Written as per the revised G Scheme syllabus prescribed by the Maharashtra State Board of Technical Education (MSBTE) w.e.f. academic year 2012-2013 Basic Science PHYSICS irst Year Diploma Semester - I irst Edition: June 2015 Salient eatures Concise content with complete coverage of revised G-scheme syllabus. Simple & lucid language. Neat, Labelled & authentic diagrams. Illustrative examples showing detailed solution of Numericals. MSBTE Theory Questions and Numericals from Winter-2006 to Summer-2015. MSBTE Question Papers of Summer-2014, Winter-2014, Summer-2015. Three Model Question Papers for practice. Important Inclusions: Additional Theory Questions, Practice Problems, Knowledge Bank and Physics behind Printed at: Repro India Ltd., Mumbai No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher. TEID : 917
PREACE Target s Basic Science Physics is compiled with an aim of shaping engineering minds of students while catering to their needs. It is a complete & thorough book designed as per the new revised G-scheme of MSBTE curriculum effective from June 2012. Each unit from the syllabus is divided into chapters bearing specific objectives in mind. The sub-topic wise classification of this book helps the students in easy comprehension. Each chapter includes the following features: Theory is provided in the form of pointers. Neat labelled diagrams have been provided wherever required. Italicized definitions are hard to miss and help students map answers easily. Illustrative Examples are provided in order to understand the application of different concepts and formulae. By introducing them after formulae, these examples enable students to gain command over formulae. An array of problems from simple to complex are included. (Examples here are similar to problems asked in previous years MSBTE Question Papers and also problems important from examination point of view) ormulae are provided for quick recap and last minute revision. MSBTE Theory Questions covered in separate section to give a clear idea of the type of questions asked. (Reference of answer to each question is provided.) MSBTE Numericals till latest year are included. Additional Theory Questions to help the student gain insight on the various levels of theory-based questions. Problems for Practice (With final answers) covers a variety of questions from simple to complex. or the enrichment of students and igniting their interest in Physics, we have ensured inclusion of: Knowledge Bank which is designed to bridge the gap between the knowledge required to understand the concept covered in syllabus but does not fall in the scope of syllabus. It acts as a base to understand the concept. Physics behind. is an effort to make students aware of real life engineering situations where physics plays prominent role or day-to-day experiences ruled by physics. MSBTE Question Papers of year 2014 and Summer-2015 are added at the end to make students familiar with the MSBTE examination pattern. A set of three Model Question Papers are designed as per MSBTE Pattern for thorough revision and to prepare the students for the final examination. rom, Publisher Best of luck to all the aspirants!
SYLLABUS Topic and Contents Hours Marks Topic 1] Properties of solids Specific Objectives: [8 Marks] Calculate the Young s Modulus of material of wire. Elasticity: Definitions of deforming force, restoring force, elasticity, plasticity, actors affecting elasticity. Stress: Tensile, Compressive, Volumetric and Shear stress. Strain: Tensile, Volumetric and Shear strain. 05 08 Elastic limit, Hooke s law. Elastic co-efficient-young s modulus, bulk modulus, modulus of rigidity and relation between them Stress-strain diagram, behavior of wire under continuously increasing load, yield point, ultimate stress, breaking stress, factor of safety, compressibility, Poisson s ratio. Topic 2] Properties of liquids Specific objectives: Determine the surface tension of the given liquid Determine the coefficient of viscosity by Stoke s method. 2.1 luid friction: [8 Marks] Pressure, pressure-depth relation (P = ρ h g), atmospheric pressure, Pascal s law, Archimedes principle. Viscous force, definition of viscosity, velocity gradient, Newton s law of viscosity, coefficient of viscosity and its SI unit. Streamline and turbulent flow with examples, critical velocity, Reynold s number and its significance. Up thrust force, terminal velocity, Stoke s law, and derivation of coefficient of viscosity by Stoke s method, effect of temperature and adulteration on viscosity of liquid. 09 12 2.2 Surface tension: [4 Marks] Cohesive and adhesive force, Laplace s molecular theory of surface tension, surface Tension: definition and unit, effect of temperature on surface tension. Angle of contact, capillarity and examples of capillary action, derivation of expression for surface tension by capillary rise method, applications of surface tension. Topic 3] Thermal properties of matter Specific objectives: Distinguish between isothermal and adiabatic process. Determine the relation between specific heats. 08 12 3.1 Modes of transformation of heat: [6 Marks] Difference between heat and temperature, definition of calorie, Absolute zero, units of temperature: C,, K, with their conversion.
