractive index of core material at core axis is maximum & it is minimum at core cladding interface. In graded index optical fiber because of different refractive indices within the core light travels with different speeds through different parts with in core path (propagation) of light through graded index optical fiber is curved (curled, coiled, helical) Multimode grades index optical fiber Core diameter of this fiber varies form 50-200µm. Because of different refractive index within core light travels with different speed longer paths are faster & shorter paths are slower. Therefore, all rays reach at same time. Page | 58  Basic Science (22102) Fig 3.5.5.c 3.5.6: Applications of optical fiber Fiber optic cable has outstanding properties they are lighter in weight, less bulky flexible and carries large data with High speed so it has many applications 1) Optical fiber in communication- Because of large bandwidth it can handle number of channels hence found large applications in communication 3) Telephone- Using optical fiber communication we can connect faster & have clear conversations 4) Internet– Optical fiber can pass large data & with faster speed. This technology is used internet cables 5) Computer networking- networking between computer in same building is made easier & faster with the use of fiber optical cables 6) Used in industrial atomization system 7) Used for signaling purpose in military 8) To observe internal organs in medical field 9) Used in cable television 10) Used in defense for confidential communication 11) Used for transmission of digital data 12) Sensors- Optical fiber sensors are used to control liquid level temperature, pressure, chemical concentration etc. in automation system example if temperature changes then refractive index of optical fiber change & this property is used in temperature sensors 13) Used as phase modulated sensors, polarized modulated sensors SOLVED EXAMPLES Examples on Refraction: Example 1 : Find the velocity of light in a glass whose refractive index is 1.5. Solution: Given: a𝜇 g = 1.5 Take va = 3  108 m/s speed of light in air vg = ? Page | 59 Basic Science (22102) va We have, a𝜇 g = v g 𝑉𝑎 3  108  vg = = 𝑎𝜇𝑔 1.5 vg = 2  108 m/s Example 2: Speed of light in diamond is 1.2  108 m/s. Calculate refractive index of diamond. Solution: Given: vd = 1.2  108 m/s Take va = 3  108 m/s a𝜇 d = ? va 3  108 We have, 𝜇 = a d vd = 1.2  108 amd = 2.5 Example 3: A light ray enters water medium making an angle of 60 with water surface. If it suffers deviation of 15 in water, calculate refractive index of water. Solution: Ray makes an angle of 60 with water surface i.e. it makes an angle of 30 with normal to water. Angle of incidence, i = 90  60 = 30 r = (30  15) = 15 Refractive index of water w.r.t. air. sin i sin 30 i.e. a𝜇 w = sin r = sin 15 amw = 1.93 Page | 60 Basic Science (22102) Example 4 : Velocity of light in water is 2.3  108 m/s. Velocity of light in glass is 2  108 m/s. Calculate (a) R.I. of glass w.r.t. water, (b) Also R.I. of water w.r.t. glass. Solution: Given: vw = 2.3  108 m/s vg = 2  108 m/s w𝜇 g = ?, g𝜇 w = ? vw 2.3  108 (a) We have, w𝜇 g = vg = 2  108  wmg = 1.15 µ vg 2  108 (b) We have, = v = g w w 2.3  108 gmw = 0.87 Example 5: The refractive index of water is 1.3. The refractive index of glass is 1.5. Find the velocity of light in water and glass. µ Solution: Given: a w = 1.3 µ a g = 1.5 Take va = 3  108 m/s vw = ? vg = ? µ va µ va We have, a w = and a g = vw vg va va  vw = m and vg = m a w a g 3  10 8 3  108 = 1.3 = 1.5 vw = 2.31  108 m/s vg = 2  108 m/s Example 6 : How long will light take in travelling a distance 500 m in water ? R.I. of water is 4/3 and velocity of light in vacuum is 3  108 m/s. Solution: Given: Distance = 500 m, aµw = 4/3, va = 3  108 m/s, t = ? µ va a w = vw va 3  108  vw = m = 4/3 a w 9 vw = 4  108 = 2.25  108 m/s Distance covered in water But vw = Time Distance 500  t = = vw 2.25  108 Page | 61 Basic Science (22102)  t = 222.22  10–8 sec Examples on critical angle: Example 7 : For silicate glass optical fiber, calculate the critical angle if refractive index of core is 1.55 and refractive index of cladding is 1.35. Solution: Given: 𝜇 core = 1.55, 𝜇 clad = 1.35, 𝜃c = ?  1.35 We have, 𝜃 c = sin1 Clad = sin1 1.55 = sin1 (0.871) = 60.57°  core   Example 8: For a typical optical fiber, refractive index of core is 1.48 and refractive index of cladding is 1.46. Calculate the critical angle required. Solution: Given: 𝜇 core = 1.48, 𝜇 clad = 1.46, 𝜃c = ?  We have, 𝜃c = sin1 Clad  core 𝜃c = sin1 (0.9864) = 80.57° E learning websites 1. www.examfear.com 2. www.khanacademy. Com 3. amrita virtual labs 4. the physics classroom phet interactive simulations Sample MCQS HEAT: SAMPLE MCQ 1. Which statement is the best example of heat energy transfer by conduction. a) Heat energy is transferred from the bottom of the top of a lake b) Heat energy is transferred from the sun to the earth c) Heat energy is transferred from the surface soil to the rocks below d) Heat energy is transferred from the earth’s surface to the upper atmosphere. 