PHYSICS VOL.1

217If is the original length and the mean radius of the experimentallrwire, then Young’s modulus of the material of the wire is given byq = F/Adl/l = 2F/ rπdl/l(i.e)q = 2 Flπrdl5.2.6 Applications of modulus of elasticityKnowledge of the modulus of elasticity of materials helps us tochoose the correct material, in right dimensions for the right application.The following examples will throw light on this.(i) Most of us would have seen a crane used for lifting and movingheavy loads. The crane has a thick metallic rope. The maximum loadthat can be lifted by the rope must be specified. This maximum loadunder any circumstances should not exceed the elastic limit of thematerial of the rope. By knowing this elastic limit and the extension perunit length of the material, the area of cross section of the wire can beevaluated. From this the radius of the wire can be calculated.(ii) While designing a bridge, one has to keep in mind the followingfactors (1) traffic load (2) weight of bridge (3) force of winds. The bridge isso designed that it should neither bend too much nor break.5.3 FluidsA fluid is a substance that can flow when external force is appliedon it. The term fluids include both liquids andgases. Though liquids and gases are termedas fluids, there are marked differences betweenthem. For example, gases are compressiblewhereas liquids are nearly incompressible. Weonly use those properties of liquids and gases,which are linked with their ability to flow,while discussing the mechanics of fluids.5.3.1 Pressure due to a liquid columnLet be the height of the liquid columnhin a cylinder of cross sectional area . If Aρ isthe density of the liquid, then weight of thehFig. 5.10 Pressure

218liquid column W is given byW = mass of liquid column × = gAh gρBy definition, pressure is the force acting per unit area.∴ Pressure= secweight of liquid columnarea of crosstion−= ρ AhgA = h gρ∴ P = h gρ5.3.2 Pascal’s lawOne of the most important facts aboutfluid pressure is that a change in pressure atone part of the liquid will be transmittedwithout any change to other parts. This wasput forward by Blaise Pascal (1623 - 1662), aFrench mathematician and physicist. This ruleis known as Pascal’s law.Pascal’s law states that if the effect ofgravity can be neglected then the pressure in afluid in equilibrium is the same everywhere.Consider any two points A and B inside the fluid. Imagine acylinder such that points A and B lie at the centre of the circularsurfaces at the top and bottom of the cylinder (Fig. 5.11). Let the fluidinside this cylinder be in equilibrium under the action of forces fromoutside the fluid. These forces act everywhere perpendicular to thesurface of the cylinder. The forces acting on the circular, top and bottomsurfaces are perpendicular to the forces acting on the cylindrical surface.Therefore the forces acting on the faces at A and B are equal andopposite and hence add to zero. As the areas of these two faces areequal, we can conclude that pressure at A is equal to pressure at B.This is the proof of Pascal’s law when the effect of gravity is not takeninto account.Pascal’s law and effect of gravityWhen gravity is taken into account, Pascal’s law is to be modified.Consider a cylindrical liquid column of height and density hρ in aABFig. 5.11 Pascal’s law inthe absence of gravity

219vessel as shown in the Fig. 5.12.If the effect of gravity is neglected, thenpressure at M will be equal to pressure at .NBut, if force due to gravity is taken into account,then they are not equal.As the liquid column is in equilibrium, theforces acting on it are balanced. The verticalforces acting are(i) Force P A1 acting vertically down on thetop surface.(ii) Weight mg of the liquid column acting vertically downwards.(iii) Force P A2 at the bottom surface acting vertically upwards.where P 1 and P 2 are the pressures at the top and bottom faces, isAthe area of cross section of the circular face and is the mass of themcylindrical liquid column.At equilibrium,P A + mg - P A = 0 12or P A + mg = P A12P = P +21mgAButm = Ahρ∴P = P + 21ρ AhgA(i.e)P = P + h g21ρThis equation proves that the pressure is the same at all pointsat the same depth. This results in another statement of Pascal’s lawwhich can be stated as change in pressure at any point in an enclosedfluid at rest is transmitted undiminished to all points in the fluid and actin all directions.5.3.3 Applications of Pascal’s law(i) Hydraulic liftAn important application of Pascal’s law is the hydraulic lift usedto lift heavy objects. A schematic diagram of a hydraulic lift is shownin the Fig. 5.13. It consists of a liquid container which has pistonsfitted into the small and large opening cylinders. If a1 and a 2 are theareas of the pistons A and B respectively, is the force applied on AFand W is the load on B, thenM P 1P 2hNFig. 5.12 Pascal’s lawand effect of gravity

220=12FW aaorW = F 21aaThis is the load that can be liftedby applying a force F on A. In the aboveequation 21aa is called mechanicaladvantage of the hydraulic lift. One cansee such a lift in many automobileservice stations.(ii) Hydraulic brakeWhen brakes are applied suddenly in a moving vehicle, there isevery chance of the vehicle to skid because the wheels are not retardeduniformly. In order to avoid this danger of skidding when the brakes areapplied, the brake mechanism must be such that each wheel is equallyand simultaneously retarded. A hydraulic brake serves this purpose. Itworks on the principle of Pascal’s law.Fig. 5.14 shows the schematic diagram of a hydraulic brake system.The brake system has a main cylinder filled with brake oil. The maincylinder is provided with a piston P which is connected to the brakeFig. 5.14 Hydraulic brakeFig. 5.13 Hydraulic liftABFWBrake PedalLever systemMain CylinderBrake oilT-tubeP 1P 2Pipe line toother wheelsS 1S 2Inner rim of the wheelP

221pedal through a lever assembly. A shaped tube is provided at theTother end of the main cylinder. The wheel cylinder having two pistonsP and P is connected to the tube. The pistons P and P are connected12T12to the brake shoes S and S respectively.12When the brake pedal is pressed, piston P is pushed due to thelever assembly operation. The pressure in the main cylinder istransmitted to P and P . The pistons P and P push the brake shoes1212away, which in turn press against the inner rim of the wheel. Thus themotion of the wheel is arrested. The area of the pistons P and P is12greater than that of P. Therefore a small force applied to the brakepedal produces a large thrust on the wheel rim.The main cylinder is connected to all the wheels of the automobilethrough pipe line for applying equal pressure to all the wheels .5.4 ViscosityLet us pour equal amounts of water and castor oil in two identicalfunnels. It is observed that water flows out of the funnel very quicklywhereas the flow of castor oil is very slow. This is because of thefrictional force acting within the liquid. This force offered by the adjacentliquid layers is known as viscous force and the phenomenon is calledviscosity.Viscosity is the property of the fluid by virtue of which it opposesrelative motion between its different layers. Both liquids and gases exhibitviscosity but liquids are much more viscous than gases.Co-efficient of viscosityConsider a liquid to flowsteadily through a pipe as shownin the Fig. 5.15. The layers ofthe liquid which are in contactwith the walls of the pipe havezero velocity. As we move towardsthe axis, the velocity of the liquidlayer increases and the centrelayer has the maximum velocity . Consider any two layers P and Qvseparated by a distance dx. Let dv be the difference in velocity betweenthe two layers.Fig. 5.15 Steady flow of a liquidP Qdxv

222The viscous force acting tangentially between the two layers ofFthe liquid is proportional to (i) area A of the layers in contact(ii) velocity gradient dvdx perpendicular to the flow of liquid.∴ F A αdvdxF = A ηdvdxwhere is the coefficient of viscosity of the liquid.ηThis is known as Newton’s law of viscous flow in fluids.If A = 1m 2and dvdx = 1s–1then F = ηThe coefficient of viscosity of a liquid is numerically equal to theviscous force acting tangentially between two layers of liquid having unitarea of contact and unit velocity gradient normal to the direction of flowof liquid.The unit of is ηN s m–2. Its dimensional formula is ML T .–1–15.4.1 Streamline flowThe flow of a liquid is said to be steady, streamline or laminar ifevery particle of the liquid follows exactly the path of its preceding particleand has the same velocity of its preceding particle at every point.Let abc be the path of flowof a liquid and v 1, v 2 and v 3be the velocities of the liquidat the points a, b and crespectively.Duringastreamline flow, all the particlesarriving at ‘a’ will have the samevelocity which is directed along the tangent at the point ‘a’. A particlev 1arriving at b will always have the same velocity v 2. This velocity v 2 mayor may not be equal to . Similarly all the particles arriving at the pointv 1c will always have the same velocity . In other words, in the streamlinev 3flow of a liquid, the velocity of every particle crossing a particular pointis the same.abcv 1v 2v 3Fig. 5.16 Steamline flow