Conduction, law of thermal conductivity, coefficient of thermal conductivity, good conductors of heat and insulators with suitable examples, applications of conduction. Convection, applications of convection. Radiation, applications of radiation. 3.2 Gas laws: [6 Marks] Gas Laws: Boyle s law, Charles law, Gay Lussac s law (Statement and mathematical equation only) Perfect gas equation (PV = RT) (No derivation), specific heat of a substance, SI unit, specific heat of gas at constant volume (C V ) specific heat of gas at constant pressure (C P ), ratio of specific heats, Mayer s relation between C P and C V, isothermal process, adiabatic process, difference between isothermal process and adiabatic process. Topic 4] Optics Specific objectives: Calculate refractive index of prism. Determine the numerical aperture of optical fiber Refraction of light: [6 Marks] Refraction of monochromatic light, Snell s law, derivation of prism formula, total internal reflection, critical angle. Optical fibre: principle, structure of optical fiber, propagation of light wave through optical fibre, derivation of numerical aperture and acceptance angle. Topic 5] Wave motion Specific objectives: Differentiate between transverse waves and longitudinal waves Derive expression for displacement, velocity and acceleration of a body executing S.H.M. 5.1 Wave motion: [6 Marks] Definition of a wave, wave motion, wave velocity, wave period, wave frequency, wavelength, vibratory motion, periodic motion, amplitude of a vibrating particle, derivation of v = n λ Simple harmonic motion (S.H.M.), examples of S.H.M., equation of S.H.M., expression of velocity and acceleration of a body executing S.H.M. Types of progressive waves: transverse and longitudinal waves with examples. 5.2 Resonance: [6 Marks] Stationary wave, formation of stationary wave, examples of stationary wave, characteristics of stationary waves, free and forced vibrations with examples. 04 06 06 12 Resonance: definition of resonance, examples of resonance, formula to calculate velocity of sound by resonance tube method. TOTAL 32 50
Target Publications Pvt. Ltd. Basic Physics Unit (.Y.Dip.Sem.-1) I: Properties of MSBTE Solids UNIT I Properties of Solids Chapter - 1 Properties of Solids Chapter-1 Properties of Solids 1.1 Introduction 1.2. Elasticity 1.2.(a) Deforming force 1.2.(b) Restoring force 1.2.(c) Elasticity 1.2.(d) Plasticity 1.2.(e) Comparison between elasticity and plasticity 1.2.(f) actors affecting elasticity 1.3 Stress 1.3.(a) Longitudinal stress 1.3.(b) Volumetric stress 1.3.(c) Shear stress 1.4. Strain 1.4.(a) Tensile strain 1.4.(b) Volumetric strain 1.4.(c) Shear strain 1.5. Elastic Limit, Hooke s law 1.5.(a) Elastic limit 1.5.(b) Hooke s law 1.6. Elastic coefficient 1.6.(a) Young s modulus 1.6.(b) Bulk modulus 1.6.(c) Compressibility 1.6.(d) Modulus of rigidity 1.6.(e) Relation between moduli of elasticity 1.6.(f) Comparision between moduli of elasticity 1.7. Poisson s ratio 1.8 Stress-strain diagram 1.8.(a) Behaviour of wire under continuously increasing load 1.8.(b) Ultimate stress 1.8.(c) actor of safety 1
Target 01 Publications Pvt. Ltd. Properties of Solids Basic Science Physics (.Y.Dip.Sem.-1) MSBTE 1.1 Introduction i. When a body is subjected to a set of balanced forces, there is no motion of the body but the force may produce change in its size, shape or both. ii. Due to applied force, some bodies undergo change in dimension (length, breadth, height, etc.) easily, while some bodies oppose any change in dimension. iii. Some bodies show permanent change in their dimension after removal of applied force, whereas some bodies preserve their original dimension, i.e., size, shape or both after removal of applied force. 1.2 Elasticity Knowledge Bank A hard, solid body with definite size and shape is said to be rigid if its dimensions do not change when a force is applied to it. In reality, no body is rigid. But for practical purposes, stone, iron, etc. are taken to be rigid bodies. A solid body is an array of atoms or molecules. These atoms and molecules are bounded together by interatomic or intermolecular forces and remain in equilibrium positions. Lattice point Crystal structure Spring-ball model for the illustration of elastic behaviour of solids 2 1.2.(a) Deforming force: i. When an external force is applied on a body, it changes the dimensions of the body. The change in size, shape or both of a body due to applied external force is called deformation.
