B. B. 6yX08I4e8. B. JI. KpU8IleHIC08.r. fl. MRKuwe8, B. n. llJaAbH08CBOPHHK3A)lAqno3JIEMEHTAPHOA0where k =0.12 mA/V, areconnected to a circuit as shown inFig. 190.Draw a diagram showinghowthe current I in the circuitdependsonthevoltageVifeft1 = 2 V, itI =5 V, G8 =7 V, andVcan change from -10Vto + 10V.543. Calculatethesensitivityof acathode-raytubetovoltage,i. e., the deflectionof the light spot onthe screen causedbyapotential difference of 1Von the control grids. The lengthof.the controlgrids is l, thedistancebetweenthemis d, thedistancefromtheendof thegridstothescreen is L, andtheacceleratingpotential difference isUe-Fig. 189 Fig. 190112PROBLEMS3-4. Magnetic Field ofaCurrent. Action ofaMagneticField on aCurrent and Moving Charges544. Determinethe dimension and magnitude of the coefficient kin the expression for the intensity of the magnetic fie-ld of asolenoidH =k 4nJ ~ . if H is measured in oerstedsand I inegs electrostatic units.Thedimensionof theoersted coincideswith that of the electricfieldintensity incgs units.545. Twowindings connectedasshowninFig. 191arewoundarounda thin iron ringwitharadius R =10ern. Thefirst wind-ing has 2,000 turns and thesecond1,000 turns. Find theinten-sityof themagneticfield inside the ring if acurrent of I =IDAflows through the windings. .546. A current I flows through an infinitely long conductorABCbent to forma right angle(Fig. 192).How manytimes will the intensityof themagneticfield changeat point Mif an infinitely long straight conductor BDissoconnected to point B that the current I branches at point Binto twoequal parts and the current in the conductor ABre-mains thesame?Note. Take into account the fact thai the intensityofamag-netic field induced at a certain point by a small element ofcurrent is perpendicular to the plane containing this elementand a radius-vector drawn from this current element to thegiven point.547. Acurrent flowsthrougha conductorarrangedin oneplaneasshowninFig. 193. Find the intensity of the magnetic fieldTNJJI C__- ...- ~ I ~ _ - - -P 0"AFig. 191 Fig. 192Fig. 193ELECTRICITY ANDMAGNETISM113IIIJJIJjIJIJI....---+-t-..J...J ~ " ' : : J ~I -Fig. 194 Fig. 195 Fig. 196at anarbitrarypoint online AB, which istheaxis of symmetry0-1 the conductor.548. Howwill amagneticpointer bepositionedif it is placedin the centre of a single-layer toroidal solenoid through whicha direct current flows?549. Acurrent I flows alongan infinite straight thin-walledpipe. Bearing in mind that the intensityof the magnetic fieldof an infinite straight conductor at a distance , fromit is pro-portional to I [r, find the intensity of the magnetic field at anarbitrary point inside the pipe.550. Rememberingthat the intensityof amagnetic field insidea long cylindrical conductor H = k 2njr, where j is the currentdensity and r is the distance fromthe conductor axis, find theintensity of thefieldat an arbitrarypoint on a longcylindricalspace inside the conductor (Fig. 194) through which a currentwith a density j flows. The axis of, the space is parallel totheaxis of the conductor and is atadistancedfromit.551. Drawthe distributionof the force lines of a magneticfield in the space of thecylindrical conductor descri bed InProblem550.552. Determine the dimension and themagnitudeof thecoef-ficient k in the expression for the force F = kHI1sincp actingfroma magnetic field on a current if Hisinoerstedsand I inegs electrostatic units.8-2042114PROBLEMS553. Will thedensityof a direct current flowing in a cylindricalconductor be constant across the entire cross section of theconductor?554. Alightning arrester is connectedtoearth byacircularcopper pipe. After lightningstrikes, it is discovered that thepipe became a circularrod.Explain the cause of thisphenomenon.555. Averygreat current is madeto flow for ashort timethrough a thickwindingof 8 solenoid. Describe thedeformationof the windingfromthe viewpoint of quality.556. The magnetic systemof a galvanometer consistsof amagnet, poleshoesAandB, andacylinder madeof soft iron(Fig. 195). The magnetic force lines in the gap betweentheshoesandthe cylinder are perpendicular tothe surface of thecylinder. Theintensityofthemagneticfieldis H. Arectangularcoil with n turns isplacedin the gaponaxisO. Thesidesofthe coil are parallel tothe diameter and thegeneratrix of thecylinder. The areaofeachturn isA. One endofa spiral springissoattached tothe axisof thecoil that when the latter ro-tates through ananglea, the deformation of thespringcreatesa rotational moment ka that tends to turnthecoil to a posi-tionof equilibrium. Determinetheanglethroughwhichthecoilwill turn if acurrent I passes through it.557. A current of I =1Aflows through a wirering with aradius R= 5emsuspendedontwoflexible conductors. Theringisplacedina homogeneous magnetic fieldwith an intensityofH = 10Oewhose forcelinesarehorizontal. What force will theringbetensionedwith?558. Awireringwith a radius R= 4ern is placed intoaheterogeneousmagnetic field whose forcelinesat the points ofintersectionwiththe ring form an angle of a = 100withanormaltothe plane of the ring(Fig. 196). The intensityof the mag-netic field actingonthe ring H=100 Oe. A current of I = 5 Aflows through the ring. What forcedoes the magnetic field actonthe ringwith?559. Arectangular circuit ABeDwithsidesa and b placedinto a homogeneous magnetic field with an intensity Hcanrevolvearoundaxis 00'(Fig. 197). Adirect current I constantlyflows through thecircuit.Determine the workperformed by the magnetic field whenthecircuit isturned through 1800if initiallythe plane of thecircuit is perpendicular to the magnetic field and arrangedas showninFiz. 197.ELECTRICITY ANDMAGNETISMolJ115aFig. 197Fig /988 *Ho0'ABHFig. /99116PROBLEMSticfieldinthe directionof sidea with a velocitye. The inten-sity of the magnetic field Hisperpendicular to the base ofthe blockwith thesidesaandc(Fig. 199). .Determine the intensity of the electric field inthe blockandthe density of the electriccharges onthe surfacesof the paral-lelepipedformedbysidesa andb.567. An uncharged metal cylinder witha radius r revolves aboutits axis inamagnetic field with an angular velocity ro. Theintensity of the magnetic field isdirectedalongthe axis of thecylinder.What should the intensity of the magnetic field be for noelectrostatic field to appear inthecylinder?3-5. Electromagnetic Induction. AlternatingCurrent568. Determine the direction of the intensity of anelectricfield inaturn placedin a magneticfield(Fig. 200) directed awayfromusina direction perpendicular tothe planeof the turn.The intensity of the magnetic field grows with time.569. Arectangular circuit ABeDmovestranslationallyin themagneticfield of a current flowing alongstraight long conductor00' (Fig. 201). Find the directionofthe current induced inthecircuit if the turn movesawayfromthe conductor.570. Anon-magnetized ironrodflies throughacoil connectedto a batteryand anammeter (Fig. 202). Drawanapproximatediagramof the changeof current inthe coil with time as therodflies through it.571. A current in a coil grows directlywithtime. What is thenature of the relation betweenthe current andtime inanothercoil inductively connectedtothe first one?Fig. 2000'B..---.OA 0aFig. 201 Fig. 202o+aB 2 EELECTRICITY ANDMAGNETISMFig. 20,1ADFig. 204tZaa F117572. Will the result of Problem571 change if an ironcore isinserted into the second coil?573. Awire ringwitha radiusr is placed intoa homogeneousmagnetic field whose intensity is perpendicular to the plane ofthe ring and changes with time according to the law H= kt.Find the intensity of the electric field in the turn574. Aringof a rectangular cross section (Fig. 203) ismadeof a material whose resistivity is p, The ring is placed ina ho-mogeneous magnetic field. The intensityof the magnetic fieldisdirected along the axis of the ringand increases directly withtime, H" = kt. Find the intensity of the current induced in thering.575. Acoil having n turns, each with an area of A, is con-nectedto a ballistic galvanometer. (Thelatter measures the quantityof electricity passing through it.) The resistance of the entirecircuit is R. First thecoil is between the poles of a magnet ina region where the magnetic field H is homogeneous and itsintensity is perpendicular to the area of theturns. Thenthecoilis placed intoa space with no .magnetic field.What quanti tyof the electricity passes through~the galvanometer? (Express the answer in cou-lombs.)576. Determine the current in the conductorsh of the circuit shown in Fig. 204 if the intensi-jty of a homogeneous magnetic field is perpen-dicular to theplaneof thedrawingand changesintime according to the lawH = kt, The resi-stance of a unit- of length ofthe conductors is r.577. The winding of a laboratory regulatingFig. 205 autotransformer iswoundaroundan ironcoreha-118Fig. 206 Fig. 207PROBLEMSving theform ofa rectangular toroid (Fig. 205). For protectionagainst eddy (Foucault) currents the core is assembledof thinironlaminas insulated from one another by a layer of varnish.This can be done in various ways:(1) byassemblingthecoreof thinrings piledon one another;(2) by rolling up a long band with a width h;(3) byassembling thecore of rectangular laminas 1X hinsize,arranging them along the radii of the cylinder. Which is the best way?578. Adirect induced current I is generatedina homogeneouscircular wirering. The variablemagneticfieldproducingthiscur-rent is perpendicular to the plane of the ring, concentratednearits axis and has an axisofsymmetry passing throughthecentreof the ring (Fig. 206).What isthe potential differencebetweenpoints Aand B? Whatis the readingof an electrometer connected to these points?579. Avariable magnetic fieldcreates a constant e.rn.f. Cina circular conductor ADBKA(seeProblem578). The resistancesof the conductors ADB, AKBand ACB(Fig. 207) are equal toR1 , R2and Ra, respectively. What current will be shown byammeter C? The magnetic field is concentratednear the axis ofthe circular conductor.580. The resistance of conductor ACB (see Problem579) isRa=O. Find the currents 11' I, and /a and the potential diffe-rence V A-VB"581. Amedical instrument usedtoextract alienparticles froman eye has theformof astrongpermanent magnet or anelectro-magnet. When brought close to the eye (without touching it) itextracts iron and steel particles (filings, chips, etc.),What current shouldflowthrough theelectromagnet toextract,without touching theeye, metal objectsmadeof non-ferromagne-tic materials (aluminium, copper, etc.)?ELECTRICITY ANDMAGNETISM119582. Awire ringsecuredontheaxis passing throughits centreandperpendicular to the forcelinesisplacedinahomogeneousmagnetic field (Fig. 208). The intensity of the field begins togrow. Find the possible positions of equilibriumof theringandshowtheposition of stableequilibrium. What willhappeniftheintensi tyof the field decreases?583. Aconductor withalengthl andmass mcanslidewith-out friction along two vertical racks AB and CDconnectedbya resistor R. Thesystemisinahomogeneousmagneticfieldwhoseintensity Hisperpendicular totheplane ofthe drawing(Fig. 209). How will themovable conductortravel inthe fieldof gravityif the resistanceof theconductor itself and the racks is negle-cted?584. A conductorwitha mass m and lengthI can move withoutfriction alongtwometallic parallel racks inahorizontal planeand connected across capacitor C. The entire systemis inahomogeneous magneticfield whose intensity H is directedupward.AforceFisapplied to the middleof the conductor perpendi-cular toit and parallel to the racks(Fig. 210).Determine the acceleration ofthe conductor if the resistanceof the racks, feedingwiresandconductor iszero. What kindsofenergywillthework of the force F be converted into?Assumethat the velocityof theconductor is zero attheinitialmoment.585. Consideringthe motionof astraight magnet inaplaneperpendicular toawire andusing the lawof conservationofenergy, provethat the fieldof a longforwardcurrent dimin-isheswith the distance fromthe wireas I/R.586. A cylindermade of anon-magneticmaterial hasNturnsof awire (solenoid) wound aroundit. TheradiusofthecylinderRA.. t-C.m.0NImHctE0NB 0Fig. 209 Fig. 210120PROBLEMSis , and its length l (r ~ I). The resistance of the wire is R.What should the voltageat theendsof thesolenoidbe for thecurrent flowing in it to increasedirectlywith time, i.e., I =kt?587. Asolenoid (see Problem586) is connected to a batterywhosee.m.I. is cC. Thekey isclosedat themomentt = O. Whatis the intensity of the current flowing through the circuit ofthe solenoid if the resistance Rof the solenoid, battery andfeeding wires is neglected?588. Calculate the work of the battery (see Problem587)during the time 1". What kindof energyis this workconvertedinto?589. Aringmade of a superconductor is placed intoa homo-geneous magnetic field whose intensitygrows fromzero to HoThe plane of the ring is perpendicular to theforce linesof thefield. Find the intensityof the inductioncurrent appearing inthe ring. The radius of the ring is , and its inductance L.590. Asuperconductive ring wi th a radius r is in. a homoge-neousmagneticfieldwithanintensity H ~ The force lines of thefield are perpendicular to the plane of the ring. There is nocurrent in the ring.Find the magnetic flux piercing the ring after the magneticfield is swi tched off.591. Find the inductance of acoil woundontothe ironcoreshown in Fig. 211. The number of turns of the coil N, thecross-sectional area A, the perimeter of the core (mediumline)I and the permeabili ty of the core l-t, are known.Note. Take into account the fact that .the intensityof themagnetic field inside the core is practically constant and canbe approximatelyexpressed by the formulaH = OAn~ I.592. Estimate approximatelythecoefficient of mutual induc-tance of the windings of a transformer. Consider the windingsFig. 211 Fig. 212ELECTRICITY ANDMAGNETISM121as coils of identical cross section. Disregard the dispersion ofthe force lines of the magnetic field.Note. Thecoefficient of mutual inductance of twocircuits istheratiobetweenthemagneticflux =k.Then,cos a+ksin acos (a-qcos cporcos a+k sin a= YI +k" cos( ~ - q FHence, k= ..r,. al_FI112. Theforces actingonthecylinder are shown in Fig. 315. Sincethecylinder does not move translationally,F1,-Fcosa=OF sin a-mg+N=OSince themaximum value of cos (a- q isunity, thenF . _ kGeun>:YI +ki0.75.MECHANICS197The forceof friction FIr =kN-Hence,F kmgcos a+k sinaThe denominator of this expressioncan be written as A sin (a+cp). whereA=Yl+k2(see ProblemIll).Therefore, the minimumforcewith which the string should be pulled IskmgFmin "Ir f 1+k2The anglealcan be found fromthe equation cosa1 +k sin al = Y1+kZ,and tana1 =k.113. The forces actingon the piston and the rear lid of the cylinder areF1=Fz=pA(Fig. 316a). Point Cof the wheel is also acted upon in a hori-zontal direction by the force F2 transmitted from the piston through thecrank gear.The sumof the moments of the forces acting on the wheel with respectto its axis is zero. (The mass ofthe wheel is neglected.) Therefore. FIrR =F2"where Ftr is the forceof friction. Since thesumof the forces which act onthe wheel is also zero, the force F3 applied to the axis by the bearings ofthe locomotive isF3 ~ FIr +F2- According to Newton's third law. the forceF.=Fsacts on the locomotive fromthe side of theaxis. Hence, the tractivereffortF=F.-FI=Fjr=pA R'In the second positionof the pistonand the crankgear, the forces we areinterested in are illustrated inFig. 316b. For thesame reason as in the pre-rvious case, Flr=F. R'Thetractiveeffort F=F.-F.=F/,=pA~ ,Ascould beexpected. the tractiveeffort is equal to the force of friction,for the latter is the only external forcethat acts on the locomotive.114. The maximumlength of the extending part of the top brick is 1/2.The centre of gravity of the two upper bricksC, isat a distanceof 1/4 fromthe edge of the second brick (Fig. 317). Therefore. the second brickmayextend by this length relative to the thirdone.The centreof gravity of the three upper bricks C3Is determined by theequality of the moments of the forces of gravity with respect to C3; namely.G( ~ -x) =2Gx. Hence x=~ , i.e. the third brick mayextend over the&- fa)Fig. 316198Fig. 31799.0Fig. 318ANSWERSAND SOLUTIONS,%,----------J(fourthoneby not morethan 1/6. Similarly it canbefound that the fourthbrickextends over the fifthoneby L/B, etc. The nature of thechange inthelength of theextending part with an increase in the number of bricks isobvious. Themaximumdistance over whichthe right-handedge of the upperbrickcanextend over the right-hand edge of the lowermost brickcanbewritten asthe seriesL=f(l+; + ~ + + + ... )When the number of the bricks is increased infinitely thissumtends toinfinity.Indeedt thesumof the series1 1 1 1 1 1 11+2+3'+4 +5+6+7+8+ ...isgreater than that of the series~ Y2,..-"'-. . . . . - - - " ' - - ~1 1 1 1 1 I 11+"2+4+4+8+8"+8+8+ ...andthe latter sumwill beinfinitelygreat if the number of terms isinfinite.Thecentreof gravityof all the bricks passes through theright-hand edgeof the lowermost brick. Equilibrium will be unstable. The givenexamplewould bepossibleifthe Earth wereflat.115. Let us inscribea right polygon into a circle witha radius, (Fig. 318).Let usthen findthe moment of the forces of gravity (with respect totheMECHANICS199axis AK) applied tothe middles ofthe sides of the polygon AB, BC,CD, DE, etc., assuming that theforce of gravityacts at right ang-lesto the drawing. Thismoment isequal to pg (ABx1+8Cx2+CDxs++DEx.