223The streamline flow is possible only as long as the velocity of thefluid does not exceed a certain value. This limiting value of velocity iscalled critical velocity.5.4.2 Turbulent flowWhen the velocity of a liquid exceeds the critical velocity, thepath and velocities of the liquid become disorderly. At this stage, theflow loses all its orderliness and is called turbulent flow. Some examplesof turbulent flow are :(i) After rising a short distance, the smooth column of smokefrom an incense stick breaks up into irregular and random patterns.(ii) The flash - flood after a heavy rain.Critical velocity of a liquid can be defined as that velocity of liquidupto which the flow is streamlined and above which its flow becomesturbulent.5.4.3 Reynold’s numberReynolds number is a pure number which determines the type offlow of a liquid through a pipe. It is denoted by N R .It is given by the formulaN R = ρ c vDηwhere vc is the critical velocity, ρ is the density,ηis the co-efficientof viscosity of the liquid and is the diameter of the pipe.DIf N R lies between 0 and 2000, the flow of a liquid is said to bestreamline. If the value of N R is above 3000, the flow is turbulent. IfN R lies between 2000 and 3000, the flow is neither streamline norturbulent, it may switch over from one type to another.Narrow tubes and highly viscous liquids tend to promote streamline motion while wider tubes and liquids of low viscosity lead totubulence.5.4.4 Stoke’s law (for highly viscous liquids)When a body falls through a highly viscous liquid, it drags thelayer of the liquid immediately in contact with it. This results in arelative motion between the different layers of the liquid. As a resultof this, the falling body experiences a viscous force . Stoke performedF

224many experiments on the motion of small spherical bodies in differentfluids and concluded that the viscous force acting on the sphericalFbody depends on(i) Coefficient of viscosity of the liquidη(ii) Radius of the sphere anda(iii) Velocity of the spherical body.vDimensionally it can be proved thatF = kηavExperimentally Stoke found thatk = 6π∴F = 6π ηavThis is Stoke’s law.5.4.5 Expression for terminal velocityConsider a metallic sphere of radius ‘ ’ andadensity to fall under gravity in a liquid of densityρσ . The viscous force acting on the metallic sphereFincreases as its velocity increases. A stage is reachedwhen the weight W of the sphere becomes equal tothe sum of the upward viscous force and the upwardFthrust due to buoyancy (Fig. 5.17). Now, there isUno net force acting on the sphere and it moves downwith a constant velocityv called terminal velocity.∴ W - F - U = O...(1)Terminal velocity of a body is defined as theconstant velocity acquired by a body while fallingthrough a viscous liquid.From (1), W = F + U...(2)According to Stoke’s law, the viscous force isFgiven by = 6Fπηav.The buoyant force = Weight of liquid displaced by the sphereU= 43π a 3σ gThe weight of the sphereW = 43πa g 3 ρWFUFig. 5.17 Spherefalling in aviscous liquid

225Substituting in equation (2),43π a 3 ρg = 6π ηav + 43π a 3 σ gor 6π ηav = 43πa ( – )g 3 ρ σ∴ v = 2 ()2ag9ρ ση −5.4.6 Experimental determination of viscosity ofhighly viscous liquidsThe coefficient of highly viscous liquid like castoroil can be determined by Stoke’s method. Theexperimental liquid is taken in a tall, wide jar. Twomarking B and C are marked as shown in Fig. 5.18.A steel ball is gently dropped in the jar.The marking B is made well below the free surface of the liquidso that by the time ball reaches B, it would have acquired terminalvelocity .vWhen the ball crosses B, a stopwatch is switched on and the timetaken to reach C is noted. If the distance BC is s, then terminaltvelocity =vst.The expression for terminal velocity isv = 2( - )ρ ση 2ag9∴ st=2 2( - )ρσ9ηagorη = 29a 2 ( - ) ρσtgsKnowing , aρ and σ, the value of of the liquid is determined.ηApplication of Stoke’s lawFalling of rain drops: When the water drops are small in size,their terminal velocities are small. Therefore they remain suspended inair in the form of clouds. But as the drops combine and grow in size,their terminal velocities increases because vαa 2. Hence they startfalling as rain.sBCFig. 5.18Experimentaldeterminationof viscosity ofhighly viscousliquid

2265.4.7 Poiseuille’s equationPoiseuille investigated the steady flow of a liquid through a capillarytube. He derived an expression for the volume of the liquid flowing persecond through the tube.Consider a liquid of co-efficient of viscosity flowing, steadilyηthrough a horizontal capillary tube of length and radius . If is thelrPpressure difference across the ends of the tube, then the volume ofVthe liquid flowing per second through the tube depends on , andηrthe pressure gradient Pl⎛⎞ ⎜⎟ ⎝⎠ .(i.e)Vα ηx yrPl⎛⎞ ⎜⎟ ⎝⎠ zV = kη xr y⎛⎞ ⎜⎟ ⎝⎠ z Pl...(1)where k is a constant of proportionality. Rewriting equation (1) interms of dimensions,[L T ] = [ML T ] [L]3 -1-1-1 xyz -1 -2ML TL⎡⎤ ⎢⎥ ⎣⎦Equating the powers of L, M and T on both sides we getx = -1, = 4 and = 1yzSubstituting in equation (1),V = k η -1 r4 1 Pl⎛⎞ ⎜⎟ ⎝⎠V = 4kPrl ηExperimentally was found to be equal to kπ8 .∴ V = 48PrlπηThis is known as Poiseuille’s equation.

2275.4.8 Determination of coefficient of viscosity of water byPoiseuille’s flow methodA capillary tube of very fine boreis connected by means of a rubber tubeto a burette kept vertically. The capillarytube is kept horizontal as shown inFig. 5.19. The burette is filled with waterand the pinch - stopper is removed.The time taken for water level to fallfrom A to B is noted. If is the volumeVbetween the two levels A and B, thenvolume of liquid flowing per second isVt. If and are the length and radiuslrof the capillary tube respectively, thenπ=η48VPrtl...(1)If is the density of the liquidρthen the initial pressure differencebetween the ends of the tube is P 1 = h g1 ρ and the final pressure differenceP 2 = h 2 ρ g. Therefore the average pressure difference during the flow ofwater is P whereP = + 122 PP= 12ρρ 2 hhgh g⎛⎞+=⎜⎟ ⎜⎟ ⎝⎠12 h+ hh=2⎡⎤ ⎢⎥ ⎣⎦ ∵Substituting in equation (1), we getVt= 4π ρh gr8lη orη = 4π ρh gr t8lV5.4.9 Viscosity - Practical applicationsThe importance of viscosity can be understood from the followingexamples.(i) The knowledge of coefficient of viscosity of organic liquids isused to determine their molecular weights.AB h 1h 2Fig. 5.19 Determination ofcoefficient ofviscosity by Poiseuille’s flow

228(ii) The knowledge of coefficient of viscosity and its variation withtemperature helps us to choose a suitable lubricant for specificmachines. In light machinery thin oils (example, lubricant oil used inclocks) with low viscosity is used. In heavy machinery, highly viscousoils (example, grease) are used.5.5 Surface tensionIntermolecular forcesThe force between two molecules of a substance is calledintermolecular force. This intermolecular force is basically electric innature. When the distance between two molecules is greater, thedistribution of charges is such that the mean distance between oppositecharges in the molecule is slightly less than the distance between theirlike charges. So a force of attraction exists. When the intermoleculardistance is less, there is overlapping of the electron clouds of themolecules resulting in a strong repulsive force.The intermolecular forces are of two types. They are (i) cohesiveforce and (ii) adhesive force.Cohesive forceCohesive force is the force of attraction between the molecules ofthe same substance. This cohesive force is very strong in solids, weakin liquids and extremely weak in gases.Adhesive forceAdhesive force is the force of attraction between the moelcules oftwo different substances. For example due to the adhesive force, inksticks to paper while writing. Fevicol, gum etc exhibit strong adhesiveproperty.Water wets glass because the cohesive force between watermolecules is less than the adhesive force between water and glassmolecules. Whereas, mercury does not wet glass because the cohesiveforce between mercury molecules is greater than the adhesive forcebetween mercury and glass molecules.Molecular range and sphere of influenceMolecular range is the maximum distance upto which a moleculecan exert force of attraction on another molecule. It is of the order of10 –9m for solids and liquids.