2. Name the method of energy transfer that requires no medium for transfer a) Conduction b) convection c) radiation d) none of above 3. Heat is measured in a) Joule b) calorie c) both and B d) joule / second 4. The amount of heat required to raise the temperature of 1 kg by 10c is called as a) Heat capacity B0 work capacity c) specific heat capacity d) energy capacity 5. For a gas, which pair of variables are inversely proportional to each other if the remaining conditions kept constant. a) P,V b) P,T c) V,T d) none of above Page | 62 Basic Science (22102) OPTICS SAMPLE MCQ 6. Snell's law states that, for any two media, the  (a) product of sin i to sin r is constant (b) ratio of sin i to sin r is constant (c) Sum of sin i and sin r is constant (d) difference of sin i and sin r is constant 7. When ray of light travels from denser medium to rarer medium and if angle of incidence is greater than critical angle, then only reflection takes place. This phenomenon is known as . (a) Total internal reflection (b) total internal refraction (c) Interference (d) diffraction 8. In step index optical fiber, the refractive index of . (a) Core is uniform throughout the fiber (b) core and cladding is same (c) Core is changing from axis to boundary (d) none of these 9. In multimode step index optical fiber, for light  (a) There are many zig-zag paths (b) there is only one zig-zag path (c) There are many curved paths (d) there is only one curved path 10. For a glass optical fiber, calculate the critical angle if refractive index of core is 1.5 and refractive index of cladding is 1.3. (a) 55.23 (b) 64.25 (c) 57.83 (d) 60.07 Answers 1 (c) 2(b) 3(a) 4(c) 5(a) 6 (b) 7 (a) 8 (a) 9 (a) 10 (d) References 1. Engineering physics R.k gaur, S.S.Gupta, Dhanpat rai Pub. 2. Text book of physics for class XI & XII (Part I, Patt –II) NCERT New delhi 3. Physics standard XI, XII 4. Controller Maharashtra state, textbook bwceau, Mumbai Self directed learning 1. Collect information on good and bad conductors of heat. 2. Find applications of gas laws in everyday life 3. Collect the information on specific heat capacity of gas used in industrial and engineering application 4. Find the types of optical fibres Suggestion for micro project 1. Demonstrate modes of heat transfer 2. Demonstrate gas laws. 3. Determine thermal conductivity of a metal bar 4. Demonstrate phenomena of TIR Page | 63 Basic Science (22102) SECTION – II [CHEMISTRY] Unit 4 CHEMICAL BONDING AND CATALYSIS Rationale: Chemical bonds are used to create number of chemical compounds that allow us to find new substances. Chemical bonds are extremely important to chemists, scientists, engineers and technologist. One of the most crucial reason behind the knowledge of chemical bonds is to understand the various chemical and physical properties associated with an element which technologist have to use in various engineering applications. Ex: Metals, Alloys, Polymer, adhesives etc. The knowledge of chemical reactions is so important because they not only change the property of the substances that participate in the reaction, but also store or release energy. Course Outcome : Apply the catalysis process in industries. Unit Outcomes: 4a. Explain the properties of given material based on the bond formation. 4b. Describe the molecular structure of given solid, liquid and gases. 4c. Describe the crystal structure of the given solids. 4d. Select the relevant catalyst for given application. Significance: Atom consists of nucleus composed of protons and neutrons and the electrons are revolving around the nucleus in the extra nuclear part of atom. In chemical bonding the electrons located in the outermost shell or orbital are involved to form different types of chemical bonds. Chemical Bonding refers to the formation of a chemical bond between two or more atoms, molecules, or ions to give rise to a chemical compound. These chemical bonds are keeping the atoms together in the resulting compound. 4.1 Electronic Theory of Valency and Chemical Bonding 4.1.1 Electronic Theory of Valency The electronic theory of valency was originated by Kossel and Lewis in 1916 and was applied by Langmuir in 1919. According to this theory every element has a tendency to acquire or to have stable electronic configuration (ns2 np6). The theory is based on the following facts: 1) The valency of an element depends on the number of electrons present in the outermost shell, such electrons are called as valency electrons. The number of valence electrons lost or gained or shared by an atom of an element in order to complete its octet and become stable is known as valency. 2) Atoms with eight electrons in the outermost shell (two in case of helium) are chemically stable. Page | 64 Basic Science (22102) 3) Atoms with less than eight electrons in valence shell are chemically active and actively take part in chemical reactions. 4) Every element has a tendency to have stable electronic configuration which is achieved by forming chemical bond. “Chemical force which holds two or more atom together is known as chemical bond”. Following are the different types of chemical bonds formed by the different types of combinations of elements: 1) Electropositive elements + Electronegative elements Ionic bond: Complete transfer of electrons from electropositive element to electronegative element. 