+ EFx"+FKxe),where pis the mass of 8 unit ofthe wirelength.o Fromthe similarity of the cor-F. 319 respondingtrianglesit canbe shown~ g . that theproductsABx1' BCx,. CDxa.etc., are equal respectivelyto AB'h,B'C' h, C' D' h, etc., whereh is the apothemof thepolygon.Therefore, the moment isequal topgh(AB' +B'C' +C'D' +D'E' +E'F' +F'K)=pgh2rIf the number ofsides increasesinfinitely, the value of hwill tend torand the moment to2r2pg.Onthe other hand, the moment is equal to theproduct of the weight of the wirenrpgand the distance x fromthe centreofgravitytoaxis AK. Therefore, 2r2pg=itrpgx,whence x=.! r,n116. Let us divide the semicircle into triangles and segments, as showninFig. 319. Thecentre of gravityof a triangle, as is known, isat the pointof intersection of its medians. In our case the centre of gravity ofeachtriangle is at a distance of ~ hfrompoint 0 (hIs theapothem). When thesides are increasedinfinitely innumber the centres of gravityof thetriangleswill lieonacircle with aradius of f r, while the areas of the segmentswi11 tend to zero.Thus, the problemconsists indetermining the centre of gravityof ase..micircle with aradius of ~ r,It followsfromthe solutionof Problem115 that x, which isthe distancebetweenthe centre of gravity of the semicircle and point 0, isequal to2 2 4x="1t3r= 3n ' 117. By applyingthe method usedtosolve Problems 115and 116, it canbe shown that the centre of gravity is at point C at a distance2 ex81n"2CO= r fromthe centre of curvature of the arc (see Fig. 45).alt8. Fromthesolutions of Problems 115, 116 and 117, It canbe shown ex4810'2that the centre of gravity isat point C at adistanceof CO==3-a- rfrompoint O.119. When the centre of gravity isdetermined, the plate with a cut-outportion canformally beconsidered asasolid one if weconsider that a se-200ANSWERSANDSOLUTIONSrnicircle with a negative mass equal inmagnitude to the massofthecut..outportion Is superposedonit.The moment of the gravity forces of the positiveandnegativemasseswithrespect to axis AB is equal to(r nr'4) 1pg 2rB----r =-r3pg2 2 3n 3if the force of gravity acts at right angles to the drawing (seeFig. 47),where p is the mass of a unit area of the plate (see thesolutionto Pro-blem 116). On the other hand, this moment is equal to the product of theweight of the plate and the distance x=OC from its centre of gravity toaxis AB.(nrl) 1Hence, xPi 2r2-2=3" r3pg.2Therefore, %=3 (4-n) r.1-6. Work and Energy120. The work of a force does not depend on themass of the body actedupon by the given force. A force of 3 kgf will perform the workW=Fh= 15 kgl-m. This work is used to increase the potential energy(5kgf-m) andthe kinetic energy (10 kgf-m) of the load.121. k=O.098J/kgfcm.122. First let us find the force with which the air presses on one of thehemispheres. Assume that its base is coveredwith a flat lid in the formofadisk with a radius R. If the air is pumped out from this vessel the forceof pressure on the flat cover will be PI =pA=prtR2. Obviously, thesamepressurewill beexerted on the hemisphere by the airt otherwise the forceswill not be mutually balanced and the vessel will perpetually move in thedirection of the greater force. The number of horses should be FI/F, sincethe other hemisphere maysimply be tied to a post. The tensioned rope willproduce the same force as the teamof horses pullingat the other side.123. The change in the momentumof a body Is equal to the impulse ofthe force of gravity. Sincethe forces actingon the stone and the Earth arethesame and act during the same time, the changes in the momenta ofthese bodies are also thesame.The change in the kinetic energy of a body is equal to the work of the. forces of gravitational attraction. The forces are equal, but thepaths traver-sed by the stone and the Earth are inversely proportional to their masses.This is why the law of conservation of energy may be written in a formwhichdisregards the change in the kinetic energy of theEarth: mt +Ep=const.wheremis themass ofthe stone and Ep the potential energyof interaction.124. Accordingto the lawof conservationof energy,m l v ~mlgh=- 2-whereml is the mass of the pile driver, h the height from which it drops,and VI its velocity before the impact.MECHANICS201Sincethe Impact isInstantaneous, theforce of resistance cannot appreciablychangethe total momentum of thesystem.Seeingthat the Impact isinelasticmlvl =(ml +m.) v.where m2 isthemass ofthepile, thevelocityofthe driver andthepileat thefirstmoment after the impact.Themechanicalenergyofthe driver andthe pile isspent onworkdonetoovercome theresistanceofthesoil F:(m+m)vl122I+(ml+m2)gs=Fswhere s isthedepthwhichthepile is driventointothesoil.Hence,ml hF= + - mtg+mlR+m2g=32,500 kg!ml m2 s125. As aresult ofthe inelastic impact, the linear velocityof theboxwiththe bullet at thefirstmoment will beequal to u=M where uisthe velocityofthe bullet. Onthebasisof the law of conservation of energy,the angleof deflection a isrelated tothevelocityvbytheexpression.(M+m) ulm2v2(M+m) L (I-cos a)g2 2(M+m)whence. aM+m.. r-v=2stn2"-my Lg126. Since the explosionisinstantaneous, theexternal horizontal forces(forcesoffriction) cannot appreciably change the total momentum ofthesystemduringtheexplosion. This momentumis zero both before anddirectlyaftertheexplosion.Therefore, mtVt +m2v2 =0.Hence, VI =_m2 V2 mlSince thecarts finallystop, their initial kineticenergies are spent on workagainst the forces of friction:m2v:-2-=kmtgsl' and vr 51Hence, .=-,and, therefore, s2=2 metres.VI 81127. Let usdenote thespeed of thebodyand the wagon after theystopmovingwithrespect toeachother byu: Accordingto the lawof conserve-tionofmomentum,(M +m) U= Mvo (1)202ANSWERSANDSOLUTIONSThe wagon loses its kinetic energy in viewof the fact that the force offriction f actlngon it performs negative work 2 2-where S is the distance travelled by the wagon.The body acquires kinetic energy because the force of friction actingonitperforms positiveworkmul2=lsHere s is the distance travelled by the body.It iseasy to see that the change inthe kinetic energy of the system _ [MU2+ mU2]=/ (S-s) (2)2 2 2is equal to the force of friction multiplied by the motionof the body rela-ttve to the cart.It follows fromequations (1) and (2) that S-s=2f(M+m) SinceS-sI. then1 2/ (M+m)Bearing inmind that f=kmg. wehave l;a.128. The combustion of the second portion increases the velocity of therocket vby Av. Since combustion is instantaneous, then accordingtothe lawof conservation of momentum,(M+m) v=M (v+t\v)+m (v-u)where mis the mass of a fuel portion, M the mass of the rocket withoutfuel, u the outflow velocity of the gases relative to the rocket.The velocity increment of the rocket Av= Zu does not depend on theveloci ty vbefore the second portion of the fuel burns. On the contrary, theincrement in the kinetic energy of the rocket (without fuel)se, M(vtliV)1wi11 be the greater, the higher isv.The maximumaltitudeoftherocket isdeterminedbythe energyit receives.For this reason the second portion of the fuel can be burnt to the greatestadvantage when the rocket attains its maximumvelocity. i. e. directly afterthe first portion is ejected. Here the greatest part of the mechanical energyproduced by the combustion of the fuel will be impartedtotherocket, whilethe mechanical energy of the combustion products will be minimum.MECHANICS203129. It will be sufficient toconsider theconsecutive combus-tion of twoportions of fuel. Letthe mass of the rocket with thefuel first beequal to M +2m.dh After combustion of the firstportion, the velocityof therock-mUl .my et v=M+m.where Ul IS thevelocity of thegases with respectto the rocket. The initial velo-,,11 city of the rocket isassumed tobezero.Fig. 320 The increment inthe velocityof the rocket after the secondportionburns is =m;;2 t where U2 is the new velocity of the gases withrespect tothe rocket.Combustionof the first portion produces the mechanical energy dEl =(M +m)tJ2 mu2= 2 +T t and of thesecondportiontheenergy(M +m) v22According to the initial condition,=Hence,2 ! m2m \ _ 2 (m2m )Ul \ 2 (M+m) +2) - U2 2M + "2Therefore, Ut >u2 ; the velocity of the gases with respect to the rocket di-minishes because the mass of the rocket decreasesas the fuel burns.130. Both. slopes may be broken into any arbitrary number of small in-clined planes with various angles of inclination. Let us consider one ofthem(Fig. 320).The work done to lift the body up such an inclined plane is equal tothe work against the forcesof gravitymgplus the work against theforcesof friction F/,lBut FIr = kmg cosex and =_0__ . Therefore, FJrt1s= kmg/1l. Thetotalcosex.work L\W = mg (L\h+k81). If weconsider all the incl ined surfaces and sumup the elementary works, the total workwill beW = =mg8l)=mgh+kmglThe workisdetermined onlybythe height ofthe mountain h andthe length lof its foot.131. The forceappl ied to the handle will beminimumif it forms arightangle with the handle. Denotingthe forcesought by F, we shall have fromthe golden rule of mechanics: 2nRF=Gh. Hence,204ANSWERSANDSOLUTIONS132. According to the definition, the efficiency f) WI'.;W2 ,where WI == GHis the workdoneto lift theload G toaheight H, andW2 is thedoneagainst the forces offriction. Since the force of friction is capable ofholdingthe load Inequilibrium, the workof thisforce cannot be lessthantheworkWI. Theminimumwork oftheforces of frictionis WI = WI- There-fore, rr50%.'133. As the manclimbs the laddertheballoon willdescend byaheight h.Therefore, the workdone bythe manwill bespent to increase his potentialenergybythe amount mg (I-h)and that of theballoonbymgh (theballoonwithout theman will be acteduponbytheliftingforce mgdirected upwards).Hence,W=mg(l-h)+mgh=mglThis result canbeobtainedat onceincalculatingthework done bythe maninthe systemrelated to the ladder.If the manclimbs with a velocityv with respect tothe ladder, hemovesat V-VI with respect to the Earth, where VI isthevelocityofthe ascendingballoon.Accordingtothe lawofconservation of momentum, (v-v.)m= Mv.,Hence,m1= M+mU134. To deliver twice asmuch water ina unit of time, avelocity twotimes greater should be impartedto the double mass ofthe water. Theworkof themotor isspent to impart akinetic energy m;2 to the water. Therefore,the power ofthe motor should beincreasedeight times.135. (1) Theworkdonetoraise thewater out of the pit isH 3 3Wi=pg2 AX4'lf=gpgAH2wherepis the density ofwater.Thework1 HW.=2"P 2Av2Is required to impart kinetic energytothe water.Thevelocity (I with which the water flows out of the pipe onto theground canbefoundfromthe ratioA= nR2ut.The tot al workisw-! 2 1 H8A3- 8 pg AH+ 16Pn2R'T2(2)In the second casethe work required toraise thewater is less thanWI byTheworkrequired to impart kineticenergyMECHANICSto the water is205FFig. 321The total work 136. It isthe simplest to solve the probleminacoordinate systemrela..ted to the escalator. Theman will walk a distance l=_.h_+ut relativesm ato the escalator, where ur isthedistance covered bytheescalator. He shouldperformthe work W=(-.h_+v-r)mgsincx, since during the ascent thesinexforce mgis applied over the distance1 andformsanangleof 900- awithit.Part of the work mgh isspent to increase the potential energyof theman, andthe remainder mgucsin a. together with the work of the motorthat drives the escalator isspent toovercomethe forcesof friction.137. The relation betweenthe elastic force and the deformation isshowninFig. 321. The workdoneto stretch (or compress) the spring by asmallamount 8Xis shown bytheareaof thehatched rectangleThe totalwork in stretching or compressingthe spring by the amount I, equal toitspotential energy Epis shownby the areaof triangleOBC: Let us recall that anexpression forthe distance coveredin uniformly ac-celerated motion can usually beobtained bysimilar reasoning.138. The man actingwithforce Fon thespringdoes theworkWI =- FL.At the same time, the floorofthe railway carriage isacted uponfromtheside of the manby the force of friction F. Theworkof this forceW2=FL.Therefore, the total work performed by the man in a coordinate systemrelatedtothe Earth iszero, inthe sameway as in a system relatedto the train.139. In the system of the train the workdoneisequal tothe potentialklienergyof the stretched spring (see Problem137) W= T' sincetheforce offriction between the man and the floorofthe railwaycarriage doesnot doworkinthis system.In thesystem relatedto the Earth,the workthe man does tostretchthespring is equal tothe productof themean force and thedistan-ce L-l,i. e., W(L-l). Themanacts onthe floor ofthecarria-ge withthesame mean force. Theklworkof this force W2 =2' L. Theo ....IC total workinthe given coordinatekltsystem W=W1+W2= T is thesame as in thesystemorthe carriage.206ANSWERSANDSOLUTiONS140. On the basis of the lawsofconservation ofmomentumandenergywecanwrite the followingequations:mtvt+ 2 2 '2 '2mtVl + m 2V2= m1Vl +m2V22 2 2 2where and v; arethe velocities of thespheres after the collision.Uponsolving thesesimultaneousequations, weobtain, (mt-m2)vl+2m2v2. d ,_(m2-mt)v2+2mtVlVt ,an V2 - -------..ml+m2mt+m2(1) If the secondspherewasat rest beforethe collision (v2=0), then(m1- m2) VI ' 2m1V1v. V2m1+m2' mt +m2If ml >m2' the firstspherecontinues to move in thesamedirectionasbeforethe collision, but at a lower velocity.If m1 ex ' 0sin2a.2gsin 2aXo= 2g164. The trajectoryof the ball takes the form of aparabolapassing througha point with the coordinates hands. Therefore (seethe solution to Pro--blem163),. h=g s2+tao as cos2aHence,gs2 n-2cos> a(tan as -h) (s sin 2a.- h cos 2a)-hy 52 +h' sin (2a-q-hwhere tan cp=h/s. The minimumvelocityVgs2 Vtlo= V = g (h+ Vhl+SI)s2+h'l.-h220" ...--", ,ANSWERSANDSOLUTIONSFig. 837Fig. 338Therefore,isattained whenm 1t h+Ys2+h2sa=..I2 +-4=arctan s arctanh2+ s2-h(Fig. 337.)165. Thecoordinatesandthe velocities of the body at any moment oftime withrespect tothereadingsystemshownin Fig. 338 are determinedbythesameequations as in Problem162.Atthemoment whenthe bodyfallsintothe water its coordinate y=- H.For this reason the durationofflight T canbefoundfrom theequationgT2-H=vosin aT-THence,Vo sin a Vsin2a.+2gHT=gSinceT > Ot we shall retain the plussign. Thedistancefrom thebank issa 2VoSID a.+vocosaV sa I +2HL=vo cos a.T 2g --g- Vosin a gThebody will beat a height habovethe water after thetimev sina Vv2sin2a+2g(H-h)'t=_gIf I h I < I H I, onlythe plussignhas a physical meaning. Whenhn,bothsolutionshaveameaning. Duringitsmotion the bodywill betwiceatthesameheight abovethe water.It isthesimplest tofindthe final velocity vwith theaidof the lawofconservationofenergy2 2mvo+ H mv2 mg=Tv=Vv:+2gHMECHANICS Fig. 339221166. Inthereading system depicted" in Fig. 339, the coordinatesof thestonearedeterminedat anymoment oftime by the followingequations:X=Vocos aigt2y=ho+vosin aJ-TAt the momentwhenthestonefalls, y=OandX=S, wheresisthedistancecoveredbythestone.Uponsolving theseequations withrespecttotheanglea, weobtaintana= (I V1+2;hO_g;:3)gs 0 0Thisexpression hasameaningwhen1+2gho _g2s2 0 Vo Vv2+2ghoPoVHence, 50 Itsmaximumvalue is smaxg gWhens is smaller, twovaluesofthe angleacorrespondto each of its va-lues, the difference between which isthe less, the nearersis toits maximumvalue.Fig. 340222ANSWERSANDSOLUTIONSTherefore, forthemaximumdistanceof Oight,v: 00 1tan 1%= V ,r-;and 1%=30"gSm"f 3187. Thecomponents ofthe velocitiesof the bodies alongxand y at anymoment oftime aredetermined as011/=VOsin al- gt ; (121/=00 sin -gt;vtx=uo cos at; and Vlx=- Vo cos a2Let ubethe velocityofthesecondbodywith respect tothe first one. Henceu,=oo sin al-gt-vosin CXs+gt=vo (sin at-sin ( 2)uJ'=vo (cos at +COSCXs)Therefore, the velocityuisequal tou=V(1%Itl%l)(/0Thebodiesmovewith respect toeachother at aconstant velocity. After thetimel' the distance betweenthemwill bes=2vocos(1%1 tlXs) "C168. Thehorizontal path of the bomb s=Yl2_h2= v cos at, where t isthe durationof fallingof the bomb. The vertical path is h=v sinI%t +(Fig. 340).Upon excludingthe time fromtheseequations, wefindtan a=- Vi .. I(VI)2+2hvt-1gs JI gs gslThe solutionwith the plussignhasameaning. The minus signcorrespondstoex < 0, i.e., the bombisdroppedwhen the dive-bomber is flying upward.169. It will beconvenient tosolvethis probleminareading systemrelatedtothe uniformlymovingvehicles.Inthis system, the highway moves backwith the speedof 0=50km/h,the vehiclesareat rest withrespect toeachother and their wheels rotate.Thelinearspeedofpointsonthe circumference ofthe wheeland tha t of thestuckstone IS alsov. Thestonewill fly the maximumdistance if it flies outwhenits speed forms anangleof450with the horizon. Let us find thisdistance. Neglectingthe fact that thestone is somewhat above the level of2 2 Ithehighwaywhenit isthrownout, weobtain 1 v sin a v =19.6met.g gres. Theminimumdistancebetweenthe vehicles should be19.6 metres.170. It will bemucheasiertosolvethe problemif theaxes of coordinatesaredirectedalongthe inclinedplaneand perpendicular toit (Fig. 341).Inthis casethe components ofthe acceleration of the ball on the axesxandy will beequal to as=g.=g sinex and a1l= g J/= - gcosa.