229Sphere of influence is a sphere drawn around a particular moleculeas centre and molecular range as radius. The central molecule exerts aforce of attraction on all the molecules lying within the sphere ofinfluence.5.5.1 Surface tension of a liquidSurface tension is the property of the free surfaceof a liquid at rest to behave like a stretched membranein order to acquire minimum surface area.Imagine a line AB in the free surface of a liquidat rest (Fig. 5.20). The force of surface tension ismeasured as the force acting per unit length on eitherside of this imaginary line AB. The force isperpendicular to the line and tangential to the liquidsurface. If F is the force acting on the length of thelline AB, then surface tension is given byT = Fl.Surface tension is defined as the force per unit length actingperpendicular on an imaginary line drawn on the liquid surface, tendingto pull the surface apart along the line. Its unit is N m –1and dimensionalformula is MT .–2Experiments to demonstrate surface tension(i) When a painting brush is dipped into water, its hair getsseparated from each other. When the brush is taken out of water, itis observed that its hair will cling together. This is because the freesurface of water films tries to contract due to surface tension.Fig. 5.21 Practical examples for surface tensionABFig. 5.20 Force ona liquid surfaceHair clings together when brush is taken outNeedle floats on water surfaceWT vTT

230(ii) When a sewing needle is gently placed on water surface, itfloats. The water surface below the needle gets depressed slightly. Theforce of surface tension acts tangentially. The vertical component ofthe force of surface tension balances the weight of the needle.5.5.2 Molecular theory of surface tensionConsider two molecules P and Q as shownin Fig. 5.22. Taking them as centres andmolecular range as radius, a sphere of influenceis drawn around them.The molecule P is attracted in all directionsequally by neighbouring molecules. Thereforenet force acting on P is zero. The molecule Q ison the free surface of the liquid. It experiencesa net downward force because the number ofmolecules in the lower half of the sphere ismore and the upper half is completely outsidethe surface of the liquid. Therefore all themolecules lying on the surface of a liquidexperience only a net downward force.If a molecule from the interior is to be brought to the surface ofthe liquid, work must be done against this downward force. This workdone on the molecule is stored as potential energy. For equilibrium, asystem must possess minimum potential energy. So, the free surface willhave minimum potential energy. The free surface of a liquid tendsto assume minimum surface area by contracting and remains in astate of tension like a stretched elasticmembrane.5.5.3 Surface energy and surface tensionThe potential energy per unit area ofthe surface film is called surface energy.Consider a metal frame ABCD in which ABis movable. The frame is dipped in a soapsolution. A film is formed which pulls ABinwards due to surface tension. If is theTsurface tension of the film and is the lengthlQPFig. 5.22 Surfacetension based onmolecular theoryDCABB /A /xlFig. 5.23 Surface energy

231of the wire AB, this inward force is given by 2 × Tl . The number 2indicates the two free surfaces of the film.If AB is moved through a small distance as shown in Fig. 5.23xto the position A B′′, then work done isW = 2TlxWork down per unit area = 2 Wlx∴ Surface energy = 22TlxlxSurface energy = TSurface energy is numerically equal to surface tension.5.5.4 Angle of contactWhen the free surface of a liquidcomes in contact with a solid, itbecomes curved at the point ofcontact. The angle between thetangent to the liquid surface at thepoint of contact of the liquid with thesolid and the solid surface inside theliquid is called angle of contact.In Fig. 5.24, QR is the tangent drawn at the point of contact Q.The angle PQR is called the angle of contact. When a liquid has concavemeniscus, the angle of contact is acute. When it has a convex meniscus,the angle of contact is obtuse.The angle of contact depends on the nature of liquid and solid incontact. For water and glass, lies between 8 and 18 . For pure waterθooand clean glass, it is very small and hence it is taken as zero. The angleof contact of mercury with glass is 138 .o5.5.5 Pressure difference across a liquid surfaceIf the free surface of a liquid is plane, then the surface tensionacts horizontally (Fig. 5.25a). It has no component perpendicular tothe horizontal surface. As a result, there is no pressure difference betweenthe liquid side and the vapour side.If the surface of the liquid is concave (Fig. 5.25b), then the resultantQPFor waterRQPRFor mercuryFig. 5.24 Angle of contact

232force due to surface tension on a molecule on the surface act verticallyRupwards. To balance this, an excess of pressure acting downward onthe concave side is necessary. On the other hand if the surface isconvex (Fig. 5.25c), the resultant acts downward and there must beRan excess of pressure on the concave side acting in the upward direction.Thus, there is always an excess of pressure on the concave side ofa curved liquid surface over the pressure on its convex side due to surfacetension.5.5.6 Excess pressure inside a liquid dropConsider a liquid drop of radius . The molecules on the surfacerof the drop experience a resultant force acting inwards due to surfacetension. Therefore, the pressure inside the drop must be greater thanthe pressure outside it. The excess of pressure inside the drop providesPa force acting outwards perpendicular to the surface, to balance theresultant force due to surface tension. Imagine the drop to be dividedinto two equal halves. Considering the equilibrium of the upperhemisphere of the drop, the upward forceon the plane face ABCD due to excesspressure is PP π r2 (Fig. 5.26).If is the surface tension of theTliquid, the force due to surface tensionacting downward along the circumferenceof the circle ABCD is T2 r.πAt equilibrium, P r = T 2 rπ2π∴ = P2 TrFig. 5.25 Excess of pressure across a liquid surfaceTT(a)TTRexcess pressure(b)TTexcess pressureR(c)ACBDTPFig. 5.26 Excess pressureinside a liquid drop

233Excess pressure inside a soap bubbleA soap bubble has two liquid surfaces in contact with air, oneinside the bubble and the other outside the bubble. Therefore the forcedue to surface tension = 2 × 2 rTπ∴ At equilibrium, P r π2= 2 × 2 rTπ(i.e)P = 4 TrThus the excess of pressure inside a drop is inversely proportionalto its radius i.e. . As Pα1rPα 1r, the pressure needed to form a verysmall bubble is high. This explains why one needs to blow hard to starta balloon growing. Once the balloon has grown, less air pressure isneeded to make it expand more.5.5.7 CapillarityThe property of surface tension gives rise to an interestingphenomenon called capillarity. When a capillary tube is dipped in water,the water rises up in the tube. The level of water in the tube is abovethe free surface of water in the beaker (capillary rise). When a capillarytube is dipped in mercury, mercury also rises in the tube. But the levelof mercury is depressed below the freesurface of mercury in the beaker(capillary fall).The rise of a liquid in a capillarytube is known as capillarity. The heighth in Fig. 5.27 indicates the capillaryrise (for water) or capillary fall (formercury).Illustrations of capillarity(i) A blotting paper absorbs ink by capillary action. The pores inthe blotting paper act as capillaries.(ii) The oil in a lamp rises up the wick through the narrow spacesbetween the threads of the wick.(iii) A sponge retains water due to capillary action.(iv) Walls get damped in rainy season due to absorption of waterby bricks.For mercury hhFor waterhhFig. 5.27 Capillary rise

2345.5.8 Surface tension by capillary rise methodLet us consider a capillary tube ofuniform bore dipped vertically in a beakercontaining water. Due to surface tension,water rises to a height in the capillary tubehas shown in Fig. 5.28. The surface tension Tof the water acts inwards and the reaction ofthe tube R outwards. R is equal to T inmagnitude but opposite in direction. Thisreaction R can be resolved into tworectangular components.(i) Horizontal component R sin acting θradially outwards(ii) Vertical component R cos θ actingupwards.The horizontal component acting allalong the circumference of the tube canceleach other whereas the vertical component balances the weight of watercolumn in the tube.Total upward force = R cos θ × circumference of the tube(i.e) = 2 Fπr R cosθorF = 2 πr T cosθ...(1)[ ∵R = T]This upward force is responsible for thecapillary rise. As the water column is inequilibrium, this force acting upwards isequal to weight of the water column actingdownwards.(i.e)F = W...(2)Now, volume of water in the tube isassumed to be made up of (i) a cylindricalwater column of height and (ii) water in thehmeniscus above the plane CD.Volume of cylindrical water column = πr h 2Volume of water in the meniscus = (Volume of cylinder of heightr and radius – (Volume of hemisphere)r)T TRRR sinR cosR cosR sinrhDCFig. 5.28 Surface tensionby capillary rise methodrrDCFig. 5.29 Liquidmeniscus