2) Electronegative elements + Electronegative elements Covalent bond: Mutual sharing of electrons between atoms of two electronegative elements. 3) Electropositive elements + electropositive elements Metallic bond. 4) Electronegative elements and hydrogen atom attached to electronegative element. Hydrogen bond Concept of valency: Valency: The valency of an element is the number of electrons its atom can lose or gain or share so as to complete its octet or duplet and become stable. Type of valency: i) Electrovalency ii) Covalency iii) Coordinate valency i) Electrovalency The number of electrons lost or gained by an atom of an element during the formation of an electrovalent bond is termed as its electrovalency. The elements which give up electrons to form positive ions have positive valency, while the elements which accept electrons to form negative ions have negative valency.  Positive electrovalency The valency obtained by the loss of valency electrons from the atom of metallic element so as to complete the outermost orbit is called as positive electrovalency. Generally metals atoms lose electrons and acquire positive charge. Examples: Table 4.1 Sr. Element Symbol Atomic Electronic Valency Positive No Number configuration electrovalency Na Na+ + e- 1 Sodium Na 11 2,8,1 +1 (2,8, 1) (2, 8) Mg Mg2+ +2e- 2 Maganesium Mg 12 2,8,2 +2 ( 2,8, 2) (2, 8) Al Al +3e- 3+ 3 Aluminium Al 13 2,8,3 +3 (2,8,3) ( 2, 8) Page | 65 Basic Science (22102)  Negative electrovalency The valency obtained by the gain of valency electrons by the atoms of non-metallic element so as to complete the outermost orbit is called as negative electrovalency. Generally non-metals gain electrons and acquire negative charge. Examples: Table 4.2 Sr. Element Symbol Atomic Electronic Valency Negative No Number configuration electrovalency Cl + e- Cl- 1 Chlorine Cl 17 2,8,7 -1 (2,8,7) (2,8,8) O +2e- O2- 2 Oxygen O 8 2,6 -2 (2,6) ( 2,8) N + 3e - N3- 3 Nitrogen N 7 2,5 -3 (2,5) (2,8) Exercise: Find out the type of valency of K (19), Ca (20), F (9) P (15), S (16) ii) Covalency: The valency obtained by mutual sharing of electrons between the atoms of similar or dissimilar elements is termed as covalency. iii) Co-ordinate valency: The valency obtained by one sided sharing of two electrons (lone pair of e-) between the atoms of dissimilar elements is termed as co-ordinate valency. 4.1.2 Types of Chemical Bonds: Definition: Chemical bond is defined as “The attractive force which holds various Constituents (atom, ions etc.) together in different chemical species is called chemical bond.” The chemical bonds are of following types. 4.1.2.a Electrovalent /Ionic Bond and its Characteristics: An ionic bond is formed due to the “electrostatic force of attraction between stable ions formed by complete transfer of electrons from atom of metallic element to atom of non-metallic element. The metallic atom loses electron to complete octet and acquire nearest inert gas configuration and form positively charged ion called as cation. The non-metallic atom gains the electrons to complete octet and acquire nearest inert gas configuration and form negatively charged ion called as anion. Positively charged cation and negatively charged anion held together by electrostatic force of attraction, the bond formed is called as ionic bond or electrovalent bond. For example: Formation of sodium chloride (NaCl) molecule 1. A molecule of sodium chloride (NaCl) consists of one atom of sodium and one atom of chlorine. It is formed by electrovalent linkage (bond). 2. Sodium atom (Atomic Number = 11) has an electronic configuration (2, 8, 1) and chlorine atom (Atomic Number = 17) has an electronic configuration (2, 8, 7). Page | 66 Basic Science (22102) 3. During the formation of sodium chloride, a) Sodium atom loses its one valency electron and acquires a +1 charge and attains a stable electronic configuration of nearest inert gas element Neon (2, 8). b) The electron lost by sodium atom is gained by chlorine atom and it acquires – 1 charge and attains a stable electronic configuration of nearest inert gas element Argon ( 2, 8, 8). 4. These two equal and oppositely charged ions (Na+ and CI- ions) which are produced, bound together by the electrostatic forces of attraction to form sodium chloride (NaCl). Thus electrovalency of sodium is + 1 and that of chlorine is – 1. For example: Formation of magnesium oxide molecule (MgO) 1. Molecules of magnesium oxide consist of one atom of magnesium and one atom of oxygen. 2. Magnesium atom (Atomic Number = 12) has electronic configuration (2, 8, 2) and oxygen (Atomic Number = 8) has electronic configuration (2, 6). 3. In the formation of Magnesium oxide molecule, a) Magnesium atom loses its two valence electrons and it acquires +2 charges and attains a stable electronic configuration as that of Neon (2,8). b) The lost electrons are accepted by oxygen atom and acquire – 2 charges and attains a stable electronic configuration as that of Neon (2,8). 4