Target Publications Pvt. Ltd. Basic Chapter Physics (.Y.Dip.Sem.-1) 01: Properties of MSBTE Solids ii. iii. iv. Since the applied force changes the configuration of the body, it is termed as deforming force. The force required to produce deformation in a body is called deforming force. When a body is deformed by applying an external force, the molecules are displaced from their original positions of stable equilibrium. Physics behind changing the ball after certain overs in a cricket match The force with which batsman hits the ball, acts as a deforming force. After certain overs, ball loses its elastic property and original shape. Hence, the players in a cricket match ask for change of ball after certain overs. 1.2.(b) Restoring force: i. Under deformed condition, every shifted molecule tries to come back to its original position. ii. Due to elastic property of the body, internal molecular forces are set up within the body, which tend to oppose the changes in size and shape of the body. iii. Restoring force is defined as internal force developed in a body, in order to regain its original size and shape after application of deforming force. iv. ormula: Applied internal force = Restoring force 1.2.(c) Elasticity: i. Under deformed condition, the molecules of a body are displaced from their original position. ii. The intermolecular distances change and restoring forces act on the molecules which bring them back to their original position. iii. This leads to the phenomenon of elasticity in material body. iv. The property by virtue of which material bodies regain their original dimensions (size, shape or both) after removal of deforming force is called elasticity. v. Elastic property is more in solid, less in liquid and least in gas. Elastic body: A body which possesses the property of elasticity i.e., the body which changes its size, shape or both when a deforming force is applied and comes back to its original position as soon as deforming force is removed is called elastic body. e.g. Steel wire, rubber band, sponge ball, etc. Perfectly elastic body: i. A body which regains its original shape, size or both completely after removal of deforming force is called perfectly elastic body. ii. Perfectly elastic bodies do not exist in nature, however some bodies are considered to be perfectly elastic. e.g. Quartz, phosphor bronze, etc. 1.2.(d) Plasticity: i. The property by virtue of which material bodies undergo permanent deformation even after the removal of external deforming forces is called plasticity. 3
Target Publications Pvt. Ltd. Basic Science Physics (.Y.Dip.Sem.-1) MSBTE 4 ii. Plastic bodies do not regain their original dimensions (size, shape or both) after removal of deforming forces. iii. These bodies can be deformed to a large extent by a small deforming force. Perfectly plastic body: i. A body which does not regain its original configuration at all on the removal of deforming force, how-so-ever small the deforming force may be, is called perfectly plastic body. ii. Ideal perfectly plastic bodies do not exist in nature, however some bodies are considered to be perfectly plastic. e.g. Wax, mud, putty, clay, plasticine, polythene plastic, etc. Note: Elastic properties of materials always lie between those of perfectly elastic and perfectly plastic bodies. 1.2.(e) Comparison between elasticity and plasticity: No. Elasticity Plasticity i. Body regains its original shape or size after removal of external force. Body does not regain its original shape or size after removal of external force. ii. External force changes the dimensions of the body temporarily. External force changes the dimensions of the body permanently. iii. Internal restoring force is set up inside the body. Internal restoring force is not set up inside the body. iv. Ratio of stress and strain remains Ratio of stress and strain does not remain constant. constant. 1.2.