235∴ Volume of water in the meniscus = (π r 2 × ) – rπ323r⎛⎞ ⎜⎟ ⎝⎠= 13 πr 3∴ Total volume of water in the tube = πr h 2 + 13 πr 3= π r 23 rh⎛⎞+⎜⎟ ⎝⎠If ρ is the density of water, then weight of water in the tube isW = π r 2ρ3 rhg⎛⎞+⎜⎟ ⎝⎠ ...(3)Substituting (1) and (3) in (2),π r 2ρ3 rhg⎛⎞+⎜⎟ ⎝⎠ = 2πrT cosθT = ρθ32rhr gcos⎛⎞+⎜⎟ ⎝⎠Since is very small, can be neglected compared to .r 3 rh∴T = θ ρ2hr gcosFor water, is small, therefore cos θθ1∴ T = ρ2hr g5.5.9 Experimental determinationof surface tension of waterby capillary rise methodA clean capillary tube of uniformbore is fixed vertically with its lowerend dipping into water taken in abeaker. A needle N is also fixed withthe capillary tube as shown in the Fig.5.30. The tube is raised or lowered untilthe tip of the needle just touches thewater surface. A travelling microscopeM is focussed on the meniscus of theFig. 5.30 Surface tensionby capillary rise methodhNM

236water in the capillary tube. The reading R corresponding to the lower1meniscus is noted. The microscope is lowered and focused on the tip ofthe needle and the corresponding reading is taken as R . The difference2between R and R gives the capillary rise .12 hThe radius of the capillary tube is determined using the travellingmicroscope. If is the density of water then the surface tension of waterρ is given by = Tρ2hr g where is the acceleration due to gravity.g5.5.10 Factors affecting surface tensionImpurities present in a liquid appreciably affect surface tension. Ahighly soluble substance like salt increases the surface tension whereassparingly soluble substances like soap decreases the surface tension.The surface tension decreases with rise in temperature. Thetemperature at which the surface tension of a liquid becomes zero iscalled critical temperature of the liquid.5.5.11 Applications of surface tension(i) During stormy weather, oil is poured into the sea around theship. As the surface tension of oil is less than that of water, it spreadson water surface. Due to the decrease in surface tension, the velocityof the waves decreases. This reduces the wrath of the waves on theship.(ii) Lubricating oils spread easily to all parts because of their lowsurface tension.(iii) Dirty clothes cannot be washed with water unless somedetergent is added to water. When detergent is added to water, one endof the hairpin shaped molecules of the detergent get attracted to waterand the other end, to molecules of the dirt. Thus the dirt is suspendedsurrounded by detergent molecules and this can be easily removed.This detergent action is due to the reduction of surface tension of waterwhen soap or detergent is added to water.(iv) Cotton dresses are preferred in summer because cotton dresseshave fine pores which act as capillaries for the sweat.

2375.6 Total energy of a liquidA liquid in motion possesses pressure energy, kinetic energy andpotential energy.(i) Pressure energyIt is the energy possessed by a liquidby virtue of its pressure.Consider a liquid of density containedρin a wide tank T having a side tube near thebottom of the tank as shown in Fig. 5.31. Africtionless piston of cross sectional area ‘ ’ais fitted to the side tube. Pressure exertedby the liquid on the piston is P = hρ gwhere is the height of liquid column above the axis of the side tube.hIf is the distance through which the piston is pushed inwards, thenxVolume of liquid pushed into the tank = ax∴ Mass of the liquid pushed into the tank = ax ρAs the tank is wide enough and a very small amount of liquid ispushed inside the tank, the height and hence the pressure may behPconsidered as constant.Work done in pushing the piston through the distance = Forcexon the piston × distance moved(i.e) W = PaxThis work done is the pressure energy of the liquid of mass axρ .∴ ∴ ∴ ∴ ∴ Pressure energy per unit mass of the liquid = ρρ PaxPax=(ii) Kinetic energyIt is the energy possessed by a liquid by virtue of its motion.If is the mass of the liquid moving with a velocity , the kineticmvenergy of the liquid = 12mv2 .Kinetic energy per unit mass = 212mvm22 v=hxTFig. 5.31 Pressure energy

238(iii) Potential energyIt is the energy possessed by a liquid by virtue of its height abovethe ground level.If is the mass of the liquid at a height from the ground level,mhthe potential energy of the liquid = mghPotential energy per unit mass = mghm = ghTotal energy of the liquid in motion = pressure energy + kineticenergy + potential energy.∴ Total energy per unit mass of the flowing liquid = 2ρ2Pv+ + gh5.6.1 Equation of continuityConsider a non-viscous liquid in streamline flow through a tubeAB of varying cross section as shown in Fig. 5.32 Let a1 and a 2 be thearea of cross section, v 1 and v 2 be thevelocity of flow of the liquid at A and Brespectively.∴ Volume of liquid entering per secondat A = a v1 1.If ρ is the density of the liquid,then mass of liquid entering per secondat A = a v1 1ρ .Similarly, mass of liquid leaving persecond at B = a v2 2ρIf there is no loss of liquid in the tube and the flow is steady, thenmass of liquid entering per second at A = mass of liquid leaving persecond at B(i.e)a v1 1ρ = a v2 2ρor a v1 1 = a v2 2i.e. av = constantThis is called as the equation of continuity. From this equationv α1a .i.e. the larger the area of cross section the smaller will be the velocityof flow of liquid and vice-versa.Fig. 5.32 Equation ofcontinuity

2395.6.2 Bernoulli’s theoremIn 1738, Daniel Bernoulliproposed a theorem for thestreamline flow of a liquid basedon the law of conservation ofenergy. According to Bernoulli’stheorem, for the streamline flow ofa non-viscous and incompressibleliquid, the sum of the pressureenergy, kinetic energy andpotential energy per unit mass is a constant.(i.e) 2ρ2Pv++ gh = constantThis equation is known as Bernoulli’s equation.Consider streamline flow of a liquid of density through a pipe ABρof varying cross section. Let P 1 and P 2 be the pressures and a 1 and a 2 ,the cross sectional areas at A and B respectively. The liquid enters Anormally with a velocity and leaves B normally with a velocity . Thev 1v 2liquid is accelerated against the force of gravity while flowing from A toB, because the height of B is greater than that of A from the groundlevel. Therefore P 1 is greater than P 2. This is maintained by an externalforce.The mass of the liquid crossing per second through any sectionmof the tube in accordance with the equation of continuity isa v1 1ρ = a v2 2ρ = mor a v1 1 = a v2 2 = ρ m = V..... (1)As a 1 > a2, v 1 < v 2The force acting on the liquid at A = P a1 1The force acting on the liquid at B = P a22Work done per second on the liquid at A = P a1 1 × v 1 = P V1Work done by the liquid at B = P a2 2 × v 2 = P 2V∴ Net work done per second on the liquid by the pressure energyin moving the liquid from A to B is = P 1V – P 2V...(2)h 1h 2BPa11Pa22Ground levelFig. 5.33 Bernoulli’s theorem

240If the mass of the liquid flowing in one second from A to B is ,mthen increase in potential energy per second of liquid from A to B ismgh2 – mgh1Increase in kinetic energy per second of the liquid= 12mv2 2 – 12mv1 2According to work-energy principle, work done per second by thepressure energy = Increase in potential energy per second + Increasein kinetic energy per second(i.e)P V – P V12= (mgh2 -mgh1)+22211122mvmv⎛⎞−⎜⎟ ⎝⎠ P V + mgh + mv11121 2= P V + mgh + mv22122 21PVm + gh +121 12 v= 2PVm + gh + 212v 2 21ρ P+ gh + 112 v1 2= 2ρ P+ gh + v2122 2ρmv⎛⎞=⎜⎟ ⎝⎠ ∵or ρ P+ gh + v122= constant...(3)This is Bernoulli’s equation. Thus the total energy of unit mass ofliquid remains constant.Dividing equation (3) by g,ρ Pg+22 vg + = constanthEach term in this equation has the dimension of length and henceis called head. ρ Pgis called pressure head, 22 vg is velocity head and ishthe gravitational head.Special case :If the liquid flows through a horizontal tube, h 1 = h 2. Thereforethere is no increase in potential energy of the liquid i.e. the gravitationalhead becomes zero.∴ equation (3) becomesρ P + 12v 2 = a constantThis is another form of Bernoulli’s equation.