Fig. 341223Fig. 342Upon the first impact with the inclined plane, the velocity of the ball willbeVo= Y2gh. The initial velocity of the ball after the first impact is Voand formsan anglea with the y-axis (Fig. 341).The distance between the points of the first and second impacts is. t211 =vosinat) + g a ,where it is the duration of flight and is determi-nedby the equationgcos2oHence, i1=2voand i1=8h sin a. The velocity of the ball at the secondgimpact can be found fromthe equationsv)x=vOX+aXil=vo sin a+g sin at l =3vo sin aVly=Voy + aytl =00 cos a,- g cos atl=- Vocos a.After the impact these velocities are equal tov 2X=vt x , andV2y=-v1YThe distance between the points of the second and third impacts is equal to. g sin 12=300 sin at2+ 2where 12 .is the time during which the ball is in flight. Since the initialvelocity alongthe y' axis is thesame as during the first impact, 12= / }.Therefore, 12= 16hsin a.Similarly, it can be shown that the distance between the next pointsla=24h sina.Consequently. 11:l2 : [3 =1:2:3, etc.224ANSWERSANDSOLUTIONS171. Themotionof the bodycanbeconsidered assuperrosftionofmove-ment alongacircumference With aradius Rinahorizonta planeandverti-cal falling. Accordingly, the velocity ofthe body v at the given momentcanberepresentedasthe geometrical sumof two components: VI =v cos adirected horizontallyand v2=v sin a. directed vertically (Fig. 342). Hereais the angle formedbythe helical lineof the groove with the horizon.Incurvilinear motion the acceleration of a bodyisequal to the geomet-rical sumof thetangenti al andnormalaccelerations. Thenormal accelerationthat corresponds tomovement along the circumference isv2cost exal n = 7[ = RThevertical motion isrectilinear, and thereforea2n =O.Thesought acceleration a = V+a:'t +where a1tand a2't are thetangential accelerations that correspondto motion along the circumferenceandalong the vertical. The total tangential acceleration Q't Is obviouslyequal to a't=V..Thevalue ofa't canbefoundby mentally developIng the surface of thecylinder with the helical grooveinto aplane. Inthis casethe groove willbecome aninclined planewith B height nhanda length ofIts base 2nRn.hApparently, sin a.=g y .h2+4n2R2Todetermine DIn' let usfind v fromthe lawof conservation of energy:mv22 Bn"nhgR .T=mghn. Consequently, () =2ghnand a1nh2+4n2R" Upon InsertingFig. 343MECHANICS225the found accelerations a'f and alninto theexpressionfor the sought accele-ration, wegetghYh?+4nIRI + 64n4lnIRsa= h2+ 4n2RI172. As usual, the motion of the ball can be considered as the' resultof summation of vertical (uniformlyaccelerated) andhorizontal (uniform)mo-tions.The simplest method of solution is to plot a diagram showing howthecoordinates of the ball alongthe horizontal depend on the time forthe limi-ting velocities 267cmls and 200cmls (Fig. 343). The lower broken line cor-responds to the maximum velocity and the upper one to the minimumvelocity. In the course of time, as can beseen from the diagram, the inde..finiteness of the ball coordinatexshown by the section of the horizontalstraight line between the linesofthediagramincreases. Thevertical hatchingin .Fig. 343shows the movement ofthe ball fromMtoN, andthehorizontalhatching-fromNtoM. The cross..hatched areas correspond to indefinitenessin the direction of the horizontal velocity.(I) The diagram shows that after the ball bounces once fromslab Nthedirectionof itshorizontal velocity will be indefinite when the durationoffalling OK< t ~ OL or t > AB(whereOK=0.15 s, OL=O.2 sandAB=0.225 s).gt2Hence, 10 em (2k- 1)q>-;- >001 > --2-:r-MECHANICS227(2)Sincethe front andrear wheels have thesame number of spokes, the wheelswill seem torevolvecounterclockwise if thespeedof thecart iskepr kfPr cpr-:r>v>-:r-2't (1)kcpR > v > kcpR _fPR't 't' 2-rSince R=1.5r, thesecondinequality canberewritten as foHows:I 5 kcpr 1.5kfPr I.SepT T > v > - ' t - - ~BothInequalities, whicharecongruent onlywhen k= 1, givethepermissiblespeeds of thecart inthe formcpr > v > 0.75q>r't 't2n .Or, since CP=6' wehave8.8m/s> v > 6.6 m/s.(2) Thespokes of therearwheel will seemtorevolveclockwise if duringthetime r the wheel turns throughtheangle ~ 2 whichsatisfies the condition(2k-l) : > Pi > (k- 1) cp (Fig. 345b). Hence, thefollowing inequalityis trueforthespeedof thecart:1 5 (2k-l)cpt 1.5 (k-l)cpr 2't >v> 'tAt thesametime inequality(1) shouldbecompliedwith. When k= 1, bothinequalities arecongruent if 0.75 cpr > v > 0.5 fPr . When k=2theyare can-'t 't'gruent if 2cpr > v >1.5cpr If k > 2 the inequalities are incongruent.T "F'II15 *D'O(h)Fig..845228ANSWERSANDSOLUTIONS6.6 m/s > v > 4.4mls17.6 mls >t1 >14.2mlsor176. The instantaneous axisof rotation(see Probtern173) passes through point C(Fig. 346). For this reason the velocityof point A relative to the block isR+,VA=V-,-.Point Bhas~ Therefore,_.......~ ~ - - ~Fig. 346R-rthe velocity vB=v --.rPoints onacircle withtheradius r whosecentre is point Chave an instantaneousvelocity equal tothat of the spool core.177. The trajectories of points A, Bandcare shownin Fig. 347. Point Bdescri-bes a curve usually called anordinary cycloid. Points AandCdescribeanelongated and a shortenedcycloids.178. The linear velocity of points on the circumference of the shaftUl=OO ~ and that of points onthe raceu2= Q~ . Since the balIs do notslip, the same instantaneous velocities will be imparted tothepoints ontheball bearing that at this moment areincontact with theshaft and therace.The instantaneous velocity of any point onthe ball can be regarded as thesumof two velocities: the velocity ofmotionof its centre Vo and the linearvelocityof rotationaround the centre. Theball will rotate withacertainangular velocity (00 (Fig. 348).Therefore,Hence,VI =vo-wo'V2=VO+OOO'1 1vO=T (vt +v2)=4 ( 2mg (Fig. 356). Therefore, thesystemwill not be in equilibrium. The right-hand weight will have a grea-ter pull.193. The direction of the acceleration coincides withthat of the resultantforce. The acceleration is directed downward when the ball is in its twoextreme upper positions Band C (Fig. 357). The accelerationwill be di-rected upward if the ball is in its extremebottom position Aand horizon-tally in positions Dand Ldetermined by the angle cx.Let us find ex. According to Newton's second law, the product of themass and thecentripetal acceleration isequal to the sum of the projectionsof the forces on the direction of the radius of rotation:mv2-1- =T - mg cosexOn the other hand, ascan beseenfromFig. 357, we have T=mg . Oncosexthe basis of the lawof conservation of energy:molT=mgl cos a,1We can find from these equations that cosa= ya' and thereforeex~ 54045'.194. Let us denote the angular velocity of therod by co at the momentwhen it passes through the vertical position. In conformity with the lawofconservationofenergy:co2t 22(miT!+m,T.)=g (I-cos a) (mITt +m2's)orMECHANICS235Fig. 357whenceFig. 358Vl=ror1=2rt 2 V m1' 2 +m2' S. a. ml'l +m2'2vs=r2=2r2sln- g 2 22 mt'2+m2'2195. The resultant of the forcesapplied tothe ball F=mg tan a. shouldbuild up a centripetal acceleration a=oo2r, where r=1 sin a. (Fig. 358).Hence,mgtan a-=moo2l sin(IThis equationhas two solutions:CXt=Ocx! =arccos IBoth solutions arevalid inthe second case: a1=0(herethe ball isinastate of unstable equilibrium) and=600Inthe first case theonlysolution is al =0.198. Let us resolve the force Facting fromthe side of the rod on theweight mintomutually perpendicular componentsT and N(Fig. 359).Let us project the forces ontoavertical and a horizontal lines and writeNewton's equations forthese directionsmooll sin cp=T sin cp- N cos q>mg=T cos cp+N sin,Let us determine T andNfrom these equationsT=m(CJ)11 sintcp+g coscp)N=m19-CJ)21coscp) sincp236Ho0'AANSWERSANDSOLUTIONSa0'Fig. 859Therefore.Fig. 360F =YT2+N2=m y g2+ro4l2sin2cp197. The forces actingonthe bead are shown inFig. 360: f is the forceof friction, mgthe weight and Nthe forceof the normal reaction.Newton's equations for the projection of the forceson a horizontal and avertical directions will have the formf sin q> 1=N cosq>=mro2l sinq>f cos q> N sinq>-mg=OThe upper sign refers to the caseshown in Fig. 360and the lower one tothe case when the force Nacts in the opposite direction. We findfromtheseequations thatf=mro2l sin2cp+mg cos q>N= (mg sinq> - mro2l sincp cos qInequilibriumf ~ kNork sincp-coscp g1 cos cP sinq> (J)andk sin gI ~ . (k .) 2 when k ~ tanq>sinq> cos cp-slnq> co198. Figure 361shows the forcesactingon the weights. Here T1 and T2are the tensions of the string. Let us write Newton's equations for the pro-j ectlons ontoa horizontal and a vertical directions.MECHANICS 237For the first weightTIsin q>-T2 sin 'i'=mooll sin,T1 cos q> - Tt COS'\1'-mg=O (1)For the second weightmro21(sinq>+ sin 'i')=T2 sin", (2)T2 cos"i'=mgUpon excluding T1 and T2 fromthe systemof equations (I) and (2), weobtain the equationsa sinq>=2 tan cp- tan",a (sin q>+sin "p) =tan",002lwhere a=-'gFromthese equationswe get 2 tan q>- tan", < tan'Pand, therefore, q> =T1 sin q>-NIcos +N1 sinq>=mgMw2(b+a) sin q>=T2 sincp+N2 cos q>; T2 cos cp-N2 sin cp=MgUpon excludingtheunknownquantities T 1. T2' Nt and N2fromthe system,We find thatg mb+M (a+b)(1) qJ=Ot and (2)cosq>= 002 mb2+M (a+b)2oFig. 361 Fig. 3620'238ANSWERSANDSOLUTIONSThe first solution istrue forany angularvelocities of rotation,and the second when 0) ~.. / mb+M (a+b) (see theV g mb2+M(a+b)2solution toProblem195).200. In the state ofequilib-riummro2x=kx, wherex isthedistance from the body totheaxis.It isthus obvious that withanyvalueof x the spring impartsthe centripetal acceleration ne-Fig. 963 cessaryfor rotation to the body.For this reasonthe latter willmove after the impetus with a constant velocity up to stop A or aslong asthelaw of proportionalitybetweenthe forceacting onthe springandits deformationis valid.201. Let us write Newton's secondlawfor asmall portion of the chainhaving the mass ~ R Aaand showninFig. 363:7R Aa(2nn)'R=2Tsin A;Sincethe angle Aa issmall, sin A; ~ A; , whence T=mln'=9.2kgf.202. Let us take a small element of the tube with the length R Aa(Fig. 364). The stretchedwalls of thetube impart an acceleration a=~ to thewaterflowing along thiselement. Accordingto Newton'sthird law, tne waterwill act onthe element ofthe tube with the force~ ~Yo--:-rn;MECHANICS243If Vo < CiloT, the hoop will stop at themoment of time 't'=m;o when rotat-Voing with the angular velocity 00=(00--. Then the hoop begins to moverwith slipping in the reverse direction. In a certain time the hoop will stopslippingand will roll without slippingtothe left withatranslational velocitywor-vov 2 (seeProblem213).IfVo > CiloT, thenafter thetime't'= m ~ C i l o elapses thehoopwill stop rotatingand will move to the right with a translational velocity v=vo-rroo. Nextthe hoop will rotate in the reverse direction, and in some time it will rollwithout slippingto the right. Its angular velocity will bevo-rwo00= 2rPractice shows that the loop will also be braked when it does not sli p,We did not get such a result since the specificrollingfrictionwas neglected.215. Since the hoops donot slip, Vo (velocity of thecentre of gravity ofthe hoops) andv (velocityofthe weight) are related by the expressionRvo=v R-rAssume that the weight lowers through the distanceh. If the systemwasatrest at the initial moment. fromthe lawof conservation ofenergy we havemv22mgh=-2-+Mvo(seeProblem207).This expressioncan be used to findthe velocit y of the weight:V2mghv= R 2m+2M (R-r)Hence, the accelerationof the weight isa mgm+2M ( RRT rThe weight lowers with the accelerationaunder the actionof the forceofgravitymgand the tensionTof the string.The sought tension T is16 *T=m(g-a)2mMg(h)'m+2M (RRr r244ANSWERSANDSOLUTIONSSince thecentreofgravity ofthe hoopmoves withanacceleration equal toa -RR under the action of theforceTandtheforce of frictionF, Newton's-rsecondlawgivesusthe followingequationforthe forceFRF=T-Ma--R-rorFMmg(h Y(2- ) Mmg('+i)m+2M( RRr r m ( 1-; r+2MThe friction force ofrest cannot exceed the value kMg. For this reasonslipping occurs whenMmg(hr(1+N--)(R )2 > kMgm+2M -R--ror