(f) actors affecting elasticity: Elasticity depends upon the material of the body, hence the structure of material greatly affects its elastic properties. Though for a given material, elasticity is constant, following factors can affect its elasticity. i. Change in Temperature: a. As the temperature of material increases, intermolecular distance increases. b. This decreases intermolecular force and in turn, internal restoring force. c. Thus, if temperature is increased, elasticity of material decreases and vice-versa. ii. e.g. When lead is cooled in liquid air, it becomes elastic. A carbon filament which is elastic at room temperature turns plastic when heated by passing current through it. d. Exception: Invar steel, an alloy of nickel (36 %) and iron (64 %) shows elasticity independent of change in temperature. Impurities: a. The impurity affects elastic properties of material to which they are added. b. Sometimes suitable impurities are deliberately added to increase binding between the lattice points (individual points that come together to constitute lattice structure) without disturbing their orientation. c. In metals, impurities are added to increase their elastic properties. e.g. Addition of potassium in gold, addition of carbon in molten steel, etc. d. On the other hand, if impurity added is less elastic than the material itself, then addition of impurity results in decrease in elastic properties of material.
Target Publications Pvt. Ltd. Basic Chapter Physics (.Y.Dip.Sem.-1) 01: Properties of MSBTE Solids iii. iv. Hammering and rolling: Operations like hammering and rolling break up crystal lattice into smaller units which increase the elastic properties of material. Annealing: a. Annealing is a process of bringing desired consistency, texture or hardness in a material by gradually heating and cooling. b. Due to annealing, smaller crystals present inside the material get aligned to form large crystal structure and softening of material is observed. c. Thus, annealing decreases elasticity of material. v. Recurring stress: If recurring (repeated) stress is applied on the same body, softening of material takes place decreasing the elasticity. 1.3 Stress i. The internal elastic restoring force per unit cross-sectional area of a body is called stress. OR Stress is defined as applied force per unit cross-sectional area of a body. Stress = Physics behind declaring bridges unsafe after long use A bridge during its use undergoes recurring stress depending upon the movement of vehicles on it. When bridge is used for long time, it loses its elastic strength and ultimately may collapse. Hence, the bridges are declared unsafe after long use. Applied force Area of cross-section Elastic restoring force = Area of cross-section ii. ormula: Stress = A iii. Unit: N/m 2 or Pa in SI system and dyne/cm 2 in CGS system. iv. Dimensions: [M 1 L 1 T 2 ] v. When an external force acts on a solid, there are three ways in which a solid may change its dimension. Depending upon the type of change, we get three types of stress. a. Longitudinal stress b. Volumetric stress c. Shearing stress 1.3.(a) Longitudinal stress: i. If deforming force produces change in length of a body (wire, beam etc.) the stress associated is called longitudinal stress. Applied force on wire ii. ormula: Longitudinal stress = Area of cross section = A = Mg 2 r where, M = mass of the wire r = radius of cross-section of wire g = acceleration due to gravity. iii. It occurs only in solids like wire, rod, beam, etc and comes into picture when one of the three dimensions viz. length, breadth or height is much greater than other two. iv. Longitudinal stress is either tensile or compressive. a. Tensile stress: Longitudinal stress produced due to increase in length of a body under a deforming force is called tensile stress. 5