2415.6.3 Application of Bernoulli’s theorem(i) Lift of an aircraft wingA section of an aircraft wingand the flow lines are shown inFig. 5.34. The orientation of the wingrelative to the flow direction causesthe flow lines to crowd together abovethe wing. This corresponds toincreased velocity in this region andhence the pressure is reduced. But below the wing, the pressure isnearly equal to the atmospheric pressure. As a result of this, the upwardforce on the underside of the wing is greater than the downward forceon the topside. Thus there is a net upward force or lift.(ii) Blowing of roofsDuring a storm, the roofs of huts ortinned roofs are blown off without anydamage to other parts of the hut. Theblowing wind creates a low pressure P1on top of the roof. The pressure P under2the roof is however greater than P . Due1to this pressure difference, the roof is liftedand blown off with the wind.(iii) Bunsen burnerIn a Bunsen burner, the gas comes out of thenozzle with high velocity. Due to this the pressure inthe stem of the burner decreases. So, air from theatmosphere rushes into the burner.(iv) Motion of two parallel boatsWhen two boats separated by a small distancerow parallel to each other along the same direction,the velocity of water between the boats becomes verylarge compared to that on the outer sides. Becauseof this, the pressure in between the two boats getsreduced. The high pressure on the outer side pushesthe boats inwards. As a result of this, the boats come closer and mayeven collide.air flowHigh velocity; Low pressureLow velocity; High pressureFig. 5.34 Lift of an aircraft wingWindP 1P 2Low PressureAirAirGasFig. 5.35 Blowing of roofsFig. 5.36 BunsenBurner

227Solved problems5.1A 50 kg mass is suspended from one end of a wire of length 4 mand diameter 3 mm whose other end is fixed. What will be theelongation of the wire? Take q = 7 × 10 N m for the material of10− 2the wire.Data : l = 4 m; d = 3 mm = 3 × 10 m; m = 50 kg; q = 7×10 N m− 310− 2Solution :q = FlAdl∴ dl = π 2-3 210Fl50 ×9.8 × 4 = r q3.14 × (1.5 ×10 ) ×7 ×10= 3.96 × 10−3 m5.2A sphere contracts in volume by 0.01% when taken to the bottomof sea 1 km deep. If the density of sea water is 10 kg m , find the3− 3bulk modulus of the material of the sphere.Data : dV = 0.01%i.e 0.01100dVV= ; h = 1 km ; = 10 kg mρ3− 3Solution : dP = 10 × 10 × 9.8 = 9.8 × 10336/ dPkdV V∴ = = 69.8 ×10 ×1000.01 = 9.8 × 10 N m10− 25.3A hydraulic automobile lift is designed to lift cars with a maximummass of 3000 kg. The area of cross section of the piston carrying−the load is 425 × 10−4 m . What maximum pressure would the 2piston have to bear?Data : m = 3000 kg, A = 425 × 10 m− 42Solution: Pressure on the piston = Weight of carArea of piston= mgA= -43000 ×9.8425 ×10= 6.92 × 10 N m5 − 25.4A square plate of 0.1 m side moves parallel to another plate with avelocity of 0.1 m s , both plates being immersed in water. If the− 1viscous force is 2 × 10 N and viscosity of water is 10− 3−3 N s m − 2 ,find their distance of separation.

228Data :Area of plate A = 0.1 × 0.1 = 0.01 m2Viscous force F = 2 × 10 N− 3Velocity dv = 0.1 m s− 1Coefficient of viscosity = 10 N s mη− 3− 2Solution : Distance dx = AdvF η= -3-310 ×0.01×0.12×10 = 5 × 10 m− 45.5Determine the velocity for air flowing through a tube of10−2 m radius. For air = 1.3 kg m and = 187 x 10 N s m .ρ− 3η− 7− 2Data : r = 10 m ; = 1.3 kg m ; = 187 × 10 N s m ; N = 2000− 2ρ− 3η − 7−2 RSolution : velocity v = ηρ R ND= -7-22000 ×187 ×101.3× 2 10×= 1.44 m s− 15.6Fine particles of sand are shaken up in water contained in a tallcylinder. If the depth of water in the cylinder is 0.3 m, calculate thesize of the largest particle of sand that can remain suspended after40 minutes. Assume density of sand = 2600 kg m and viscosity of− 3water = 10 N s m .− 3− 2Data : s = 0.3 m, t = 40 minutes = 40 × 60 s, = 2600 kg mρ− 3Solution: Let us assume that the sand particles are spherical inshape and are of different size.Let r be the radius of the largest particle.Terminal velocity v = 0.340 60× = 1.25 × 10 m s− 4− 1Radius r= ησ ρ9v2( -)g= -3-49×10 ×1.25 ×102 (2600 - 1000) 9.8= 5.989 × 10 m− 6

2295.7A circular wire loop of 0.03 m radius is rested on the surface of aliquid and then raised. The pull required is 0.003 kg wt greaterthan the force acting after the film breaks. Find the surface tensionof the liquid.Solution: The additional pull F of 0.003 kg wt is the force due tosurface tension.∴Force due to surface tension,F = T × length of ring in contact with liquid(i.e) F = T × 2 × 2 r = 4 Trππ(i.e) 4 Tr = Fπ∴ π 4 Tr = 0.003 × 9.81or T = ××× 0.003 9.8143.140.03= 0.078 N m− 15.8Calculate the diameter of a capillary tube in which mercury isdepressed by 2.219 mm. Given T for mercury is 0.54 N m , angle− 1of contact is 140 and density of mercury is 13600 kg mo− 3Data :h = 2.219 × 10−−3 m; T = 0.54 N m ; = 140 ;− 1θoρ = 13600 kg m− 3Solution : hr g = 2T cos ρθ∴ r = 2 cos ρ Thgθ= −××−××× 320.54 cos 140( 2.219 10 ) 13600 9.8o= 2.79 × 10 m− 3Diameter = 2r = 2 × 2.79 × 10 m = 5.58 mm− 35.9Calculate the energy required to split a water drop of radius1 × 10 m into one thousand million droplets of same size. Surface− 3tension of water = 0.072 N m− 1Data :Radius of big drop R = 1 × 10 m− 3Number of drops n = 10 × 10 = 10 ; T = 0.072 N m369-1Solution : Let r be the radius of droplet.

230Volume of 10 drops = Volume of big drop910 × 9π343r = π343R 10 r = R = (10 )9 33−3 3(10 r) = (10 )3 3-3 3r = − 33 1010 = 10−6 mIncrease in surface area ds = 10 × 4 r 4 R9π2− π 2(i.e) ds = 4 [ 10 × (10 ) (10 ) ] = 4 [10 10 ] mπ9−6 2−−3 2π− 3−− 62∴ds = 0.01254 m2Work done W = T.ds = 0.072 × 0.01254 = 9.034 × 10−4 J5.10Calculate the minimum pressure required to force the blood fromthe heart to the top of the head (a vertical distance of 0.5 m). Givendensity of blood = 1040 kg m . Neglect friction.− 3Data : h h = 0.5 m , = 1040 kg m , P P = ?2−1ρ− 31−2Solution : According to Bernoulli’s theoremP P = g(h h ) + 1−2ρ2−112ρ (v v )2 2−1 2If v = v , then21P P = 1−2gρ(h h )2−1P P = 1040 × 9.8 (0.5)1−2P P = 5.096 × 10 N m1−23− 2

231Self evaluation(The questions and problems given in this self evaluation are only samples.In the same way any question and problem could be framed from the textmatter. Students must be prepared to answer any question and problemfrom the text matter, not only from the self evaluation.)5.1If the length of the wire and mass suspended are doubled in a Young’smodulus experiment, then, Young’s modulus of the wire(a) remains unchanged(b) becomes double(c) becomes four times(d) becomes sixteen times5.2For a perfect rigid body, Young’s modulus is(a) zero(b) infinity(c) 1(d) –15.3Two wires of the same radii and material have their lengths in theratio 1 : 2. If these are stretched by the same force, the strainsproduced in the two wires will be in the ratio(a) 1 : 4(b) 1 : 2(c) 2 : 1(d) 1 : 15.4If the temperature of a liquid is raised, then its surface tension is(a) decreased(b) increased(c) does not change(d) equal to viscosity5.5The excess of pressure inside two soap bubbles of diameters in theratio 2 : 1 is(a) 1 : 4(b) 2 : 1(c) 1 : 2(d) 4 : 15.6A square frame of side l is dipped in a soap solution. When the frameis taken out, a soap film is formed. The force on the frame due tosurface tension T of the soap solution is(a) 8 Tl(b) 4 Tl(c) 10 Tl(d) 12 Tl