l+!..Rr216. The centreofgravity ofthespool will not bedisplacedif the ten-sion ofthe threadsatisfiestheequationT=Mgsina.Let usfind theaccelerationof theweight withamass mto determinethetension ofthethread T. Let the weight lower througha distanceh. Since,accordingtothe initial condition, thecentreofgravity of thespoolshouldremainat rest, the changeinthe potential energyis equal tomgh. Ifvisthevelocitywithwhichtheweight havinga mass mmoves, thevelocityofthepoints onthespool at the distanceRfrom the axis of rotation isvR .rHence, the kineticenergyofthesystemismv" Mv2RIE"=T+-2-fiIt follows fromthe lawofconservationofenergythat(m+M v; ==mghorV2mghv= R2m+M-rlMECHANICS245Therefore, theaccelerationoftheweight isa= mgRIm+M-;rIfwe know theacceleration 0that(AI +A2)2 >4"'1AIHence,At+A22AIA2 Il 'l-2- >Al +12 ,I.e., "'I >"'II313. Thequantity ofheat suppliedbytheheater into thewater throughthe panbottomisAQ=7f(T-Tt) A=mrwhere T1 istheboilingpoint ofthe water and, isthespecificheat ofva-porization.Therefore,23. Prope'rtiesof Gases314. Theremovable capactsasa pump and under it ararefied spaceisformed that sucks out theink. The orificeserves to maintainaconstant pres-sureunderthecap.315. Assuming that thetemperatureremains constant, let usapply Boyle'slawtothe volume ofair abovethemercury:(POI-PI) (1-748mm)=(P02-P2) (1-736mm)whence1=764mm.316. Inaposition of eqnilibrium f-6--F=O, where f isthe force ofexpulsion equal toyh1A (here y isthespecificweight ofthewatert andhlthe height oftheair columninthe tubeaftersubmergence). Inour case theforce of expulsionIs built up by the differenceof pressureon the solderedendof thetube frombelowandlromabove:f=ptA-(lJo+'Vh)A, wherePiis theair pressure inthe tubeafter submergence. AccordingtoBoyle'slaw,PolA=PlhiA.It follows fromthis systemofequationsthatE=: [y (po+yh)2+4Poyl.:.....(po+yh)-G=8.65gf18-2042274PA ---------------------BANSWERSANDSOLUTIONS1/ Fig. 395317. First. the pressurepofthe air will decrease approximately isother-mally owingtothe dropof the level of the water inthe vessel. This willcontinue until the total pressure at the level of the lower endofthe tubebecomes equal tothe atmospheric pressurePo; i, e., p+pgh=Po. wherehisthe height ofthe water column in the vessel above the level ofthe lowerendof the tube. Fromthis moment onair bubbleswill begin to pass into thevessel. The pressure at the level ofthe lower endofthe tubewill remainequal tothe atmospheric pressure, while the air pressure P=Po-pghwillgrowlinearly with adropinthewater level. Thewater will flowout fromthe vesselat aconstant velocity.Therelation betweenp and QisshowninFig. 395. The negligible fluctu-ations of pressure whenseparate bubbles passinarenot showninFig. 395.318. Whenthe air isbeingpumpedout ofthe vessel, the pressure in thevessel after onedoublestrokewill become equal toPt After theo V)2seconddoublestrokePtV=pz(V+00)and, consequently, Pz=Po (v +00 etc.After n doublestrokesthe pressure inthe vessel will bep'=Po (V:vJ"When air is being delivered into the vessel. after n doublestrokesthepressure will be_ ,+Ponvo_{( V )n+nuo}p-p ---Po -- -V V+vo VHerep > Po at anyn, sinceduring deliverythepumpduring eachdoublestrokesucksinair with a pressure(Jo, and duringevacuation (10 of the air ispumpedout at pressures belowPo.319. ApplyingBoyle's lawtothetwo volumes ofgasinthe closedtube,we obtainL-I (L-l )P-2- A=Pl -2--A1AL-l (L-l)p -2-A=tJ2 APl=P2+VlHEAT. MOLECULARPHYSICS275Here p isthe pressurein the tube placed horizontally, Pl and P2 are thepressuresin the lower and upper endsof the tube placed vertically whenitsends areclosed, y is the specificweight of the mercury, Aisthe cross-sectio-nal areaofthe tube.Hence, the initial pressureinthe tube isp=y.!- (l!. _&1)2 III 10Here. forthe sake of brevity we have designated L 21 by 10,If oneendof the horizontal tube isopened', the pressureofthe gas Jn thetube will becomeequal tothe atmospheric pressure.According toBoyle's law. ploA =yH11 A (hereH isthe atmospheric pres-sure), and, therefore, 1- 2H &l 10The mercurycolumn will shift through the distanceL\11=lo- lt =l!JL[2H 2H I L\l 10For the mercurynot to flow out of the tube, the followingcondition isrequired V\Vhenthe upper end of the vertical tube isopenedploA ="1 (H +1)l2AHence,tlI2= 10-122(:+1)[27- ( - +2]The mercurywill not flowout ofthe tube. if