Target Publications Pvt. Ltd. Basic Science Physics (.Y.Dip.Sem.-1) MSBTE L L+ L L x Physics V 6 (a) A cylindrical body under tensile stress elongates by L (b) Shearing stress on a cylinder deforming it by an angle b. Compressive stress: Longitudinal stress produced due to decrease in length of a body under a deforming force is called compressive stress. 1.3.(b) Volumetric or volume stress or bulk stress: i. If a deforming force produces change in volume of a body, the stress associated with it is called volume stress. ii. This stress can be produced in solid, liquids and gases. iii. Let a gas balloon with original volume V be compressed by P additional pressure P on all sides. Due to applied additional pressure P, volume of gas balloon V V P decreases by V. Thus, new volume of gas in balloon = V V P iv. If original pressure is P and new pressure applied is P + P then increase in pressure P tries to restore the body back to its original dimensions. v. P generates an internal restoring force which acts on the walls of the balloon. vi. Applied force = Internal restoring force = P A vii. Applied force ormula: Volume stress = = A P = P Area A 1.3.(c) Shearing (shear) or tangential stress: (c) A body subjected to shearing stress (d) A solid body under a stress normal to the surface at every point i. If deforming force produces change in shape of a body, the stress associated with it is called shearing stress. OR The restoring force per unit area developed due to applied tangential force is called shearing stress or tangential stress. T T S S ii. When a tangential force is applied on the surface QRSV of a cube as shown in the figure U U V x V x then position of surface changes at an angle without change in volume of the cube. h In the figure, PQRSTUVW = original shape of cube W R PQRS T U V W = new shape of cube in P Q deformed state.
Target Publications Pvt. Ltd. Basic Chapter Physics (.Y.Dip.Sem.-1) 01: Properties of MSBTE Solids iii. = tangential force TT = SS = UU = VV = x = lateral displacement of edge ormula: Shearing stress = Tangentialforce = Area A Illustrative Examples: Example 1 A wire of length 3 m and a uniform area of cross section 1.5 10 6 m 2 is stretched by a force of 50 N. Calculate the stress on the wire. Solution: Given: L = 3 m, A = 1.5 10 6 m 2, = 50 N To find: Stress ormula: Stress = A Calculation: rom formula, 50 Stress = = 3.33 10 7 N/m 2 6 1.5 10 Ans: The stress on the wire is 3.33 10 7 N/m 2. Example 2 ind the maximum weight that can be attached to a copper wire of area of cross-section 2.4 mm 2 if breaking stress for copper is 5 10 8 N/m 2. Solution: Given: A = 2.4 mm 2 = 2.4 10 6 m 2, Breaking stress = 5 10 8 N/m 2 To find: Weight () ormula: Stress = A Calculation: In this case, weight attached is the force acting on the wire. The wire can bear the weight attached to it upto specific limit depending on its breaking stress. If the weight attached exceeds this limit, the wire will break. rom formula, = Stress A = 5 10 8 2.4 10 6 = 1200 N Ans: The maximum weight that can be attached to the wire is 1200 N. 1.4 Strain i. The change in dimension per unit original dimension of a body is called strain. OR Strain is the ratio of change in dimension to the original dimensions of the material. ii. Change in dimension ormula: Strain = Original dimension iii. Strain is the ratio of two similar physical quantities, hence it has no unit and dimension. iv. Strain is classified into three types: a. Longitudinal strain b. Volume strain (Bulk strain) c. Shearing strain (Tangential strain) 7
Target Publications Pvt. Ltd. Basic Science Physics (.Y.Dip.Sem.-1) MSBTE 1.4.(a) Longitudinal strain: i. The ratio of the change in length (l) to the original length (L) is called longitudinal strain. ii. In the figure, L = original length of wire L = L + l = deformed length due to load Mg Longitudinal strain = L L L l L = L L l = L O L A Initial length O L l B L = Deformed length iii. ormula: Longitudinal strain = L l Mg iv. Longitudinal stress is either tensile or compressive. a. Tensile strain: Longitudinal strain produced due to increase in length of a body under a deforming force is called tensile strain. b. Compressive strain: Longitudinal strain produced due to decrease in length of a body under a deforming force is called compressive strain. v. It is observed in solid metal wires. vi. It has no units and dimensions. Illustrative Examples: Example 1 A cylindrical bar of length 0.2 m deforms to 2 mm. What will be the strain developed in the bar? Solution: Given: l = 2 mm = 0.002 m, L = 0.2 m To find: Strain l ormula: Strain = L 8 Calculation: rom formula, Strain = 0.002 0.2 = 0.01 Ans: The strain developed in the bar will be 0.01. Example 2 A material wire elongates by 4% of its original length when loaded. Calculate tensile strain for the wire. Solution: 4 Given: l = 4 % of L = 100 L To find: Tensile strain