2325.7The rain drops falling from the sky neither hit us hard nor makeholes on the ground because they move with(a) constant acceleration(b) variable acceleration(c) variable speed(d) constant velocity5.8Two hail stones whose radii are in the ratio of 1 : 2 fall from a heightof 50 km. Their terminal velocities are in the ratio of(a) 1 : 9(b) 9 : 1(c) 4 : 1(d) 1 : 45.9Water flows through a horizontal pipe of varying cross section at the−rate of 0.2 m s . The velocity of water at a point where the area of3 − 1cross section of the pipe is 0.01 m is−2(a) 2 ms− 1(b) 20 ms− 1(c) 200 ms− 1(d) 0.2 ms− 15.10An object entering Earth’s atmosphere at a high velocity catches firedue to(a) viscosity of air(b) the high heat content of atmosphere(c) pressure of certain gases (d) high force of g.5.11Define : i) elastic body ii) plastic body iii) stress iv) strain v) elasticlimit vi) restoring force5.12State Hooke’s law.5.13Explain the three moduli of elasticity.5.14Describe Searle’s Experiment.5.15Which is more elastic, rubber or steel? Support your answer.5.16State and prove Pascal’s law without considering the effect of gravity.5.17Taking gravity into account, explain Pascal’s law.5.18Explain the principle, construction and working of hydraulic brakes.5.19What is Reynold’s number?5.20What is critical velocity of a liquid?5.21Why aeroplanes and cars have streamline shape?5.22Describe an experiment to determine viscosity of a liquid.

2335.23What is terminal velocity?5.24Explain Stoke’s law.5.25Derive an expression for terminal velocity of a small sphere fallingthrough a viscous liquid.5.26Define cohesive force and adhesive force. Give examples.5.27Define i) molecular range ii) sphere of influence iii) surface tension.5.28Explain surface tension on the basis of molecular theory.5.29Establish the relation between surface tension and surface energy.5.30Give four examples of practical application of surface tension.5.31How do insects run on the surface of water?5.32Why hot water is preferred to cold water for washing clothes?5.33Derive an expression for the total energy per unit mass of a flowingliquid.5.34State and prove Bernoulli’s theorem.5.35Why the blood pressure in humans is greater at the feet than at thebrain?5.36Why two holes are made to empty an oil tin?5.37A person standing near a speeding train has a danger of fallingtowards the train. Why?5.38Why a small bubble rises slowly through a liquid whereas the biggerbubble rises rapidly?Problems5.39A wire of diameter 2.5 mm is stretched by a force of 980 N. If theYoung’s modulus of the wire is 12.5 × 10 N m , find the percentage10− 2increase in the length of the wire.5.40Two wires are made of same material. The length of the first wire ishalf of the second wire and its diameter is double that of secondwire. If equal loads are applied on both the wires, find the ratio ofincrease in their lengths.5.41The diameter of a brass rod is 4 mm. Calculate the stress and strainwhen it is stretched by 0.25% of its length. Find the force exerted.Given q = 9.2 × 10 N m for brass.10− 2

2345.42Calculate the volume change of a solid copper cube, 40 mm on eachside, when subjected to a pressure of 2 ×10 Pa. Bulk modulus of7copper is 1.25 × 10 N m .11− 25.43In a hydraulic lift, the piston P has a diameter of 50 cm and that of2P is 10 cm. What is the force on P when 1 N of force is applied on12P ?15.44Calculate the mass of water flowing in 10 minutes through a tube ofradius 10 m and length 1 m having a constant pressure of 0.2 m of−2 water. Assume coefficient of viscosity of water = 9 × 10 N s m− 4− 2and g = 9.8 m s .− 25.45A liquid flows through a pipe of 10 m radius and 0.1 m length− 3under a pressure of 10 Pa. If the coefficient of viscosity of the liquid3is 1.25 × 10 N s m , calculate the rate of flow and the speed of the− 3− 2liquid coming out of the pipe.5.46For cylindrical pipes, Reynold’s number is nearly 2000. If the diameterof a pipe is 2 cm and water flows through it, determine the velocity ofthe flow. Take for water = 10 N s m .η− 3− 25.47In a Poiseuille’s flow experiment, the following are noted.i) Volume of liquid discharged per minute = 15 × 10 m− 63ii) Head of liquid=0.30 miii) Length of tube=0.25 miv) Diameter=2 × 10 m− 3v) Density of liquid=2300 kg m .− 3Calculate the coefficient of viscosity.5.48An air bubble of 0.01 m radius raises steadily at a speed of5 × 10 m s through a liquid of density 800 kg m . Find the− 3− 1− 3coefficient of viscosity of the liquid. Neglect the density of air.5.49Calculate the viscous force on a ball of radius 1 mm moving througha liquid of viscosity 0.2 N s m at a speed of 0.07 m s .− 2− 15.50A U shaped wire is dipped in soap solution. The thin soap film formedbetween the wire and a slider supports a weight of 1.5 × 10 N. If− 2the length of the slider is 30 cm, calculate the surface tension of thefilm.

2355.51Calculate the force required to remove a flat circular plate of radius0.02 m from the surface of water. Assume surface tension of wateris 0.07 N m .− 15.52Find the work done in blowing up a soap bubble from an initial surfacearea of 0.5 × 10 m to an area 1.1 × 10 m . The surface tension of− 42− 42soap solution is 0.03 N m .− 15.53Determine the height to which water will rise in a capillarytube of 0.5 × 10 m diameter. Given for water, surface tension− 3is 0.074 N m .− 15.54A capillary tube of inner diameter 4 mm stands vertically in a bowlof mercury. The density of mercury is 13,500 kg m and its surface− 3tension is 0.544 N m . If the level of mercury in the tube is 2.33 mm− 1below the level outside, find the angle of contact of mercury withglass.5.55A capillary tube of inner radius 5 × 10 m is dipped in water of− 4surface tension 0.075 N m . To what height is the water raised by− 1the capillary action above the water level outside. Calculate the weightof water column in the tube.5.56What amount of energy will be liberated if 1000 droplets of water,each of diameter 10 m, coalesce to form a big drop. Surface tension− 8of water is 0.075 N m .− 15.57Water flows through a horizontal pipe of varying cross section. If the−pressure of water equals 2 × 10 m of mercury where the velocity of−2 flow is 32 × 10 m s find the pressure at another point, where the− 2− 1velocity of flow is 40 × 10 m s .− 2− 1

236Answers5.1(a)5.2(b)5.3(d)5.4(a)5.5(c)5.6(a)5.7(d)5.8(d)5.9(b)5.10 (a)5.390.16 %5.401 : 85.412.3 × 10 N m , 0.0025, 2.89 × 10 N8 -23 5.42-1.024 × 10 m-8 35.4325 N5.445.13 × 10 kg35.453.14 × 10 m s , 1 m s-63 -1-15.460.1 ms-15.474.25 × 10 N s m-2-25.4834.84 N s m-25.492.63 × 10 N-45.502.5 × 10 N m-2-15.518.8 × 10 N-35.521.8 × 10 J-65.536.04 × 10 m-25.54124 36’o5.553.04 × 10 m, 2.35 × 10 N-2-45.562.12 × 10-14 J5.572636.8 N m-2

1Mathematical Notes(Not for examination)LogarithmIn physics, a student is expected to do the calculation by usinglogarithm tables. The logarithm of any number to a given base is thepower to which the base must be raised in order to obtain the number.For example, we know that 2 raised to power 3 is equal to 8 (i.e) 2 = 8. In3the logarithm form this fact is stated as the logarithm of 8 to the base 2 isequal to 3. (i.e.) log2 8 = 3.In general, if a = N x, then log N xa = .We use “common logarithm” for calculation purposes. Commonlogarithm of a number is the power to which 10 must be raised in order toobtain that number. The base 10 is usually not mentioned. In other words,when base is not mentioned, it is understood as base of 10.For doing calculations with log tables, the following formulae shouldbe kept in mind.(i)Product formula: log mn = log m + log n(ii)Quotient formula: log mn = log m log n−(iii)Power formula: log m = n log mn(iv)Base changing formula: log m = log m × log ba b aLogarithm of a number consists of two parts called characteristicand Mantissa. The integral part of the logarithm of a number afterexpressing the decimal part as a positive is called characteristic. Thepositive decimal part is called Mantissa.To find the characteristic of a number(i)The characteristic of a number greater than one or equal toone is lesser by one (i.e) (n − 1) than the number of digits (n)present to the left of the decimal point in a given number.(ii)The characteristic of a number less than one is a negativenumber whose numerical value is more by one i.e. (n+1) than