Whenthe lowerend isopenedploA =Y (H-I) 13AwhenceM3= 10- 13= 2 [27- ( --2]The followingconditionshould be satisfied to prevent the mercury columnfrombeing forcedout of the tube 1//-4 (H -1)2 1+2 (H -I)t1lr 12+ 118 *.276ANSWERSANDSOLUTIONS320. Since for one gramme-moleculeof any. gas at P= I atm and T =2370Kwehave V,.=22.4lltres, then for one moleThis constant is usually denoted byRandcalled the universal gas constant.The values of Rin the various systems of units are:R=0.848kgl-rn/mole- deg=8.3X 107erg/mole- deg= 1.986cal/rnole-deg321. At a fixed pressure and temperature the volume occupied by the gasis proportional to its mass. Avolume ofVfA- corresponds to one gramme-mo-lecule anda volume of V to an arbitrary mass m. Obviously. VIi =Vwhere I..tis the molecular weight expressed in grammes.Upon insertingthis expression into the equation of state for onegramme-molecule. we havepV=!!!...RTJ.I.

....

JrFig. 397/1."'

--.:.---.-.,IFig. 396322. If the attraction between the molecules suddenly disappeared. thepressure shou ld increase. Toprove this, let us mentallysingleout two layersJ and IJ inside a fluid (Fig. 396). The molecules penetrating fromlayer Iinto layer I J owingto thermal motioncollidewith the molecules in layerII, and as a result this layer is acted upon by the pressure forcesPt thatdepend on the temperature. The forcesof attractionact on layer I I fromtheside of the molecules in layer I in the oppositedirection. Theresultant pres-sure of layer I on layer I J P=Pt- Pi, where Pi is the pressure'caused bythe internal forces of attraction. WhenPIdisappears, the pressuregrows.323. If the forces of attraction betweenthe molecules disappeared the water wouldbeconverted into an ideal gas. The pressurecan befound fromthe equation of stateofan idea 1 gas:mRT 1,370atmHEAT. MOLECULARPHYSICS271324. Let us separate a cylindrical volume of the gasin direct contactwith the wall (Fig. 397). Theforces actingontheside surfaceof the cylin..der aremutually balanced. Since the volume isinequilibrium, the pressureonthe gas fromthesideof the wall should always beequal tothe pressureonthe .other baseof the cylinder fromtheside of thegas. We canconclude,on the basisof Newton's third law, that the pressureofthe gas onthewallis equal tothe pressure inside the vessel.325. The pressureinthe gas dependsonthe forces of interaction betweenthe molecules (see Problem 322). The forces of mutual interaction of themolecules andof the molecules with the wa11 are different, however. Hencethe pressuresinside the gas and at the walls of the vessel(see Problem324)canbeidentical only if the concentrations are different.326. Since the vol ume is constant~ = . ~ , orP2-Pl=T2- T1= O.004PI T 1 PI T 1Hence,T =T2-T1=2500K1 0.004327. FromArchimedes' law, mg+G=yV, whereyisthe specific weightofwaterandVisthe volume of the sphere. The equation of state gives(Po+ yh) V=!!!:.... RT~UpondeletingVfromtheseequations, wefindthatm GJ-t (Po +yh) ~ 0.666gVRT - fJ-g (Po +yh)andequilibriumwill beunstable.328. Whenthe tube ishorizontal, the devicecannot beusedas athermo-meter, sincethe pressures exerted onthe dropfromthe right and fromtheleft will bebalanced at any temperature.If the tube is placed vertically, the pressureof the gas inthe lower ballwill behigher than inthe ufper onebyaconstant magnitude. Ifthevolumeisthesame, the pressurewil growwith arise in thetemperaturethefaster,thehigher isthe initial pressure. Tomaintaina constant difference of thepressures inthe balls, the dropwill begintomoveupward, and inthiscasethe devicecanbeemployedas a thermometer.329. Sincethe masses of the gas are thesame inbothendsandthepistonis inequilibrium,Hence,T2 =!.!. T1 =3300KVIApplyingBoyle's lawtothe volume of the gaswhosetemperature does notchange, we obtainp=PoVo=l. 05atmVI278ANSWERSANDSOLUTIONS330. Whentheexternal conditions are the same, equal volumesofvariousgases contain an equal number of molecules (Avogadro's law). Therefore,Vl:V,:Vs:V.=Nt:N2:Ns:Nc, whereVi isthe volumeofa gas and N,. thenumber of moleculesof this gas.The mass of a certain amount of agas is proportional to the nurnberofitsmolecul es and themolecular weight ofthegas:ml: m2: ma:m. = N1""'1: N2""2: N8""3: N.J.t.Ontheother hand, denotingtherelative volume of this gas in per centVby nj=V100%, wehave. .. VI. V2 Vs Vc N1.N2.N3 N.ntn2 .nan,=VVV V=7jN'N!iIf thecomposition of air inpercent is described byni= mi 100% (com-mpositionbyweight), we canobtain fromthe previous ratios thatn ' 'n' n' 'n' _m1.m2.m3, m4_N1Jll . Na""2 .Naf.ls . N.Jl4_nII. n II. nII. in u1 2 3 - m m m m - N N . N N - lr1' 2r2' 3r3 4r4Hence,, n, n.,.,..n1Jll+nlJ.L 2 +naJla+n.J1. I IRememberingthat n; +n; +n; +n; = 100per cent, we obtain, nilJi1OO%n1nl""'l +n2J.t2 -l- nal-'a +n.J.t4Therefore,n;==75.52%; n;=23.15%; n;=1.28%;331. For each gas, the equation of state can be written as follows:PIV=.!!!.L RTJ.ll