Target Publications Pvt. Ltd. Basic Chapter Physics (.Y.Dip.Sem.-1) 01: Properties of MSBTE Solids ormula: Tensile strain = L l Calculation: rom formula, 4L /100 4 Tensile strain = = L 100 = 0.04 Ans: Tensile strain for the wire is 0.04. 1.4.(b) Volumetric or volume strain or bulk strain: i. The ratio of change in volume per unit original volume of a body is called volume strain. V ii. ormula: Volume strain = V where, V = original volume of gas, V = change in volume iii. It has no units and dimensions. Note: In this case, change in volume is without change in shape of body. Illustrative Example: Calculate the magnitude of volume strain if volume changes by 2% of original volume. Solution: 2 Given: V = 2% V = 100 V To find: Magnitude of volume strain V ormula: Volume strain = V Calculation: rom formula, 2 V Volume strain = 100 = 0.02 V Ans: Magnitude of volume strain is 0.02. 1.4.(c) Shearing (shear) or tangential strain: i. The ratio of lateral displacement of any layer to its perpendicular distance from fixed layer is called shearing (shear) strain. Relative displacement Shearing strain = Distance from fixed layer ii. In the figure, base of the block AB is fixed. A tangential force is applied to the upper surface so that lateral displacement of the D x Q C x P face is x. h h In right angled ADQ, Shearing strain = x = tan, h where is called shear angle. If is very small, then tan Shearing strain = = x h A B 9
Target Publications Pvt. Ltd. Basic Science Physics (.Y.Dip.Sem.-1) MSBTE iii. ormula: Shearing strain = x h iv. It has no dimensions but expressed in radian. v. Shear strain is possible only for solid bodies. If the deforming force is applied parallel to the liquid surface, it will begin to flow in the direction of applied force. e.g. On twisting a wire or rod, shearing strain is produced in it. Illustrative Example: or a cube each of side 8 cm, the upper face is displaced by 1.2 cm due to a force of 1 N. Calculate shearing stress and shearing strain. Solution: Given: = 1 N, h = 8 cm = 8 10 2 m, A = h 2 = 64 10 4 m 2, x = 1.2 cm = 12 10 3 m To find: Shearing stress, shearing strain ormulae: i. Shearing stress = ii. Shearing strain = x A h Calculation: rom formula (i), 1 Shearing stress = = 156.25 N/m 4 64 10 rom formula (ii), 3 12 10 Shearing strain = = 0.15 2 8 10 Ans: Shearing stress is 156.25 N/m and shearing strain is 0.15. 1.5 Elastic limit, Hooke s law 1.5.(a) Elastic limit: i. The maximum stress to which an elastic body can be subjected without causing permanent deformation is called as elastic limit. ii. The deformation produced in an elastic body depends upon the magnitude of deforming force. If the deforming force is small or large, then deformation produced will also be small or large. iii. If the deforming force is removed, body regains its original shape and size. iv. However, if we go on increasing the deforming force, a stage is reached, where the body does not regain its original dimensions, even after the removal of deforming force. The body is said to acquire permanent deformation called set. v. When such a stage is reached, we say that body is stretched beyond the elastic limit. 10 1.5.(b) Hooke s law: i. Statement: Within the elastic limit, the stress developed in a body is directly proportional to the strain produced in it. ii. Mathematically, Stress Strain iii. ormula: Stress = E (strain) where, E is constant of proportionality called modulus of elasticity. iv. Value of E depends upon the nature of material of the body and the manner in which the body is deformed. Stress O Strain