2the number of zeroes ( ) between the decimal point and thenfirst significant figure of the number.ExampleNumberCharacteristic5678.93567.89256.78915.678900.5678910.05678920.00567893To find the Mantissa of a numberWe have to find out the Mantissa from the logarithm table. Theposition of a decimal point is immaterial for finding the Mantissa.(i.e) log 39, log 0.39, log 0.039 all have same Mantissa. We use the followingprocedure for finding the Mantissa.(i)For finding the Mantissa of log 56.78, the decimal point is ignored.We get 5678. It can be noted that the first two digits from the leftform 56, the third digit is 7 and the fourth is 8.(ii) In the log tables, proceed in the row 56 and in this row find thenumber written under the column headed by the third digit 7.(i.e) 7536. To this number the mean difference written underthe fourth digit 8 in the same row is added (i.e)7536 + 6 = 7542. Hence logarithm of 56.78 is 1.7542. 1 is thecharacteristic and 0.7542 is the Mantissa.(iii) To find out the Mantissa of 567, find the number in the rowheaded by 56 and under the column 7. It is 7536. Hence thelogarithm of 567 is 2. 7536. Here 2 is the characteristic and0.7536 is the Mantissa.(iv) To find out the Mantissa of 56, find the number in the rowheaded by 56 and under the column 0. It is 7482. Hence thelogarithm of 56 is 1.7482. Here 1 is the characteristic and 0.7482is the Mantissa.(v) To find out the Mantissa of 5, find the number in the row headedby 50 and under the column 0. It is 6990. Hence the logarithm

3of 5 is 0.6990. Here 0 is the characteristic and 0.6990 is theMantissa.AntilogarithmTo find out the antilogarithm of a number, we use the decimal partof a number and read the antilogarithm table in the same manneras in the case of logarithm.(i)If the characteristic is n, then the decimal point is fixed after(n+1)th digit.(ii) If the characteristic is n, then add (n 1) zeroes to the left side−and then fix the decimal point.(iii) In general if the characteristic is n or n, then fix the decimalpoint right side of the first digit and multiply the whole numberby 10 or 10 .n− nExampleNumberAntilogarithm0.93288.567 or 8.567 × 1001.932885.67 or 8.567 × 1012.9328856.7 or 8.567 × 1023.93288567.0 or 8.567 × 1031.93280.8567 or 8.567 × 10− 12.93280.08567 or 8.567 × 10− 23.93280.008567 or 8.567 × 10− 3EXERCISE - 11.Expand by using logarithm formula(i)T = 2 πlg(ii)v =e2gR(iii)q = 2mglπrx(iv)log 2e2.Multiply .5670 by 353.Divide .6990 by 234.Evaluate using logarithm(i)2×22×64007×7918.4(ii)39.8×6370×10

4(iii)-23-22×7.35×109.8×10 ×8.5×10(iv)2 π 0.5245Some commonly used formulae of algebra(i)(a+b)2= a + 2ab + b22(ii)(a b)−2= a 2ab + b2−2(iii)(a+b+c) = a + b + c + 2ab + 2bc + 2ca2222(iv)(a+b)3= a + b + 3a b + 3b a3322(v)(a b)−3= a b 3a b + 3ab3−3−22Quadratic equationAn algebraic equation in the form ax + bx + c = 20 is called quadraticequation. Here a is the coefficient of , b is the coefficient of and c is thex 2xconstant. The solution of the quadratic equation isx = −±− 242bbacaBinomial theoremThe theorem states that (1 + x) n = 1 + nx + − (1)2! nn x +2 −− (1)(2)3!nnn x + 3... where is less than 1 and n is any number. If nxis a positive integer the expansion will have (n+1) terms and if n isnegative or fraction, the expansion will have infinite terms.Factorial 2 = 2! = 2 × 1Factorial 3 = 3! = 3 × 2 × 1Factorial n = n! = n(n − 1) (n − 2) ....If is very small, then the terms with higher powers of can bexxneglected.(i.e)(1 + ) = 1 + nx nx(1 + ) = 1 nx − n−x(1 ) = 1 n−x n−x(1 ) = 1 + n−x − nxEXERCISE- 21. Find the value of x in 4 + 5 2 = 0x 2x−2. Expand Binomially (i) 21hR−⎡⎤ +⎢⎥ ⎣⎦(ii) (1 2 )−x 3

5TrigonometryLet the line AC moves in anticlockwise direction from the initialposition AB. The amount of revolution that the moving line makes with itsinitial position is called angle. From the figure = θCAB. The angle ismeasured with degree and radian. Radian is the angle subtended at thecentre of a circle by an arc of the circle, whose length is equal to the radiusof the circle.1 radian = 57 17 45o′″1 right angle = /2 radianπ1 = 60 (sixty minutes). 1 = 60 (sixty seconds) ο′′″Triangle laws of sine and cosinea = b + c – 2bc cos 222αsinsinsinabcαβγ== Trigonometrical ratios (T ratios)− − − − −Consider the line OA making anangle in anticlockwise direction withθOX.From A, draw the perpendicularAB to OX.The longest side of the right angledtriangle, OA is called hypotenuse. The sideAB is called perpendicular or opposite side.The side OB is called base or adjacent side.1.Sine of angle = θsin θ = perpendicularhypotenuse2.Cosine of angle = θcos = θbasehypotenuse3.Tangent of angle = θtan = θperpendicularbase4.Cotangent of = θcot = θbaseperpendicularCBAθθBAXOX ′YY ′)))γαβabc)

65.Secant of = θsec = θhypotenusebase6.Cosecant of = θcosec = θhypotenuseperpendicularSign of trigonometrical ratiosII quadrantI quadrantsin and θcosecθAll positiveonly positiveIII quadrantIV quadranttan and θcotθcos and θsec only θonly positivepositiveT ratios of allied angles− − − − −− θ , 90o − θ, 90 + , 180o θo − θ, 180 + , 270o θo − θ, 270 + are calledo θallied angles to the angle . The allied angles are always integral multiplesθof 90 .o1. (a) sin ( ) = −θ−sin θ(b) cos ( ) = −θcosθ(c) tan ( ) = −θ−tanθ2. (a) sin (90 ) = o− θcosθ(b) cos (90 ) = o− θsinθ(c) tan (90 ) = o− θcotθ3. (a) sin (90 + ) = oθcosθ(b) cos (90 ) = o+ θ−sinθ(c) tan (90 + ) = –oθ cot θ4. (a) sin (180 – ) = oθsinθ(b) cos (180 ) = o− θ−cosθ(c) tan (180 – ) = –oθ tanθ5. (a) sin (180 ) = –o+ θ sinθ(b) cos (180 ) = – o+ θcosθ(c) tan (180 ) = o+ θtanθ6. (a) sin (270 – ) = – cos oθθ(b) cos (270 ) = o− θ−sinθ(c) tan (270 – ) = oθcotθ7. (a) sin (270 + ) = – oθcosθ(b) cos (270 ) = o+ θsinθ(c) tan (270 + )= – oθcotθ

7T ratios of some standard angles − − − − −Angle0 o30o45o60o90o120o180osinθ01212321320cosθ13212120–12–1tanθ01313∞– 30Some trigonometric formulae1.sin (A + B) = sin A cos B + cos A sin B2.cos (A + B) = cos A cos B sin A sin B−3.sin (A – B) = sin A cos B – cos A sin B4.cos (A – B) = cos A cos B sin A sin B+5.sin A + sin B = 2 sin 2 AB+ cos 2AB −6.sin A – sin B = 2 cos 2 AB+ sin 2AB −7.cos A + cos B = 2 cos 2 AB+ cos 2AB −8.cos A – cos B = 2 sin 2 AB+ sin 2AB −9.sin 2A = 2 sin A cos A = 22tan A1tan +A10.2 sin A cos B = sin (A + B) + sin (A – B)11.2 cos A sin B = sin (A + B) – sin (A – B)12.2 sin A sin B = cos (A – B) – cos (A + B)13.2 cos A cos B = cos (A + B) + cos (A – B)14.cos 2 A = 1 – 2 sin A2Differential calculusLet y be the function of x(i.e)y = f( )x .....(1)The function y depends on variable . If the variable is changed toxxx + x, then the function is also changed to y + y∆∆