fJ2

""3 RTf.14Hence,(PI +P2+P3+P4)V=RTf.tl ""2 f.La fJ4Onthe other hpnd, for amixtureof gases pVRT t where m=ml +f.1+ m2+m3+tn4and Il. is thesought molecular weight.HEAT. MOLECULARPHYSICS279I/otm1 ---- !0.5Z/lZ6 ------4.3IIIJZ J 4 .J{litFig. 398According to Dalton's law, p=p,+p!+Ps+P,. Therefore,11= ml+mS+ma+mn ~ + n ; + n ~ + n ~ =28.966. ! ! ! ! . . + ~ + ~ + ~ !!i+ n; +n ~ +n;""'1 fi! J,Ls J,L. ftl fl.' fJ.af.t.where n i = ~ 100%is the composition of the air inper cent by weight.mTheresul t obtained inthe previous problemallows usto find.... fromtheknowncomposi tion of the air. by volumef..L ....l nt +""2n2+""sna+J.t.n. = 28.966nl +n2+nS+n.332. Onthe basis ofClapeyron's equation,J1 = mRT=pRT=72g/molepV pThesought formula is C&H12(oneof the pentane isomers).333. When the gas is compressed in a heat .. impermeableenvelope, theworkperformed by the external forceis spent toincreasetheinternal energyof the gas. and its temperature increases. The pressure in the gas will in-creasebothowingtoareduction inits volumeandan increaseinits tempe..rature. In isothermal compression the pressure rises onlyowingto a reduc-tion inthevolume.Therefore. the pressure will increase more in'the first case than inthesecond.334. Adiagramof p versusVisshownin Fig. 398. The greatest workequal to the hatched area inFig. 398 is performed during the isothermalprocess(1.. 2). .The temperature doesnot change onsection 1-2, and is halved on sec-tion 2-3. After this the temperature rises. and, T,=T1whenV4=4lit.335. Line 1-2 is an isobaric line(Fig. 399). Thegas is heated at acon-stant pressure, absorbing heat.280p1.....---+---....vANS,WERSANDSOLUTIONSLine 2-3 is anisochoric line. Thegasiscooled at a constant volume,thepressure dropsand heat is Iibera...ted.Line3..l is an isothermal line. Thevolume of the gas diminishes at aconstant temperature. Thepressureri ...sese Thegasis not heated, althoughitissubjected to thework of externalforces. Hence, the gasrejects heat onthissection.338. The amountof heat liberatedperhour upon combustion of the meth-ane isFig. 399 Q_90PVof.'1- RTwhere 11=16g/moleisthemass of one mole ofthegasandT=t+273 == 2840Kis its temperature. The amount ofheat receivedby the water inone hourisnDIQ'=Tvpc (tt-tl ) 3,600where p=1 g/cms,is the density of the water andc= 1 cal/deg- gisthespecificheat.Accordingtothecondition,QQ: ='1=0.6Upon solvingthese simultaneous equations, we find thatt t+qopVoJ.LTJ - 930r2 =1 900nD2vpcRT = -\J337. Intheinitial state plV= ~ . RTl' where J.tl isthe molecularweight~ 1ofthe ozone. Inthe final state, P2V =mRT2' where .... 2 is themolecular1-12weight of theoxygen.The heat balance equationgives~ q =Cvm(T2 - TI)J.Ll J.L2Upon solvingthesesimultaneous equations, we find thatP'I.=-q...-+J.LI=lOPI CvTt ""2338. Inview of thelinear dependence of pressure on volumewe can write:p==aV+b.HEAT. MOLECULARPHYSICSrI \I\I\I\\I\I\I\,\I\I\\I\va~ O : G IfIIFig. 400281The constantsaandb can be found from theconditionof theproblem:a=PVl - Pv2 :!!:-0.5atm/lit1- 2b plV1- plV2S!!E 20 atmVI-V'Upon inserting the expression for p intothe equation of stateof anidealmgaspV=-RT=const T, we find thatJ.LaVI+bV=const T (1)The relation between T and V (see Fig. 400) isaparabola. The curvereaches its maximumat Vmax= - :a 2E 20lit when therootsof quadraticequation(1) coincide. HerebPmax=aVmax+b=2 e:10 atmTherefore,T _PmaxVmaxJ.L~ 4900KmQX- mR -339. The energyof 8 unit volumeof gas "1 =CTp, where p Is the density of theair. According to the equation of state of an Ideal gas, Pi =mB (B is a constant).Sincep = ~ , then pT= ; . Therefore, U l = ~ P is determined onlybythepressure. The energy ofall the air intheroomisalsodetermi nedonlybythepressure. The pressureinthe roomis equal totheatmospheric pressure282ANSWERSANDSOLUTIONSFig. 401anddoes not changewhentheair is heated. For this reasontheenergyoftheair inthe room alsodoes not change. As the air isheated, someof itflows outsidethroughcracks, andthis ensuresa constancyofenergydespitetheheating. Theenergywouldincreasewithheatingonlyina hermeticallyclosed room.340. On the basis of the equation ofstate. thesought mass of theairwill beA _ f.LpV T2 - T1 ~ 13kum------ gR T1T2341.Let thetube first benear thebottominastateof stableequilibrium.Uponheating, theair pressureinthe tube and, correspondingly. the forceofexpulsionincrease. At acertain temperature TIthe tubebegins torise to thesurface. Sincethe pressure ofthe water graduallydecreases upwardsfromthebottom, thevolume oftheair inthe tubeand, therefore. the force ofexpul-sioncontinue to increase. The tube will quickly reachthe surfaceofthewater. Upona further increase inthe temperature, the tubewill beat thesurface. If the temperature lowers, the tubewill not sinkat T1. because ithas a great reserve of buoyancy caused byanappreciableincreaseintheforce of expulsionas the tuberises. It isonly whenT2 < T1 that the tubebegins tosink. Herethe forceofexpulsionwill dropbecause, as submergen-ce continues, the air in the tubewill occupyasmaller volume. Thetubewill reachthebottomveryquickly.Therelation between the depth of submergence h ofthe tubeand thetemperatureT isshowninFig. 401.Thetubewill always be at thebottomwhen T < T2 andat thesurfacewhenT > T1. If T2 < T < Tit thetubewill beeither at thebottomoratthe surface, dependingontheprevioustemperatures.342.Thegas expands at a certain constant pressurepbuilt upby thepiston. Thework W=p (V2-VI). where ViandV2 arethe Initial andfinalvolumesof the gas. Byusingtheequation of state. let usexpressthe pro-duct pVthroughthetemperatureT. Then.W = ~ R (T2- Td~ 33.9kgf-mLLHEAT. MOLECULARPHYSICS283343. Theheat imparted tothe gas is usedtoheat it and performmecha..nical work. Accordingtothe lawofconservationofenergy,2-4. Propertiesof Liquids344. It ismoredifficult to compress a litre of air, sincemoreworkhastobedoneinthis case.Water has a small compressibility, and asmall reduction involumeisrequired toincreasethe pressureinside it tothree atmospheres.345. Amaximumthermometer can be made as follows. A small unweUablefreelymoving body is placed inside the tube ofahorizontal thermometer(Fig. 402). The position of the bodywill showthe maximumtemperature,since the body will movealongthe tube when the liquid expands andwillremain inplace whentheliquid inthe tube iscompressed.To make a minimum thermometer, a wettable body should be placedinsidethe liquid inthe tube.346. Whenanelasticrubber filmisstretched, theforce of tension dependsonthe amount of deformation of the film. Theforce of surfacetension isdeterminedonly by the properties of the liquid anddoesnot change withanincreaseof its surface.347. The surface tension of pure petrol is less than that of petrol inwhichgrease is dissolved. For this reasonthe petrol applied tothe edgeswill contract thespot towards thecentre. If the spot itself iswetted, itwillspreadover the fabric.348. Capillaries of the type shownin Fig. 403 forminacompact surfacelayer ofsoil. They convergetowards thetop, and thewater in themrisestothe surface, fromwhich it is intensivelyevaporated. Harrowing destroys thisstructure of the capillaries and the moisture isretained in the soil longer.349. Leather contains a great number ofcapillaries. Adropofa wettingliquid insideacapillaryhaving aconstant cross sectionwill bein equllib-rium. When the liquid is heated, the surfacetensiondiminishes and theliquid isdrawntowards the cold part of the capillary. Thegreasewill bedrawninto the leather if it is heated outside.350. Thegreasemelts, and capillary forces carry it to thesurfaceof thecold fabric placed under the clothing(seeProblem349).351. Theendof the pieceof woodinthe shade iscolder, and the capll-taryforcesmovethe water inth is direction.352. The hydrostatic pressure should bebalanced bythe capillary pres-4 ~sure: pgh={f' Hence, h=30em.353. Thefollowingforcesact verticallyon sectionabed of thefilm: weight,surfacetension Fab applied to line ab and surface tension Fcdapplied tocd.Fig. 402284ANSWERSANDSOLUTIONSEquilibriumispossibleonlyif Fallisgreater than Fcd by anamountequal tothe weight of the sectionof the film being considered.Thedifference betweentheforcesof surface tension can be explai-nedby the differencein the concent-ration of the soap inthe surfacelayersof the film.354. The force of expulsion ba-lancesthe weight of the cubemgandtheforce of surface tension 4aa.Fig. 403 i.e., a2xpg= mg+ 4aa, wherex isthe sought distance. Therefore,mg+4aa.~ 2 3x a2pg -. emTheforces ofsurfacetension introduceacorrectionofabout 0.1em.355. Thewater risestoaheight h= 2a. Thepotential energyof thewa-pgrter colurnnisEp=mgh =2na;22 pg4n(2Theforces ofsurfacetensionperformthe work W=2nrah=--. One halfPiof this work goes to increase the potential energy, and the other half toevolveheat. Hence,2na2Q=-pg356. The pressureinsidethe liquid at a point that isat aheight h aboveacertain level is lessthan the pressureat this level bypgh. The pressureiszeroat the level of the liquid inthe vessel. Therefore, the pressure atthe height his negative(the liquidis stretched) and isequal top= -pgh.357. The forces of attr action acting on a molecule in thesurfacelayerfrom all theother molecules produce a resultant directed downward. Theclosest neighbours, however, exert a force of repulsionon the molecule whichistherefore inequi librium.Owingto the forcesof attractionand repulsion, the density of the liquidis smaller in thesurfacelayer than inside. Indeed, molecule/ (Fig. 404) isacted upon by the force of repulsion frommolecule 2 andthe forces ofattract ionfromall the other molecules(3, 4, ...). Molecule 2 isacteduponby the forces of repulsion from3and/ and theforces of attractionfrom themoleculesinthe deep layers. As a result, distance /-2 should begreaterthan 2-3, etc.This course of reasoning is quite approximate (thermal motion, etc., isdisregarded), but neverthelessit gives a qualitativelycorrect result.An increase inthe surfaceof the liquid causes newsections of therare-fied surfacelayer toappear. Hereworkshouldbeperformed against thefor-cesofattractionbetweenthe molecules. It is this workthat constitutes thesurfaceenergy.HEAT. MOLECULARPHYSICS285358. Therequired pressure shouldexceedthe atmospheric pressure byanamount that canbalance the hydrostatic pressureofthewatercolumn andthe capillary pressureintheair bubblewitharadius r,Theexcess pressure is p+pgh+2a =4,840dyne/ems,r359.Sinceinthis casep gh < 2a, , thewater rises tothetopendof thertube. Themeniscus will bea part ofa spherical segment (Fig. 405): Theradiusofcurvature ofthesegment isdeterminedfromtheconditionthat theforces of surface tension balance the weight of the water column:2Jtra cosq:> =rtr2hpg.rhpaHence. cosp= 20.'. It is obvious fromFig. 405 that the radius ofcur-r 2avatureof the segment R=--=h-=O.74mm.cos q> gp360. When the tube is opened. aconvex meniscus ofthesame shapeason the top is formed at its lower end. For this reason the lengthofthewater columnremaininginthetube will be 2h if I ~ h. and1+h if t ; h.361. (1) The forces of surface tensioncanretainawatercolumnwithaheight not over hinthis capillary tube. Therefore. thewater will flowoutofthe tube.(2) Thewater does not flowout.Themeniscus is convex. andwillbea hemisphere for anabsolutelywettingliquid.(3) Thewater does not flowout.Themeniscus isconvexandis lesscurved than inthesecond case.(4)Thewater does not flowout.ThemeniscusisOat.----------------@-------@--------------@-------------@-------Fig. 404 Fig. 405------286

--_.....---_...ANSWERSANDSOLUT IONS

-..------ - - -(aj-------. -#-- .r.__ ...- --.---(b)Fig. 406(5) Thewater does not flowout. Themeniscus is concave.362. The pressure p inside the soap-bubble witharadius Rexceeds theatmospheric pressure by the amount ofthedouble capillarypressure, sincethebubblefilmis double: P=Po+ Thepressureinside thebubblewitharadiusRtogether withthe pressureof the section of the film between the bubbles shouldbalancethe pressureinside thesmaller bubble. Therefore,+:: =where u, isthera-dius of curvature of sectionAB. Hence, Rx= RRr -rAt any point of contact theforces of surface tensionba lanceeachotherandaremutuallyequal. This is possible onlywhenthe angles between themare equal to120.363. Accordingto the lawof conservationof energy, thecrosswill notrotate. The components of theforces of surface tension arebalancedbytheforces of hydrostatic 'pressure, since the hydrostatic pressure of thewater.higherthanthe level inthevessel is negative(see Problem356).364. If thebodies arewettedbywater, its surface will take the formshowninFig. 4000. Betweenthematches, abovelevel MN, thewater istensionedby the capillary forces, and thepressure inside thewater is lessthantheatmospheric pressure. Thematches will beattracted toward eachother, sincetheyaresubjectedtotheatmospheric pressureontheir sides.Forunwettedmatches, the formofthesurface is shown inFig. 40Sb. Thepressure betweE'n the matches is equal to the atmospheric pressureandisgreaterthanthe latter onthesides belowlevel M N.

HEAT. MOLECULARPHYSICS287 ------::-.!L- -_-=-=(a) --- w- -----(h)Fig. 407In the last case two various forms of the surface correspondtothewettingangles when the matches approach each other (Fig. 407). Oneofthem, howe-ver, cannot be obtained in practice (Fig. 407a). The pressure at level KLshould be the same everywhere. In particular, the pressure of columns ABand CDof different height should be the same. But this is impossible, sincethe positionof the columncan be soselected that their surfaces areidenticaJinform. In this case the additional pressure of the surface forces will be thesame, and' the hydrostatic pressure different. Asa result, when the matchesapproach each other, the surface of the water between themwill tend toassurne a horizontal form(Fig. 407b). In this case, as can beseenfromthefigure, the pressure between the matches at level MNis equal to the atmos-pheric pressure. The pressure exerted fromthe left on the first match is alsoequal to the atmospheric pressure below level MN. The pressure actingon the second match fromthe right is less than the atmospheric pressureabove level JWN. Asa result, the matches will be repulsed.2-5.Mutual Conversion of Liquids and Solids385. Water will freezeat zero only in the presenceof centres of crystalli-zation. Any insoluble particles can serve as such centres. When the massof the water is great, it will always contain at least onesuch centre. Thiswill beenoughfor all the water to freeze. If the mass of the water is divi-ded into very finedrops, centres of crystallization will be present only inacomparativelysmall number of the drops, and only they will freeze.288ANSWERSANDSOLUTIONSc386. The water and the icereceive about thesame amount ofheat inaunit oftime, sincethedifferencebetween thetemperatures ofthe water andthe air in the room isapproximatelythesame asthat fortheice andtheair. In15 minutes thewater receives 200 calories. Therefore, theice receives8,000 caloriesintenhours. Hence, H=80cal/g.367. 0=2,464m/s.368. Theheat balanceequations have theformQ1=mtcIL\#+CAtAtQ2=mlclT+mtH+mlc2T+C&twhere ml and Cl are themassandheat capacityof theice, CIs theheatcapacity of the calorimeter, c2is the heat capacity of the water, and

Hence,Q(C2 HI) Q1-+--+-- 22c12 150 lidCI At H :z: ca eg"C;T-T+C;-369. The amount of heat that canbeliberatedbythewaterwhenit iscooledto OC is 4,000cal. Heating ofthe ice toOCrequires12,000cal.Therefore, the ice canbeheatedonlybytheheat liberatedwhenthewaterfreezes. .One hundred grammes of water should be frozen toproduce thelacking8,000 calories.Asaresult, thecalorimeter will containamixtureof 500gof water and500 gofice at atemperatureofOC.370. The final temperature of the contents inthevessel isO=OC. Theheat balanceequationhasthe formm1c1(tt-8)=mtcl (8-t2)+ (m2- ma) Hwhere m1Is the sought mass of the vessel and c, istheheat capacityoftheice. Therefore,m,Ct (9-tt )+ (mt - ms) H 200ml = CI (tl - 6) g371. (1)The sought mass of the ice mcanbefound fromtheequationmH=Mc (-t). Hence, m= 100 g.(2) The heatbalance equation can bewrittenin this case as MH=Me (- t).Hence, t=-80C.372. The melting point oftheice compressedto 1,200atmwill dropbyAt =8.8 C. Theice will melt until it cools to -8.8 C. TheamountofheatQ=mtHis absorbed, where m\ isthemass ofthemeltedice andHisthespecific heat of fusion. Fromtheheat balance equationm1Hwherec is theheat capacityof the ice.Hence,cmtuoc:m1 =n -5.6gHEAT. MOLECULARPHYSICS2892-6. Elasticity and Strength373. F AE (R-r) 60kgf.r 374. When the rodwith fastenedends is heated by t degrees, it developsanelastic forceFequal, according toHooke's law, toF =AEAI = AEaJIwhereEis the modulus ofelasticityof steel and (X is its coefficient of ther-rna I expansion.If one of the rod ends is gradually released, the length ofthe rod. willincrease by Al=lat. The force will decreaselinearlyfromFtozeroand itsaverage magnitudewill beF/2.F IThesought work W=2Al =2AEla2t2375. The tensionof the wireT=2 It follows from Hooke's lawsinaAithat T =2f EA.Since Al=2 (_/__1) , thenT I-cosa. At small an-cos ex cosex. Sin agles, sin a a, and cos a=1-2 sinS e; 1- Bearingthis inmind, weobtainVMgcx= AE376. The rod heated by At would extend by Al=loaAt ina free state,where Lois the initial length of the rod. Tofit the heated rod betweenthewalls, it should becompressedby 111. Inconformity with Hooke's law,IFAl=EATherefore, F =EAaAt =110 kgf.377. When the rods are heated in a free state, their total length willincreasebyAl=All+Al,=(al'l+a21z)t. .Compression by the sameamount will reduce the lengths of the rodsbyandAI;, whereAI; +This requires the forceF - ElAAI' EzAA1;- '1 t 12Uponsolving this systemofequations, wefindthatF allt +asll At

E) 2The rods will act uponeachother with this force.19-2042290ANSWERSANDSOLUTIONS378. It is obvious from cons iderations of symmetry that the wires willelongate equally. Let us denote this elongation byOn the basis ofHooke's law, the tension of a steel wireAEsand of a copper oneAlFe=-l- AEcIt follows that the ratiobetween the tensions is equal totheratiobetweenhe respective Young's moduli .Fe Ee1p;= E$ ="2Inequilibrium2Fc+Fs=mg.Therefore, Fc ="'4g=25 kgf and Fs =2Fc =50 kgf.379. On the basis of Hooke's law, Fe=-l- AcEcand Fi=-l- AiE;FIt follows that =2.Thus, of the load are resisted by the concrete and one-third bythe iron.FI380. The compressive forceFshortens the tube by AE and the tensileFI c CforceFextends the bolt by AE s sThe sum + isequal to the motionof the nut alongthe bolt: AsEsAcEcHence,F -!!:.. AsEsAcEc- I AsEs+AcEe381. Since the coefficient of linear thermal expansionof copper acIs grea-ter thanthat of steel as,. the increase intemperaturewill leadtocompressionof the copper plate and tension of thesteel ones. In viewof symmetry, therelative elongationsof all the three plates are thesame. Denoting thecomp-ressive force acting on the copper plate fromthe sides of the steel platesby F, we shall have for the relative elongationof thecopper plate: =F=act- AE .cEither steel plate is subjected to the tensile force F/2 fromthe side ofthe copper one. Upon equating the relative elongation of the plates, weobtain:HEAT. MOLECULARPHYSICS2918 Fig. 408On upper nutOn lower nutO=Fg oFFpllf----.-,HenceF _2AEcEs (Xc-as) t- 2Es+Ecmvl382. Whenthe ring rotates, the tension T = 2-appears InIt (see Prob-nrtern 201). For a thin ring m=2nrAp, where Ais thecross sectionof thering. Therefore. ~ =pv.Hence. the maximum velocity v=V ~ Q ! ; ~ l m/s.:$83. Initially, an elastic force Po acts oneachnut fromthe sideof theextended bolt. .The load G~ F0 cannot increasethe length of the part of the bolt bet-weenthe nuts and change its tension. For this reasontheforce acting ontheupper nut from the side of the blockwill not change as longas G~ Fe-The lower nut isacted uponbythe force F0 fromthe sideof thetop partof the bolt and by the forceGfromthe bottompart. Sincethe nut is inequilibrium, the force exerted onit fromthe block is F=Fo-G. Thus theaction ofthe load G~ F0 consistsonly inreducingthe pressureof thelowernut onthe block.When G > FOIthe length of the bolt will increaseand theforce actingon the lowernut fromthe sideof the blockwill disappear. Theupper nutwill beacted onbythe forceG.Therelationbetweenthe forces actingonthe nuts and the weight of theloadGisshownin Fig. 408. .27. Properties ofVapours384. Thecalorimeter will contain 142 gofwater and 108g of vapour ata temperature 1000C.385. By itself, water vapour cr steamis invisible. Wecanobserve onlyasmall cloud ofthe finest drops appearingafter condensation. Whenthe gasburner is switched off, thestreams ofheated air that previously enveloped19 ':292ANSWERSANDSOLUTIONSthe kettle disappear, and the steamcoming out of the kettle iscooled andcondenses.386. Onthe basisof theequationof stateofanideal gas p= = ;';. .If the pressure is expressed in mm Hg and the volume inm3, then R:::;::760X 0.0224 mm Hgms= 273 deg.mole273Therefore,p=1.06 PT' At temperatures near room temperature,g/ms.387. It seems at first sight that the equationofstateof an ideal gas can-not give values of the densityor specificvolumeofsaturated vapours closeto the actual ones. But this is not so. If wecalculatethe densityof a va-pourbythe formula p= = and compare the values obtained withthose in Table 2 (p. 85), we shall observe good agreement.This isexplained as. follows. The pressureof an i deal gas grows in directproportion to the temperature at a constant volume of the gas and, there-fore, at a constant density. The relation between the pressure of saturatedvapours and the temperature depicted in Fig. 146 corresponds to a constantvolume of a saturated vapour and the liquid which it is in equilibriumwith. As the temperature increases, the densityof the vapeur grows, sincethe liquid partially transforms intoa vapour. Anappreciable increase in themass of the vapour corresponds toasmall change in the volume it occupies.The pressure-densityratiobecomesapproximately proportional to the tempe-rature, as with an ideal gas.The Clapeyron-Mendeleyev equation mainly gives a correct relationshipbetween p, Vand Tfor water vapour up to the valuesof these parametersthat correspond to the beginning of condensation. This equation, however,cannot describe the processof transitionof 8 vapour into a liquid and indi-catethe values of p, 1> and Tat which this transition begins.388. At 300Cthe pressure of saturated vapours p=31.82mmHg. Accor-ding to the equationof state of an ideal gas,V=.!!!. RT e: 296 lit1-1 p389. When the temperature gradually increases, the pressure of the watervapours in the roommay be consideredconstant.Thevapourpressure p=corresponds to ahumidity of Wo= 10 percent, where Po= 12.79 mm Hg is the pressure of the saturated vapours-at 15 C. At a temperatureof 25Cthe pressure of thesaturated vapours isPI =23.76mmHg. For this reason the sought relative humidity isw=LPI PJ'390. According to the conditions of the problem, the relative humidityoutside and in the roomis close to 100 per cent. The pressure of saturatedwater vapours outside, however, ismuch smaller than in the room, becausethe temperature of the air in the roomishigher and much time is requiredtoequalize the pressures owing to penetrationof the vapoursoutsidethroughHEAT. MOLECULARPHYSICS293slits. Therefore, if the windowis opened, the vapours will quickly flowoutfromthe roomand the washingwill be dried faster.391. (1) The water levels will become the same as in communicatingvessels. The water vapours in the left-hand vessel will partlycondense. andsome water will evaporate in the right-hand vessel.(2) Thelevels will become the same because the vapours will flowfrom onevessel into the other.At a given temperature the pressure of saturated vapours is identical inboth vessels at the surface of the water and will decrease at the same ratewith height. For this reason the pressure ofthe vapoursinthevessels at thesame level is different. This causes the vapour to flowover and condense inthe vessel with the lower water level.392. When t2=30 C, the pressure of the vapours is equal to the pressureP20 of saturated vapours (P20=31.8 mmHg) only if the air pressureis10 at.Upon isothermal reduction of the air pressure to one-tenth, the volume ofthe air will increase ten times. Hence, at atmospheric pressure and atempe-rature of 30 C, the pressure of the water vapour isp=3.18mmHg. It fol-lows from the Clapeyron equation that at a temperature of 11=10C, theTvapour pressure PI =PT:' where TI =283 Kand T1=303 K.The sought relative humidity isw=Pt 1000/0 =P.. TTl 1000/0~ 32.60/0Po Po 2where Po=9.2 mm Hg is the pressureof saturated vapours at t1= 10C393. The pressure p=6.5mmHg is the pressure of saturatedwater va-pours at t=5 C. Asharp drop of the pressure shows that all thewater hasbeen converted into vapour. The volume of the vapour pumpedout until thewater is evaporatedcompletely is V=3,600litres.On the basis of the Clapeyron-Mendeleyevequationof state, the soughtmass of thewater isv .m=PR/ !!!!S. 23.4 g394. An amount of heat Q1=mcAt =3,000cal is required to heat thewater to 100C. Therefore, Q2=Q- QI =2,760cal will bespent for vapourformation. The amount of the water converted into vapour is ml =Q=J=5.1g.rInconformity with the equation of state of an ideal gas, this vapourwilloccupy a volume of V=m1 RT. Upon neglecting the reductionof the YO-J.L P .lume occupied by the water, wecan findthe height which the piston is rai-Vsedto: h=;r = 17 ern.CHAPTER3ELECTRICITYANDMAGNETISM3-1.ElectrostaticsQ2395. F=7=918kgf.The forceis verygreat. It isimpossibletoimpart a charge of one coulombto a small bodysince the electrostatic forces ofrepulsion aresohigh thatthe charge cannot beretained onthe body.396. The balls will bearrangedat the cornersofanequilateral trianglewith aside ~ 3 1 . The force acting fromany two balls on the third is4Q2=12 Y3'The ball will beinequilibriumif tan a,=f- (where a=300). Hence,mgIQ=2 Ym g ~ 100CGSQ_397. Since the threads do not deflect fromthe vertical, thecoulombianforce of repulsion is balanced bythe force ofattractionbetween the ballsinconformitywith the lawof gravitation.Therefore, in a vacuumQ2 p2V.-;:2=Y7andinkerosene (taking intoaccount theresults ofProblem230)Q2 (p-PO)2 ViErr2= 'V r 2where Vis thevolumeof theballs.Hence,P=Po yer~ 2.74 g/cm8Ye,.-l398. Theconditionsof equilibriumof thesuspendedball givethe follow-ingequations forthe twocasesbeingconsidered:T- QQs Y2 01 sin(Xt- 2a2X -2-=T +QQs Y2 QQs 01 cos al 2a2X -2--7-mg=T QQs V2 02 srn (X,2- 2a2X -2-=T2cosa2+QQs-QQsx V2-mg=Oa22a22ELECTRICITYANDMAGNETISMFig. 409 Fig. 410295whereT1 and T2 are the tensions of the thread, al andthe angles ofdeflectionof the thread, +QandQthe charges of the fixedballs, +Qsthe charge of the suspended ball, and mg is the weight of the suspendedball (Fig. 409).Upon excludingthe unknowns fromthe above simultaneous equations, wegetcot a1-cot a2=cotai-cot 2al =2 (V2-1)whence,cot al =2 (2 Y2 - 1) V35-16 Y2Thus, C&J. =756' and a,=1552' when mg > ( 1- -':2 ), andal=8204' anda2=164OB'when mg < (1- -':2 ).399. In uniformmotion the drop isacted upon by theforceof gravity G.the expulsive forceof the air (Archimedean force) F, the force oftheelectro-static fieldeEand the forceof friction againsttheair kv=k. All the for-cesare balanced. Therefore,G-F-eEO+k..!..=Ot1