8∴ y + y = f ( + ∆x∆ x ) ....(2)Subtracting equation (1) from (2)∆y = f ( + x∆ x ) – f ( )x dividing on both sides by ∆ x, we get(x) f( ) = yfxxxx∆+ ∆ −∆∆Taking limits on both sides of equation, when x approaches zero, ∆we get00(x) f( )xxyfxxLtLtxx∆→∆→∆+ ∆ − ⎛⎞ =⎜⎟∆∆ ⎝⎠In calculus 0 xy Ltx∆→∆∆ is denoted by dydx and is called differentiation ofy with respect to x.The differentiation of a function with respect to a variable meansthe instantaneous rate of change of the function with respect to thevariable.Some theorems and formulae1.ddx ( ) = 0 , if c is a constant.c2.If y = c u, where is a constant and is a function of thencuxdydx = ddx (cu) = c dudx3.If y = w±± uv where , and are functions of thenu vwx =(uv w) ±±dyddxdx= ±± dudvdwdxdxdx4.If y = x n, where n is the real number then1 =(x) n x−= nndyddxdx5.If y = uv where and are functions of thenuvx =( ) u=+dyddvduuvvdxdxdxdx

96.If is a function of , then yxdy = dydx . dx7.() e xxdedx=8.1(log )edxdxx =9.dd θ (sin ) = θcosθ10.dd θ (cos ) = – θsinθ11.If is a trigonometrical function of and is the function of , thenyθθtddt (sin ) = θcosθ ddtθ12.If is a trigonometrical function of and is the function of , thenyθθtddt(cos θ) = – sinθ ddtθEXERCISE - 31.If = ysin 3 find θθ dyd2.If = yx5/7 find dydx3.If = y2 1x find dydx4.If = 4 + 3 + 2, find yx 3x 2dydx5.Differentiate : (i) a + b + cx 2x6.If s = 2t – 5t + 4t – 2, 32 find the position ( ), velocity sdsdt⎛⎞ ⎜⎟ ⎝⎠ andacceleration ⎛⎞ ⎜⎟ ⎝⎠ dvdt of the particle at the end of 2 seconds.IntegrationIt is the reverse process of differentiation. In other words integrationis the process of finding a function whose derivative is given. The integralof a function with respect to is given by yx∫y dx. Integration is representedby the elongated S. The letter S represents the summation of all differentialparts.

10Indefinite integralWe know that32() 3x dxdx=32(x4) 3xddx+= 32(x) 3xdcdx+=The result in the above three equations is the same. Hence thequestion arises as to which of the above results is the integral of 3 . Tox 2overcome this difficulty the integral of 3 is taken as ( + c), where c isx 2x 3an arbitrary constant and can have any value. It is called the constant ofintegration and is indefinite. The integral containing c, (i.e) ( + c) isx 3called indefinite integral. In practice ‘c’ is generally not written, though itis always implied.Some important formulae(1)dx = x∫() 1 dxdx=∵(2)11 nnxxdxn+⎛⎞ =⎜⎟+⎝⎠∫11 ndxdx n+⎛⎞ ⎜⎟+⎝⎠ ∵= x n(3)=∫∫ cu dx c u dx where c is a constant(4)()v w dxvdx±±=±±∫∫∫∫ uu dxw dx(5)∫xe1dx = logx(6) = e∫xxedx(7)cos dsinθ θθ =∫(8)sin dcosθ θ θ =−∫Definite integralsWhen a function is integrated between a lower limit and an upperlimit, it is called a definite integral.

11[] ()dx() = ( ) f( )bbaafxfxf ba′=− ∫ is a definite integral. Here a and b arelower and upper limits of the variable .xEXERCISE – 41.Integrate the following with respect to x(i) 4x 3(ii) 2 1x(iii) 3 + 7 - 4x 2x(iv) 2/7 57x(v) 3 2x −(vi) 12 + 6x 2x2.Evaluate(i) 322 ∫xdx(ii) 41 ∫xdx(iii) 42 ∫xdx(iv) /2/2cos dππθ θ−∫ANSWERSExercise - 11.(i)log 2 + log 3.14 + 12 log l - 12 log g(ii)12 (log 2 + log g + log R)(iii)log m + log g + log l – log 3.14 – 2 log r – log x(iv)0.69312.14.70103.2.84954.(i) 5.080(ii) 7.9 × 103(iii) 1.764 × 10–4(iv) 2.836 × 10–1

12Exercise - 2(1)5578 −±(2) (i) 2h 1R −(ii) 1 – 6xExercise - 3(1) 3 cos 3θ(2) 2/757 x −(3) 3 2x −(4) 12 + 6x 2x(5) 2ax + b(6) 2, 8, 14Exercise - 41.(i) x 4(ii) 1x −(iii) 32 74x2 xx+−(iv) x5/7(v) 2 1x(vi) 4x 3 + 3x 22.(i) 193(ii) 143(iii) 6(iv) 2

237Proof for Lami’s theoremOADBCRPQProof for Lami’s theoremLet forces , → →P Q and →R acting at a point be in equilibrium. LetOOA and OB(=AD) represent the forces and →P→Q in magnitude anddirection. By the parallelogram law of forces OD will represent theresultant of the forces and →P→Q . Since the forces are in equilibrium DOwill represent the third force R.In the triangle OAD, using law of sines, = = sinsinsinOAADODODAAODOADFrom Fig. 2.35,∠ODA = ∠BOD = 180 o− ∠BOC∠AOD = 180 o− ∠AOC∠OAD = 180 o− ∠AOBTherefore,osin (180 - )OABOC∠ = osin (180 - )ADAOC∠ = osin (180 - )ODAOB∠(i.e) = sinsinOAADBOCAOC∠∠ = sin ODAOB∠If ∠BOC = , α∠AOC= , β∠AOB= γ==sinsinsinPQRα βγ which proves Lami’s theorem.ANNEXURE(NOT FOR EXAMINATION)

2381. Moment of inertia of a thin uniform rod(i) About an axis passing through itscentre of gravity and perpendicularto its lengthConsider a thin uniform rod AB ofmass M and length as shown inlFig. 1. Its mass per unit length willbe Ml. Let, YY′ be the axis passingthrough the centre of gravity of theGrod (and perpendicular to thelength AB).Consider a small element of lengthdx of the rod at a distance from G.xThe mass of the element= mass per unit length × length of the element = Ml× dx...(1)The moment of inertia of the element dx about the axis YY′ is,dI= (mass) × (distance)2= ()2 xMdxl⎛⎞ ⎜⎟ ⎝⎠...(2)Therefore the moment of inertia of the whole rod about YY ′ is obtainedby integrating equation (2) within the limits – l2 to + l2 .ICG= /22/2+−⎛⎞ ⎜⎟ ⎝⎠ ∫ llMdx xl= /22/2+−∫ llMxdxlICG= /23/23+−⎛⎞ ⎜⎟ ⎝⎠ llMxl = 33322Mlll⎡⎤ ⎛⎞⎛⎞−−⎢⎥ ⎜⎟ ⎜⎟⎝⎠⎝⎠ ⎥ ⎣⎢⎦= 333 l88⎡⎤+⎢⎥ ⎣⎦ Mll = 32l3l8⎡⎤ ⎢⎥ ⎣⎦ MICG = 3212l12 = MlMl...(3)Y 1/Y 1Y /YxdxBGAl2Fig 1 Moment of inertia ofa thin uniform rod

239(ii) About an axis passing through the end and perpendicular to itslengthThe moment of inertia about a parallel axis IY Y1 1′ passing throughone end can be obtained by using parallel axes theoremA∴I = ICG + ⎛⎞⎜⎟⎝⎠ = 222lMlMlM+21242Ml I3 =2 Moment of inertia of a thin circular ring(i) About an axis passing through its centre andperpendicular to its planeLet us consider a thin ring of mass M andradius with as centre, as shown in Fig. 2. As theROring is thin, each particle of the ring is at a distanceR from the axis XOY passing through O andperpendicular to the plane of the ring.For a particle of mass m on the ring, its momentof inertia about the axis XOY is mR2. Therefore the moment of inertia ofthe ring about the axis is,I = mR = Σ 2 ()mΣR = MR 22(ii) About its diameterAB and CD are the diameters of the ringperpendicular to each other (Fig. 3). Since, the ringis symmetrical about any diameter, its moment ofinertia about AB will be equal to that about CD. Letit be I .d If is the moment of inertia of the ring aboutIan axis passing through the centre andperpendicular to its plane then applyingperpendicular axes theorem,∴ I = + = IdIdMR2 (or) = Id12MR2XYROFig 2 Moment ofInertia of a ringABCODEFRFig 3 Moment ofinertia of a ringabout its diameter