t;G-F-k!..=Otwhereeis the charge of the drop, Ethe intensityof the electric field, andsthe distance covered by the drop.296ANSWERSANDSOLUTIONSUpon solving the equations, we gett=2t lt2tl-t9Fig. 411400. It can, if weuse the phenome-non of electrostatic induction. Bring aconductor on an insulated support upto the charged body and connect theconductor to the earthforashort time.The conductorwill retain a chargeop-posite in sign to the given one, whilethe like charge will pass into theearth.The chargecan be removed fromthe conductor by. introducing the latterintoa metallicspace. The operation may be repeated many times with acharge of any magnitude.Electrostatic machines operate on a similar principle.401. The energy is produced by the mechanical work that has to be per-formed in moving the conductor from the oppositely charged body to thebody .that accumulates the charge.402. They can, if the chargeof one ball ismuch greater thanthat of theother. The forces of attraction caused by the induced charges mayexceedtheforces of repulsion.403. Since Q ~ q, the interaction between the separate elements ofthe ringcan be neglected. Let us take a small element oftheringwithalengthRA.a,(Fig. 410). From the side of the charge Q it is acted upon by the forceIJ.F =~ ~ q where t1q =q:na.. The tension forces of the ringTbalance IJ.F.Fromthe conditionofequilibrium, and remembering that AeJ is small. wehavet1F=2T sin (IJ.;) rsThesought force is the tension T =: : . ~ . 404.Let us consider thecase of oppositechargesQ1 > 0andQ, WI' Besides, the heat Qis libe-rated in the conductor.According to the lawof conservationof energy, however, thetotal amountof energy in the balls should be the same in both cases. Since the work WIand, correspondingly, W 2' is the potential energy of the second ball in thefield of the first one in the first and second cases, thenW1 +Wet =W2 +Q+We22 2where Wcl = +is the intrinsic energy of the balls before connection2 q2and We2= ;, + 2, is the intrinsic energy of the ballsafter the chargesareredistributed (see Problem422).The energy liberated as heat is(qt -q2)2 (J.. _-!-)Q= Wel- We2 +W1- W2 4 r I426. Assume that the radius of the envelope increases by6, which maybe an infinitely small quantity. Theexpanding force will performthe workW=4nR2f6, wheref is the force per unit ofarea. Thiswork is done at theexpense of a reduction in the electrostatic energy. First the electrostatic304ANSWERSANDSOLUTIONSFig. 416energy is 2QR2, andafter expansion Q22 (R+6)ThechangeintheenergyQ2 QI QI 62R 2(R+6) 2isequal totheworkW, I. e.,

2R(R+6)Taking into account the fact that 6canbeinfinitelysmall, we obtain thefollowing expres-sionforthe force:f- Q2 -2'"'1'1'2-8nRt- ".vHere, (1=is thedensityof theelectricity, Le., thechargeper unitof area.Thesought forcecanalsobefounddirectly. Let usconsidera small areaaonthesphere(Fig. 416).Let usfindthe intensity1oftheelectricfield createdon thearea beingconsidered by all the chargesexcept theonesontheareaitself. Tointro-duce definiteness, let usconsiderthecasewhenthespherecarries apositivecharge.Let us denote by 2 the intensity oftheelectric field created by thecharges on the area itself. Since theresulting intensityiszeroinside thesphere, then1 =2QThe resulting intensity on the sphere E1+EJ=W' and, therefore,Q21 =W=4no. Hence, 1=2na.To find the force that acts fromall the chargesoutside t heareaoftthechargesonthearea, the intensityE1shouldbe multipliedby themagnitudeof theelectricchargeof thearea aa:F=E1oa=2nosaTheforce perunit ofenvelope areawill bef=21tCJ1427. For the charge qtobeinequilibrium, the charges -Qshouldbeat equal distances afrom it (Fig. 417). The. sumof the forces actingonthecharge - Qis alsozero:QS Qq4a2 -Qi'=OHence, q= . The distanceamayhave anyvalue. Equilibriumisun-stable since when the charge -Qisshiftedalong001to the left overadistancex, the force ofattractionQ2Fg=4 (a+x)1ELECTR ICITY ANDMAGNETISM305acting fromthe side of the charge q is less than the force ofrepulsionQ2FQ=(2a+x)2and the charge -Qmoves still farther fromthe positionofequilibrium.When the charge -Qis shifted along001over a distance x towards thecharge q, then Fq > FQ for x y2'~ 1.4.695. The angle of incidenceof the rayonface Be isequal tothe soughtangle cx. For the ray tobecompletely reflected from face Be, the angle exshould exceedthe limitingone. .Therefore, sin a > ::' wherenzis the refraction indexof water.Hence, a > 62030'.696. This phenomenonis nothing but a mirage frequently observed in de-serts.The hot layer of air indirect contact with the asphalt has a smaller re-fraction indexthan the layers above. Total internal reflectionoccursand theasphalt seemstoreflect the light just as well as the surfaceof water.697. Let us divide the plate into many plates sothinthat their refractionindexcan beassumed constant within the limits ofeach plate(Fig. 503).Assumethat the beamenters the plate froma medium with a refractionindexofno and leavesit for a mediumwith a refraction indexof na390ANSWERSANDSOLUTIONSFig. 503Then.accordingtothe lawofrefraction,sin a ntsin ~ =nosin p n'--=-siny nlsin'V n"sin~ = n 'sinq> n2sin, = n(n)sin~ nasinx=n2Upon multiplyingtheseequations we"getsin a nsSiiiX=noHence, the angle at whichthe beamleavesthe platex=arcsin (:: ~ i n a)depends onlyonthe angleof incidence ofthe beamonthe plateandontherefractionindices ofthemediaon both sides of the plate. Inparticular, ifng=no. thenx=a..GEOMETRICALOPTICSFig. 504n"NFig. 505F391Generally speaking, the angle O. at which the beamis inclined tothever-tical is related to the refraction index nat any point on the plate by theratio n sin 8=const=nosin a.. If the refraction index reaches a value ofn =nosin ex anywhere insidethe plate, full internal reflection will takeplace.In this case the beamwill leave the plate for the mediumat the sameangleex at which it entered the plate (Fig. 504).698. The minimumamount of water determined by the level x (Fig. 505)can be found fromthe triangle MNF. We have NF=x-b=xtanr. Fromthe lawof refraction. sini51nr=--nTherefore,bx=-t--t-a-n-,since i=45 and 11= .The amount of water required is V=xa2=::: 43.2 litres,699. The man's eyes are reached by rays coming in a narrow beam fromanarbitrary point Con the bottom. They seem to the eye as issu ingfrompoint C' (Fig. 506). Sincedi anddr are very small, we can write: cos r cos Byequatingthe values of ABfrom triangles ABDand ABD', we have-!!-.s.. f1.tcos2rcos2l392ANSWERSANDSOLUTIONSaFig. 506Usingthe lawofrefraction, we can find the ratio ~ ~ . Indeed,s ~ n i =nand sin (i +ai) =nsin, sin (, +ar)Remembering that ~ i and ~ r aresmall, wehavesi n~ i ~ 8i, s i n ~ , === ~ " andcos L\i === cos~ , =:: tTherefore,the last equationcanberewrittenassini +cos i Ai=n sin ,+n cos r-br